Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$ Solve to find the general value of $x$:
$$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$
My Attempt:
$$5(1-\sin^2 (x))-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$5-5\sin^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$2\sin^2 (x)+4\sin (x)\cos (x)=3$$
| Since $\sin^2(x)+\cos^2(x)=1$
$5\cos^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=\sin^2(x)+\cos^2(x)+\sin^2(x)+\cos^2(x)$
Let $\cos(x)=a, \sin(x)=b$
$5a^2-4ab+3b^2=2a^2+2b^2$
$3a^2-4ab+b^2=0$
$(3a-b)(a-b)=0$
$3a=b$ or $a=b$, which means $3\cos(x)=\sin(x)$ or $\cos(x)=\sin(x)$
$\frac{\sin(x)}{\cos(x)}=\tan(x)=3$ or $\frac{\sin(x)}{\cos(x)}=\tan(x)=1$
You may go from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing that $\lim_{x \to -\infty}(x+{1\over x}) = -\infty$ My attempt. I' not sure about this, but I think the definition for this limit will be:
$\forall M <0 \space \exists N < 0, \space x < N \Rightarrow f(x) < M$
We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1 \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$. Now, since ${1\over N^2} + 1 > 1$ and $N < 0$, we have that $N({1\over N^2} + 1 ) < N$. Thefore, $x({1\over x^2} + 1) < N$, which mean that we will choose N = M in our proof.
Proof:
Pick N = M. Since $x < N \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)< N = M$, as required.
Is this correct?
| You wanted to evaluate the limit from first principles, so
Yes, you want to prove :
$\forall M <0,\space \exists N < 0, \space x < N \Rightarrow f(x) < M$
"We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1$" This is correct.
$\Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$.
This is wrong! Consider $x=-0.5,N=-0.4,x<N$, but
$\Rightarrow x({1\over x^2 } + 1)=-2.5 > N({1 \over N^2} + 1)=-2.9!$
You are making an invalid comparison here.
Instead you can simply do this:
Proof:
Pick N = M. Since $x < N,x^2>xM$,
$x^2+1>x^2>xM$, since obviously $x\neq 0$,
$f(x)=x+\frac{1}{x}<x<M$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$
My Attempt
$$
x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\
2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\
\frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\
\frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy}
$$
How do I proceed further and find the derivative ?
| Hint:
Squaring both sides of $$x\sqrt{1+y}=-y\sqrt{1+x}$$
$$x^2(1+y)=y^2(1+x)$$
$$\iff0=(x-y)(x+y+xy)$$
But $x\sqrt{1+y}=-y\sqrt{1+x}\implies x,y$ are of opposite sign, hence $x\ne y$ unless $x=y=0$
Otherwise, $$y(1+x)=-x\implies y=\dfrac{1-(x+1)}{1+x}=\dfrac1{1+x}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number theory question involving Wilson's theorem. How to solve?? Given p is a prime number greater than 2, and
$ 1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{p-1} = \frac{N}{p-1}$
how do I show, $ p | N $ ???
The previous part of this question had me factor $ x^{p-1} -1$ mod $p$. Which I think is just plainly $(x-1) ... (x-(p-1))$
| Working $\bmod{p}$:
$$
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1}\\
\equiv 1 + 2^{-1} + 3^{-1} + \cdots + (p-1)^{-1}\\
\equiv 1 + 2 + \cdots + (p-1) \ \text{ reordered}
$$
which you have shown is $0\pmod{p}$ when you factored $x^{p-1} -1\pmod{p}$.
So if $\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p-1} =\frac{N}{p-1}$ then necessarily $N(p-1)^{-1} = 0 \pmod{p}$ so $p$ divides $N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with this trigonometric equation Here is the equation:
$$\sin^5(x) = \frac {\cos(x)}4$$
The answers should be between: 0° and 360°
I really tried and I dont even think it is possible to solve algebraically, maybe using numerical methods or other kind of black magic but that is very advanced for my grade. I am in high school.
| \begin{align}
\sin^5x &= \frac{\cos x}4
,\\
(\sin^5x)^2 &= \left(\frac{\cos x}4\right)^2
,\\
\sin^{10}x &= \tfrac1{16}{\cos^2 x}
,\\
(\sin^{2}x)^5 &= \tfrac1{16}(1-\sin^2 x)
,
\end{align}
\begin{align}
16(\sin^{2}x)^5+\sin^2 x-1&=0
,\\
(2\sin^2x-1)
(8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1)
&=0
,\\
\cos2x\,(8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1)
&=0.
\end{align}
\begin{align}
8\sin^8x+4\sin^6x+2\sin^4x+\sin^2x+1&>0\quad \forall x\in\mathbb{R}
,
\end{align}
\begin{align}
\cos2x&=0
,\\
2x&=\tfrac\pi2+\pi k,\quad k=0,1,\dots
,\\
x&=\tfrac\pi4+\tfrac{\pi k}2,\quad k=0,1,\dots
.
\end{align}
For $x$ restricted to the interval $[0,2\pi]$,
we have four solutions
\begin{align}
x_1&=\tfrac\pi4=45^\circ
,\\
x_2&=\tfrac\pi4+\tfrac\pi2=\tfrac{3\pi}4=135^\circ
.\\
x_3&=\tfrac\pi4+\pi=\tfrac{5\pi}4=225^\circ
.\\
x_4&=\tfrac\pi4+\tfrac{3\pi}2=\tfrac{7\pi}4=315^\circ
.\\
\end{align}
Substitution of $x_1,\dots,x_4$ into original equation
shows that only solutions $x_1,x_3$ are valid:
\begin{align}
\sin^5x_1 &=\tfrac{\sqrt2}8
,\\
\frac{\cos x_1}4&=\tfrac{\sqrt2}8
,\\
\sin^5x_2 &=\tfrac{\sqrt2}8
,\\
\frac{\cos x_2}4&=-\tfrac{\sqrt2}8
,\\
\sin^5x_3 &=-\tfrac{\sqrt2}8
,\\
\frac{\cos x_3}4&=-\tfrac{\sqrt2}8
,\\
\sin^5x_4 &=-\tfrac{\sqrt2}8
,\\
\frac{\cos x_4}4&=\tfrac{\sqrt2}8
.
\end{align}
As @Malcolm noted, two extra solutions were
introduced by the squaring the original equation,
that's why the final check of the found roots is essential.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by Induction $\frac{n+1}{2n} = \left(1-\frac{1}{2^2}\right)\dotsc \left(1-\frac{1}{n^2}\right)$ proof by induction. I'm pretty close to done with this proof but got stuck at the end
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{n^2}\right) = \frac {n+1}{2n}$$
for all integers $n \geq 2$
*
*basis step
$$1-\frac{1}{2^2} = .75 \qquad \frac{2+1}{2\times2} = 3/4= .75$$
*set $n = k$ assumed to be true
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$
*set $n = k+1$
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1}$$
from step 2 we know that
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dotsc \left(1-\frac{1}{k^2}\right) = \frac {k+1}{2k}$$
so we are going to plug that in
$$\frac {k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2k+1} $$
now to get both sides equal we multiply by $2k$ to get common denominators
$$\frac{k+1}{2k} \times {\left(1-{1\over(k+1)^2}\right) \times 2k \over 2k}$$ This is the left hand side
now to combine it
$${\left((k+1) \left(1- {1\over(k+1)^2}\right) \times 2k \right)\over 2k}$$ - this is where I got stuck...
