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Find the identity under a given binary operation I have two problems quite similar. The first:
In $\mathbb{Z}_8$ find the identity of the following commutative operation:
$$\overline{a}\cdot\overline{c}=\overline{a}+\overline{c}+2\overline{a}\overline{c}$$
I say:
$$\overline{a}\cdot\overline{i}=\overline{a}+\overline{i}+2\overline{a}\overline{i} = \overline{a}$$
Since $\overline{a}$ is always cancelable in $\mathbb{Z}_9$ I can write:
$$\overline{i}+2\overline{a}\overline{i} = \overline{0}$$
$$\overline{i}(\overline{1}+2\overline{a}) = \overline{0}$$
so $i=0$ whatever $\overline{a}$.
Second question:
In $\mathbb{Z}_9\times\mathbb{Z}_9$ find the identity of the following
commutative operation: $$(\overline{a}, \overline{b})\cdot
(\overline{c}, \overline{d})= (\overline{a}+ \overline{c},
\overline{8}\overline{b}\overline{d})$$
So starting from:
$$(\overline{a}, \overline{b})\cdot
(\overline{e_1}, \overline{e_2})= (\overline{a}+ \overline{e_1},
\overline{8}\overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$
that is:
$$\overline{a}+\overline{e_1}=\overline{a}\qquad (1)$$
$$\overline{8}\overline{b}\overline{e_2}=\overline{b}\qquad (2)$$
In (1) there's always cancellable element for $\overline{a}$ since $\overline{-a}$ is always present in $\mathbb{Z}_9$.
In (2) I should multiply both member for $\overline{8^{-1}}$ and $\overline{b^{-1}}$ to know exactly $\overline{e_2}$.
This happens only if both number are invertible. $\overline{8}$ is easily to demonstrate that it's invertible, cause $gcd(8,9)=1$.
But what about $b$.
| So since you want $$(\overline{a},\overline{b})=(\overline{a}+\overline{e_1},\overline{8}\overline{b}\overline{e_2})$$
You must have $\overline{a}=\overline{a}+\overline{e_1}$, meaning $\overline{e_1}=\overline{0}$, and from the second entry we see that
$\overline{b}=\overline{b}(\overline{8}\overline{e_2})$; since this holds for any $\overline{b}$ we must have $\overline{8}\overline{e_2}=1$, and as you said $\overline{8}$ has a multiplicative inverse, so you have that $\overline{e_2}=\overline{8}^{-1}(1)=\overline{8}^{-1}=\overline{8}$ since $8\cdot 8=64=1$ modulo $9$. So your answer is $\overline{e_1}=\overline{0}$ and $\overline{e_2}=\overline{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric identity, possible error I need to prove the following trigonometric identity:
$$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?
| Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.
Loosely,
*
*Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
*Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$\pi \cdot \text{any}$") preserves "sin" and "cos" (and possibly affects sign)
*Negating the argument preserves "sin" and "cos" (and possibly affects sign)
Since squaring eliminates sign considerations, we have, simply:
$$\begin{align}
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{odd} \right) &= \mathrm{cotrig}^2\theta \\
\mathrm{trig}^2\left( \pm \; \theta \pm \frac{\pi}{2} \text{even} \right) &= \mathrm{trig}^2\left( \pm \; \theta \pm \pi \cdot \text{any} \right) = \mathrm{trig}^2\theta
\end{align}$$
where each "$\pm$" is independent, "any" means (of course) "any integer", and "trig" can in fact be any of the six trig functions.
This makes pretty quick work of the simplification process ...
$$\begin{align}
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\left(\alpha-4\pi\right)} - \cot^2\left(\alpha-11\pi\right) \; \sin^2\left(-\alpha-\frac{13\pi}{2}\right) &\stackrel{?}{=} \sin^2\alpha \\[1em]
\frac{\sin^2\left(\frac{5\pi}{6}-\alpha\right)}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha &\stackrel{?}{=} \sin^2\alpha
\end{align}$$
... right up to the point at which the process shudders to a halt.
Given the nature of all the other terms (and @Adam's comment that sum and difference identities are not allowed), I suspect that "$\frac{5\pi}{6}$" is a typo of "$\frac{5\pi}{2}$", which would get us a little further ...
$$\frac{\cos^2\alpha}{\cos^2\alpha} - \cot^2\alpha \; \cos^2\alpha = 1 - \cot^2\alpha \;\cos^2\alpha \stackrel{?}{=} \sin^2\alpha$$
... but we hit another snag. Could it be that "$\sin^2\left(-\alpha-\frac{13\pi}{2}\right)$" is a typo of "$\cos^2(...)$"? If so, then that factor should've simplified to "$\sin^2\alpha$", and we'd have
$$1 - \cot^2\alpha \;\sin^2\alpha = 1 - \cos^2\alpha = \sin^2\alpha$$
as desired.
(It's also possible that, instead of a sin-cos typo, "$\cot$" is a typo for "$\tan$", but it seems like that would be an easier one for the OP to notice.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$ Compute
$$\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$$
I did
$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x = \lim_{x\to\infty} \exp(x\cdot \ln(\frac{x-2}{x+2})) = \exp( \lim_{x\to\infty} x\cdot \ln(\frac{x-2}{x+2}))$$
But how do I continue? The hint is to use L Hopital's Rule. I tried changing to
$$\exp(\lim_{x\to\infty} \frac{\ln(x-2)-\ln(x+2)}{1/x})$$
This is
$$(\infty - \infty )/0 = 0/0$$
But I find that I can keep differentiating?
| A nitpick: $\infty-\infty$ is not 0! It's undefined. Your limit is of the form $0/0$ though.
You can apply L'H'ôpital from the start if you like: $\lim\limits_{x\rightarrow\infty}{x-2\over x+2} =1$, and $\ln 1=0$.
So
$$
\lim_{x\rightarrow\infty} \Bigl(x \ln{x-2\over x+2} \Bigr)
=\lim_{x\rightarrow\infty} {\ln{x-2\over x+2}\over1/x}
=\lim_{x\rightarrow\infty} {{x+2\over x-2}\cdot{1(x+2)-1(x-2)\over (x+2)^2} \over- 1/x^2 }
=\lim_{x\rightarrow\infty} {{-4x^2\over (x+2) (x-2)} }=-4.
$$
(use L'Hopital again to evaluate the limit on the right hand side if you like).
So,
$$\lim_{x\rightarrow\infty}\Bigl({x-2\over x+2}\Bigr)^x
=e^{ \lim\limits_{x\rightarrow\infty}\bigl(x\ln{x-2\over x+2}\bigr)}=e^{-4}.
$$
To answer more directly, L'Hôpital applied to $$\lim_{x\rightarrow\infty}{\ln(x-2)-\ln(x+2)\over 1/x}$$ gives you
$$\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}.$$
Now simplify:
$$
{{1\over x-2}-{1\over x+2}\over- 1/x^2}
=-x^2\Bigl({1\over x-2}-{1\over x+2}\Bigr)
= {-4x^2\over (x+2)(x-2)}.
$$
So, using L'Hôpital's rule again
$$
\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}
=\lim_{x\rightarrow\infty} {-4x^2\over (x+2)(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over (x+2)+(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over2x}=-4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
If z is one of the fifth roots of unity, not 1... If z is one of the fifth roots of unity, not 1, show that:
$1+z+z^2+z^3+z^4=0$
Which wasn't too bad, but the next part is killing me: show that:
$z-z^2+z^3-z^4=2i(sin(2\pi/5)-sin(\pi/5))$
Can anyone help? Thanks!
| For the first part you should note that by the formula for the geometric series,
$$1 + z + z^2 + z^3 + z^4 = \frac{z^5 - 1}{z- 1}.$$
So if you put $z = e^{2\pi i/5}$ into $z^5 - 1$ you get $(e^{2\pi i/5})^5 =1$ so that $1 + z + \ldots z^4 = 0$. You can see also that if you put $z = (e^{2\pi i/5})^n$ for $n=2,3,4$ you see that $z^5 - 1$ is always zero.
For the second part you should be able to see that if $z = e^{2\pi i/5}$, then $$z - z^4 = 2i\sin(2\pi/5).$$
The reason is if you draw a pentagon in the complex plane with vertices at the points $$1, e^{2\pi i/5}, e^{4\pi i/5}, e^{6\pi i/5}, e^{8\pi i/5}$$
then the vertices $e^{2\pi i /5}$ and $e^{8 \pi i/5}$ have equal real part so subtracting one from the other gives the result above. You should be familiar with de Moivre's formula for this.
Now similarly if you calculate $z - z^4$ you should get
$$z^3 - z^2 = 2i\sin(6 \pi/5).$$
But then $2i\sin(6 \pi/5) = -2i\sin(\pi/5)$ so adding this together with that found for $z - z^4$ gives
$$z +z^3 - z^2 -z^4 = 2i(\sin(2 \pi/5) - \sin(\pi/5)).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$? I'm trying to evaluate the integral $\displaystyle\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$.
My book explains that to evaluate integrals of the form $\displaystyle\int_0^\infty x^\alpha R(x)dx$, with real $\alpha\in(0,1)$ and $R(x)$ a rational function, one first starts with a substitution $x=t^2$, to transform the integral to
$$
2\int_0^\infty t^{2\alpha+1} R(t^2) dt.
$$
It then observes that
$$
\int_{-\infty}^\infty z^{2\alpha+1}R(z^2)dz=\int_0^\infty(z^{2\alpha+1}+(-z)^{2\alpha+1})R(z^2)dz.
$$
Since $(-z)^{2\alpha}=e^{2\pi i\alpha}z^{2\alpha}$, the integral then equals
$$
(1-e^{2\pi i\alpha})\int_0^\infty z^{2\alpha+1}R(z^2)dz.
$$
How are you suppose to apply the residue theorem to this integral if the integrand is not a rational function?
In my case, I have
$$
\int_0^\infty x^{1/3}R(x)dx=2\int_0^\infty t^{5/3}R(t^2)dt.
$$
Also
$$
\int_0^\infty z^{5/3} R(z^2)dz=\frac{1}{1-e^{(2\pi i)/3}}\int_{-\infty}^{\infty}\frac{z^{5/3}}{1+z^4}dz.
$$
I don't know what to do after that, since this last integrand still has a fractional power in the numerator. How does this work? Thanks.
(This is part (g) of #3 on page 161 of Ahlfors' Complex Analysis, part of some self-study.)
| Here is the simplest I could get without introducing other variables or making strange substitutions. Everything is in complex on the main branch ($0 \le arg(z) < 2\pi$).
First we calculate a bound change :
$$\begin{eqnarray*}
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz &=& \int_{-\infty}^0 \frac{z^{1/3}}{1+z^2} dz + \int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz\\
&=& \int_{0}^\infty \left( \frac{z^{1/3}}{1+z^2} + \frac{(-z)^{1/3}}{1+(-z)^2}\right) dz \\
&=& \int_{0}^\infty \frac{z^{1/3} + (e^{i\pi}z)^{1/3}}{1+z^2} dz \\
&=& (1+e^{i\frac{\pi}{3}})\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz \\
\end{eqnarray*}$$
So we have (1)
$$\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz = \frac{1}{1+e^{i\frac{\pi}{3}}} \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz
$$
Now we only have to find $\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz$. Because $1+z^2 = (z-i)(z+i)$, the only singularity in the upper half-plane is $i$ and by the Residue Theorem we immediatly get
$$
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = 2 \pi i \text{Res}_i \text{ with } \text{Res}_i = \lim_{z \to i} (z-i) \frac{e^{log(z)/3}}{(z-i)(z+i)} = \frac{e^{i\frac{\pi}{6}}}{2i}
$$
So we have (2)
$$
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = \pi e^{i\frac{\pi}{6}}
$$
Finally, by (1) and (2) we get
$$
\int_{0}^\infty\frac{z^{1/3}}{1+z^2} dz = \frac{\pi e^{i\frac{\pi}{6}}}{1+e^{i\frac{\pi}{3}}} = \frac{\pi}{\sqrt{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
How does one solve the differential equation $y'=\frac {y} {3x-y} $ ? I need to find a solution to the differential following equation:
$y'=\frac {y} {3x-y} $.
I tried to use use some kind of substitution, but I didn't manage to solve it.
Any suggestion\help?
Thanks a lot!
| $$ y' = \frac{y}{3x - y} = \frac{1}{\frac{3x}{y} - 1} $$
$$ \frac{y}{x} = v \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = x \frac{dv}{dx} + v $$
$$ x \frac{dv}{dx} + v = \frac{1}{\frac{3}{v} - 1} = \frac{v}{3 - v} $$
$$ x \frac{dv}{dx} = \frac{v}{3-v} - v = \frac{v(v-2)}{3-v} $$
$$ \frac{(3-v)dv}{v(v-2)} = \frac{dx}{x} $$
$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = \int \frac{dx}{x} = \ln x + c $$
$$ \frac{(3-v)}{v(v-2)} = \frac{A}{v} + \frac{B}{v-2} $$
$$ \Rightarrow A(v-2) + Bv = 3 - v $$
$$ v = 2 \Rightarrow B = \frac{1}{2} $$
$$ v = 0 \Rightarrow A = -\frac{3}{2} $$
$$ \Rightarrow \int \frac{(3-v)}{v(v-2)} \ dv = -\frac{3}{2} \int \frac{1}{v} \ dv + \frac{1}{2} \int \frac{1}{v-2} \ dv = \frac{1}{2} \ln(v-2) - \frac{3}{2} \ln v = \ln x + c $$
$$ v = \frac{y}{x} \Rightarrow \frac{1}{2} \ln \left(\frac{y}{x}-2 \right) - \frac{3}{2} (\ln y - \ln x) = \ln x + c $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/123764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Exponential Equation with mistaken result I'm on my math book studying exponential equations, and I got stuck on this Problem:
What is sum of the roots of the equation:
$$\frac{16^x + 64}{5} = 4^x + 4$$
I decided to changed: $4^x$ by $m$, so I got: $$\frac{m^2 + 64}{5} = m + 4$$
working on it I've got: $m^2 - 5m + 44 = 0$
but solving this equation the roots were: $x = \frac{5 \pm \sqrt{151i}}{2}$
which isn't even close from the possible answers: 1, 3, 8, 16 or 20.
What's the mistake ? thanks in advance;
| $$
\begin{align}
\frac{16^x+64}{5}&=4^x+4\\
16^x+64&=5\cdot 4^x+20\\
16^x-5\cdot 4^x&=-44\\
y^2-5y+44&=0 \quad \text{for } y=4^x\\
y&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
4^x&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
x&=\log_{4}\left( \frac{1}{2}\left(5\pm i\sqrt{151}\right)\right)\\
\sum(x)&=\log_{4}\left( \frac{1}{2}\left(5+ i\sqrt{151}\right)\right)+\log_{4}\left( \frac{1}{2}\left(5- i\sqrt{151}\right)\right)\\
&=2\Re\left(x\right)\\
&\approx 2\cdot 1.89209=3.78418
\end{align}
$$
This is the logical answer I arrived at. Is there an error in the problem itself?
It seems that this becomes a very complex problem when it is taken to the complex plane. (I have omitted those answers as a result, since I do not understand them.) Perhaps someone more enlightened than myself would elaborate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the limit $\lim\limits _{ x\to \infty } \left( \sqrt { x^2+3x } -\sqrt { x^2+x } \right) $? When I try to do this type of indeterminations I reach to this point:
$\lim\limits_{ x\to \infty } \dfrac { 2x }{ \sqrt { x^ 2 +3x } +\sqrt { x^2 +x } } $
but I don't know how to continue. Thanks.
| $\lim\limits_{ x\to \infty }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x } } } $
$\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3}{x}}+\sqrt{ 1+\frac{1}{x}}}}=\frac {2}{\large{\sqrt{ 1+0}+\sqrt{ 1+0}}}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to expand $\cos nx$ with $\cos x$? Multiple Angle Identities:
How to expand $\cos nx$ with $\cos x$, such as
$$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$
See a list of trigonometric identities in english/ chinese
| \begin{equation}
\text{You can use De Moivre's identity:}
\end{equation}
\begin{equation}
\text{Let's Call:}\\\\
\end{equation}
\begin{equation}
\mathrm{z=\cos x+i \sin x}\\
\mathrm{\frac{1}{z}=\cos x-i \sin x}\\
\end{equation}
\begin{equation}
\text{Now, addind both equations together, we get:}\\
\end{equation}
\begin{equation}
\mathrm{2\cos x=z+\frac{1}{z}}\\
\text{And we know that:}\\
\end{equation}
\begin{equation}
\mathrm{z^n=(cis x)^n=cis~nx}\\
\end{equation}
\begin{equation}
\text{So:}\\
\mathrm{2\cos x=z+\frac{1}{z}}\Rightarrow\\
\mathrm{1024\cos^{10} x=\left (z+\frac{1}{z} \right )^{10}}\\
\end{equation}
\begin{equation}
\text{Expanding the RHS:}\\
\end{equation}
\begin{equation}
\mathrm{1024\cos^{10} x=z^{10}+\frac{1}{z^{10}}+10\left (z^{8}+\frac{1}{z^{8}} \right )+45\left (z^{6}+\frac{1}{z^{6}} \right )+120\left (z^{4}+\frac{1}{z^{4}} \right )+210\left (z^{2}+\frac{1}{z^{2}} \right )+252}\\
\end{equation}
\begin{equation}
\mathrm{1024\cos^{10} x=2\cos 10x+20\cos 8x+90\cos 6x+240\cos 4x+420\cos 2x+252}\\
\end{equation}
\begin{equation}
\boxed{\boxed{\mathrm{\therefore\cos^{10} x=\frac{1}{512}\cos 10x+\frac{5}{256}\cos 8x+\frac{45}{512}\cos 6x+\frac{15}{64}\cos 4x+\frac{105}{256}\cos 2x+\frac{63}{256}}}}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Number of positive solutions? What is the number of positive solutions to
$$ (x^{1000} + 1)(1 + x^2 + x^4 + \cdots + x^{998}) = 1000x^{999}? $$
I tried to solve it.
First I used by using sum of Geometric Progression.
Then the equation becomes too complicated and is in the power of 1998.
How can I get the number of positive solutions with that equation?
Thanks in advance.
| Consider$$(x^{m} + 1)(1 + x^2 + x^4 + \cdots + x^{m-2}) = mx^{m-1}?$$
by multiplying $x^2-1$ in tow sides of equation,we have
$$(x^{m} + 1)(x^m-1)=mx^{m-1}(x^2-1)$$,
Last we have
$$x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1 = 0 \tag{#}$$
Factorization
$$\begin{align*}
&x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1\\
&=(x^{2m}-1)-mx^{m-1}(x^2-1)\\
&=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2})-mx^{m-1}(x^2-1)\\
&=(x^2-1)(1+x^2+x^4+\cdots x^{2m-2}-mx^{m-1})\\
\end{align*}$$
since $1+x^2+x^4+\cdots x^{2m-2}\geq mx^{m-1}$(algebraic average is great than geometry average $\forall x>0$)if and only if $1=x^2= \cdots$ ,then$$1+x^2+x^4+\cdots x^{2m-2}= mx^{m-1}$$
so the root of equation(#)is $1$ and -$1$,but sine we first multiply with$(x^2-1)$,
so maybe there are some extraneous roots in it,substitute $1$ and$-1$ for original equation,we conclude that the root of original equation is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Euclidean algorithm as a linear combination substitution and simplifying This problem is from my discrete mathematics textbook.