I wanted to cancel the $2k$'s on the top and bottom but then there would be no way to get it to equal the $(k+1)+1\over2(k+1)$ because the denominator required a $2k$. Any help is appreciated thank you.
| Why not multiply the whole thing by $2k\cdot (k+1)^2\cdot 2(k+1)$ and see if you get an identity (after correcting the denominator of the RHS) ...
| {
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Proving the congruence $((p-1)\ /\ 2)!^2 \equiv (p-1)\ (\textrm{mod}\ p)$ $$\left(\frac{p-1}{2}\right)!^2 \equiv (-1)\ (\textrm{mod}\ p)$$ Given $p$ is a prime congruent to $1$ modulo $4$.
I know there's a way to do this with Wilson's theorem but wondering if my proof is right.
We know since $p \equiv 1\ mod\ 4$, $p$ is of the form $4n + 1$
We can rewrite $x = \frac{p-1}{2}!$ as $\frac{p-1}{2} \cdot \frac{p-1}{2} - 1 \cdot \frac{p-1}{2} -2\ \cdot\ ...\ \cdot1 = \frac{p-1}{2} \cdot \frac{p-3}{2} \cdot \frac{p-5}{2}\ \cdot\ ...\ \cdot1 = \frac{4n+1-1}{2} \cdot \frac{4n+1-3}{2} \cdot \frac{4n+1-5}{2}\ \cdot\ ...\ \cdot1 = \frac{4n}{2} \cdot \frac{4n-2}{2} \cdot \frac{4n-4}{2}\ \cdot\ ...\ \cdot 1 = 2n\cdot (2n-1) \cdot (2n -2)\ \cdot\ ...\ \cdot1$
Now taking $x^2 = 4n^2((2n-1) \cdot (2n -2)\ \cdot\ ...\ \cdot1)^2$
Let $m = n^2((2n-1) \cdot (2n -2)\ \cdot\ ...\ \cdot1)^2$
We can write $x^2$ as $4m$.
$-1\ mod\ p $ is of the form $4n$ because we have $p$ of the form $4n + 1$ and $p - 1$ is then of the form $4n$.
So $x^2 \equiv -1\ mod\ p$
| If $p \equiv 1 \bmod 4$, then $\left( \frac{p-1}{2} \right)^2 = 1 \times 2 \times \cdots \frac{p-1}{2} \times (-1) \times (-2) \times \cdots \times \left(-\frac{p-1}{2}\right)$, because the RHS has an even number of negative terms. The RHS covers every nonzero equivalence class modulo $p$, so we are left with $$(p-1)! \equiv -1 \pmod p$$ which is true because only $1$ and $-1$ are their own multiplicative inverses modulo $p$, so every other term in the product for $(p-1)!$ cancels.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating infinite series with $a^{(2^n)} + 1$ form denominator Evaluate:
$$\sum_{n = 0}^\infty \frac{2^n}{a^ \left( 2^n \right) + 1 } $$
This doesn't seem to have factors in the denominator. I tried stuff like expanding this series and then multiplying by a factors like $ \frac{1}{x-1} $ and then shifting the terms, subtracting the series.
(Edit: Sorry for not telling that a $ \gt $ 1).
| Like all sums ever, this sum telescopes: we can write
$$
\frac1{a+1} = \frac1{a-1} - \frac2{a^2-1}
$$
and more generally
$$
\frac{2^n}{a^{2^n}+1} = \frac{2^n}{a^{2^n}-1} - \frac{2^{n+1}}{a^{2^{n+1}}-1}
$$
and therefore
$$
\sum_{n=0}^{N} \frac{2^n}{a^{2^n}+1} = \sum_{n=0}^N \left(\frac{2^n}{a^{2^n}-1} - \frac{2^{n+1}}{a^{2^{n+1}}-1}\right) = \frac1{a-1} - \frac{2^{N+1}}{a^{2^{N+1}}-1}.
$$
If $\lim_{n \to \infty} \frac{n}{a^n-1} = 0$, which happens when $|a|>1$, then the infinite sum simplifies to $\frac1{a-1}$.
Okay, but seriously, what's the deal with telescoping sums?
If you don't notice the identities above, I don't blame you: Mathematica doesn't either, and in fact Mathematica can't tell that this sum is $1$ when $a=2$ unless you ask it to approximate the sum by numerical methods.
However, telescoping identities appear from partial sums, and so if you don't know what to do, the partial sums might be a good place to look. Here, summing the $n=0,1,2$ terms gives us
$$
\frac{a^6+2 a^5+3 a^4+4 a^3+5 a^2+6 a+7}{(a+1) \left(a^2+1\right) \left(a^4+1\right)}
$$
(you might not want to expand the numerator but you should be able to see that this is the denominator) and now it's much less of a stretch to spot that this expression becomes much nicer when we multiply top and bottom by $a-1$. If we do, we get
$$
\frac{a^7+a^6+a^5+a^4+a^3+a^2+a-7}{a^8-1} = \frac1{a-1}-\frac{8}{a^8-1}
$$
which directly leads to the telescoping solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731886",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}
$$
$$
2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}
$$
$$
2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)
$$
I can find the limit of $g$, but what to do with the other parts?
| Using the identity
$$
\frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}=\frac{2k-1}{2^k}\tag1
$$
the sum can be written as a telescoping series
$$
\begin{align}
\sum_{k=1}^\infty\frac{2k-1}{2^k}
&=\sum_{k=1}^\infty\left(\frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}\right)\\
&=3\tag2
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?
How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?
I took $x$ as common from each bracket making $x^{190}$ but I don't understand what to do next. Please help.
| Hint. Note that $\sum_{k=1}^{20} k=210$ and
$$(x-1)(x^2 - 2)(x^3-3)\cdots(x^{20} - 20)=x^{210}
\left(1-\frac{1}{x}\right)
\left(1-\frac{2}{x^2}\right)
\left(1-\frac{3}{x^3}\right)\cdots\left(1-\frac{20}{x^{20}}\right).$$
Now consider the integer partitions with distinct parts of $210-203=7$:
$$7,\quad 6+1,\quad 5+2,\quad 4+3,\quad 4+2+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736573",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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The square of the solution of the equation... The square of the solution of the equation
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$
$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$
$\implies x^2 = 2$
I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks!
| The formula $(a-b)^2=a^2+b^2-2ab$ is used.