I'm trying to find $\gcd(420,66)$
I compute
$$\begin{align*}
420 &= 6 \times 66 + 24\\
66 &= 2\times 24 + 18\\
24 &= 1 \times 18 + 6\\
18 &= 3 \times 6 + 0
\end{align*}$$
then I rewrite the equation
$$\begin{align*}
6 &= 24 - 1 \times 18\\
18 &= 66 - 2 \times 24\\
24 &= 420 - 6 \times 66\\
\end{align*}$$
Now I try to perform substitutions which give me
$$\begin{align*}
6 &= 24 -1 \times 18\\
& = 24-1 (66 - 2 \times 24)\\
&= 3 \times 24 -66
\end{align*}$$
My question is how do you transition from
$$ 24-1 (66 - 2 \times 24)$$
to
$$3 \times 24 -66$$
I just can't wrap my head around this part. Maybe I'm way over thinking this step.
Any help is appreciated thanks!
| Distribute, reorder, associate:
$$\begin{align*}
24 - 1(66-2\times 24) &= 24 -1(66) -1(-2\times 24)\\
&= 24 - 66 +2\times 24\\
&= 24 + 2\times 24 - 66\\
&= (1+2)24 - 66\\
&= 3\times 24 - 66.
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you explain the concept of logarithm to a five year old? Okay, I understand that it cannot be explained to a 5 year old. But how do you explain the logarithm to primary school students?
| A logarithm is how much bigger or smaller a number is than another
number when you combine numbers using multiplication instead
of addition.
For instance, to see how much bigger $3$ is than $2$ in this way:
$$
\begin{array}{|cccccccccccccccc|}
\hline
2^1 & \cdot & 2^2 & \color{red}{2^3} & \cdot & 2^4 & \cdot & 2^5 & 2^6 & \cdot & 2^7 & \cdot & \color{red}{2^8} & 2^9 & \cdot & 2^{10} \\
\hline
2 & \cdot & 4 & \color{red}{8} & \cdot & 16 & \cdot & 32 & 64 & \cdot & 128 & \cdot & \color{red}{256} & 512 & \cdot & 1024 \\
\cdot & 3 & \cdot & \cdot & \color{red}{9} & \cdot & 27 & \cdot & \cdot & 81 & \cdot & \color{red}{243} & \cdot & \cdot & 729 & \cdot \\
\hline
\cdot & 3^1 & \cdot & \cdot & \color{red}{3^2} & \cdot & 3^3 & \cdot & \cdot & 3^4 & \cdot & \color{red}{3^5} & \cdot & \cdot & 3^6 & \cdot \\
\hline
\end{array}
$$
This shows that
$$
1.5 = \frac{\color{red}{3}}{\color{red}{2}} < \log_2{3} < \frac{\color{red}{8}}{\color{red}{5}} = 1.6.
$$
You can measure $\log_2{3}$ as accurately as you like by comparing bigger powers
of $2$ and $3$.
(After this, either you end up with one very confused five-year-old,
or else you've got a budding Eudoxus on your hands.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "94",
"answer_count": 21,
"answer_id": 16
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Help finding integral: $\int \frac{dx}{x\sqrt{1 + x + x^2}}$ Could someone help me with finding this integral
$$\int \frac{dx}{x\sqrt{1 + x + x^2}}$$
or give a hint on how to solve it.
Thanks in advance
| $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{x\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$.$\;\;$ Now let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$, $dx=\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta$
to get $\displaystyle\int\frac{1}{(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2})(\frac{\sqrt{3}}{2}\sec\theta)}\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta=\int\frac{\sec\theta}{\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}}d\theta=\int\frac{1}{\frac{\sqrt{3}}{2}\sin\theta-\frac{1}{2}\cos\theta}d\theta$
$\;\;=\displaystyle\int\frac{1}{-\cos(\theta+\frac{\pi}{3})}d\theta=-\int\sec\big(\theta+\frac{\pi}{3}\big)d\theta=-\ln\left|\sec\big(\theta+\frac{\pi}{3}\big)+\tan\big(\theta+\frac{\pi}{3}\big)\right|+C$
$\;\;=-\displaystyle\ln\left|\frac{-2\sqrt{x^2+x+1}}{\sqrt{3}x}-\frac{x+2}{\sqrt{3}x}\right|+C=-\ln\left|\frac{2\sqrt{x^2+x+1}+x+2}{\sqrt{3}x}\right|+C$
$\;\;\displaystyle=\ln\left|\frac{\sqrt{3}x}{2\sqrt{x^2+x+1}+x+2}\right|+C=\ln|x|-\ln\left(2\sqrt{x^2+x+1}+x+2\right)+C$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}+\sqrt{2}i)^{-2011}
=\cos{(-2011\theta)} + i \sin{(-2011\theta)} = e^{i(-2011)\theta}$$
Is it correct?
The given answer is
...
$$\arg{z} = \frac{3\pi}{4}, \qquad z=2e^{i \frac{3\pi}{4}}$$
$$LHS
= (\frac{1}{z})^{2011}=2^{-2011}e^{-2011(\frac{3\pi}{4})i}
= 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
= 2^{-2011}e^{-\frac {\pi i}{4}}
= 2^{-2011} \color{blue}{\frac{1 - i}{\sqrt{2}}}
= ...
$$
Why is the red $1508\pi i$ removed in the following step?
How do I get the blue $\frac{1 - i}{\sqrt{2}}$ from the prev step?
| Since everyone has been answering about the blue part, I will answer about the red part:
\begin{eqnarray*} 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
&=& 2^{-2011}e^{-(\color{red}{1508\pi i})} \cdot e^{\left( \frac{\pi i }{4}\right)} \\
&=& 2^{-2011}{\left(e^{(\color{red}{2\pi i})}\right)}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \\
&=& 2^{-2011}{(\color{red}{1})}^{-\frac{1508}{2}} \cdot e^{\left( \frac{\pi i }{4}\right)} \hspace{5mm} \text{(Using the fact that $e^{2 \pi i } = 1$)} \\
&=& 2^{-2011} e^{\left( \frac{\pi i }{4}\right)}. \end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Applying difference of cubes to cube roots I am stumped as to why this application of the difference of cubes is valid...
I am rationalizing the denominator. I don't understand the reasoning of why the difference of cubes formula is applicable to cubed roots, removing the root one gets an exponent of $a^{1/3}$ - I know how to simplify this expression, but I am hoping someone can help me along with the logic.
$$\frac{1}{\sqrt[3]{a}-\sqrt[3]{b}}.$$
| You use the the formula
$$(x-y)(x^2+xy+y^2) = x^3-y^3$$
with $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$. You already have one of the factors on the left hand side, so you multiply by the other factor (and cancel it out).
If you have
$$\frac{1}{x-y}$$
then you can transform it into
$$\frac{1}{x-y} = \frac{x^2+xy+y^2}{(x-y)(x^2+xy+y^2)} = \frac{x^2+xy+y^2}{x^3-y^3}.$$
That is, you want to multiply the numerator and denominator by
$$\left( \sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}\right).$$
(Just like to rationalize $$\frac{1}{\sqrt{a}-\sqrt{b}}$$you use the formula
$$(x-y)(x+y)=x^2-y^2$$
with $x=\sqrt{a}$ and $y=\sqrt{b}$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A highschool factoring problem $x+y+z=0$
$x^3+y^3+z^3=9$
$x^5+y^5+z^5=30$
$xy+yz+zx=?$
I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?
| We have the Newton-Girard identities $$x^3+y^3+z^3=(x+y+z)^3+3xyz-3(x+y+z)(xy+xz+yz)$$ and $$\begin{split}x^5+y^5+z^5=&(x+y+z)^5-5(x+y+z)^3 (xy+xz+yz)+\\5(x+y+z)&(xy+xz+yz)^2-5xyz(xy+xz+yz)+5xyz(x+y+z)^2\end{split}$$
Replacing all instances of $x+y+z$ with $0$, we have the simultaneous equations
$$\begin{align*}
3xyz&=9\\
-5xyz(xy+xz+yz)&=30
\end{align*}$$
You should now be able to solve for what you need.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?
| Just in case you may be interested. More generally you can get a similar expression for such sums of powers of inverses. They basically come from the factorization
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \cdots + a + 1)$$
which can be proved by multiplying out the right hand side and then most of the terms cancel out and you'll be left with the expression on the left.
Then from this factorization, by dividing by $a - 1$ you get
$$\frac{a^n - 1}{a - 1} = a^{n-1} + a^{n-2} + \cdots + a + 1$$
Now by dividing by $a^n$ you get
$$
\frac{a^n - 1}{a^{n}(a - 1)} = \frac{a^{n-1}}{a^n} + \frac{a^{n-2}}{a^n} + \cdots + \frac{a}{a^n} + \frac{1}{a^n}
$$
$$
= \frac{1}{a} + \frac{1}{a^2} + \cdots \frac{1}{a^{n-1}} + \frac{1}{a^n}
$$
So that you have
$$
\boxed{
\displaystyle\frac{a^n - 1}{a^{n}(a - 1)} = \frac{1}{a} + \frac{1}{a^2} + \cdots \frac{1}{a^{n-1}} + \frac{1}{a^n}
}
$$
Therefore the expression you want to prove by induction is obtained just by setting $a = 2$, but now you can give different values to $a$ and get similar expressions.
| {
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"url": "https://math.stackexchange.com/questions/141126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Highest power of a prime $p$ dividing $N!$ How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?
This is being done to reduce abstract duplicates. See
Coping with *abstract* duplicate questions. and List of Generalizations of Common Questions for more details.
| Largest power of a prime dividing $N!$
In general, the highest power of a prime $p$ dividing $N!$ is given by
$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.
Number of zeros at the end of $N!$
The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.
To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.
Note that $\alpha_5 < \alpha_2$ whenever $N \geq 2$. (Why?)
The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$
If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.
Note that there will be
1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 \mathrel\| N$
2. A jump of $2$ zeroes going from $(N-1)!$ to $N!$ if $5^2 \mathrel\| N$
3. A jump of $3$ zeroes going from $(N-1)!$ to $N!$ if $5^3 \mathrel\| N$ and in general
4. A jump of $k$ zeroes going from $(N-1)!$ to $N!$ if $5^k \mathrel\| N$
where $a \mathrel\| b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.
Largest power of a prime dividing other related products
In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.
For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots \times (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.
Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p(N!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.
Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p(N!) - s_p(r!) - s_p((N-r)!)$$ where $s_p(N!)$ is defined above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "70",
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Trouble manipulating a log expression This question sort of follows on from question Functions with logarithmic integrals. The book presents an example of integrating a function whose integral is logarithmic:
$$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$
$$= -\frac{1}{3}\ln{A|4 - 3x|}$$
$$= \frac{1}{3}\ln{\frac{A}{|4 - 3x|}}$$
I'm having trouble seeing how the final step is reached. My approach is to separate the logarithm of the product to the addition of separate logs then distribute the minus:
$$-\frac{1}{3}\ln{A|4 - 3x|} = -\frac{1}{3}(\ln{A} + \ln{|4 - 3x|}) = \frac{1}{3}(-\ln{A} - \ln{|4 - 3x|})$$
The I use the property that log minus another log is the log of the first divided by the second to get this:
$$\frac{1}{3}(-\ln{\frac{A}{|4-3x|}})$$
But I still have a minus that the example in the book doesn't have. Could someone help me with this please? Apologies for asking another question so soon.
| From $-\dfrac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) $ we don't
get $\dfrac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }$, because
$$\begin{equation*}
-\frac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) \neq \frac{1}{3}
\ln \frac{A}{\left\vert 4-3x\right\vert }.
\end{equation*}$$
However if we write the constant of integration $C$ as $C=\frac{1}{3}\ln A$, we get the final result, as follows:
$$\begin{eqnarray*}
\int \frac{1}{4-3x}dx &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +C \\
&=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +\frac{1}{3}\ln A \\
&=&\frac{1}{3}\left( -\ln \left\vert 4-3x\right\vert +\ln A\right) \\
&=&\frac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }.
\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an integral I am having trouble with this integral:
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$$
One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.
| You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$$
that becomes
$$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\sqrt{24}}^1\frac{1}{\sqrt{y^2+1}}dy.$$
This gives finally
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\sqrt{48} - \sqrt{26} + {\rm arcsinh}\left(\frac{1}{\sqrt{24}}\right) - {\rm arcsinh}(1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/141897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$. In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that:
If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$?
I'm not familiar with these kind of calculations, so I'd like to see, if my answer is correct:
*
*Let $n=3k+2$ so then
$2^{3k+2}+3\equiv 2^{3k+2}-4 \equiv 4\left(2^{3k}-1\right)\equiv \phantom{4\cdot } 0 \pmod 7$.
*Reading a bit about Modular arithmetics, I found that
$
a_1 \equiv b_1 \pmod N $ and $ a_2 \equiv b_2 \pmod N
$
then
$
a_1 a_2 \equiv b_1 b_2 \pmod n$, if $a_i$ and $b_i$ are integers.
*Since we have $4\equiv 4 \pmod 7$, I conclude that $2^{3k}-1 \equiv 0 \pmod 7$.
*Finally we use that
$
2^{3n}-1=(2^3-1)\cdot \left(1+2^3+2^{2\cdot 3}+2^{3\cdot 3}+\cdots+2^{(k-1)3}\right)
$
and are done.
Are there alternative ways to prove it?
| $$\begin{align*}
2^{3k+2}+3&\equiv 2^{3k+2}-4\\
&\equiv 4\left(2^{3k}-1\right)\\
&\equiv 4\left((2^3)^k-1\right)\\
&\equiv 4\left(1^k-1\right)\\
&\equiv 4\cdot0\\
&\equiv 0 \pmod 7\;,
\end{align*}$$
since $8\equiv1\pmod7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
| If $(n+1)((n+1)^2+8)=(n+1)(n^2+2n+9)$ then if $(n+1)= 0\ mod\ 3$, we're done. If not, than $(n+1)=1\mod 3$ or $(n+1)= 2\mod 3$. So if $n+1=1\mod 3 $ then $n=0\mod 3$ so $3|(n^2+2n+9)$, and if $n+1=2\mod 3$ then $n^2=n=1\mod 3$ now let $n^2=3k+1$ and $n=3l+1$ so we have $(n^2+2n+9)=(3k+1+6l+2+9)=3(k+2l+1+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
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Show $\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh}(t)}{t}dt=\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du$ How would I show (or explain) that
$$\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh} t}{t}dt,$$
$$\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du,$$
and
$$-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv$$
are all equivalent, without evaluating any of the integrals?
| I'd like to give a suggestion for the equality
$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=-\int_{1/3}^{1/2} \frac{\log v}{1-v^2}\text{ d}v$
The idea is to rewrite $\displaystyle \text{arctanh } t=\int_0^t \frac{1}{1-s^2}\text{ d}s$, so that we have
$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=\int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t$
Now we can change the order of integration (note that the region isn't "simple"):
$\displaystyle \int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t=\int_{0}^{1/3} \int_{1/3}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s+\int_{1/3}^{1/2} \int_{s}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s$
It is very easy to explicitly evaluate several of the terms now:
$\begin{align*}
&=\displaystyle \frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)-\log s}{1-s^2}\text{ d}s \\
&=\frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)}{1-s^2}\text{ d}s - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\
&=\frac{1}{2} \log(3/2) \log(2)-\frac{1}{2} \log(3/2) \log(2) - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\
&=- \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s
\end{align*}$
This is precisely the desired term!
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio and number theory The question goes as follows:
Let $K$ be a three digit number such that the ratio of the number to the sum
of its digit is least. What is the difference between the hundreds and
the tens digits of $K$?
Now I was able to do this question by trial and error, assuming hundredth digit place to be 1 and unit as well as tens digit to be 9
So the number is 199, but I am not able to do it logically, any way to do it?
| Doing it in a single function, just for fun:
For $x \in \mathbb{Z}$ we want to minimize the ratio $\frac{100a+10b+c}{a+b+c}$ where $ a,b,c\in\mathbb{Z}$ and $100a+10b+c=x$ We can rewrite the ratio as: $$ f(x)=\frac{100\left\lfloor\frac{x}{100}\right\rfloor+10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor+\left\lfloor x-100\left\lfloor\frac{x}{100}\right\rfloor-10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor\right\rfloor}{\left\lfloor\frac{x}{100}\right\rfloor+\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor+\left\lfloor x-100\left\lfloor\frac{x}{100}\right\rfloor-10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rfloor}{10}\right\rfloor\right\rfloor}$$
Which, when graphed, gives:
which clearly is minimized at 199 (or 1-9 if the $x$ is allowed to have two leading $0$s)
| {
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"timestamp": "2023-03-29T00:00:00",
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Long division in integration by partial fractions I am trying to figure out what my book did, I can't make sense of the example.
"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write
$$\int \frac{x^3 + x}{x -1} dx = \int \left(x^2 + x + 2 + \frac{2}{x-1}\right)dx = \frac{x^3}{3} + \frac{x^2}{2} + 2x + 2\ln|x-1| + C$$
I am mostly concerned with the transformation of the problem by long division I think.
I attempt to do this on my own.
$(x+1)$ and $(x^3 + x)$ inside the long division bracket
I am left with $x^2 - 1$ on top and a leftover -1
This is not in their answer, I do not know how they did that.
| You did not do the long division correctly.
x^2 + x + 2
_________________________
x - 1 | x^3 + x
- x^3 + x^2
-----------
+ x^2 + x
- x^2 + x
--------
2x
-2x + 2
-------
+ 2
So the quotient is $x^2 + x + 2$, and the remainder is $2$.
You can verify this by doing the product and adding the remainder:
$$(x-1)(x^2+x+2) = x^3 + x^2 + 2x - x^2 - x -2 = x^3 + x - 2$$
so
$$(x-1)(x^2+x+2) + 2 = x^3 + x -2 + 2 = x^3 + x.$$
Whereas you claim a quotient of $x^2-1$ and a remainder of $-1$, which would give
$$(x-1)(x^2-1) -1 = x^3 -x - x^2 + 1 -1 = x^3 - x^2 - x \neq x^3 + x.$$
(Even if you tried with $x+1$ instead fo $x-1$, your answer is still incorrect, since
$$(x+1)(x^2-1)-1 = x^3 - x + x^2 -1 -1 = x^3 + x^2 - x -2 \neq x^3+x.$$
If you divide by $x+1$ correctly, you'll get a quotient of $x^2-x+2$ and a remainder of $-2$,
$$(x+1)(x^2-x+2)-2 = x^3 -x^2 +2x +x^2 -x + 2 -2 = x^3 +x$$
which is the correct total.)
| {
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"timestamp": "2023-03-29T00:00:00",
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$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
| As David Mitra commented, one may factor using polynomial division:
$\frac {x^3 - 8} {x-2} = (x^2+2x+4)$
To solve this, we can quickly complete the square as follows:
Notice that we are setting $x^2 + 2x+4 = 0$
$(x+1)^2 + 3 = 0$
$(x+1)^2 = -3$
$x+1 = \pm i \sqrt{3}$
$x = -1 \pm i \sqrt{3}$ and we have our answer. I find this slightly faster than using the quadratic formula.
| {
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Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$
$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$
Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought?
| I agree with the answer given by Gerry Myerson. The only detail left out is, is to complete the square on the terms: $x^2-2ax = (x-a)^2 - a^2$ and the same for: $y^2-2by = (y-b)^2 - b^2$
adding these 2 equations and equal to zero then gives the equation in Cartesian form as shown above by Gerry Myerson. I am just adding in the details for those that may not realize.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x^2 + 2y^2 - 6x + 4y + 7 = 0$, find center, foci, vertex/vertices So the equation is:
$$ x^2 + 2y^2 - 6x + 4y + 7 = 0 $$
Find the coordinates of the center, the foci, and the
vertex or vertices.
What I did was put the equation in the form:
$$
\frac{(x-3)^2}{4}+ \frac{(y+1)^2}{4} = 1
$$
Now based on that, I said the center is at $(3,-1)$, the foci is at ~+- 2.45 (since $c = \sqrt {a^2 + b^2}$ ). so the coordinates of that are $(3+2.45,-1)$ and $(3-2.45,-1)$ and the vertices are $(1,-1)$ and $(5,-1)$. I also went ahead and found the asymptote, which is just done by setting the equation to $0$, correct?
| The equation should be $$\frac{(x-3)^2}{4}+\frac{(y+1)^2}{2}=1.$$ You've correctly identified the center and vertices. The focal length should be $\sqrt{a^2-b^2}$, not $\sqrt{a^2+b^2}$. Ellipses don't have asymptotes, you're thinking of hyperbolae.
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial fractions for geometric probability-generating function wrong Let $X\sim \text{Geo}(1/4), Y\sim \text{Geo}(1/2)$ be given. First I have to compute $\mathbb{E}[z^{X+Y}]$:
$$\mathbb{E}[z^{X+Y}]=\mathbb{E}[z^{X}]\cdot\mathbb{E}[z^{Y}]=\frac{\frac{1}{4}z}{1-\left(1-\frac{1}{4}\right)z}\cdot\frac{\frac{1}{2}z}{1-\left(1-\frac{1}{2}\right)z}=\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}.$$
The next task is to find $\alpha,\beta,\gamma\in\mathbb{R}$ such that
$$\mathbb{E}[z^{X+Y}]=\alpha+\frac{\beta}{1-\frac{1}{2}z}+\frac{\gamma}{1-\frac{3}{4}z}.$$
Here is my procedure (which seems to be wrong...):
$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
$$\frac{10z-8}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=2\cdot\frac{5z-4}{(z-2)(3z-4)}=\frac{A}{z-2}+\frac{B}{3z-4}$$
$$\frac{A(3z-4)+B(z-2)}{(z-2)(3z-4)}=\frac{(3A+B)z+(-4A-2B)}{(z-2)(3z-4)}$$
$$\left.\begin{cases}
3A+B&=5 \\ -4A-2B&=-4
\end{cases}\right\}\implies A=3,\;B=-4\implies 2A=6,\;2B=-8$$
$$\rightsquigarrow \mathbb{E}[z^{X+Y}]=1+\frac{6}{z-2}-\frac{8}{3z-4}=1-\frac{3}{1-\frac{1}{2}z}+\frac{2}{1-\frac{3}{4}z}\implies\alpha=1,\;\beta=-3,\;\gamma=2$$
However it seems that the values for $\alpha,\beta,\gamma$ should each be just a third of my result! Can anyone explain me, what I did wrong?
| Due to anons hint (thanks!) I found the mistake.
$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{\frac{10}{3}z-\frac{8}{3}}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
This leads to $\alpha=1/3,\;\beta=-1,\;\gamma=2/3$ which is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$ So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$
I divided by cos(x) and I got
$$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$
But I'm stuck here.
I tried to substitute $t=\cos(x)$
$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$
Any help would be greatly appreciated.
| $$
\begin{aligned}
& \int \frac{\sin x}{3 \cos ^{3} x+\sin ^{2} x \cos x} d x \\
=& \int \frac{\sin x d x}{\cos x\left(3 \cos ^{2} x+\sin ^{2} x\right)} \\
=& \int \frac{d(\cos x)}{\cos x\left(2 \cos ^{2} x+1\right)} \\
=& \int\left(\frac{1}{\cos x}-\frac{2 \cos x}{2 \cos ^{2} x+1}\right) d(\cos x) \\
=& \ln (\cos x)-\frac{1}{2} \int \frac{d\left(2\cos ^{2} x+1\right)}{2 \cos ^{2} x+1} \\
=& \ln |\cos x|-\frac{1}{2} \ln \left(2 \cos ^{2} x+1\right)+C
\end{aligned}
$$
| {
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What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recursive method.)
| I thought I might post a little more elaborate version of Henning's hint (see his comment).
$$\begin{align}
1&=2^0\\
10&=2^1\\
100&=2^2\\
1000&=2^3\\
\vdots&=\vdots\\
10\dots0&=2^n\\
\hline
11\dots1&=2^0+2^1+\dots+2^n\\
1&=1\\
\hline
100\dots0&=2^0+2^1+\dots+2^n+1=2^{n+1}
\end{align}$$
Hence $2^0+2^1+\dots+2^n=2^{n+1}-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
| The best answer is Byron's, but it's possible to show by induction that the characteristic polynomial of $M$ is $\big(\lambda - \frac{n+1}2\!\big)\big(\lambda - \frac 1 2\!\big)^{n-1}$, which means that the $M$ is (orthogonally) conjugate to a diagonal matrix with these (positive) eigenvalues along the diagonal.
| {
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"timestamp": "2023-03-29T00:00:00",
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A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
| Essentially, you want to prove that (multiplying throughout by $4$) $$c^2 -2cn -4c +n^2 + 8n \leq n^2 -2n +5$$
i.e. $$c^2 - 2cn -4c + 10n \leq 5$$
Since $n \geq c \geq 5$, we have that $(10-2c)n \leq (10-2c)c$ (since $10-2c \leq 0$). Hence, $$c^2 -4c + (10-2c)n \leq c^2 - 4c +(10-2c)c = -c^2 + 6c = c(6-c) = 9 - (c-3)^2$$ Note that the function is a decreasing function for $c \geq 3$. Hence, it is a decreasing function for $c \geq 5$ as well. Hence, the maximum occurs at $c=5$, which gives us that $c(6-c) \leq 5 (6-5) = 5$, which is what we wanted.
Putting these together, we get that $$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1$$
| {
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| Directly by Sophie Germain's Identity or:
$$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$$
After splitting the initial fraction we get:
$$ \int \frac{1}{x^4 +1} \ dx = \int \frac{\frac{x}{2\sqrt2}+\frac{1}{2}}{x^2+\sqrt2x+1} \ dx+\int \frac{\frac{-x}{2\sqrt2}+\frac{1}{2}}{x^2-\sqrt2x+1} \ dx=$$
$$ \frac{\sqrt2}{8} \int \frac{2x+2\sqrt2}{x^2+\sqrt2x+1} dx-\frac{\sqrt2}{8} \int \frac{2x-2\sqrt2}{x^2-\sqrt2x+1} dx=$$
$$\frac{\sqrt2}{8}\int \frac{2x+\sqrt2}{x^2+\sqrt2x+1} dx+\frac{1}{4} \int \frac{1}{x^2+\sqrt2x+1} dx-\frac{\sqrt2}{8}\int \frac{2x-\sqrt2}{x^2-\sqrt2x+1} dx+\frac{1}{4} \int \frac{1}{x^2-\sqrt2x+1} dx=$$
$$\frac{\sqrt2}{8}\left( \int \frac{2x+\sqrt2}{x^2+\sqrt2x+1} dx -\int \frac{2x-\sqrt2}{x^2-\sqrt2x+1} dx \right)+$$
$$\frac{\sqrt2}{4} \left( \int \frac{\sqrt2}{(\sqrt2x+1)^2+1} dx+\int \frac{\sqrt2}{(\sqrt2x-1)^2+1} dx \right)=$$
$$\frac{\sqrt2}{8} \left(\ln(x^2+\sqrt2x+1)-\ln(x^2-\sqrt2x+1) \right) +\frac{\sqrt2}{4} \left(\arctan(\sqrt2x+1)+ \arctan(\sqrt2x-1)\right)+C$$
$$=\frac{\sqrt2}{8} \ln\frac{(x^2+x\sqrt2+1)}{(x^2-x\sqrt2+1)}+\frac{\sqrt2}{4}\arctan\frac{x\sqrt2}{1-x^2}+C.$$
Q.E.D.
| {
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Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
| $$
\left.\frac{(x+2)^2}{5}\right|_0^1 = \frac{(1+2)^2}{5} - \frac{(0+2)^2}{5} = \frac{9}{5} - \frac{4}{5} = \frac{5}{5} = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$ It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$$
Thanks.
| The case of cube roots is probably more interesting than square roots; namely, simplifying
$$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{12}+\sqrt[3]{16}}+\frac{1}{\sqrt[3]{16}+\sqrt[3]{20}+\sqrt[3]{25}}. \tag{$\circ$}$$
To evaluate this, as sos440 did in the comments, one notes the structure of the denominators are apparent as $a^2+ab+b^2$, which appears in the factorization in a difference of cubes, $b^3-a^3$. More generally, this sort of homogeneous polynomial results from using the geometric series formula on the common ratio $b/a$, but I digress. It is clear that we have
$$\frac{1}{a^2+ab+b^2}=\frac{b~-~a}{b^3-a^3} \tag{$\bullet$}$$
as the individual terms in $(\circ)$, with $b=\sqrt[3]{k+1}$ and $a=\sqrt[3]{k}$, for $k=2,3,4$. The denominators will all simply be $1$, and the three terms (the numerators) subsequently telescope:
$$\big(\sqrt[3]3-\sqrt[3]2\big)+\big(\sqrt[3]4-\sqrt[3]3\big)+\big(\sqrt[3]5-\sqrt[3]4\big)=\sqrt[3]5-\sqrt[3]2. \tag{$\square$}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.
| According to Maple, your integral is $\dfrac{\pi^2}{8} - \dfrac{10}{9}$, so your inequality becomes $\pi < \sqrt{89}/{3}$. In fact, an antiderivative is
$$F(x) = \dfrac{x^3 \ln(x)}{3} - \dfrac{x^3}{9} + x \ln(x) - x - \dfrac{\ln(x) \ln(x+1)}{2} - \dfrac{\text{dilog}(x)+\text{dilog}(x+1)}{2}$$
More generally, for $p > -1$ $$\int_0^1 \dfrac{x^p \ln(x)}{x^2-1}\ dx = \dfrac{\Psi(1,(p+1)/2)}{4}$$
where for even integers $p=2n$, $$\Psi(1,n+1/2) = \sum_{k=n+1}^\infty \dfrac{4}{(2k-1)^2}$$
while for odd integers $p=2n-1$,
$$\Psi(1,n) = \sum_{k=n}^{\infty} \dfrac{1}{k^2}$$
| {
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Is this proof that $\sqrt 2$ is irrational correct? Suppose $\sqrt 2$ were rational. Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.
Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.
Squaring both sides,
$2 = \frac 21 = \frac{a^2}{b^2}$.
Lowest terms representations of rational numbers are unique, so we have $a^2 = 2$ and $b^2=1$.
But there is no such integer $a$, and therefore we have a contradiction and $\sqrt 2$ is irrational.
I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound.
| Hhmmm ... Here is a short one.
We all know that $\sqrt{2}$ is a root of $x^2-2 = 0$
Apply Rational Root Theorem and you're done
It goes something like this : if $\sqrt{2} = \dfrac{p}{q}$, then $q=1$ and $p$ is a factor of $2$. Hence the only possible rational root of $x^2-2=0$ would be $-2,-1,1,-2$.
A quick check shows that this is not the case. Therefore $\sqrt{2}$ is irrational.
A bit further now : using the same Theorem, shiw that $\sqrt{n^2+1}$ is irrational.
| {
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Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot get this result. Below is my solution of the problem. I've checked it many times, must be something very obvious that I'm missing:
(original image)
Euler Substituion
$\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}$
Let $\sqrt{x^2+x+1}=t-x$.
$x^2+x+1=t^2-2xt+x^2$
$x(1+2t)=t^2-1\implies x=\dfrac{t^2-1}{1+2t}$
$dx=\left(\dfrac{t^2-1}{1+2t}\right)'dt=\dfrac{2t(1+2t)-(t^2-1)2}{(1+2t)^2}=\dfrac{2t+4t^2-2t^2+2}{(1+2t)^2}=\dfrac{2(t^2+t+1)}{(1+2t)^2}$
$\sqrt{x^2+x+1}=t-x=t-\dfrac{t^2-1}{1+2t}=\dfrac{t^2+t+1}{1+2t}$
$\implies\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}=2\int\frac{\frac{t^2+t+1}{(1+2t)^2}\;dt}{\frac{t^2-1}{1+2t}\cdot\frac{t^2+t+1}{1+2t}}=2\int \frac{1}{t^2-1}\,dt$
$\dfrac{1}{t^2-1}=\dfrac{1}{(t+1)(t-1)}=\dfrac{A}{t+1}+\dfrac{B}{t-1}\implies \begin{eqnarray}&&At-A+Bt+B=1\\&&A+B=0\implies A=-B\\ &&B-A=1\implies B=\frac{1}{2},A=-\frac{1}{2}\end{eqnarray}$
$\implies \displaystyle 2\int \frac{1}{2}\frac{1}{2t-1}\,dt-2\int\frac{1}{2}\frac{1}{t+1}\,dt=\int\frac{1}{t-1}\,dt-\int\frac{1}{t+1}\,dt=$
$=\ln|t-1|-\ln|t+1|=\ln\left|\dfrac{t-1}{t+1}\right|$
$t-x=\sqrt{x^2+x+1}\implies t=\sqrt{x^2+x+1}+x$
$\implies \ln\left|\dfrac{t-1}{t+1}\right|=\ln\left|\dfrac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|$
I'll appreciate any help.
Thanks in advance!
| You may use substitution x=1/t, or Euler substitution \sqrt{x^2+x+1}=tx-1
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving a complex integral I need help solving an integral from John Conway book.
Lets $\alpha$ complex number different from 1 find integral $$\int\frac{dx}{1-2\alpha\cos{x}+{\alpha}^2}$$ from 0 to $2\pi$ in unit circle $$(z-\alpha)^{-1}(z-\frac{1}{\alpha})^{-1}$$
| Substitute: $z = e^{i x}$ then:
$$\cos x = \frac{1}{2} \left( z + \frac{1}{z} \right), \; dx = \frac{1}{i} \cdot \frac{dz}{z}$$
and we can rewrite the integral as:
$$ i \int_{|z|=1} \frac{dz}{a (z-a)(z - \frac{1}{a})}$$
There are two cases:
*
*$|a| < 1$ then only $z=a$ is a pole inside a circle and the residue is:
$${\rm res}_{z=a} \frac{1}{a (z-a)(z - \frac{1}{a})} = \lim_{z \to a} \frac{z-a}{a (z-a)(z - \frac{1}{a})} = \frac{1}{a^2 - 1}$$
hence the result is:
$$2 \pi i \cdot \frac{i}{a^2 - 1} = \frac{2 \pi}{1 - a^2}$$
*$|a| > 1$ similarly only $1/a$ lies inside the circle and the residue is:
$${\rm res}_{z=1/a} \frac{1}{a (z-a)(z - \frac{1}{a})} = \lim_{z \to 1/a} \frac{z-\frac{1}{a}}{a (z-a)(z - \frac{1}{a})} = \frac{1}{1 - a^2}$$
hence the result is:
$$2 \pi i \cdot \frac{i}{1 - a^2} = \frac{2 \pi}{a^2 - 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following? ${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ?
A. ${21 \choose 5}$
B. ${20 \choose 5}-{11 \choose 4}$
C. ${21 \choose 5}-{10 \choose 5}$
D. ${20 \choose 4}$
Please give me a hint. I'm unable to group the terms.
By brute force, I'm getting ${21 \choose 5}-{10 \choose 5}$
| The same thing, combinatorially. We want to choose $5$ positive integers from the first $21$. This can be done in $\binom{21}{5}$ ways.
We count the same thing in a different way. If the biggest chosen number is $21$, the rest can be chosen in $\binom{20}{4}$ ways. If the biggest is $20$, the rest can be chosen in $\binom{19}{4}$ ways. If the biggest is $19$, the rest can be chosen in $\binom{18}{4}$ ways. And so on, until if the biggest is $5$, the rest can be chosen in $\binom{4}{4}$ ways.
We conclude that
$$\binom{20}{4}+\binom{19}{4}+\binom{18}{4}+\cdots+\binom{10}{4}+\binom{9}{4}+\cdots +\binom{4}{4}=\binom{21}{5}. \tag{$1$}$$
The same reasoning shows that
$$\binom{9}{4}+\binom{8}{4}+\cdots+\binom{4}{4}=\binom{10}{5}. \tag{$2$}$$
Now subtract $(2)$ from $(1)$.
| {
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"url": "https://math.stackexchange.com/questions/165044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Power (Laurent) Series of $\coth(x)$ I need some help to prove that the power series of $\coth x$ is:
$$\frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + O(x^5) \ \ \ \ \ $$
I don't know how to derive this, should I divide the expansion of $\cosh(x)$ by the expansion of $\sinh(x)$? (I've tried but without good results :( )
Or I have to use residue calculus?
Anyone can suggest me a link where I can find a detailed explanation of this expansion?
Thanks.
| $$ \begin{eqnarray}
\coth(x) &=& \frac{\cosh(x)}{\sinh(x)} = \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{x + \frac{x^3}{6} + \frac{x^5}{120} + \mathcal{o}\left(x^5\right)} = \frac{1}{x} \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{1 + \frac{x^2}{6} + \frac{x^4}{120} + \mathcal{o}\left(x^4\right)} \\ &=&
\frac{1}{x} \left( 1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right) \right) \left( 1 - \frac{x^2}{6} + \frac{7 x^4}{360} + \mathcal{o}\left(x^4\right) \right) \\ &=&
\frac{1}{x} \left( 1 + \frac{x^2}{3} - \frac{x^4}{45} + \mathcal{o}\left(x^4\right) \right) = \frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + \mathcal{o}\left(x^3\right)
\end{eqnarray}
$$
where the reciprocation and multiplication of series used:
$$
\frac{1}{1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right)} = 1 -a x^2 + \left( a^2-b \right) x^4 + \mathcal{o}\left(x^4\right)
$$
$$
\left( 1 + a x^2 + b x^4 + \mathcal{o}\left(x^4\right) \right) \left( 1 + c x^2 + d x^4 \mathcal{o}\left(x^4\right) \right) = 1 + \left(a+c\right) x^2 + \left(b + d + a c\right) x^4 + \mathcal{o}\left(x^4\right)
$$
The result for the reciprocation is obtained using the geometric series:
$$
\frac{1}{1-w} = 1 + w + w^2 + \mathcal{o}(w^2)
$$
Now substitute in the above $w = a x^2 + b x^4 + \mathcal{o}(x^4)$, and use $$w^2 = \left( a x^2 + b x^4 + \mathcal{o}(x^4) \right)^2 = a^2 x^4 + \mathcal{o}(x^4)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a good way to solve for the inverse of $(u^2-u+4)$? I'm having trouble calculating the inverse of a polynomial. Consider the polynomial $f(x)=x^3+3x-2$, which is irreducible over $\mathbb{Q}$, as it has no rational roots. So $\mathbb{Q}[x]/(f(x))\simeq \mathbb{Q}[u]$ is a field.