Here, $a=\sqrt{8+3\sqrt7}$ and $b=\sqrt{8-3\sqrt7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739227",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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How Do I Find Tangent and Normal Lines for Polar Equations? Given the equation $r=1+\sin(\theta)$, I am trying to compute:
(a) $\frac{dy}{dx}$
(b) the equation of the tangent and normal lines to the
curve at the indicated θ–value (which is $\frac\pi6$ in this case).
Computing $\frac{dy}{dx}$ is easy, since it involves simply plugging everything into the equation $\frac{(f'(\theta)\sin(\theta) + f(\theta)\cos(\theta)) }{ (f'(\theta)\cos(\theta) - f(\theta)\sin(\theta)}$.
When I used this equation, I got a $\frac{dy}{dx}$ of $\frac{\cos(\theta)(2\sin(\theta)+1) }{ \cos^2(\theta)-\sin(\theta)-\sin^2(\theta)}$.
At this point, I am stuck. I can't figure out how to compute the tangent and normal lines from here and I am getting an indeterminate form $\frac{\sqrt3} 0$ whenever I plug in the value of $\frac\pi6$.
Could someone please assist me in finding these equations? Thanks in advance!
| $$\frac{dy}{dx}=\frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}
{\frac{dr}{d\theta} \cos \theta - r \sin \theta}$$
For $r = 1+\sin \theta, \quad \dfrac{dr}{d\theta} = \cos \theta$. So
\begin{align}
\frac{dy}{dx}
&=\frac{\cos \theta \sin \theta + r \cos \theta}
{\cos^2 \theta - r \sin \theta} \\
&=\frac{\sin \theta + r}
{\cos \theta - r \tan \theta} \\
\end{align}
At $\theta_0 = \frac \pi 6, \quad
\sin \theta_0 = \frac 12, \quad
\cos \theta_0 = \frac{\sqrt{3}}{2},
\quad \quad r_0 = \frac 32$
$\quad x_0 = r_0 \cos \theta_0 = \frac{3\sqrt 3}{4},
\quad y_0 = r_0 \sin \theta_0 = \frac 34$.
So $\displaystyle m =\frac{dy}{dx}
=\frac{\frac 12 + \frac 32}
{\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}}
= undefined$
In other words, the tangent line is a vertical line.
So the tangent line at $\theta_0 = \frac \pi 6$ would be
\begin{align}
x &= x_0 \\
r \cos \theta &= \frac{3\sqrt 3}{4} \\
r &= \frac{3\sqrt 3}{4 \cos \theta}
\end{align}
The equation of the normal line would be computed using $y = y_0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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What is $x$, if $\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots = 1$?
Let
$$S_{n}=\cot ^{-1} \left(3x+\frac{2}{x}\right)+\cot ^{-1} \left(6x+\frac{2}{x}\right)+\cot ^{-1} \left(10x+\frac{2}{x}\right)+\cdots \quad\text{($n$ terms)}$$
where $x>0$. If $\lim _{n \to \infty} S_{n}=1$, then find the value of $x$.
Can the given series be converted to telescopic series? I converted in into $\tan^{-1}$ but in every term $x$ is in numerator? Could someone please given some hint?
| Arguing loosly,
$\begin{array}\\
S_n
&=\sum_{i=1}^{n} arccot(\frac{(i+1)(i+2)x}{2}+\frac{2}{x})\\
&\approx \frac{2}{x}\sum_{i=1}^{n} \dfrac1{(i+1)(i+2)}
\qquad\text{since }arccot(x) \approx \frac1{x} \text{ for large } x \\
&=\frac{2}{x}\sum_{i=1}^{n} (\dfrac1{i+1}-\dfrac1{i+2})\\
&=\frac{2}{x}(\frac12-\frac1{n+2})
\\
&=\frac1{x}-\frac{2}{x(n+2)})
\\
\end{array}
$
So if $S_n \to 1$,
must have $x = 1$.
It would be entertaining
if this sloppy argument
was correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of solutions to a polynomial I need to find the number of solution to the polynomial
$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)=0$$
Clearly, $x=1$ and $y=0$ is a solution. How to prove it has no other solutions or more solutions?
| \begin{align*}
&\;(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)\\
=&\;\frac{1}{2}(x-2y-1)^2+\frac{1}{2}(4x-3y-4)^2+\frac{1}{2}[(x-2y-1)+(4x-3y-4)]^2\\
=&\;0
\end{align*}
if and only if $x-2y-1=4x-3y-4=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When am I allowed to remove a term by dividing 0 by the term It seems that you are allowed to remove terms from an equation if there is a 0 on the other side of the equals sign:
$$
3x = 0
$$
$$
\frac{3x}{3} = \frac{0}{3}
$$
$$
x = 0
$$
However, you can't just do whatever you want:
$$
3x = 0
$$
$$
\frac{3x}{x} = \frac{0}{x}
$$
$$
3 \ne 0
$$
$$
3x^2 + 2x - 4 = 0
$$
$$
\frac{3x^2 + 2x - 4}{3x^2 + 2x - 4} = \frac{0}{3x^2 + 2x - 4}
$$
$$
1 \ne 0
$$
What exactly is the rule here? Is it simply unsafe only when the denominator would contain a variable or be 0? When is dividing 0 by something to remove that something from the equation allowed/safe, if ever?
| You can't ever divide by something that is zero. And if something might be zero, you must consider that it might not be to divide by the thing you are dividing.
Suppose you had $x^2 +x -6 = (x+2)(x-3) = 0$
But you are allowed to try. If you do try and get a valid result you can state the result is valid but only if the thing you were dividing by was not $0$.
So I can say if $x-3 \ne 0$ then I can do the following
$\frac {(x+2)(x-3)}{x-3}=\frac {0}{x-3} $ (!!!IF!!! $x -3 \ne 0$)
$x+2 = 0$
$x=-2$.
So I can conclude $x = 2$ IF $x -3 \ne 0$.
But if I do that I MUST immediately also consider the case that will happen if $x-3$ DOES equal $0$.
So I must conclude.
$x = -2$ if $x-3 \ne 0$ and if $x -3$ does equal $0$ then $x = 3$.
So either $x = -2$ or $x = 3$.
However if I try to divide and don't get a valid result, then we can conlcude we weren't allowed to do that.
Example: Let $(x^2 + 2)(x-3) = 0$
And I try: IF $x - 3 \ne 0$ then
$\frac {(x^2 + 2)(x-3)}{x-3} =\frac 0{x-3}$
$x^2 + 2 = 0$
so $x^2 = -2$. .... but that's impossible.
Then I must conclude, hmmm, I guess $x - 3\ne 0$ was false, and instead $x -3 = 0$ so $x = 3$.
Note: that is exactly the result we would have gotten if we had divide by $x^2 + 2$ instead.
Ex: Now we know $x^2 \ge 0$ so $x^2 + 2 \ge 2$ so $x^2 + 2 \ne 0$. So we can do
$\frac {(x^2 +2)(x-3)}{x^2 + 2} = \frac 0{x^2 + 2}$
$x -3 = 0$
$x = 3$.