How would I calculate $(u^2-u+4)^{-1}$ as a polynomial? Calling this $p(u)$, I tried solving for it by writing
$$
p(x)(x^2-x+4)=q(x)f(x)+1
$$
but I don't know what $q(x)$ is. How can I write down $(u^2-u+4)^{-1}$ explicitly? Thanks.
| Let's calculate the $\gcd$ of the two polynomials $x^3+3x-2$ and $x^2-x+4$ in $\mathbb{Q}[x]$. Using the Euclidean algorithm,
$$x^3+3x-2=x(x^2-x+4)+(x^2-x-2)$$
$$x^2-x+4=(x^2-x-2)+6$$
Because $2$ is a unit in $\mathbb{Q}[x]$, we see that the polynomials are relatively prime. Now, building back up, we see that
$$6=(x^2-x+4)-(x^2-x-2)$$
$$6=(x^2-x+4)-[(x^3+3x-2)-x(x^2-x+4)]$$
$$6=(-1)(x^3+3x-2)+(1+x)(x^2-x+4)$$
$$1=\left(-\tfrac{1}{6}\right)(x^3+3x-2)+\left(\tfrac{1+x}{6}\right)(x^2-x+4)$$
Plugging in $u$, which is a root of $x^3+3x-2$, we see that
$$1=\left(\tfrac{1+u}{6}\right)(u^2-u+4)$$
so that $\tfrac{1+u}{6}$ is the multiplicative inverse of $u^2-u+4$ in $\mathbb{Q}(u)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to prove $p$ divides $a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots + b^{p - 2}$ when $p$ is prime, $a, b \in \mathbb{Z}$ and $a,b \lt p$? If $p$ is a prime number and $a, b \in \mathbb{Z}$ such that $a,b \lt p$, then how could we prove that $p$ divides
$\left(a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots + b^{p - 2}\right)$?
| We have
$$(a-b)(a^{p-2} + a^{p-3}b + \ldots + ab^{p-3} + b^{p-2}) = a^{p-1} - b^{p-1}.$$
If $p\nmid a$ and $p\nmid b$ then the right side is congruent to $1-1 = 0$ by Fermat's little theorem, i.e $p$ divides the product on the left. If furthermore $a \not\equiv b \pmod p$ then $p$ must already divide $a^{p-2} + a^{p-3}b + \ldots + ab^{p-3} + b^{p-2}$.
The statement becomes false if you do not assume $a \not\equiv b \pmod p$, e.g. for $a=b=1$ the sum equals $p-1$ which is not divisible by $p$.
Also, if $p| a$ (w.l.o.g.) the sum is congruent to $b^{p-2}$ which is only divisible by $p$ if $p|b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.
| Completion of the square yields the following.
$$4x^2 - 6x + 9 = 4(x^2 - 3/2 x) + 9 = 4(x^2-3/2x + 9/16) + 27/4 = 4(x-3/4)^2 + 27/4.$$
This definitively shows your residual quadratic can have no real roots, since its graph never goes below the line $y = 27/4.$ This representation will allow you to find the complex ones easily, if you so wish.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$
| Hint 1: For all $x$, $\sin^2x+\cos^2x=1$
Hint 2: For $at^2+bt+c=0$ the solutions are $t_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Finding remainders of large number This question was asked in a competitive exam.
Find the remainder of dividing $6^{83}+ 8^{83}$ by $49$
Are there any theorems/rules to compute the answer ?
| Using Euler's Totient theorem,
As $(6,49)=1$ and $\phi(49)=42$
$6^{42}\equiv 1\pmod {49}\implies 6^{42.k}\equiv1\pmod {49}$ where is any integer.
$\implies 6^{84}\equiv 1\pmod{49}\implies 6^{83}\equiv 6^{-1}\pmod{49}$
By observation, $6\times 8=48\equiv -1\pmod{49}\implies 6\times (49-8)\equiv 1\pmod{49}\implies 41\equiv 6^{-1}\pmod{49}$
Similarly, $8^{84}\equiv 1\pmod{49}\implies 8^{83}\equiv 8^{-1}\pmod{49}$ as $(8,49)=1$
As, $6\times 8=48\equiv -1\pmod{49}$, so $(49-6)\times 8\equiv 1\pmod {49}\implies 43\equiv 8^{-1}\pmod{49}$
So, the remainder be $43+41=84\equiv 35\pmod{49}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor oscillates between -1, 0 and 1.
Here is my attempt to prove that it is divergent:
Suppose that the sequence is convergent. This should lead to a contradiction. If it is convergent, then, for every $\epsilon>0$, there is an $N>0$ such that if $n>N$, then $\left|\dfrac{n}{n+1}\sin\dfrac{n\pi}{2} -L\right|<\epsilon$.
In particular, for $\epsilon = \frac{1}{2}$:
$$-\frac{1}{2}<\frac{n}{n+1}\sin\frac{n\pi}{2} -L<\frac{1}{2} \text{ for every integer }n>N$$
Notice that the sequence of values of $\sin\dfrac{n\pi}{2}$ is $1, 0, -1, 0, 1, 0, -1, \cdots$ for $n=1, 2, 3, 4, 5, 6, 7, \cdots$. If $n = 1, 5, 9 \cdots$, $\sin\frac{n\pi}{2}$ is $1$, so, if we choose a particular $n$ from this list of values, the value of $a_n$ is $\dfrac{n}{n+1}$. In this case, $\sin\dfrac{(n+2)\pi}{2}$ will be $-1$, therefore $a_{n+2} = -\dfrac{n+2}{n+3}$. So, we have that:
$$-\dfrac{1}{2}<\dfrac{n}{n+1}-L<\dfrac{1}{2} \text{ and } -\dfrac{1}{2}<-\dfrac{n+2}{n+3}-L<\dfrac{1}{2}$$
Rearranging the terms:
$$\dfrac{1}{2}>-\dfrac{n}{n+1}+L>-\dfrac{1}{2} \text{ and } \dfrac{1}{2}>\dfrac{n+2}{n+3}+L>-\dfrac{1}{2}$$
$$\dfrac{n}{n+1}+\dfrac{1}{2}>L>\dfrac{n}{n+1}-\dfrac{1}{2} \text{ and } \dfrac{1}{2}-\dfrac{n+2}{n+3}>L>-\dfrac{n+2}{n+3}-\dfrac{1}{2}$$
The first inequality says that $L$ is positive, but the second inequality says that $L$ is negative; if this is correct so far, then I found a contradiction, proving that the sequence is divergent.
My question is: is this correct or is there some inconsistency?
| Here is another approach: If the sequence converges, then the difference between two successive terms $|a_{n+1}-a_n|$ must go to zero. So, if you can show that there are pairs of successive terms that do not go to zero, then the sequence diverges.
My suggestion is to consider pairs $a_{1+4k}, a_{2+4k}$, where $k$ is a positive integer. What is $|a_{1+4k} - a_{2+4k}|$, and does it converge to zero?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Computing a sum of binomial coefficients: $\sum_{i=0}^m \binom{N-i}{m-i}$ Does anyone know a better expression than the current one for this sum?
$$
\sum_{i=0}^m \binom{N-i}{m-i}, \quad 0 \le m \le N.
$$
It would help me compute a lot of things and make equations a lot cleaner in the context where it appears (as some asymptotic expression of the coefficients of a polynomial defined over cyclic graphs). Perhaps the context doesn't help much though.
For instance, if $N = m$, the sum is $N+1$, and for $N = m+1$, this sum is $N + (N-1) + \dots + 1 = N(N+1)/2$. But otherwise I don't know how to compute it.
| First note that $\dbinom{n-i}{m-i} = \dbinom{n-i}{n-m}$. Hence, you have
$$\dbinom{n-m}{n-m}+\dbinom{n-m+1}{n-m}+\cdots+\dbinom{n-1}{n-m}+\dbinom{n}{n-m}$$
Let $n-m = k$. Hence, we want to find an expression for
$$\dbinom{k}k+\dbinom{k+1}{k}+\cdots+\dbinom{n-1}{k}+\dbinom{n}{k}$$
From the Pascal's triangle, we have
$\hskip2.5in$
\begin{align}
\color{red}{\dbinom{n+1}{k+1}} & = \color{blue}{\dbinom{n}{k}} + \color{red}{\dbinom{n}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{red}{\dbinom{n-1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \color{red}{\dbinom{n-2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{red}{\dbinom{k+2}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{red}{\dbinom{k+1}{k+1}}\\
& = \color{blue}{\dbinom{n}{k}} + \color{blue}{\dbinom{n-1}{k}} + \color{blue}{\dbinom{n-2}{k}} + \cdots \color{blue}{\dbinom{k+2}{k}} + \color{blue}{\dbinom{k+1}{k}} + \color{blue}{\dbinom{k}{k}}
\end{align}
Hence, your result is $\dbinom{n+1}{n-m+1} = \dbinom{n+1}m$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find coefficient of $x^ny^m$ how can I find the coefficient of $x^ny^m$ of following series
$$\frac{\log\big(\frac{1}{1-xy}\big)}{(1-x)(1-y)(1-xy)}$$
and
$$ \frac{\log(\frac{1}{1-x})}{(1-x)(1-y)(1-xy)}$$
where $1 \leq m \leq n$
| Use $\log\left(\frac{1}{1-x y}\right) = -\log(1- xy) = - \sum_{n=1}^\infty \frac{1}{n} x^n y^n$, and
$$
-\frac{\log(1-x y)}{1-x y} = \sum_{n=1}^\infty x^n y^n \sum_{m=1}^n \frac{1}{n} = \sum_{n=1}^\infty x^n y^n H_n
$$
Then use
$$
\frac{ \sum_{n \geqslant 1, m\geqslant 1} c_{n,m} x^n y^m}{(1-x)(1-y)} =
\sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^n \sum_{\ell=1}^m c_{k,\ell}
$$
In our case $c_{k,\ell} = H_k \delta_{k,\ell}$, and $\sum_{k=1}^n \sum_{\ell=1}^m H_k \delta_{k,\ell} = \sum_{k=1}^{\min(n,m)} H_k$
$$
-\frac{\log(1-x y)}{1-x y} \cdot \frac{1}{(1-x)(1-y)} = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \sum_{k=1}^{\min(n,m)} H_k = \sum_{n=1}^\infty \sum_{m=1}^\infty x^n y^m \left( \left(\min(n,m) +1\right)\left( H_{\min(n,m)+1}-1\right) \right)
$$
For the second series, we start with
$$
[x]^n \frac{-\log(1-x)}{1-x} = H_n \quad [x]^m \frac{1}{1-x y} = y^m
$$
Therefore
$$
[x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} = \sum_{k=0}^{n-1} H_{n-k} y^k
$$
Then
$$
[y]^m [x]^n \frac{-\log(1-x)}{1-x} \cdot \frac{1}{1-x y} \cdot \frac{1}{1-y} = \sum_{k=0}^{n-1}H_{n-k} \mathbf{1}\left( k \leqslant m \right) = \sum_{k=0}^{\min(m, n-1)}H_{n-k}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$ So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither.
| Yet another approach :
Let us first analyze the sum till infinity. Let $$ S= \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ $$\Rightarrow S=(\frac{1}{2^2}+\frac{1}{4^2}+ \cdots\infty) +(\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty ) $$$$\Rightarrow S= \frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\infty)+ S' $$ Where $$S'=\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty $$ $$\Rightarrow S=\frac{1}{4}(1+S)+S'$$ $$\Rightarrow 3S=4S'+1........ Eqn(1)$$ Now examine the following inequality $$ (\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{4^2}-\frac{1}{5^2})+ \cdots \infty > 0$$ $$\Rightarrow \frac{1}{2^2}+\frac{1}{4^2}+\cdots > \frac{1}{3^2}+\frac{1}{5^2}+ \cdots$$ $$\Rightarrow \frac{1}{4}(1+\frac{1}{2^2}+\cdots) >\frac{1}{3^2}+\frac{1}{5^2}+ \cdots $$ $$\Rightarrow \frac{1}{4}(1+S)> S'$$ $$\Rightarrow (1+S)> 4S'......Eqn(2)$$ From Equation 1 and 2 we get $$ 1+S> 3S-1$$ $$\Rightarrow 1> S$$ Which shows $$ 1> \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$
| {
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"url": "https://math.stackexchange.com/questions/174828",
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Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$
I am trying to find all solutions to
(1) $y^3 = x^2 + x + 1$, where $x,y$ are integers $> 1$
I have attempted to do this using...I think they are called 'quadratic integers'. It would be great if someone could verify the steps and suggest simplifications to this approach. I am also wondering whether my use of Mathematica invalidates this approach.
My exploration is based on a proof I read that $x^3 + y^3 = z^3$ has no non-trivial integer solutions. This proof uses the ring Z[W] where $W = \frac{(-1 + \sqrt{-3})}{2}$. I don't understand most of this proof, or what a ring is, but I get the general idea. The questions I have about my attempted approach are
*
*Is it valid?
*How could it be simplified?
Solution:
Let $w = (-1 + \sqrt{-3})/2$. (Somehow, this can be considered an "integer" even though it doesn't look anything like one!)
Now $x^3 - 1 = (x-1)(x-w)(x-w^2)$ so that, $(x^3 - 1)/(x-1) = x^2 + x + 1 = (x-w)(x-w^2)$. Hence
$y^3 = x^2 + x + 1 = (x-w)(x-w^2).$
Since $x-w, x-w^2$ are coprime up to units (so I have read) both are "cubes". Letting $u$ be one of the 6 units in Z[w], we can say
$x-w = u(a+bw)^3 = u(c + dw)$ where
$c = a^3 + b^3 - 3ab^2, d = 3ab(a-b)$
Unfortunately, the wretched units complicate matters. There are 6 units hence 6 cases, as follows:
1) $1(c+dw) = c + dw$
2) $-1(c+dw) = -c + -dw$
3) $w(c+dw) = -d + (c-d)w$
4) $-w(c+dw) = d + (d-c)w$
5) $-w^2(c+dw) = c-d + cw$
6) $w^2(c+dw) = d-c + -cw$
Fortunately, the first two cases can be eliminated. For example, if $u = 1$ then $x-w = c+dw$ so that $d = -1 = 3ab(a-b).$ But this is not possible for integers $a,b$. The same reasoning applies to $u = -1$.
For the rest I rely on a program called Mathematica, which perhaps invalidates my reasoning, as you will see.
We attack case 5. Here
$x = c-d = a^3 + b^3 - 3a^2b$, and $c = a^3 + b^3 - 3ab^2 = -1.$
According to Mathematica the only integer solutions to $c = -1$ are
$(a,b) = (3,2), (1,1), (0,-1), (-1,0), (-1,-3), (-2,1).$
Plugging these into $x = c-d$ we find that no value of x that is greater than 1. So case 5 is eliminated, as is 6 by similar reasoning.
Examining case 4 we see that $d-c = -(a^3 + b^3 - 3a^2*b) = -1$ with solutions
$(-2,-3), (-1,-1), (-1,2), (0,1), (1,0), (3,1).$
Plugging these values into $x = d = 3ab(a-b)$ yields only one significant value, namely $x = 18$ (e.g. (a,b)=(3,1) . The same result is given by case 4. Hence the only solution to (1) is $7^3 = 18^2 + 18 + 1$
However, I'm unsure this approach is valid because I don't know how Mathematica found solutions to expressions such as $a^3 + b^3 - 3ab^2=-1$. These seem more difficult than the original question of $y^3 = x^2 + x + 1$, although I note that Mathematica could not solve the latter.
| I have not verified your work. This is just to say that $y^3=x^2+x+1$ is the equation of an elliptic curve (with $j$-invariant equal to $0$). Using Sage or Magma, you can find that this elliptic curve has rank $1$ and trivial torsion subgroup. The group of rational points is generated by the point $P=(0,1)$, and the rest of the rational points are all the multiples of $P$:
\begin{align*} & \vdots \\
-3P &=\left(-\frac{703}{216},\frac{73}{36}\right)\\
-2P &=\left(18 , 7 \right)\\
-P &=\left(-1,1\right)\\
0 & = \infty\\
P &=\left(0,1\right)\\
2P &=\left(-19 , 7\right)\\
3P &=\left(\frac{487}{216},\frac{73}{36}\right)\\
&\vdots
\end{align*}
Now one can use the rest of the theory of elliptic curves to show that the only integral points are $(18,7)$, $(-1,1)$, $(0,1)$, and $(-19,7)$. So the only solution $(x,y)$ with $x,y>1$ is $(18,7)$ as you found.
| {
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"source": "stackexchange",
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Power series expansion I recently had a problem. I know how to evaluate power series but I cannot seem to find an expansion for $\sqrt{x+1}$.
I've tried differentiating it, in order to bring it in reciprocal form but that didn't help. Due to the presence of square root, I cannot change it in the form of $1/(x+1)$.
Kindly help.
| Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:
$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$
Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.
Finding $a_1$: Differentiate both sides of the expansion. This gives
$$\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}\;\;=\;\;a_{1}\;+\;2a_{2}x\;+\;3a_{3}x^2\;+\;4a_{4}x^3\;+\;5a_{5}x^4\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{1}=\frac{1}{2}$.
Finding $a_2$: Differentiate 2-times both sides of the expansion. This gives
$$\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{3}{2}}\;\;=\;\;2a_{2}\;+\;(2)(3)a_{3}x\;+\;(3)(4)a_{4}x^2\;+\;(4)(5)a_{5}x^3\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{2}=-\frac{1}{8}$.
Finding $a_3$: Differentiate 3-times both sides of the expansion. This gives
$$\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{5}{2}}\;\;=\;\;(2)(3)a_{3}\;+\;(2)(3)(4)a_{4}x\;+\;(3)(4)(5)a_{5}x^2\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{3}=\frac{1}{16}$.
Finding $a_4$: Differentiate 4-times both sides of the expansion. This gives
$$\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{7}{2}}\;\;=\;\;(2)(3)(4)a_{4}\;+\;(2)(3)(4)(5)a_{5}x\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{4}=-\frac{5}{128}$.
Finding $a_5$: Differentiate 5-times both sides of the expansion. This gives
$$\left(-\frac{7}{2}\right)\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(x+1\right)^{-\frac{9}{2}}\;\;=\;\;(2)(3)(4)(5)a_{5}\;+ ...$$
Pugging in $x=0$ on both sides leads to $a_{5}=\frac{7}{256}$.
Keep going to get as many coefficients as you want. If you keep careful track of the numbers without reducing the fractional expressions for the coefficients, you can easily determine a pattern (a pattern that can be proved by mathematical induction if you're so inclined).
| {
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"timestamp": "2023-03-29T00:00:00",
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Commutative matrix
*
*If two matrices $A$ and $B$ are commutative then all rules for real numbers $a$ and $b$ apply for the matrices?
For example, if $AB=BA$ then:
$(A+B)^2=A^2 + 2AB + B^2$
$A^3 - B^3 = (A-B)(A^2+AB+B^2)$
and so on...
*If the matrix $A$ is invertible then is $A^m A^n = A^{(m+n)}$, where $m,n$ are integers?
| *
*"Do all rules for real numbers apply to the matrix?"
If by all rules for real numbers, you mean finite factorization laws like in your two examples, then yes. How might we prove such a thing? Let's consider $(A + B)^2$ and $(A+B)^3$.
$(A + B)^2 = A^2 + AB + BA + B^2$, and as $AB = BA$ we can write this as $A^2 + 2AB + B^2$. Similarly, once we write out $(A^2 + 2AB + B^2)(A + B)$, we can simply commute the matrices to get that $(A+B)^3 = a^3 + 3A^2 B + 3AB^2 + B^3$, and so on.
If by all rules for real numbers, you actually mean all rules for real numbers, then the answer is no. For example, it's not true that a matrix A satisfies the trichotomy, $A > 0, A = 0,$ or $A<0$.
*"If matrix $A$ is invertible, then is $A^m A^n = A^{m+n}$ for $m,n \in \mathbb{Z}$?"
Let's look at a case. Suppose $m = 2, n = -3$. Then $A^2 A^{-3}$ makes sense. And $A^2A^{-3} = A(AA^{-1})A^{-2} = (AA^{-1})A = A$. Do you see how this proof might be expanded? In fact, for a general matrix $B$, $B^m B^n = B^{m +n}$ if $m,n > 0$, so the important detail here is whether or not $A^{-1}$ makes sense to write down.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$ Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$
Prove the inequality
$$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$
Thanks :)
| Let $S=a^2bc+b^2cd+c^2da+d^2ab$. We can easily find that:
$$S-(ac+bd)(ab+cd)=-bd(a-c)(b-d);$$
$$S-(bc+ad)(bd+ac)=ac(a-c)(b-d)$$
which implies $$S\le \max\{(ac+bd)(ab+cd),(bc+ad)(bd+ac)\}.$$
By AG mean inequality:
\begin{align*}
(ac+bd)(ab+cd)&\le \left(\frac{(ac+bd)+(ab+cd)}{2}\right)^2\\
{}&=\frac{(a+d)^2(b+c)^2}{4}\\
{}&\le \frac{1}{4}\left[\left(\frac{a+d+b+c}{2}\right)^2\right]^2\\
{}&=4
\end{align*}
Similarly, we have
$$(bc+ad)(bd+ac)\le 4$$.
Thus we have $S\le 4$.
| {
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"url": "https://math.stackexchange.com/questions/184266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
I want to find out the angle for the expression $a^3 + b^3 = c^3$. like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.
| Expanding on Brian M. Scott's answer, since
$\cos\theta=\frac{a^2+b^2-c^2}{2ab}$
and $c^3 = a^3+b^3$,
$\cos\theta=\frac{a^2+b^2-(a^3+b^3)^{2/3}}{2ab}
= \frac{1+(b/a)^2-(1+(b/a)^3)^{2/3}}{2}
=\frac{1+r^2-(1+r^3)^{2/3}}{2}
$
where $r = b/a$.
The derivative of the numerator
is $2r-(2/3)(3r^2)(1+r^3)^{-1/3} = 2r - 2r^2(1+r^3)^{-1/3}$
which is never 0 (else $(1+r^3)^{1/3} = r$)
so is always positive (since its value at 1 is
$2-2/2^{1/3} > 0$).
For large r,
$(1+r^3)^{2/3} = r^2 (1+r^{-3})^{2/3}
\approx r^2(1 + (2/3)r^{-3})
= r^2 + 2/(3r)
$
so
$\cos\theta \approx \frac{1+r^2 - (r^2 + 2/(3r))}{2}
= \frac{1-2/(3r)}{2}
$
which tends to 1/2 for large $r$.
This may have an error, since I would expect it to go to 1,
but I have to go now, so this this is it.
| {
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"url": "https://math.stackexchange.com/questions/184352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding the sum of this alternating series with factorial denominator. What is the sum of this series?
$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$
| Hint: We have
$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots.$$
Multiply both sides by $x$ and differentiate.
| {
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"url": "https://math.stackexchange.com/questions/185915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that
$$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$
Thanks :)
| Using cauchy Schwarz or AM-QM we have that $$LHS \leq \sqrt{3\sum_{cyc}\frac{2a}{b+c}}$$ It suffices to prove $$\sum_{cyc}\frac{2a}{b+c}\leq \sum_{cyc}\frac ab$$ By homogeneity we may suppose $a+b+c=1$.
Clearing out denominators this reduces to show $$2\sum_{cyc}a(a+b)(a+c)abc\leq \sum_{cyc}a^2c(a+b)(b+c)(c+a)$$ which is equivalent to $$0\leq\sum_{cyc} a^2c(a+c)(a+b)(b+c-2bc)$$ which is true by AM-GM and the fact that $a, b, c\leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Deriving the addition formula for the lemniscate functions from a total differential equation The lemniscate of Bernoulli $C$ is a plane curve defined as follows.
Let $a > 0$ be a real number.
Let $F_1 = (a, 0)$ and $F_2 = (-a, 0)$ be two points of $\mathbb{R}^2$.
Let $C = \{P \in \mathbb{R}^2; PF_1\cdot PF_2 = a^2\}$.
Then the equation of $C$ in the polar coordinates is:
$r^2 = 2a^2\cos 2\theta$
Let $P$ be a point of $C$ in the first quadrant.
Let $u$ be the arc length between $O = (0, 0)$ and $P$.
Then, by this question,
$u = \int_{0}^{r} \frac{2a^2dr}{\sqrt{4a^4 - r^4}}$
Let $2a^2 = 1$ and $x = r$.
Then
$u = \int_{0}^{x} \frac{dx}{\sqrt{1 - x^4}}$
$u = u(x)$ is defined in $0 \le x \le 1$.
However, the above integral can be defined on $[-1, 1]$.
So we extend the domain of $u(x)$ to $[-1, 1]$ by the above integral.
Since $\frac{1}{\sqrt{1 - x^4}}$ is invariant under the substitution $x \rightarrow -x$,
$u(-x) = -u(x)$ for every $x \in [-1, 1]$.
Since $u'(x) = \frac{1}{\sqrt{1 - x^4}} > 0$ on $(-1, 1)$, $u(x)$ is strctly increasing on $[-1, 1]$.
Hence there exists the inverse function of $u(x)$. We denote the inverse function of $u(x)$ by $s(u)$. We call $s(u)$ lemniscate sine.
Since arcsin $x = \int_{0}^{x} \frac{dx}{\sqrt{1 - x^2}}$, $s(u)$ is analogous to $\sin u$.
We denote $u(1) = \int_{0}^{1} \frac{dx}{\sqrt{1 - x^4}}$ by $\omega$.
$s(u)$ is defined on $[-\omega, \omega]$.
$\omega$ corresponds to $\frac{\pi}{2}$ in the analogy of $s(u)$ with $\sin u$.
Since $u(-x) = -u(x)$, $s(-u) = -s(u)$
We define a function $c(u)$ by $c(u) = s(\omega - u)$ and call it lemniscate cosine.
$c(u)$ is defined on $[0, 2\omega]$.
Pursuing the analogy with $\sin u$ and being motivated by this question, we consider the following total differential equation.
$$\frac{dx}{\sqrt{1 - x^4}} + \frac{dy}{\sqrt{1 - y^4}} = 0$$
Let $u = \int_{0}^{x}\frac{dx}{\sqrt{1 - x^4}}$
Then $x = s(u)$
Let $v = \int_{0}^{y}\frac{dy}{\sqrt{1 - y^4}}$
Then $y = s(v)$
Let $c$ be a constant.
Then $u + v = c$ is a solution of this equation.
Then we get
$$s(u + v) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$$
Substituting $u = \omega$, $v = -u$, we get $x = s(\omega) = 1, y = s(-u) = -s(u)$.
Hence $s(\omega - u) = \frac{\sqrt{1 - y^4}}{1+y^2} = \sqrt{\frac{1 - y^2}{1 + y^2}}$
Hence $c(u) = \sqrt{\frac{1 - s^2(u)}{1 + s^2(u)}}$
Hence
$$s(u+v) = \frac{s(u)c(v) + s(v)c(u)}{1 - s(u)s(v)c(u)c(v)}$$
Since $c(u+v) = s(\omega - u - v) = s((\omega - u) + (-v))$,
$$c(u+v) = \frac{c(u)c(v) - s(u)s(v)}{1 + s(u)s(v)c(u)c(v)}$$
Remark
Since $c(u) = \sqrt{\frac{1 - s^2(u)}{1 + s^2(u)}}$ and $s(-u) = -s(u)$, $c(-u) = c(u)$.
My question
How do we prove the following equation?
$$s(u + v) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$$
| We follow the method of my answer to this question.
Let $u = \int_{0}^{x}\frac{dx}{\sqrt{1 - x^4}}$.
Then $x = s(u)$.
Let $v = \int_{0}^{y}\frac{dy}{\sqrt{1 - y^4}}$.
Then $y = s(v)$.
Let $c$ be a constant.
$u + v = c$ is a solution of the equation:
$$\frac{dx}{\sqrt{1 - x^4}} + \frac{dy}{\sqrt{1 - y^4}} = 0$$
It suffices to prove that $s(c) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$.
Since $v = c - u$, the right hand side is a function of $u$.
We write this function by $\phi(u)$.
Namely, $$\phi(u) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$$
Let us compute $\frac{d\phi}{du}$.
$\frac{dx}{du} = 1/\frac{du}{dx} = \sqrt{1 - x^4}$
$\frac{dy}{du} = -\frac{dy}{dv} = -1/\frac{dv}{dy} = -\sqrt{1 - y^4}$
$\frac{d^2x}{du^2}= \frac{d\sqrt{1 - x^4}}{du}\cdot\frac{dx}{du} = \frac{-2x^3}{\sqrt{1 - x^4}} \sqrt{1 - x^4} = -2x^3$
$\frac{d^2y}{du^2}= \frac{d^2y}{dv^2} = -2y^3$
Hence
$\frac{d\phi}{du} = \frac{d}{du}(\frac{-x\frac{dy}{du} + y\frac{dx}{du}}{1 + x^2y^2})$
Let $f = -x\frac{dy}{du} + y\frac{dx}{du}$
Let $g = 1 + x^2y^2$
Then $\frac{d}{du}(\frac{f}{g}) = \frac{\frac{df}{du}}{g} - \frac{f\frac{dg}{du}}{g^2}$
$\frac{df}{du} = -\frac{dx}{du}\frac{dy}{du} - x\frac{d^2y}{du^2} + \frac{dy}{du}\frac{dx}{du} + y\frac{d^2x}{du^2} = y\frac{d^2x}{du^2} - x\frac{d^2y}{du^2} = 2(xy^3 - yx^3)$
$\frac{dg}{du} = 2xy^2\frac{dx}{du} + 2x^2y\frac{dy}{du} = 2xy(y\frac{dx}{du} + x\frac{dy}{du})$
$f\frac{dg}{du} = 2xy(y^2(\frac{dx}{du})^2 - x^2(\frac{dy}{du})^2) = 2xy(y^2(1 - x^4) - x^2(1 - y^4))$
Hence
$\frac{d}{du}(\frac{f}{g}) = \frac{2(xy^3 - yx^3)}{1 + x^2y^2} - \frac{2xy(y^2(1 - x^4) - x^2(1 - y^4)}{(1 + x^2y^2)^2} = \frac{2(1 + x^2y^2)(xy^3 - yx^3) - 2xy(y^2(1 - x^4) - x^2(1 - y^4)}{(1 + x^2y^2)^2} = 0$
Hence
$\frac{d\phi}{du} = 0$
Hence $\phi(u)$ is constant.
Hence $\phi(u) = \phi(0) = y = s(v) = s(c)$ as desired
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ How would you evaluate the following series?
$$\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $$
Thanks.
| Here's another approach.
First, note that
$$\begin{eqnarray*}
\sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2}
&<& \sum_{k=n^2+1}^\infty \frac{n}{k^2} \\
&\le& n\int_{n^2}^\infty \frac{dx}{x^2} \\
&=& \frac{1}{n}.
\end{eqnarray*}$$
We also need the partial fraction expansion of $\coth x$,
$$\begin{eqnarray*}
\coth x &=& \lim_{N\to\infty} \sum_{k=-N}^N \frac{1}{x-i k \pi} \\
&=& \frac{1}{x} + \sum_{k=1}^\infty \frac{2x}{x^2+k^2\pi^2}.
\end{eqnarray*}$$
Then we find
$$\begin{eqnarray*}
\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2}
&=& \lim_{n\to\infty}\left(
\sum_{k=1}^\infty \frac{n}{n^2+k^2} - \sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2}
\right) \\
&=& \lim_{n\to\infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2} \\
&=& \lim_{n\to\infty} \left(\frac{\pi}{2}\coth n\pi - \frac{1}{2n}\right) \\
&=& \frac{\pi}{2}.
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/190966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 2
} |
Prove that there is no value of the integers $x,y,z$ satisfied the equation: $19^x + 5^y + 1980z = 1975^{4^{30}} + 2010$ Prove that there is no value of the integers $x,y,z$ satisfying the equation: $$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $$
The equation $1975^{4^{30}}$ is like a double exponent :(
thanks again, it's a little hard to translate into english when my english not so well :(
| $ 19^x + 5^y =- 1980z + 1975^{4^{30}}+ 2010 $ an integer.
If, at least one of $x,y$ is $<0$,the LHS=$(19^x + 5^y)$ is a fraction.
If $x,y≥0$,
$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $
$\implies 19^x=1975^{4^{30}}+ 2010-1980z-5^y$
Observe that $(10a+5)^n$ leaves remainder $5$ when divided by $10$ ,where $n$ is positive integer, $a$ is non-negative integer.
The RHS is divisible by $5$ if $y>0$, but $(5,19)=1$, so $5∤19^x$(the LHS).
If $y=0, 19^x + 1= 1975^{4^{30}}+ 2010-1980z≡5\pmod {10} $
But $19^x≡±1\pmod {10}\implies 19^x + 1≡0$ or $2\pmod {10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A binary quadratic form and an ideal of an order of a quadratic number field Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$.
We say $D = b^2 - 4ac$ is the discriminant of $F$.
If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive.
Is the following proposition true?
If yes, how do we prove it?
Proposition
Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
By this question, there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$.
Let $ax^2 + bxy + cy^2$ be a binary quadratic form whose discriminant is $D$.
Then $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$.
Moreover, $I$ is invertible if and only if $ax^2 + bxy + cy^2$ is primitive.
| Proof that I is an ideal of $R$
By this question, $R = \mathbb{Z} + \mathbb{Z}\frac{(D + \sqrt D)}{2}$.
Hence it suffices to show that $a\frac{(D + \sqrt D)}{2} \in I$ and $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} \in I$.
$a\frac{(D + \sqrt D)}{2} = \frac{(aD + a\sqrt D)}{2} = \frac{(aD + ab + a(-b + \sqrt D))}{2} = a\frac{(D + b)}{2} + a\frac{(-b + \sqrt D))}{2}$
Since $D \equiv b^2$ (mod $4$), $D \equiv b^2 \equiv b$ (mod $2$).
Hence $\frac{(D + b)}{2}$ is an integer.
Hence $a\frac{(D + \sqrt D)}{2} \in I$.
$(-b + \sqrt D)(D + \sqrt D) = -bD - b\sqrt D + D\sqrt D + D
= -bD + D + (D - b)\sqrt D = D - b^2 + (D - b)(-b + \sqrt D)$.
Hence
$\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} = \frac{(D - b^2)}{4} + \frac{(D - b)}{2}\frac{(-b + \sqrt D)}{2}$.
Since $D \equiv b^2$ (mod $4$) and $D \equiv b$ (mod $2$),
$\frac{(D - b^2)}{4}$ and $\frac{(D - b)}{2}$ are integers.
Hence $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} \in I$.
QED
Lemma 1
Let $K$ be a quadratic number field.
Let $R$ be an order of $K$, $D$ its discriminant.
Then $R = \{\frac{(x + y\sqrt D)}{2} |\ x \in \mathbb{Z}, y \in \mathbb{Z}, x \equiv yD$ (mod $2)\}$.
Proof:
By this question, every element $\alpha \in R$ can be writen as $\alpha = u + y\frac{(D + \sqrt D)}{2}, u, y \in \mathbb{Z}$.
Hence $\alpha = \frac{(2u + yD + y\sqrt D)}{2} = \frac{(x + y\sqrt D)}{2}$, where $x = 2u + yD$.
QED
Lemma 2
Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).
By this question, there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$.
Let $ax^2 + bxy + cy^2$ be an integral binary quadratic form whose discriminant is $D$.
By what we have proved in the above, $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$.
Let $(R : I) = \{ z \in K |\ zI \subset R \}$.
Then $(R : I) = \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$.
Proof(based on Cohen's A course in computational algebraic number theory):
Since $D$ is non-square, $a \ne 0$.
Let $z \in (R : I)$
Then $za \in R$.
By Lemma 1, there exist integers such that $za = \frac{(x + y\sqrt D)}{2}$ and $x \equiv yD$ (mod $2$). Then $z = \frac{(x + y\sqrt D)}{2a}$
$z\frac{(-b + \sqrt{D})}{2} = \frac{(x + y\sqrt D)}{2a}\frac{(-b + \sqrt{D})}{2}
= \frac{(-bx + x\sqrt D - by\sqrt D + yD)}{4a} = \frac{(-bx + yD + (x - by)\sqrt D)}{4a} \in R$.
By Lemma 1, $x \equiv by$ (mod $2a$).
Hence there exists an integer $u$ such that $x = by + 2au$.
Then $z = \frac{(x + y\sqrt D)}{2a} = \frac{(2au + y(b + √D))}{2a} = u + y\frac{(b + \sqrt D)}{2a}$. Hence $z \in \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$.
Hence $(R : I) \subset \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$.
It remains to prove the opposite inclusion.
Let $\gamma = \frac{(b + \sqrt D)}{2a}$.
Since $\mathbb{Z} \subset (R : I)$, it suffices to prove that $\gamma \in (R : I)$.
Since $D \equiv b$ (mod $2$), $\gamma a = \frac{(b + \sqrt D)}{2} \in R$ by Lemma 1.