Now one nifty thing about $a*b = 0$.
If you have that, then you know without a doubt, that one of $a$ or $b$ (or possibly both) is equal to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the Taylor series of $1\over 1+x-2x^2$ centered at $0$ My attempt:
I know that the Taylor series for $f(x) = {1\over 1-x}$ is $\sum_{k=0}^\infty x^k$. We can rewrite $1\over 1+x-2x^2$ as ${1 \over 1 - (-x + 2x^2)} = f(-x+2x^2) = \sum_{k=0}^\infty (-x+2x^2)^k$. However, I was told that this is wrong by my TA, but I'm not sure why. Can anyone explain why this is wrong? What will be the correct answer?
| $$\frac{1}{1 + x - 2x^{2}} = \frac{1}{(1 - x)(1 + 2x)}$$
You can use partial fractions to split this up into two rational functions:
$$\frac{1}{(1 - x)(1 + 2x)} = \frac{A}{1 - x} + \frac{B}{1 + 2x}$$
Some work yields $A = 1/3, B = 2/3$. So:
$$\frac{1}{1 + x - 2x^{2}} = \frac{1/3}{1 - x} + \frac{2/3}{1 - (-2x)} = \frac{1}{3} \cdot \sum_{n = 0}^{\infty}x^n + \frac{2}{3} \cdot \sum_{n = 0}^{\infty}(-2x)^{n}$$
And this simplifies to:
$$\sum_{n = 0}^{\infty} \left(\frac{1 + 2(-2)^{n}}{3}\right)x^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$
then divide the numerator and the denominator by $$x-1$$
But I got again $$\frac{0}{0}$$
| Dividing numerator and denominator by $x-1$ should give $$\frac{9(x^4+x^3+x^2+x+1)-5(x^8+\cdots+x+1)}{(x^{13}+\cdots+1)-(x^8+\cdots+1)-(x^4+\cdots+1)}$$As you say, this gives $0/0$. We can write it as $$\require{cancel}\frac{-5x^8-5x^7-5x^6-5x^5+4x^4+4x^3+4x^2+4x+4}{x^{13}+x^{12}+x^{11}+x^{10}+x^9-x^4-x^3-x^2-x-1}\\=\frac{\cancel{(x-1)}(-5x^7-10x^6-15x^5-20x^4-16x^3-12x^2-8x-4)}{\cancel{(x-1)}(x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+5x^7+5x^6+5x^5+5x^4+4x^3+3x^2+2x+1)}$$Cancel the $x-1$'s and take $x=1$. This gives: $$\frac{-90}{45}=-2$$
We chose to factor the $x-1$ since $1$ was a root of both the numerator and denominator, so $x-1$ must be a factor (and it was contributing the problematic $0$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this limit solution $-∞$? Is this correct?
$$\lim_{x\to\infty} x\left(\frac{\sqrt{x^2+6x+10}}{x^2}-1\right)=-∞$$
Please help me, I don't have the solution and I want to know if I solved it right.
| I would write
$$\frac{1}{x}\left(\frac{\sqrt{x^2+6x+10}-x^2)(\sqrt{x^2+6x+10}+x^2)}{\sqrt{x^2+6x+10}+x^2}\right)$$
and you will get $$\frac{x\left(\frac{1}{x^2}+\frac{6}{x^3}+\frac{10}{x^4}-1\right)}{\sqrt{1+\frac{6}{x}+\frac{10}{x^2}}+1}$$.
This tends to $-\infty$ if $x$ tends to $\infty$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$
How do I continue from here?
| The recursion $10S_n=S_{n-1}+3n$ with $S_0=0$ implies
$$S_n=a\left(\left(1\over10\right)^n-1\right)+bn$$
for some $a$ and $b$. From $S_1={3\over10}$ and $S_2={63\over100}$ we have
$$\begin{align}
{3\over10}&=a\left({1\over10}-1\right)+b\\
{63\over100}&=a\left({1\over100}-1\right)+2b
\end{align}$$
or
$$\begin{align}
9a-10b&=-3\\
99a-200b&=-63
\end{align}$$
which solves to $b=1/3$ and $a=1/27$, so
$$S_n={1\over27}\left(\left(1\over10\right)^n-1\right)+{1\over3}n$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the function is differentiable at $0$? Is the function given by
$\displaystyle f(x) = \begin{cases} \dfrac{1}{x\log(2)} - \dfrac{1}{2^x -1}, \quad &x \neq 0, \\ \dfrac{1}{2} , &x = 0 \end{cases}$
differentiable at $0$ ?
My attempt : $f'(0) = \frac {f(x) -f(0)}{x-0}$
$$\lim _{x\to 0}f'(0) = \lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0\log(2)} - \frac{1}{2^0 -1}}{x- 0}$$
$$\lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0} - \frac{1}{0}}{x- 0}$$
Now I am not able to proceed further.
Any hints /solution will be appreciated.
Thanks in advance
| Yes, it is.
Let us start with the function$$\begin{array}{rccc}g\colon&\mathbb{R}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\frac1{\log2}-\frac x{2^x-1}&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$Since the Taylor series of $\frac{2^x-1}x$ centered at $0$ is$$\log(2)+\frac{\log^2(2)}{2!}x+\frac{\log^3(2)}{3!}x^2+\cdots,$$the Taylor series of $\frac x{2^x-1}$ at $0$ begins with$$\frac1{\log 2}-\frac x2+\frac{\log 2}{12}x^2+\cdots$$Therefore, the Taylor series of $g$ centered at $0$ begins with$$\frac x2-\frac{\log 2}{12}x^2+\cdots$$and your function is $\frac{g(x)}x$.
| {
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"url": "https://math.stackexchange.com/questions/2753893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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product of $f(x)f(y)f(z)\geq1$ for $f(x)=ax^2+bx+c$ with $xyz=1$ and $a+b+c=1$. Let $a,b,c,x,y,z$ be positive reals with $xyz=1$ and $a+b+c=1$.
Let $f(x)=ax^2+bx+c$. Prove $f(x)f(y)f(z)\geq 1$.
Im stuck, I tried a bunch of things but the best ive managed to obtain is:
$f(x)f(y)f(z)\geq 27\sqrt[9]{abc}$ which is useless.
| By Holder:
$$(ax^2+bx+c)(ay^2+by+c)(az^2+bz+c)\geq\left(a\sqrt[3]{x^2y^2z^2}+b\sqrt[3]{xyz}+c\right)^3=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$ This is a question from a past Olympiad paper:
Three positive real numbers $a, b, c$ satisfy the following constraint:
$a+b+c = 1$. Show that the following inequality is true:
$\left(\dfrac{1}{a} + \dfrac{1}{bc}\right) \left(\dfrac{1}{b} + \dfrac{1}{ca}\right)\left(\dfrac{1}{c} + \dfrac{1}{ab}\right) \geq 1728$.
Starting from the L.H.S, I end up with:
$\dfrac{abc(abc+a^2 + b^2 + c^2 + 1) + a^2b^2 + a^2c^2+b^2c^2}{(abc)^3}$.