On the other hand, $\gamma\frac{(-b + \sqrt{D})}{2a}
= \frac{(D - b^2)}{4a} = -c \in R$.
Hence $\gamma I \subset R$.
Hence $\gamma \in (R : I)$.
QED
Proof that $I$ is invertible if and only if $ax^2 + bxy + cy^2$ is primitive
The idea of the following proof is borrowed from Cohen's A course in computational algebraic number theory.
We use the following notation.
Let $x_1, \cdots, x_n$ be a finite sequence of elements of $K$.
We denote by $[x_1, \cdots, x_n]$ the subgroup of $K$ generated by $x_1, \cdots, x_n$.
By this question, $I$ is invertible if and only if $I(R : I) = R$.
By Lemma 2, $(R : I) = [1, \frac{(b + \sqrt D)}{2a}]$.
Hence $I(R : I) = [a, \frac{(-b + \sqrt D)}{2}][1, \frac{(b + \sqrt D)}{2a}]=
[a, \frac{(b + \sqrt D)}{2}, \frac{(-b + \sqrt D)}{2}, \frac{(D - b^2)}{4a}]=[a, \frac{(b + \sqrt D)}{2}, \frac{(-b + \sqrt D)}{2}, c]=[a, b, c, \frac{(-b + \sqrt D)}{2}]=$ [gcd$(a, b, c), \frac{(-b + \sqrt D)}{2}]$.
Hence if $I(R : I) = R$, gcd$(a, b, c) = 1$.
Conversely suppose gcd$(a, b, c) = 1$.
Since $D \equiv b$ (mod $2$), $\frac{(-b - D)}{2}$ is an integer.
Hence
$I(R : I) = [$gcd$(a, b, c), \frac{(-b + \sqrt D)}{2}]
= [1, \frac{(-b + \sqrt D)}{2}]
= [1, \frac{(-b - D)}{2} + \frac{(D + \sqrt D)}{2}]
= [1, \frac{(D + √D)}{2}] = R$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
| There are no real solutions, nor any complex solutions if you use the principal branch of the square root. Squaring both sides and simplifying gives you $20 \sqrt{x-4} = -92$.
EDIT: More generally, for any $a, b \ge 0$, $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$. Since
$(x+4) - (x-4) = 8$, the most $\sqrt{x+4} - \sqrt{x-4}$ can be is $\sqrt{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Homework Calculus line with circle Find an expression for the function whose graph consist of the line segment from the point $(1,-3)$ to the point $(5,7)$ together with the top of the circle with the center at $(8,7)$.
I don't understand what it means by together with a circle.
$$y = -\frac{5}{2}x + \frac{39}{2}$$
| I will assume that the circle has radius equal to $3$. If that is the case, then the circle is defined by
$C := \{ (x,y) \in \mathbb{R}^2 \mid (x-8)^2 + (y-7)^2 = 3^2\}$.
The equation that defines the circle can be rewritten in the form
$(y-7)^2 = 9 - (x-8)^2$
and, taking the square root of both sides, we obtain
$y = 7 \pm \sqrt{9 - (x-8)^2}$.
Since we want the top half of the circle, we have $y = 7 + \sqrt{9 - (x-8)^2}$.
The line that connects points $(1,-3)$ and $(5,7)$ is given by the equation
$y + 3 = \frac{10}{4} (x-1)$
or, equivalently,
$y = \displaystyle\frac{5}{2} x - \frac{11}{2}$.
Finally, we can introduce a function $f : [1,11] \to \mathbb{R}$ defined by
$f (x) = \left\{\begin{array}{cl} \displaystyle\frac{5}{2} x - \frac{11}{2}, &\quad{} x \in [1,5]\\ 7 + \sqrt{9 - (x-8)^2}, &\quad{} x \in [5,11]\end{array}\right.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A question with the sequence $e_{n}=\left(1+\frac{1}{n}\right)^{n}$ Prove that
$a$) the following sequence is increasing
$$e_{n}=\left(1+\frac{1}{n}\right)^{n},\quad n\ge1;$$
$b$) the inequality below holds
$$e_{n} \leq3,\quad n\ge1.$$
| In order to prove that the given sequence is strictly increasing, we are to demonstrate $e_{n+1} > e_n$:
\[
\bigg(1+ \dfrac{1}{n+1}\bigg)^{n+1} > \bigg(1 + \dfrac{1}{n} \bigg)^n.
\]
Let's rewrite the inequality above as:
\[ \bigg( \dfrac{1 + \dfrac{1}{n+1}}{ 1 + \dfrac{1}{n}} \bigg)^n > \dfrac{1}{1 + \dfrac{1}{n+1}}.
\]
The right-hand side equals
\[ \dfrac{1}{1 + \dfrac{1}{n+1}} = \dfrac{n+1}{n+2} = 1 - \dfrac{1}{n+2}.
\]
Now, let's focus on the left-hand side:
\[ \bigg( \dfrac{1 + \dfrac{1}{n+1}}{ 1 + \dfrac{1}{n}} \bigg)^n = \bigg( \dfrac{(n+2)/(n+1)}{(n+1)/n} \bigg)^n = \bigg( \dfrac{n(n+2)}{(n+1)^2} \bigg)^n = \bigg( 1 - \dfrac{1}{(n+1)^2}\bigg)^n.
\]
By the Bernoulli's inequality, the following holds:
\[ \bigg( 1 - \dfrac{1}{(n+1)^2}\bigg)^n \geq 1 - \dfrac{n}{(n+1)^2}
\]
Now it's purely technical to show the desired inequality
\[ 1 - \dfrac{n}{(n+1)^2} > 1 - \dfrac{1}{n+2},
\]
because
\[ \dfrac{n}{(n+1)^2} < \dfrac{1}{n+2}.
\]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$
$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$
$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$
$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$
$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$
$$ 4 \cos x = 2 \sqrt 3 $$
$$ \cos x = \frac{2 \sqrt 3}{4} $$
$$ \cos x = \frac{\sqrt 3}{2} $$
The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:
$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$
$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$
Since I earlier on exponentiated both sides I have to check my solutions:
$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$
$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$
Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.
$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$
Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...
| There is a standard method for solving equations of the form:
$$
A \sin x + B \cos x = C
$$
Divide both sides by $\sqrt{A^2 + B^2}$:
$$
\frac{A}{\sqrt{A^2 + B^2}} \sin x + \frac{B}{\sqrt{A^2 + B^2}} \cos x = \frac{C}{\sqrt{A^2 + B^2}}
$$
Find $\theta \in [0, 2\pi)$ so that:
$$
\sin \theta = \frac{B}{\sqrt{A^2 + B^2}} \\
\cos \theta = \frac{A}{\sqrt{A^2 + B^2}} \\
$$
And $\phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ so that:
$$
\sin \phi = \frac{C}{\sqrt{A^2 + B^2}}
$$
If you cannot find such a $\phi$, then the equation doesn't have any solutions. (For example, if $\frac{C}{\sqrt{A^2 + B^2}} > 1$.
Thus:
$$
\cos \theta \sin x + \sin \theta \cos x = \sin \phi
$$
Using the angle sum identity, we have:
$$
\sin(x + \theta) = \sin \phi
$$
Therefore:
\begin{align*}
x_1 &= \phi - \theta + 2 \pi n \\
x_2 &= \pi - \phi - \theta + 2 \pi n
\end{align*}
(Where $n \in \mathbb{Z}$)
Now, let's apply this method to your question. We have:
$$
A = 1, B = \sqrt{3}, C = 1 \\
\sqrt{A^2 + B^2} = 2 \\
\sin \theta = \frac{\sqrt{3}}{2},
\cos \theta = \frac{1}{2}, \theta = \frac{\pi}{3} \\
\sin \phi = \frac{1}{2}, \phi = \frac{\pi}{6}
$$
Thus:
$$
x_1 = -\frac{\pi}{6} + 2 \pi n \\
x_2 = \frac{\pi}{2} + 2 \pi n \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/201399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
How to solve and see resolution of $13^{53} \pmod 7$ using Fermat little Theorem? How to solve and see resolution of $13^{53} \pmod 7$ using Fermat little Theorem? Using Fermat's Little Theorem, I know it gives me 6 as an answer to this problem..., but why? How is the resolution?
Thanks,
| Fermat's Little Theorem says $x^7\equiv x\pmod{7}$. When $x\not\equiv0\pmod{7}$, we can divide by $x$ to get
$$
x^6\equiv1\pmod{7}
$$
In the case of $13^{53}$, we get that $13^{53}=13^{6\cdot8+5}=\left(13^6\right)^8\cdot13^5\equiv1^8\cdot(-1)^5\equiv-1\pmod{7}$ since $13^6\equiv1\pmod{7}$ and $13\equiv-1\pmod{7}$.
Of course, since $13\equiv-1\pmod{7}$, we get that $13^{53}\equiv(-1)^{53}\equiv-1\pmod{7}$.
In any case, $13^{53}\equiv-1\equiv6\pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| Substition
$$x=4\tan y \Rightarrow dx=\frac{4}{\cos^2 y} dy$$
$$\tan y=\frac{x}{4} \Rightarrow y=\arctan\frac{x}{4}$$
$$$\int\frac{1}{(16+x^2)^2}dx=\int\frac{1}{(4^2+x^2)^2}dx$=$\int\frac{\frac{4}{\cos^2 y}}{(16+(4\tan y)^2)^2}dy$=$\int\frac{\frac{4}{\cos^2 y}}{(16+16\tan^2 y)^2}dy$$
$$= \int\frac{\frac{4}{\cos^2 y}}{16^2(1+\tan^2 y)^2}dy $$
$$=\int\frac{\frac{4}{\cos^2 y}}{\frac{256}{(\cos^2 y)^2}}$$
$$=\int\frac{\frac{4}{\cos^2 y}}{\frac{256}{\cos^4 y}}$$
$$=\frac{1}{64} \int\frac{\cos^4 y dy}{\cos^2 }$$
$$=\frac{1}{64} \int{\cos^2 y dy}$$
$$=\frac{1}{64}(\frac{1}{2}\sin y\cos y+\frac{1}{2}y)$$
$$=\frac{1}{128}(\frac{1}{\tan y+\cot y}+y)$$
$$=\frac{1}{128}(\frac{1}{\tan y+\frac{1}{\tan y}}+y)$$
$$=\frac{1}{128}(\frac{\tan y}{1+\tan^2 y}+y)$=$\frac{1}{128}(\frac{4x}{16+x^2}+\arctan\frac{x}{4})+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/204961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find an ellipse whose length is the same as the outer rim of the monkey saddle
Given the monkey saddle $z=x^3-3xy^2$ over the unit circle $x^2+y^2
\leq 1$, find an ellipse whose length is the same as the length of the
outer edge of the monkey saddle.
I've already found a parameterization for the monkey saddle in cylindrical coordinates:
$x=r\cos \theta$
$y = r\sin\theta$
$z = r^3\cos3\theta$
And I've found the area of the monkey saddle in the region: $\frac{\pi}{6}[3\sqrt{10}+log(3+\sqrt{10})]$
And I know that the arc length of the ellipse $x^2/a^2+y^2/b^2=1$ is $\int_0^{2\pi}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta} d\theta}$
| The outer edge of the monkey saddle is at $r=1$, and thus $(x,y,z)=(\cos\theta,\sin\theta,\cos3\theta)$. The total arc length is
$$
\begin{align}
\int\mathrm ds
&=
\int\sqrt{\mathrm dx^2+\mathrm dy^2+\mathrm dz^2}
\\
&=
\int_0^{2\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dy}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dz}{\mathrm d\theta}\right)^2}\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{\sin^2\theta+\cos^2\theta+9\sin^2(3\theta)}\,\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{1+9\sin^2(3\theta)}\,\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{1+9\sin^2\theta}\,\mathrm d\theta\;.
\\
&=
\int_0^{2\pi}\sqrt{\cos^2\theta+10\sin^2\theta}\,\mathrm d\theta\;.
\end{align}
$$
What you wrote in the last line of the question is actually the arc length of an ellipse, not its area (as you can tell just from the units of length). Thus an ellipse with semi-axes $1$ and $\sqrt{10}$ has the same length as the outer edge of the monkey saddle.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving an equality of matrices I am asked to prove that $$\begin{pmatrix} \\ A & B\\ C &D\end{pmatrix}^{-1}=\begin{pmatrix} M & -MBD^{-1} \\ -D^{-1}CM & D^{-1}+D^{-1}CMBD^{-1} \end{pmatrix}$$
Where $M=(A-BD^{-1}C)^{-1}$.
Unfortunately, I have no idea what to do about it.
| I guess rather than verifying the inverse stated in the assignment, you should derive it. Let $$
\begin{pmatrix} A & B \cr C & D \end{pmatrix}^{-1} = \begin{pmatrix} U & V \cr W & X \end{pmatrix}
$$
We have:
$$
\begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix} = \begin{pmatrix} A & B \cr C & D \end{pmatrix} \cdot \begin{pmatrix} U & V \cr W & X \end{pmatrix} = \begin{pmatrix} A U + B W & AV + B X \cr CU + D W & CV + DX \end{pmatrix}
$$
Thus $AV = -BX$ and $CU = -DW$, giving $X = -B^{-1} A V$ and $W=-D^{-1} C U$. Substituting back into block-diagonals:
$$
A U - B D^{-1} C U = \left(A - B D^{-1} C U\right) U= 1 \qquad C V-D B^{-1} A V = \left(C - D B^{-1} A \right) V= 1
$$
Hence
$$
U = \left(A - B D^{-1} C\right)^{-1} \qquad
V = \left(C - D B^{-1} A\right)^{-1}
$$
Now $$U B D^{-1} = \left(A - B D^{-1} C\right)^{-1} B D^{-1} = \left( D B^{-1} \left(A - B D^{-1} C\right)\right)^{-1} = -V $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Triangle problem I have got one simple task to prove:
We have got a triangle $\triangle XYZ$.
Then we create points $A,B,C$ on $XY, YZ, ZX$ respectively, such that $XA = AB = BZ$ and $CZ = AY = AC$.
How to prove that $XY = \frac{XZ + YZ}{2}$?
| Choosing coordinates
Based on the variables $a$, $b$ and $c$ we define coordinates for your points like this, without loss of generality:
\begin{align*}
X &= \begin{pmatrix}-1\\0\end{pmatrix} &
Y &= \begin{pmatrix}1\\0\end{pmatrix} &
Z &= \begin{pmatrix}b\\c\end{pmatrix} \\
A &= \begin{pmatrix}a\\0\end{pmatrix} &
B &= Z + (1+a)\frac{Y-Z}{\lVert Y-Z\rVert} &
C &= Z + (1-a)\frac{X-Z}{\lVert X-Z\rVert}
\end{align*}
This choice of coordinates already ensures that $AX = BZ = (1+a)$ and $AY = CZ = (1-a)$.
Some computer algebra
Now you have to ensure that $AB$ and $AC$ play along as well, which leads to two equations that my computer algebra system initially writes as
\begin{alignat*}{2}
\lVert A-B\rVert &\,=\,&
\sqrt{{\left| \frac{{\left(a + 1\right)} c}{\sqrt{c^{2} + {\left| b - 1
\right|}^{2}}} - c \right|}^{2} + {\left| \frac{{\left(b - 1\right)}
{\left(a + 1\right)}}{\sqrt{c^{2} + {\left| b - 1 \right|}^{2}}} + a - b
\right|}^{2}} &= a + 1 \\
\lVert A-C\rVert &\,=\,&
\sqrt{{\left| -\frac{{\left(a - 1\right)}
c}{\sqrt{c^{2} + {\left| b + 1 \right|}^{2}}} - c \right|}^{2} + {\left|
-\frac{{\left(b + 1\right)} {\left(a - 1\right)}}{\sqrt{c^{2} + {\left|
b + 1 \right|}^{2}}} + a - b \right|}^{2}} &= -a + 1
\end{alignat*}
Now you square the equations, multiply by the common denominator, put the remaining root on one side and the rest on the other, square again, and eventually solve for $a$ and $c$ depending on $b$. There will be multiple solutions, most of them complex, and one degenerate with $c=0$ which I'll discuss below. The remaining solutions only differ in signs, so we can write them as one using $\pm$.
The expression for $a$ is quite complex, but the one for $b$ is simply
$$c = \pm\frac12\sqrt{12-3b^2}$$
which you can write as
$$3b^2 + 4c^2 - 12 = 0\;.\tag{1}$$
It's an ellipse
Our aim is to prove the equation from your question,
$$XY=\frac{XZ+YZ}{2}\tag{2}$$
As Arthur pointed out in his comment, this condition can be interpreted as a condition that $Z$ lies on a certain ellipse with $X$ and $Y$ as its focal points. There are many such ellipses, so you still need one more point on the ellipse to specify which one you actually want. In our coordinate setup, $(2,0)^T$ is an example of a point which satisfies your equation and therefore has to lie on the ellipse.
Equation $(1)$ does indeed describe an ellipse. I don't feel like computing its focal points from the equation just now. Instead, observe that from the form of this equation (i.e. only pure squared monomials and constant term, no bare $x$, $y$ or mixed $xy$), the ellipse has its center of symmetry at the origin, and its axes coincide with your coordinate axes. By simply inserting coordinates into the equation, you can check that the following points lie on this ellipse:
$$ \begin{pmatrix}\pm2\\0\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}0\\\pm\sqrt3\end{pmatrix} $$
Both of these pairs of symmetric points satisfy $(2)$: the first can be easily checked as all points are on a line, whereas the second corresponds to the equilateral case. For an ellipse whose axes coincide with the coordinate axes, the above points already define both radii, and therefore uniquey define the ellipse.
To conclude, we know that all points satisfying $(1)$ lie on an ellipse, we know that $(2)$ describes an ellipse, and we know that these two ellipses are the same. As all non-degenerate configurations satisfying the initial length equalities will lead to $(1)$, we now have proven $(2)$ to hold.
Degenerate situation
The equations stated above also have one solution
$$a=b\quad c=0\;.$$
This means that you can also satisfy your length equalities by choosing $A=Z$, so that your whole configuration collapses to a single line. In that case, $(2)$ does not necessarily hold, as the following example demonstrates:
\begin{align*}
X &= \begin{pmatrix}-2\\0\end{pmatrix} &
C &= \begin{pmatrix}-1\\0\end{pmatrix} &
A=Z &= \begin{pmatrix}0\\0\end{pmatrix} &
Y &= \begin{pmatrix}1\\0\end{pmatrix} &
B &= \begin{pmatrix}2\\0\end{pmatrix}
\end{align*}
This leads to
\begin{align*}
XA=AB=BZ &= 2 \\
CZ=AY=AC &= 1
\end{align*}
but
$$ XY = 3 \neq \frac32 = \frac{2+1}2 = \frac{XZ + YZ}{2}\;. $$
So you should require that your triangle is non-degenerate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/208824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Sum of a nearly classic series Assuming we know that : $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$$
How do you find the sum of a series where all terms are in this one ?
For instance, how do you prove that ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$
| Observe that:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$
and
$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$
therefore
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} & = & \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\
& = & \frac{\pi^2}{6} - \frac{\pi^2}{24} \\
& = & \frac{\pi^2}{8}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Analytic Geometry: Circle Given is a line with parametric equation:
$ x = 2 \lambda $
$ y = 1-\lambda $
Find out for which values of $\lambda$ the line is inside the circle of $x^2+4x+y^2-6x+5=0$
My attempt at solving this:
$x^2+4x+y^2-6x+5=0$
$x^2-2x+y^2+5=0$
$ (x-1)^2 -1 + y^2+5=0$
$ (x-1)^2 + y^2 =-4$
And that's where I'm stuck, this isn't correct. Can anyone point my in the right direction?
| I will assume that we have a typo, and $x^2+4x+y^2-6x+5=0$ is intended to be $x^2+4x+y^2-6y+5=0$. Two reasons for this: (i) it would be strange to have two "$x$" terms and (ii) the given equation is not satisfied by any (real) pair $(x,y)$.
The modified equation can be rewritten as $(x+2)^2+(y-3)^2=8$, a circle with centre $(-2,3)$ and radius $\sqrt{8}$.
So we want the distance from $(2\lambda,1-\lambda)$ to $(-2,3)$ to be $\lt\sqrt{8}$.
Equivalently, we want to solve the inequality
$$(2\lambda+2)^2+(1-\lambda-3)^2\lt 8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/213704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help! Integrate using substitution method. I need help integrating the following function:
$$\int\frac{2x+5}{\sqrt{16-6x-x^2}}dx$$
| If you let $u=16-6x-x^2$, then $du=(-6-2x)dx=-(2x+6)~dx$. Now you can rewrite your integral:
$$\begin{align*}
\int\frac{2x+5}{\sqrt{16-6x-x^2}}dx&=\int\frac{(2x+6)-1}{\sqrt{16-6x-x^2}}dx\\
&=-\int\frac{du}{\sqrt{u}}-\int\frac1{\sqrt{16-6x-x^2}}dx\;.
\end{align*}$$
The first of these two integrals is an easy power rule integration. For the second, start by completing the square in the denominator:
$$\begin{align*}
\int\frac1{\sqrt{16-6x-x^2}}dx&=\int\frac1{\sqrt{-(x^2+6x-16)}}dx\\
&=\int\frac1{\sqrt{-\left((x+3)^2-25\right)}}dx\\
&=\int\frac1{\sqrt{25-(x+3)^2}}dx\;.
\end{align*}$$
Now make the trig substitution $x+3=5\sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| The following argument is based on the suggestions by vesszabo. (I do not restrict the functions to the interval $[0, \, 0.1]$ as vesszabo did. That would erroneous because we are looking for the limit at $0$.) It is a rigorous argument and it does avoid using Taylor's Theorem ... but it is still hokey. The choice to use the functions $f$ and $g$ is guided by Taylor's Theorem.
Demonstration
$f$ is a function defined by
\begin{equation*}
f(x) = \sin{x} - x + \frac{1}{6} \, x^{3} .
\end{equation*}
This function is differentiable, and
\begin{equation*}
f^{\prime}(x) = \cos{x} - 1 + \frac{1}{2} \, x^{2} .
\end{equation*}
Likewise, this derivative is differentiable, and
\begin{equation*}
f^{\prime\prime}(x) = -\sin{x} + x .
\end{equation*}
$f(0) = f^{\prime}(0) = 0$, but since $f^{\prime\prime}(x) > 0$ for every positive, real number $x$, $f^{\prime}(x)$ is an increasing function, and since $f^{\prime}(0) = 0$, $f^{\prime}(x) > 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $f$ is an increasing function. $f(0) = 0$, and so $f(x) > 0$ for every positive, real number $x$.
$g$ is another twice-differentiable function defined by
\begin{equation*}
g(x) = \sin{x} - x + \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} = f(x) - \frac{1}{120} x^{5} .
\end{equation*}
For every positive, real number $x$,
\begin{equation*}
g^{\prime\prime}(x)
= f^{\prime\prime}(x) - \frac{1}{6} x^{3}
= - f(x)
< 0 .
\end{equation*}
According to a corollary to the Mean Value Theorem, $g^{\prime}$ is a decreasing function. $g^{\prime}(0) = 0$, and so $g^{\prime}(x) < 0$ for every positive, real number $x$. According to a corollary to the Mean Value Theorem, $g$ is a decreasing function. $g(0) = 0$, and so $g(x) < 0$ for every positive, real number $x$.
So, for every positive, real number $x$,
\begin{equation*}
\frac{1}{6} \, x^{3} - \frac{1}{120} x^{5}
< x - \sin{x}
< \frac{1}{6} \, x^{3} ,
\end{equation*}
and
\begin{equation*}
\frac{1}{6} - \frac{1}{120} x^{2}
< \frac{x - \sin{x}}{x^{3}}
< \frac{1}{6} .
\end{equation*}
According to the Squeeze Theorem,
\begin{equation*}
\lim_{x\to0^{+}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
$f$ and $g$ are odd functions. So, for every negative, real number $x$,
\begin{equation*}
\frac{1}{6} \, x^{3}
< x - \sin{x}
< \frac{1}{6} \, x^{3} - \frac{1}{120} x^{5} ,
\end{equation*}
and
\begin{equation*}
\frac{1}{6} - \frac{1}{120} x^{2}
< \frac{x - \sin{x}}{x^{3}}
< \frac{1}{6} .
\end{equation*}
Again, according to the Squeeze Theorem,
\begin{equation*}
\lim_{x\to0^{-}} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
A function that has the same left-sided and right-sided limits at $0$ has a limit at $0$.
\begin{equation*}
\lim_{x\to0} \frac{x - \sin{x}}{x^{3}} = \frac{1}{6} .
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/217081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
SVD and linear least squares problem Edit:
I've actually found an error:
Instead of full SVD I had to use, "economy size" SVD, where $U$ has only first $n$ columns, and $\Sigma$ becomes a square matrix. I also forgot to take the transpose of $V$, that's why I was getting wrong numbers. SO, the primary question is solved :)
But it would be great if someone could answer the bonus question.
I need to solve the linear least squares problem $\min_{x}\|Ax - b\|_{2}^{2}$ given the matrix
$A = \begin{pmatrix}
1&1 \\
0&1 \\
-1&0
\end{pmatrix} $ and vector $b = \begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix} $
The normal equations method $A^{T}Ax = A^{T}b$ and the QR-decomposition $Rx = Q^{T}b$
give the same result: $x = \begin{pmatrix}
-0.667 \\
1.333
\end{pmatrix}$
However, I have some questions about the SVD method.
My lecture notes say that the solution to the LLS is
$$x = V\Sigma^{-1}U^{T}b$$
But $$A_{32} = U_{33}\Sigma_{32}V_{22}^{T}$$
Therefore, $\Sigma$ is an $3\times2$ matrix, and the inverse is not defined for it.
I though $\Sigma^{-1}$ could mean $\Sigma^{+}$ (pseudo inverse), then the dimensions agree.
But the solution is $x = \begin{pmatrix}
1.333 \\
0.667
\end{pmatrix}$ which looks quite close to the first solution, but it doesn't give the least value.
Thank you for your time! I really hope someone could help me figure it out.
As a bonus question, why is it written $\min_{x}\|Ax - b\|_{2}^{2}$ ?
Isn't $\min_{x}\|Ax - b\|_{2}^{2} \Leftarrow\Rightarrow \min_{x}\|Ax - b\|_{2}$, because norm is always $\geq0$?
| Problem definition
$$
\begin{align}
\mathbf{A} x &= b\\
\left(
\begin{array}{rr}
1 & 1 \\
0 & 1 \\
-1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{c}
x_{1} \\
x_{2}
\end{array}
\right)
&=
\left(
\begin{array}{c}
1 \\
1 \\
1 \\
\end{array}
\right)
\end{align}
$$
Normal equations
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b\\
\left(
\begin{array}{cc}
2 & 1 \\
1 & 2 \\
\end{array}
\right)
\left(
\begin{array}{c}
x_{1} \\
x_{2}
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
2
\end{array}
\right)
\end{align}
$$
Given the inverse matrix $\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} = \frac{1}{3}\left(
\begin{array}{rr}
2 & -1 \\
-1 & 2 \\
\end{array}
\right)$, the solution is
$$
x_{LS} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*}b = \frac{1}{3}
\left(
\begin{array}{r}
-2 \\
4
\end{array}
\right)
$$
SVD
The singular value decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{U}\ \Sigma \ \mathbf{V}^{*} \\
&=
\left(
\begin{array}{ccc}
\frac{1}{\sqrt{6}}
\left[
\begin{array}{r}
2 \\
1 \\
-1 \\
\end{array}
\right] &
\frac{1}{\sqrt{2}}
\left[
\begin{array}{r}
0 \\
1 \\
1 \\
\end{array}
\right] &
\frac{1}{\sqrt{3}}
\left[
\begin{array}{r}
1 \\
1 \\
1 \\
\end{array}
\right] &
\end{array}
\right)
%
\left(
\begin{array}{cc}
\sqrt{3} & 0 \\
0 & 1 \\
0 & 0 \\
\end{array}
\right)
%
\frac{1}{\sqrt{2}}
\left(
\begin{array}{rr}
1 & 1 \\
-1 & 1
\end{array}
\right)
\end{align}
$$
The pseudoinverse solution provides the same answer
$$
x_{LS} = \mathbf{A}^{\dagger} b = \mathbf{V}\ \Sigma^{\dagger} \mathbf{U}^{*} b =
\frac{1}{3}
\left(
\begin{array}{rrr}
1 & -1 & -2 \\
1 & 2 & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
1 \\
1 \\
1 \\
\end{array}
\right)
=
\frac{1}{3}
\left(
\begin{array}{r}
-2 \\
4
\end{array}
\right)
$$
Bonus answer: You can show that when the inverse to the product matrix exists, it must be the pseudoinverse. Start with the basic definitions
$$
\mathbf{A}= \mathbf{U} \ \Sigma \ \mathbf{V}^{*}, \quad
\mathbf{A}^{*}= \mathbf{V} \ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*}, \quad
\mathbf{A}^{\dagger} = \mathbf{V} \ \Sigma^{\dagger} \mathbf{U}^{*}.
$$
To get there, use the following identities
$$
\Sigma = \left(
\begin{array}{cc}
\sqrt{3} & 0 \\
0 & 1 \\
0 & 0 \\
\end{array}
\right), \quad
%
\Sigma^{\mathrm{T}} = \left(
\begin{array}{ccc}
\sqrt{3} & 0 & 0\\
0 & 1 & 0
\end{array}
\right), \quad
%
\Sigma^{\dagger} =
\left(
\begin{array}{ccc}
\frac{1}{\sqrt{3}} & 0 & 0\\
0 & 1 & 0
\end{array}
\right).
$$
Given $\mathbf{S} =
\left(
\begin{array}{cc}
\sqrt{3} & 0 \\
0 & 1
\end{array}
\right),$
$$
\Sigma =
\left(
\begin{array}{c}
\mathbf{S} \\
0
\end{array}
\right)_{3\times 2}, \quad
%
\Sigma^{\mathrm{T}} =
\left(
\begin{array}{cc}
\mathbf{S} & 0
\end{array}
\right)_{2\times 3}, \quad
%
\Sigma^{\dagger} =
\left(
\begin{array}{cc}
\mathbf{S}^{-1} & 0
\end{array}
\right)_{2\times 3}
$$
When the normal equations matrix product exists, it must be the same as the pseudoinverse:
$$
\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} =
\left( \mathbf{V}\ \mathbf{S}^{2}\ \mathbf{V}^{*} \right)^{-1} \mathbf{V} \ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*} =
\left( \mathbf{V}\ \mathbf{S}^{-2}\ \mathbf{V}^{*} \right) \mathbf{V}\ \Sigma^{\mathrm{T}}\ \mathbf{U}^{*} =
\mathbf{V}\ \Sigma^{\dagger}\ \mathbf{U}^{*} = \mathbf{A}^{\dagger}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the imaginary roots of polynomials I'm looking for a simple way to calculate roots in the complex numbers.
I'm having the polynomial $2x^2-x+2$
I know that the result is $1/4-$($\sqrt{15}/4)i$.
However I'm not aware of an easy way to get there.
| $$2x^2 - x +2 = 2\left(x^2 - \dfrac{x}2 + 1 \right) = 2\left(\underbrace{x^2 - 2 \times x \times \dfrac14 + \left(\dfrac14 \right)^2}_{a^2 - 2ab + b^2 = (a-b)^2} - \left(\dfrac14 \right)^2 + 1 \right) = 2 \left( \left(x- \dfrac14 \right)^2 + \dfrac{15}{16}\right)$$
Setting the above to zero, we get that
$$\left(x- \dfrac14 \right)^2 + \dfrac{15}{16} = 0 \implies \left(x- \dfrac14 \right)^2 = \dfrac{-15}{16} \implies x - \dfrac14 = \pm\dfrac{i\sqrt{15}}{4}$$
Hence, $$x = \dfrac{1 \pm i \sqrt{15}}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help with Trigonometry homework - prove an identity I need to prove the following identity:
$\sin^2 2\alpha-\sin^2 \alpha = \sin 3\alpha \sin \alpha$
What I have tried, is to work on each side of the identity. I have started with the left side:
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\sin 2\alpha+\sin \alpha)\\
&=(2\sin\alpha \cos\alpha-\sin \alpha)(2\sin \alpha \cos \alpha+\sin \alpha)\\
&=\sin\alpha(2\cos\alpha-1)\sin \alpha(2\cos \alpha+1)\\
&=\sin^2 \alpha(2\cos^2 \alpha-1)\\
&=\sin^2\alpha \cos 2\alpha
\end{align}
Then, I moved on to the right side of the identity:
\begin{align}
\sin 3\alpha \sin \alpha&=\sin(2\alpha+\alpha) \sin \alpha\\
&=\sin \alpha(\sin 2\alpha \cos \alpha + \sin \alpha \cos 2\alpha)\\
&=\sin \alpha[2 \sin \alpha \cos^2 \alpha+\sin \alpha(\cos^2 \alpha-\sin ^2 \alpha)]\\
&=\sin^2 \alpha(3\cos^2\alpha-\sin^2\alpha)\\
&=\sin^2\alpha(4\cos^2\alpha+1)
\end{align}
I'm not sure how to continue. Any tips?
| You made an error in each side of the identity. Both sides simplify to
$$\sin^2 \alpha(4\cos^2 \alpha-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to the equation
$$2^a + 3^b = 5^c$$
Of course $(1,\space 1, \space 1)$ is a solution.
|
case : $b>2$ and $c>2$
A calculus with computer, modulo $N=2372625=5^3\times 19 \times 37 \times 3^3$, in $H=\mathbb Z/N\mathbb Z $ give
$\text{card}(<2>)=900$, $\text{card}(3^3\times <3>)=900$, $\text{card}(5^3\times<5>)=36$.
And $(<2>+(3^3\times <3>)) \cap 5^3\times <5>=\emptyset$
with $1\in <g>$ the subset of $H$ generated by $g$ with times ($\times$)
So if $b>2$ and $c>2$, there are not solutions.
case : $b\leq 2$ and $c>2$
Here we choose $N=5^3\times (2^{25}-1)$
so $(<2>+\{1,3,9\}) \cap 5^3\times <5>=\emptyset$
So if $b\leq 2$ and $c>2$ no solution.
case : $c \leq 2$ three solutions
Then there are only 3 solutions, not any more.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Can $11^{13}-1$ be divided exactly by 6? Can $11^{13}-1$ be divided exactly by 6?
My solution:
$$11^2 \equiv 1 \pmod 6$$
$$11^{12} \equiv 1 \pmod 6$$
$$11^{13} \equiv 5 \pmod 6 $$
Hence, $(11^{13}-\mathbf{5})$ can be divided exactly by 6. However, according to the solution on my book, （$11^{13}-\mathbf{1}$）can be divide exactly by 6. What's wrong?
| It is already not divisible by $ 3 $; notice that
\begin{align}
11^{13} - 1 &\equiv (-1)^{13} - 1 \\
&\equiv -1 - 1 \\
&\equiv -2 \\
&\equiv 1 \\
&\not\equiv 0 \, (\text{mod} \, 3).
\end{align}
Note $ 11 \equiv 2 \equiv -1 \, (\text{mod} \, 3) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \frac{5A}{6B} = \frac{9C}{14D}?$ If not what am I doing wrong? What is the right method?
| When you mix up a m/n ratio of A/B to get C, then $m+n$ units of C will contain m units of A and n units of B, or $(m+n)C = mA+nB$. I prefer to express this as $C = \frac{m}{m+n}A + \frac{n}{m+n} B$. (Note that the sum of the fractions is exactly 1, so I can work with just the fraction of A, knowing that the fraction of B is just one minus A's fraction.)
This gives the two equations $C = \frac{1}{5}A + \frac{4}{5} B$, $D = \frac{4}{7}A + \frac{3}{7} B$
Now I mix up paint E with a fraction $x$ of C and fraction $1-x$ of D (that would be a ratio $\frac{x}{1-x}$ of C/D). This gives me $E = x C + (1-x)D$, and we also want $E=\frac{5}{11}A + \frac{6}{11} B$. Equating the two formula and and comparing the quantities of A, B will give me a formula for $x$. In fact, we need only look at the fraction of A in E: This gives the formula
$$x \frac{1}{5} + (1-x)\frac{4}{7} = \frac{5}{11}$$
Working through the tedious details gives $x = \frac{45}{143}$.
To validate the answer, we take
$$\frac{45}{143} C + \frac{98}{143}D = \frac{9(A+4B)}{143} C + \frac{14(4A+3B)}{143}D = \frac{5}{11} A + \frac{6}{11}B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{last term}]\\
&=\frac{n(1^3+n^3)}{2}
\end{align}$$
What am I doing wrong?
| Let's take the suggested hint, and consider $$(k+1)^{4}-k^{4}=4k^{3}+6k^{2}+4k+1$$
Summing up both sides from $k=1$ to $n$. Notice that $$\sum_{k=1}^{n}[(k+1)^{4}-k^{4}]=[2^{4}-1^{4}]+[3^{4}-2^{4}]+\ldots+[n^{4}-(n-1)^{4}]+[(n+1)^{4}-n^{4}]$$
Cancelling, we get $(n+1)^{4}-1$. So altogether,
$$(n+1)^{4}-1=4\sum_{k=1}^{n}k^{3}+6\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$
Assuming you already know the results for $\sum1$, $\sum k$, $\sum k^{2}$, we can substitute them in:
$$n^{4}+4n^{3}+6n^{2}+4n+1-1=4S+n(n+1)(2n+1)+2n(n+1)+n$$
$$[n^{4}+4n^{3}+6n^{2}+4n]-[2n^{3}+3n^{2}+n]-[2n^{2}+2n]-[n]=4S$$
Thus, simplifying further
$$n^{4}+2n^{3}+n^{2}=n^{2}(n+1)^{2}=4S \implies \sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4} \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/239909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why isn't it a Hilbert space Let $X$ be the vector space of all the continuous complex-valued functions on $[0,1]$. Then $X$ has an inner product $$(f,g) = \int_0^1 f(t)\overline{g(t)} dt$$ to make it an inner product space. But this is not a Hilbert space.