From the numerator I suspect $a+b+c$ will be factorised, but this is the furthest I have got to.
Using the hint given by timon92:
$\dfrac{1}{a} + \dfrac{1}{bc} = \dfrac{(a+b)(a+c)}{abc}$
and likewise for other two, I end up with:
$(a+b)^2(b+c)^2(a+c)^2 \geq 1728(abc)^3$
$\left((a+b)(b+c)(a+c)\right)^\frac{2}{3} \geq 8abc$
Many thanks in advance.
| Another way, is to use a more general Holder’s inequality:
$$LHS \geqslant \left( \frac1{\sqrt[3]{abc}}+\frac1{\sqrt[3]{a^2b^2c^2}}\right)^3\geqslant (3+9)^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{(k+1)(k+2)}=\frac{1}{n+2}$
Deduce that $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\dfrac{1}{n+2}$
We know $\dfrac{1}{n+2}=\displaystyle \int_{0}^{1}t^{n+2-1}dt$
Now $\displaystyle {n \choose 0}\dfrac{1}{1\cdot2}-{n \choose 1}\dfrac{1}{2\cdot3}+{n \choose 2}\dfrac{1}{3\cdot4}+...(-1)^n{n \choose n}\dfrac{1}{(n+1)\cdot(n+2)}=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)(k+2)}\bigg)=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}-\dfrac{1}{k+2}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+1)}\bigg)-\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\bigg(\dfrac{1}{(k+2)}\bigg)$
$=\displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k\int_{0}^1 t^{k+1-1}dt \space - \space \displaystyle \sum_{k=0}^{n}{n\choose k}(-1)^k \int_{0}^1 t^{k+2-1}dt$
$=\displaystyle \int_{0}^1 t^{1-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt- \int_{0}^1 t^{2-1}\bigg(\sum_{k=0}^{n}{n\choose k}(-t)^k\bigg) dt$
$=\displaystyle \int_{0}^1 t^{1-1}(1-t)^{n}dt \space - \space \int_{0}^1 t^{2-1}(1-t)^{n}dt$
$=\displaystyle \large \beta(1,n+1)$-$\displaystyle \large\beta(2,n+1)$
$=\dfrac{\Gamma(1)\Gamma(n+1)}{\Gamma(1+n+1)}-\dfrac{\Gamma(2)\Gamma(n+1)}{\Gamma(2+n+1)}= \underbrace{\dfrac{1}{n+2}}_{\text{which is exactly what I proved}}$
PS @Chappers Thankyou all users for correcting one nasty mistake.
| $$\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2} = \int_{0}^{1} x^k (1-x)\,dx $$
hence
$$\begin{eqnarray*} \sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{(k+1)(k+2)} &=& \int_{0}^{1}(1-x)\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx\\ &=& \int_{0}^{1}(1-x)^{n+1}\,dx = \int_{0}^{1} x^{n+1}\,dx = \frac{1}{n+2}.\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If the range of $f(x)= \frac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\frac 56, \frac 32\right]$ then what is $c$?
If the range of the function $f(x)= \dfrac{x^2+x+c}{x^2+2x+c}, x\in \mathbb R$ is $\left[\dfrac 56, \dfrac 32\right]$ then $c$ is equal to?
Attempt:
$y= \dfrac{x^2+x+c}{x^2+2x+c}$
For real values of $x$, $\Delta \ge 0$
$\implies 4y^2(1-4c) +1-4y(1-2c) - 4c \ge 0$
What do I do next? I am really unable to understand the concept to be followed after this. Could someone explain that?
The answer is:
$c= 4$
| $$f'(x)=\frac{x^2-c}{(x^2+2x+c)^2}=0 \implies x=\pm \sqrt c$$
$$f(\pm \sqrt c)\in \{3/2, 5/6\}$$
$$f(\sqrt{c})=1-\frac{1}{2(\sqrt{c}+1)} = 5/6\implies c=4$$
$$f(-\sqrt{c})=1+\frac{1}{2(\sqrt{c}-1)} =3/2 \implies c=4 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ Find the sum of the series $\sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}$ using the Taylor series of $e^{x+2 }$.
Answer:
$$
e^{x+2}=1+(x+2)+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\ldots
$$
Integrating, we get
$$
\int e^{x+2} dx=x+\frac{(x+2)^2}{2!}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\ldots
$$
Again integrating, we get
$$
\int e^{x+2} dx=\frac{x^2}{2}+\frac{(x+2)^3}{3!}+\frac{(x+2)^4}{4!}+\frac{(x+2)^5}{5!}+\ldots\\
\Rightarrow \int e^{x+2} dx=\frac{x^2}{2}+(x+2)^3 \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}\\
\Rightarrow \sum_{n=0}^{\infty} \frac{(x+2)^n}{(n+3)!}=\frac{\int e^{x+2} dx-\frac{x^2}{2}}{(x+2)^3}
$$
Is this the sum of the series ?
| As $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
Set $n+3=m$
$$\sum_{n=0}^\infty\dfrac{(x+2)^n}{(n+3)!}=\sum_{m=3}^\infty\dfrac{(x+2)^{m-3}}{m!}=\dfrac1{(x+2)^3}\sum_{m=3}^\infty\dfrac{(x+2)^m}{m!}$$
Now, $$\sum_{m=3}^\infty\dfrac{(x+2)^m}{m!}=e^{x+2}-\sum_{m=0}^2\dfrac{(x+2)^m}{m!}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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find the minimum value given constraints
Let $z$ be a complex number such that $\dfrac{z-i}{z-1}$ is purely imaginary. Find the minimum value of $|z-(2+2i)|$.
Source: ISI 2017 BMATH UGA
$$$$
Attempt:
Since $\dfrac{z-i}{z-1}$ is purely imaginary,
$$$$$$\dfrac{z-i}{z-1}+\overline{\left(\dfrac{z-i}{z-1}\right)}=0$$
This reduces to
$$|z|^2=\Re(z)+\Im(z)$$
This represents the locus of $z$ on the Argand Plane. The minimum value of $|z-(2+2i)|$ will be the shortest distance between any point $z$ lying on $|z|^2=\Re(z)+\Im(z)$ and the point $(2,2)$ on the Argand Plane.
$$$$Unable to recognize the locus represented by $|z|^2=\Re(z)+\Im(z)$.
How to identify locus represented by $|z|^2=\Re(z)+\Im(z)$ without reducing to Cartesian Coordinates?
| We can just use the triangle inequality and following facts:
*
*$|z-(2+2i)| \geq ||z| - 2\sqrt{2}|$
*$\frac{z-i}{z-1}= ia \Rightarrow z = \frac{i-ia}{1-ia} \Rightarrow |z| = \frac{|1-a|}{\sqrt{1+a^2}}$
*$|z|^2 = \frac{(1-a)^2}{1+a^2} \leq 2$, because
$$\frac{(1-a)^2}{1+a^2} \leq 2 \Leftrightarrow a^2-2a+1 \leq 2a^2 + 2 \Leftrightarrow 0 \leq a^2 + 2a + 1 = (a+1)^2$$
All together yields
$$|z-(2+2i)| \geq ||z| - 2\sqrt{2}| \geq |\sqrt{2} - 2\sqrt{2}| = \sqrt{2} \mbox{ where equality is reached for } a = -1$$
Indeed for $z = \frac{i-i(-1)}{1-i(-1)} = \frac{2i}{1+i} = 1+i$
$$|z - 2(1+i)| = |1+i - 2(1+i)| = |1+i| = \sqrt{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x \in \mathbb{Q}$, $\frac{x^2}{2} < 2$ then $\exists a \in \mathbb{Q}$, $a>0$ where $\frac{x^2}{2}There is this demonstration in my calculus book that I can't understand:
I want to prove that:
If $x \in \mathbb{Q}$, $\frac{x^2}{2} < 2$ then $\exists a \in \mathbb{Q}$, $a>0$ where $\frac{x^2}{2}<a^2<2$.
Dem:
If $\frac{x^2}{2} \leq 1$ then $a = \frac{11}{10}$ satisfies the proposition.
If $y = \frac{x^2}{2} >1$, let's take an $n \in \mathbb{N}$ where
$(1+\frac{1}{n})^2<y$ and $(1+\frac{1}{n})^2 <\frac{2}{y} $
We can find this $n$, just see that
$(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
If we take an $n$ where $1+\frac{3}{n} < y^* \implies n>\frac{3}{y^*-1}$ with $y^* = min\{y,\frac{2}{y}\}$ then $(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $.
If $u = 1+\frac{1}{n}$ where $n>\frac{3}{y^*-1}$. Then there will be a term $v$ in the progression $u^2, u^4, u^6,...,u^{2k},...$
where $\frac{x^2}{2} < v <2$
Let $m \in \mathbb{N}$ where $\frac{x^2}{2}<u^{2m}<2$ then $ a = u^k$
Here are my questions:
Question 1:
How we can prove the inequality
$(1+\frac{1}{n})^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
I tried to use induction, but it did not work for me.
Solution: $\frac{1}{n^2} \leq \frac{1}{n} \implies 1 + \frac{2}{n} + \frac{1}{n^2} \leq 1 + \frac{3}{n} $
Question 2:
Why if $u = 1+\frac{1}{n}$ where $n>\frac{3}{y^*-1}$. Then there will be a term $v$ in the progression $u^2, u^4, u^6,...,u^{2k},...$
where $\frac{x^2}{2} < v <2$.
| For Question 2:
The idea is that $u^2,u^4,u^6,\ldots$ is a geometric progression with ratio $r$ greater than $1$ and with:
$$r=u^2= \left(1+\frac{1}{n} \right)^2 < \frac{2}{y}=\frac{2}{\frac{x^2}{2}}$$
and with initial term:
$$u^2 < y=\frac{x^2}{2}$$
To prove it let us suppose that there is not $k$ such that $u^{2k} \in (x^2/2,2)$.
As $u^2 <x^2/2$ and $(u^{2k})$ is an increasing sequence going to $+\infty$ (as the ration is $>1$) there exists $k_0$ such that:
$$u^{2k_0} \leq \frac{x^2}{2} < 2 \leq u^{2(k_0+1)} $$
so:
$$\frac{u^{2(k_0+1)}}{u^{2k_0}} \geq \frac{2}{\frac{x^2}{2}}$$
i.e:
$$\left(1+\frac{1}{n} \right)^2 \geq \frac{2}{y}$$
which is in contradiction with the choice of $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Change of basis matrix for polynomial Given the transformation $T : \mathbb{P}_3 \rightarrow \mathbb{P}_4$ defined as $T(\mathbf{p}(t)) = (-t)\mathbf{p}(t)$
where $\mathbf{p}(t) = t^3 + 6t - 2$
I am first supposed to find $T$, which should be easy enough and gives: $$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t$$
Now, what I can't do is to find the matrix $M$ relative to the bases $B = \{1, t, t^2, t^3\}$ and $C = \{1, t, t^2, t^3, t^4\}$
Looking in my book and other threads here, It seemed like the appropriate way is to insert every element of $B$ into $T$ and set up a matrix with the coefficients.
$$T(\mathbf{p}(1)) = -(1)^4 - 6(1)^2 + 2(1) = -5$$
$$T(\mathbf{p}(t)) = -t^4 - 6t^2 + 2t $$
$$T(\mathbf{p}(t^2)) = -(t^2)^4 - 6(t^2)^2 + 2(t^2) = -t^8 - 6t^4 + 2t^2$$
$$T(\mathbf{p}(t^3)) = -(t^3)^4 - 6(t^3)^2 + 2(t^3) = -t^{12} - 6t^6 + 2t^3$$
which gives
\begin{equation}
M =
\begin{bmatrix}
-5 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & -6 & 2 & 0\\
0 & 0 & 0 & 2\\
0 & -1 & -6 & 0
\end{bmatrix}
\end{equation}
Happy with any helps or hints, thanks
| Since $T(t^n)=-t^{n+1}$ i would say it is:
\begin{equation}
M =
\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation}
Then
\begin{equation}
\begin{bmatrix}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
6 \\
-2
\end{bmatrix}=
\begin{bmatrix}
-1 \\
0 \\
-6 \\
2\\
0
\end{bmatrix}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761585",
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"source": "stackexchange",
"question_score": "2",
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} |
If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$ Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that
$\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$
I have tried bounding them up (assuming that $a\le b\le c$), many inequalities (AM-GM, QM-AM, Cauchy) but nothing has worked. I know that equality is achieved when one of $x,y,z$ is $0$ and the other two are $1$ but that doesn't help.
| The hint.
By Holder
$$\left(\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\right)^2\sum_{cyc}(x^2+y^2)(x^2+y^2+3z^2+2(\sqrt2-1)xy)^3\geq$$
$$\geq\left(\sum_{cyc}(5x^2+2(\sqrt2-1)xy)\right)^3.$$
Thus, it remains to prove that
$$(x+y+z)^2\left(\sum_{cyc}(5x^2+2(\sqrt2-1)xy)\right)^3\geq$$
$$\geq2(9+4\sqrt2)\sum_{cyc}(x^2+y^2)(x^2+y^2+3z^2+2(\sqrt2-1)xy)^3,$$
which is true by $uvw$ (https://math.stackexchange.com/tags/uvw/info).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find smallest integer $n$ so that $\sqrt{\frac{1^2+2^2+3^2+...+n^2}n}$ is an integer
Find smallest positive integer $n$ (not equal to $1$) so that $\frac{1^2+2^2+3^2+...+n^2}n$ is a perfect square.