Why isn't is complete? Which Cauchy sequence in it is not convergent?
Thanks.
| Consider the sequence of continuous functions $f_n$ defined by $$
f_n(x) = \left\{ \begin{array}{rl}
0 &\mbox{ if $x\in \Big[0,\frac{1}{2}-\frac{1}{n}\Big]$} \\
\frac{nx}{2}-\frac{n}{4}+\frac{1}{2} &\mbox{ if $x\in\Big[\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n}\Big]$}\\
1 &\mbox{ if $x\in\Big[\frac{1}{2}+\frac{1}{n},1\Big]$}
\end{array} \right.
$$
Let $f(x)=0$ if $x\in[0,\frac{1}{2}]$ and $f(x)=1$ if $x\in (\frac{1}{2},1]$. Im gonna prove that $f_n\rightarrow f$ in your norm. In fact, we have \begin{eqnarray}
\int_0^1|f_n-f|^2 &=& \int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}+\frac{1}{n}}( \frac{nx}{2}-\frac{n}{4}+\frac{1}{2}-f)^2 \nonumber \\
&=& \int_{\frac{1}{2}-\frac{1}{n}}^{\frac{1}{2}} (\frac{nx}{2}-\frac{n}{4}+\frac{1}{2})^2+\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}(\frac{nx}{2}-\frac{n}{4}-\frac{1}{2})^2 \nonumber \\
&=& \frac{1}{12n}+\frac{7}{12n}\\
&\rightarrow& 0
\end{eqnarray}
So the sequence $f_n$ converges to a function $f$ not continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/241933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Finding sum of a series: difference of cubes I am trying to find sum of the infinite series:
$$1^{3}-2^{3}+3^{3}-4^{3}+5^{3}-6^{3} + \ldots$$
I tried to solve it by subtracting sum of even cubes from odd, but that solves only half of the numbers.
Any input is appreciated.
Thank you all (especially Marvis, Manzonni, Gottfried, Mhenni for long and descriptive answers) for you answer. It's my first visit to math.stackexchange.com. Very impressed with all the help. :)
Now, I would like to know how Marvis got to partial sum formula:
$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 = \begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$
I tried to replicate it, but can't get to it.
| Using the identities
$$
\sum_{k=0}^\infty\binom{k}{n}x^k=\frac{x^n}{(1-x)^{n+1}}
$$
and
$$
k^3=\binom{k}{1}+6\binom{k}{2}+6\binom{k}{3}
$$
yields
$$
\sum_{k=1}^\infty k^3x^{k-1}=\frac{1}{(1-x)^2}+6\frac{x}{(1-x)^3}+6\frac{x^2}{(1-x)^4}
$$
Letting $x\to-1$, gives
$$
\begin{align}
\sum_{k=0}^\infty(-1)^{k-1}k^3
&=\frac14-6\cdot\frac18+6\cdot\frac1{16}\\
&=-\frac18
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/242878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Function expansion help I want to write a function, $f(k,a,b)$, I made, in terms of combinations of the fractional part function, $$ j\left\{\frac{c \ }{d}k\right\},$$ where $c,d,$ and $j$ are any integers.
The function is as follows: $f(k,a,b)=1$ if $k\equiv b$ mod a
and $f(k,a,b)=0$, if it is not
I need a general method for finding the expansion of this function in terms of the fractional part function for any given coprime integers $a,b$.
An example of one is $$ f(k,6,1)= -\{k/6 \}+\{k/2\}+\{2k/3\} $$
Such that if, $k\equiv 1$ mod 6 , then $f(k,6,1)=1$, if not $f(k,6,1)=0$.
So I need a general method for finding the expansion of $f(k,a,b)$ in terms of the fractional part operator.
| Recall that the fractional part function $\{x\}$ is defined as the unique real number $r$ with $0\le r<1$ such that $x-r$ is an integer. Recall also that the floor function $\lfloor x\rfloor$ is the unique integer $m$ such that $m\le x<m+1$. Finally, recall that these two functions are related by the identity $x=\lfloor x\rfloor+\{x\}$.
If $a=0$, then your condition for $f(k)$ to be $1$ is $k=0n+b=b$, so we may write:
$$f(k)=\left\lfloor\frac{1}{1+|k-b|}\right\rfloor=\frac{1}{1+|k-b|}-\left\{\frac{1}{1+|k-b|}\right\}.$$
This works because for all $k$, $|k-b|\ge 0$, and $|k-b|=0$ exactly when $k=b$. Thus whenever $k\ne b$, $1+|k-b|>1$ and $0<\frac{1}{1+|k-b|}<1$, so $\left\lfloor\frac{1}{1+|k-b|}\right\rfloor=0$, and if $k=b$, then $1+|k-b|=1$ and $\frac{1}{1+|k-b|}=1$, so $\left\lfloor\frac{1}{1+|k-b|}\right\rfloor=1$.
Otherwise, the condiation for $f(k)$ to be $1$ is $k=an+b$, which is equivalent to $k-b=an$ or $\frac{k-b}{a}=n$. That is, $f(k)=1$ precisely when $\frac{k-b}{a}$ is an integer. Therefore we may write:
$$f(k)=1-\left(\left\{\frac{k-b}{a}\right\}+\left\{-\frac{k-b}{a}\right\}\right).$$
This works because for all $x$, if $\{x\}>0$, then $0\le 1-\{x\}<1$ and $-x-(1-\{x\})=-x+\{x\}-1=-(x-\{x\})-1=-\lfloor x\rfloor-1$, which is an integer, so we must have $\{-x\}=1-\{x\}$ and thus $\{x\}+\{-x\}=1$. On the other hand, if $\{x\}=0$, then $x$ is an integer, so $-x$ is also an integer and we must have $\{-x\}=0$, so $\{x\}+\{-x\}=0$. Therefore $\{x\}+\{-x\}$ is $0$ if $x$ is an integer and $1$ otherwise, so $1-\left(\{x\}+\{-x\}\right)$ is $1$ if $x$ is an integer and $0$ otherwise. We may then substitute $\frac{k-b}{a}$ for $x$ and get the desired function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/244648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?
| In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$
Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$
$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
Applying sine law in $\triangle ABC,$ $$\frac{AC}{\sin 42^\circ}=\frac{AB}{\sin 96^\circ}\implies \frac{AC}{AB}=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\cos 48^\circ}{2\sin 48^\circ \cos 48^\circ}$$ (applying $\sin 2A=2\sin A\cos A$)
So, $$AC=\frac{AB}{2\sin48^\circ} \implies AC=AD$$
So, $\angle ACD=\angle ADC=\frac{(180-24)^\circ}2=78^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
Does $x_{n+2} = (x_{n+1} + x_{n})/2$ converge? Let $0 < a < b < \infty$. Define $x_1=a$, $x_2=b$, and $x_{n+2} = \frac{x_n + x_{n+1}}{2}$ for $n \geq 1$. Does $\{x_n\}$ converge? If so, to what limit?
| $$x_3 = \dfrac{a+b}2$$ $$x_4 = \dfrac{a+3b}4$$ $$x_5 = \dfrac{3a+5b}8$$ $$x_6 = \dfrac{5a+11b}{16}$$ $$x_7 = \dfrac{11a+21b}{32}$$
In general, (by induction) $$x_n = \dfrac{J_{n-2} a + J_{n-1}b}{2^{n-2}}$$ where $J_n$'s are Jacobsthal sequence i.e. they satisfy the recurrence
$$J_{n+1} = J_n + 2J_{n-1}$$ with $J_0 = 0$ and $J_1 = 1$. Once we have this recurrence, it is not hard (by induction) to find that
$$J_n = \dfrac{2^n - (-1)^n}3$$
Hence, $$x_n = \dfrac{(2^{n-2} - (-1)^{n-2})a + (2^{n-1} - (-1)^{n-1})b}{3 \cdot 2^{n-2}} \to \dfrac{a + 2b}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/246278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factorization of polynomial I was just asked to factor $x^3+1$. I came to $(x^2-x+1)(x+1)$ after a while, and I was wondering, whether there is a good method to quicky factor such of polynomials.
| You are usually asked to remember $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^3+a^3=(x+a)(x^2-ax+a^2)$. In general,
$$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1})$$
If $$y=x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1}$$
$$xy=x^n+ax^{n-1}+...+a^{n-2}x^2+a^{n-1}x$$
$$-ay=-ax^{n-1}-a^2x^{n-2}-...-a^{n-1}x-a^n$$
Add $xy$ and $-ay$ and all the terms will cancel out except for $x^n-a^n$, showing
$$xy-ay=(x-a)y=x^n-a^n$$
For your equation, $n=3$ and $a=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/246780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral by partial fractions
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)
factor out the 5, and use partial fraction
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$.
$A\left(x-5\right) + B = x$ Then $B-5A$ has to be zero and $A$ has to be 1.
Resulting in
$$ 5 \left[\int \frac{1}{x-5} + \frac{5}{\left(x-5\right)^2}\, \mathrm{d}x \right]$$
$$ \Rightarrow 5 \left[ \ln \vert x - 5 \vert -\frac{5}{x-5}\right] + C$$
However, this approach doesn't give the answer in the book.
Book's Answer
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right) + C $$
The value should be 30, according to the book.
| Distribute:
$$
\frac{5}{x-5}((x-5)\ln|x-5|-x)=5\left(\frac{x-5}{x-5}\ln|x-5|-\frac{x}{x-5}\right).
$$
Then
$$
\frac{x-5}{x-5}=1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:
If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?
The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$
| The square of $\sqrt{x} - {1 \over \sqrt{x}}$ is $x - 2 + {1 \over x}$. In this case $x = 5 + 2\sqrt{6}$, whose reciprocal is seen to be $5 - 2\sqrt{6}$ by rationalizing the denominator.
So
$$x - 2 + {1 \over x} = (5 + 2\sqrt{6}) - 2 + (5 - 2\sqrt{6})$$
$$= 8$$
So your answer is $\sqrt{8} = 2\sqrt{2}$. (You take the positive square root since $x > 1$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Problem when integrating $e^x / x$. I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?
So, I'm trying to determine: $$ \int{\frac{e^x}{x}} \, dx $$
Integrate by parts, where $u = 1/x$, and $v \, ' = e^x$. Then $u \, ' = - 1/x^2$, and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx$$
Integrate by parts again, $u = 1/x^2$, $v \, ' = e^x$, so that $u \, ' = -2/x^3$ and $v=e^x$. So,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx$$
Repeat this process ad infinitum to get,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) $$
Expanding this gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots $$
And factoring that gives,
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( 1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
Now, considering the series itself, the ratio between the $n^{th}$ term and the $(n-1)^{th}$ term = $\Large \frac{n}{x}$. Eventually, $n$ will be larger than $x$, so the ratio between successive terms will be positive, so (assuming $x$ is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty $$
| There is no elementary antiderivative for you this function.
You can take a look at here: http://en.wikipedia.org/wiki/Exponential_integral
What you have calculated here:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( 1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
is even something like a taylorexpansion of the integral at $x=\infty$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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$f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$?
The polynomial is allowed to have repeated roots.
| [There was a mistake in the first version of this answer that caused one solution to go missing.]
The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have
$$
\begin{align}
a&=-a-b-c\;,\\
b&=ab+bc+ca\;,\\
c&=-abc\;.
\end{align}
$$
If $c=0$, this becomes
$$
\begin{align}
a&=-a-b\;,\\
b&=ab\;,\\
\end{align}
$$
and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$.
If $c\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields
$$
c=-2a+\frac1a\;,
$$
and then the second equation becomes
$$
-\frac1a=-1+2-\frac1{a^2}-2a^2+1\;.
$$
Multiplying through by $a^2$ yields
$$
2a^4-2a^2-a+1=0\;.
$$
The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields
$$
2a^3+2a^2-1=0\;,
$$
which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$.
Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\prod_{x=2}^\infty\frac{x^4-1}{x^4+1}$ Difficult question from some test somewhere (I forget).
$$\prod_{x=2}^\infty\frac{x^4-1}{x^4+1}$$
$x$ is, of course, an integer.
| Write
$$\frac{x^4 - 1}{x^4+1} = \frac{(x-a_1)\ldots(x-a_4)}{(x-b_1)\ldots(x-b_4)}$$
where $a_j$ are the roots of $x^4-1$ and $b_j$ are the roots of $x^4+1$.
Then the partial product
$$ \prod_{x=2}^n \frac{x^4 - 1}{x^4+1} = \frac{\Gamma(n+1-a_1)\ldots \Gamma(n+1-a_4) \Gamma(2-b_1) \ldots \Gamma(2-b_n)}{\Gamma(2-a_1) \ldots \Gamma(2-a_4) \Gamma(n+1-b_1) \ldots \Gamma(n+1-b_4)}$$
Now (carefully) take the limit as $n \to \infty$.
| {
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| We will do this using only divisibility.
Implying that you have this knowledge.
$(a+b,a^2-ab+b^2)=\{1\;\text{and}\;3\}$ if $(a,b)=1$
Calling $d$ the $(a+b,a^2-ab+b^2)$
$$(a+b,a^2-ab+b^2)=d\Longrightarrow d\mid a+b\;\;\text{and}\;\;d\mid a^2-ab+b^2$$$$a^2-ab+b^2=a^2-ab+b^2\underbrace{+3ab-3ab}_{=0})=a^2+2ab+b^2-3ab=(a+b)^2-3ab$$If $d\mid a+b\Longrightarrow d\mid (a+b)^2$ as $d\mid (a+b)^2-3ab$ For these data we have $d\mid3ab$
There are three possibilities
I) $d\mid 3\Longrightarrow d=1$ or $d=3$
II) $d\mid a$ Above we see that $d\mid a+b$ this implies that $d\mid b\Longrightarrow d\mid(a,b)\Longrightarrow d\mid1\Longrightarrow d=1$
III) $d\mid b$ Above we see that $d\mid a+b$ this implies that $d\mid a\Longrightarrow d\mid(a,b)\Longrightarrow d\mid1\Longrightarrow d=1$
Thus, we see that the unique possibilities for $d$ is $1$ or $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Series of positive terms I want to show that $\displaystyle 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots$ converges. I know that by using D'Alembert ratio test I easily show that this series converges but I am doing in this way:
\begin{align*}
s_{n}&=1+\frac{1}{2!}+\frac{1}{4!}+\cdots+\frac{1}{2(n-1)!}\\
&<1+\frac{1}{2}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2n-3}}(\because \frac{1}{k!}\leq\frac{1}{2^{k-1}},\forall k\geq 2)\\
&=1+\frac{1}{2}[1+\frac{1}{2^2}+\frac{1}{2^4}+\cdots+\frac{1}{2^{2n-4}}]
\end{align*}But as $n\to\infty$ the right hand side of the above equation becomes $$ 1+\frac{1}{2}.\frac{1}{1-\frac{1}{4}}=\frac{5}{3}.$$
Hence we have $s_n\leq \frac{5}{3}$. So the given positive term series is such that $(s_n)$ is bounded above hence convergent. Am O right or doing some mistake? Please suggest me!
| Maybe you can compare it to exp(1). You would consider the term of your series to be
a2n = 1/(2n!) ,
a2n+1 = 0.
Obviously for each k, ak<=1/(k!) and therefore your series converges and its sum is less than exp(1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find $1/x^3 + 1/y^3$? If I am given, $x + y = a$ and $xy = b$, how would I find the value of $\dfrac1{x^3} + \dfrac1{y^3}$?
| \begin{align}
\dfrac1{x^3} + \dfrac1{y^3} & = \dfrac{x^3+y^3}{(xy)^3} = \dfrac{\left(x+y \right)\left(x^2+y^2-xy \right)}{(xy)^3}\\
& = \dfrac{\left(x+y \right)\left(\left(x+y \right)^2-3xy \right)}{(xy)^3} = \dfrac{a\left(a^2-3b \right)}{b^3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Variance formulae I know that the variance formula is
$$\sigma^2 = \frac{
\left( x_1 - \bar{x} \right) ^2 +
\left( x_2 - \bar{x} \right) ^2 +
\dots +
\left( x_n - \bar{x} \right) ^2
}{n}$$
Where $\sigma^2$ is the variance; $x_1,\ x_2,\ \dots,\ x_n$ are the statistical data, and $n$ is the number of data.
My question is: how can I expand that formula to get this equivalent one:
$$ \sigma^2 = \frac{x_1^2 + x_2^2 + \dots + x^2_n}{n} -\bar{x}^2$$
?
| Each of the $(x_i - \bar x)^2$ terms expands into $x_i^2 - 2x_i\bar x + \bar x^2$.
Since the sum of all $x_i$ is also equal to $n \bar x$ (by definition), we get that the sum of all the $2x_i\bar x$ is actually $2n\bar x^2$.
Then, the equation becomes: $\displaystyle \sigma^2 = \frac{x_1^2 + x_2^2 + \cdots + x_n^2 - 2n\bar x^2 + n\bar x^2}{n}$
which is $\displaystyle \frac{x_1^2 + x_2^2 + \cdots + x_n^2 - n\bar x^2}{n}$ or $\displaystyle \frac{x_1^2 + x_2^2 + \cdots + x_n^2}{n} - \bar x^2$, as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can I resolve rational indefinite integral? $$\int\left( {20.56\over x^2-1.27}+x^{55}\right) dx$$
I came to something like this
$$20.56 \int {1\over (x-1)^2 - \frac{27}{100} }~dx + {x^{56}\over56} + \text{Constant}$$
Please can you help me to resolve this?
| Your integral is of the form
$$I = \int \left(\dfrac{a}{x^2 - b^2} + x^n \right)dx$$
Hence,
\begin{align}
I & = \int \left(\dfrac{a}{2b}\dfrac1{x - b} - \dfrac{a}{2b}\dfrac1{x + b} + x^n \right)dx \\
& = \dfrac{a}{2b} \log(\vert x-b \vert) - \dfrac{a}{2b} \log(\vert x+b \vert) + \dfrac{x^{n+1}}{n+1} + \text{constant}\\
& = \dfrac{a}{2b} \log\left(\left \vert \dfrac{x-b}{x+b} \right \vert \right) + \dfrac{x^{n+1}}{n+1} + \text{constant}
\end{align}
In your case, $a = 20.56$, $b = \sqrt{1.27}$ and $n = 55$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/260168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
| By AM-GM inequality,
$$\frac{a^2}{a+b} + \frac{a+b}{4} \ge a$$
Add up the similar inequalities obtained by cyclic substitution, you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Distinct natural numbers such that $ab=cd=a+b+c+d-3$ Find the distinct natural numbers $a,b,c,d$ who satisfying $ab=cd=a+b+c+d-3$.
| Find natural numbers $(a,b)$ such that:
$c$ = $\dfrac{1}{2}(a b-\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3 )$
and $d$ = $\dfrac{1}{2}$ $(a b+\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3)$
are natural numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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