I tried to use : $1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}6$
That gives us $\frac{1^2+2^2+3^2+...+n^2}n=\frac{(n+1)(2n+1)}6$
I tried to brute-force it using python, and I got $n=337$, but I should probably use modular arithmetic (or something else).
| You want to solve
$$(n+1)(2n+1)=6m^2.$$
This is
$$16n^2+24n+8=48m^2$$
or
$$(4n+3)^2-48m^2=1.$$
This is basically Pell's equation $x^2-48y^2=1$, but with the extra
stipulation $x\equiv3\pmod 4$. The positive solutions of this
Pell equation are given by
$$x_k+y_k\sqrt{48}=(7+\sqrt{48})^k.$$
The solutions with $x_k\equiv 3\pmod 4$ are those with $k$ odd. The
first is $x_1=7$ giving $n=1$. The next is $x_3=1351$ giving $n=337$.
The next is $x_5=262087$ giving $n=65521$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
show this $\prod(a^3+b^3+ab)^2\ge (\sum a^2b)^3( \sum ab^2)^3$ let $a,b,c>0$ show that :$$(a^3+b^3+abc)^2(b^3+c^3+abc)^2(c^3+a^3+abc)^2-(a^2b+b^2c+c^2a)^3\cdot(ab^2+bc^2+ca^2)^3\ge 0$$Maybe there's some kind of identity in there.so I use wolfampha Calculated the results and found no identity. and the results are very complex. Maybe this inequality has a good solution.such as this following identity
$$(x+y+z)(xy+yz+zx)=(x+y)(y+z)(z+x)+xyz$$
| We need to prove that
$$\prod_{cyc}\left(\frac{a^2}{bc}+\frac{b^2}{ac}+1\right)^2\geq\left(\sum_{cyc}\frac{a}{b}\right)^3\left(\sum_{cyc}\frac{a}{c}\right)^3.$$
Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Hence, $xyz=1$ and we need to prove that
$$\prod_{cyc}\left(\frac{x}{z}+\frac{y}{x}+1\right)^2\geq(x+y+z)^3(xy+xz+yz)^3$$ or
$$\prod_{cyc}(x^2+yz+zx)^2\geq(x+y+z)^3(xy+xz+yz)^3.$$
Now, since by AM-GM
$$\sum_{cyc}x^4y^2=\frac{1}{6}\sum_{cyc}(4x^4y^2+y^4z^2+z^4x^2)\geq\sum_{cyc}x^3y^2z,$$ we obtain:
$$\prod_{cyc}(x^2+yz+zx)=\sum_{cyc}(x^4y^2+x^3y^3+x^4yz+2x^2y^2z+3x^3z^2y+x^2y^2z^2)\geq$$
$$\geq\sum_{cyc}(x^3y^3+x^4yz+3x^3y^2z+3x^3z^2y+x^2y^2z^2).$$
Thus, it's enough to prove that
$$\left(\sum_{cyc}(x^3y^3+x^4yz+3x^3y^2z+3x^3z^2y+x^2y^2z^2)\right)^2\geq(x+y+z)^3(xy+xz+yz)^3xyz.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$(27v^6-27uv^2w^3+3w^6+27u^3w^3-27uv^2w^3+3w^6+27uv^2w^3-9w^6+3w^6)^2\geq$$
$$\geq729u^3v^6w^3$$ or $f(u)\geq0,$ where
$$f(u)=(v^6-uv^2w^3+u^3w^3)^2-u^3v^6w^3.$$
But $$f'(u)=2(v^6-uv^2w^3+u^3w^3)w^3(3u^2-v^2)-3u^2v^6w^3=$$
$$=2uw^6(u^2-v^2)(3u^2-v^2)+v^6w^3(3u^2-2v^2)\geq0,$$
which says that $f$ increases.
Thus, it's enough to prove $f(u)\geq0$ for a minimal value of $u$,
which happens for equality case of two variables.
Since the last inequality is symmetric and homogeneous, we can assume $y=z=1$,
which gives
$$(x^4+8x^3+9x^2+8x+1)^2\geq(2x^2+5x+2)^3x.$$
Let $x^2+1=2tx$.
Hence, by AM-GM $t\geq1$ and we need to prove that
$$(4t^2+16t+7)^2\geq(4t+5)^3$$ or
$$(t-1)(4t^3+20t^2+38t+19)\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A System of Simultaneous Pell Equations
Are $0$ and $\pm1$ the only integer solutions for which both $\sqrt{24n^2+1}$ and $\sqrt{48n^2+1}$ are simultaneously integers ?
Whilst pondering upon the Biblical concept of the Jubilee year, I couldn't help but notice that a time period of $7^2=49$ years, apart from approximating half of a decimal century of $10^2=100$ years, also lies conspicuously close to a third of a duodecimal “century” of $12^2=144$ years. In other words, I was left with solving the system of Diophantine equations $$\dfrac{x^2}3+1~=~y^2~=~\dfrac{z^2}2-1,$$ which, after a bit of basic number theory, boiled down to the system of Pell equations described above. Then, a Mathematica search of depth up to $10^4$ into the $($ periodic $)$ continued fraction expansion of $~\lim\limits_{n\to\infty}\sqrt{\dfrac{48n^2+1}{24n^2+1}}~=~\sqrt2~$ failed to reveal any other solutions, save for the ones mentioned earlier, implying that any other possible values of n possess at least $3,828$ digits.
*
*A $2007$ paper by Mihai Cipu and Maurice Mignotte shows that there might be at most one such positive solution, for $n>1.$
*In the same year, the $34^{th}$ volume of the Annales Mathematicae et Informaticae, published by the Eszterhazy Karoly University of Applied Sciences, contains a relevant paper by Laszlo Szalay on the resolution of simultaneous Pell equations.
| Here a general solution of this problem using Fermat's method of infinite descent (the simplest version of it currently very sophisticated).
From
$$24n^2+1=x^2\\48n^2+1=y^2$$ we have, since $x$ and $y$ are odd and $x\lt y$
$$(x+2h)^2=y^2\Rightarrow h^2+hx-6n^2=0\Rightarrow h= \frac{-x\pm\sqrt{x^2+24n^2}}{2}=$$
Because of $24n^2=x^2-1$ the radicand becomes $2x^2-1$ and must be equal to $t^2$ so we have $$2x^2-1=t^2\iff1+t^2=2x^2$$
From identity $$(r^2-s^2+2rs)^2+(r^2-s^2-2rs)^2=2(r^2+s^2)^2$$ comes the general solution of the equation $$X^2+Y^2=2Z^2..........(*) $$ it follows, because $(1,t,x)$ is a solution of $(*)$
$$\begin{cases}r^2-s^2+2rs=t \\r^2-s^2-2rs=1\iff2r^2=1+(r+s)^2 \\r^2+s^2=x\Rightarrow r\text{ and }s\lt x \end{cases}$$
This shows another solution $(1,r+s,r)$ of $(*)$ with $r\text{ and }s\lt x$ so we have $$\begin{cases}r_1^2-s_1^2+2r_1s_1=r+s \\r_1^2-s_1^2-2r_1s_1=1\iff2r_1^2=1+(r_1+s_1)^2 \\r_1^2+s_1^2=r\Rightarrow r_1\text{ and } s_1\lt r\end{cases} $$ Again another solution of $(*)$ with $r_1\text{ and }s_1\lt r$ and the procedure can be iterated how many times you want.
By descent the proof is done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Order of eigenvalues for converting quadratic equations In my textbook there is a way of converting a quadratic equation to its canonical form using eigenvalues. I'm going to give you a rough description of the method:
The quadratic equation is of the form:
$a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_12xy+2a_{23}yz+2a_{31}xz+2a_{14}x+a_{24}y+a_{34}z+a_{44}=0$
Which can be express as following
$\begin{pmatrix}x & y & z\end{pmatrix}$$\begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$+a_{44}=0$
Let $A=\begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$
because A is symmetric there exists an orthogonal matrix $P$ such that
$P^{-1}AP=$ $\begin{pmatrix}
t_{1} & 0 & 0\\
0 & t_{2} & 0\\
0 & 0 & t_{3}
\end{pmatrix}$, $t_{1},t_{2},t_{3}$ the eigenvalues of $A$
Now let $B=PB'$, $B'=$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$
then $\begin{pmatrix}x' & y' & z'\end{pmatrix}$$\begin{pmatrix}
t_{1} & 0 & 0\\
0 & t_{2} & 0\\
0 & 0 & t_{3}
\end{pmatrix}$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+a_{44}=0$$\Leftrightarrow$
$t_{1}x'^2+t_2y'^2+t_3z'^2$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+a_{44}=0$
My question now is the following, is there a specific order which determines which eigenvalue is number 1,2 etc? Because if for example an eigenvalue is zero this means that one of the $x^2,y^2,z^2$ is zeroed as well. How do you determine what eigenvalue to asign to each of the $x^2,y^2,z^2$?
| The order of the eigenvalues is determined by how you want the figure to be oriented in space. To give you an example, consider a simple case: ellipse in 2D ($a_{33} = a_{23} = a_{31} = a_{34} = 0$) centered at the origin ($a_{14} = a_{24} = 0$ and $a_{44} = -1$) with major axis along the $x$ axis ($a_{11} = 1, a_{22} = 1 / b^2$, $a_{12} = 0$, $b<1$), that is
$$
x^2 + \frac{y^2}{b^2} = 1 \tag{1}
$$
In this case
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 1/b^2 \end{pmatrix} \tag{2}
$$
This is a trivial example in the sense that the matrix is already diagonal,
$$
P = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}, ~~ T = \begin{pmatrix}1 & 0 \\ 0 & 1/b^2 \end{pmatrix} \tag{3}
$$
But here's the idea, image that you decide to organize your eigenvalues as
$$
P_{\rm new} = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, ~~ T_{\rm new} = \begin{pmatrix}1/b^2 & 0 \\ 0 & 1 \end{pmatrix} \tag{4}
$$
In this case your coordinate system is defined as
$$
\begin{pmatrix}x' \\ y' \end{pmatrix} = P_{\rm new} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}y \\ x \end{pmatrix} \tag{5}
$$
So that the ellipse is written as
$$
y'^2 + \frac{x'^2}{b^2} = 1 \tag{6}
$$
Which is now an ellipse with major axis along the $y'$ axis! so changing the order of the eigenvalues we change the orientation of the figure in space
| {
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Solving inverse function problems The following is my (likely completely incorrect) attempt.
$$f(x)= 3x + 5, \qquad \qquad g(x)= 1 - 2x$$
*
*Show that $f{^{-1}}(2) < g(-3)$
$$\frac{1}{3\cdot 2+5} < (1-2\cdot(-3))\implies\frac{1}{11} < 7$$
*Find the value of $a$ such that $f(2a) = g{^{-1}}(a)-2$
$$6a+5 = \frac{1}{-1-2a}$$
$$6a = \frac{1}{4-2a}$$
$$8a = \frac{1}{4}\implies a = \frac{1}{32}$$
*Solve $\frac{f(x)}{2} + \frac{g(x)}{3} = -23$
$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$
| You are correct that you have incorrectly answered $(1)$. Note that $f^{-1}$ most likely denotes the inverse of $f$, not its reciprocal.
Since $$f(x)-5=3x\implies x=\frac{f(x)-5}3$$ the inverse function of $f$ is $$f^{-1}(x)=\frac{x-5}3$$ so $$f^{-1}(2)=\frac{2-5}3=-1<g(-3)$$ as required.
| {
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Prove $3^{2n+1} + 2^{n+2} = 7m$ I'm trying to prove that $3^{2n+1} + 2^{n+2}$ is a multiple of 7 by using induction.
So I started to prove it for $n=1$: $3^{2(1)+1}+2^{1+2}=3^3+2^3=27+8=35=7(5)$.
Next, try to prove that the statement being true $n=k$ implies it being true for $n=k+1$. Thus:
$3^{2(k+1)+1}+2^{(k+1)+2} = 3^{2k+3}+2^{k+3} = (3)(3)(3^{2k+1})+(2)(2^{k+2}) = 9(3^{2k+1})+2(2^{k+2})$
I feel like I'm almost there, if I could've factor 9 and 2 somehow I could say that $3^{2k+1}+2^{k+2}=7m$ for some integer $m$, but I can't find a way to do it. What am I missing? Or did I do a blunder somewhere along the road?
Thanks in advance.
| Alt. hint (no induction): $3\;\cdot 9^n+ 4 \cdot 2^n=3\cdot (7 + 2)^n+4 \cdot 2^n=3 \cdot 7 \cdot (\ldots)+ (3 + 4) \cdot 2^n\,$.
| {
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Surface integral using Gauss theorem The exercise consists in determining
$$\iint_Y \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S$$
where
$$\mathbf{F}= (x^2 yz + x \sin z , x^2 + y(1 - \sin z ), x + y - xy z^2)$$
and $Y$ is the part of the conical surface $4x^2 + (y-1)^2 = z^2$ which lies between $z = 1$ and $z = 2$. The normal vector points away from the $z$-axis.
I have tried to solve it using Gauss theorem, and I think this is the correct approach. Of course, the surface is not closed, and I need to add a "lid" and "bottom" to $Y$ to use Gauss theorem in the first place. Let $\sigma$ denote the lid, $\gamma$ the bottom and $K$ the whole surface. This yields
$$\iint_Y \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S + \iint_{\sigma} \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S + \iint_{\gamma} \mathbf{F} \cdot \mathbf{N} \ \mathrm{d}S = \iiint_K \nabla \cdot \mathbf{F} \ \mathrm{d}x\mathrm{d}y\mathrm{d}z$$
where $\nabla \cdot \mathbf{F}$ is the divergence, which can be determined and it is $\nabla\cdot\mathbf{F} = 1$. This produces the triple integral
$$\iiint_{4x^2 + (y-1)^2 \leq z^2} \ \mathrm{d}x\mathrm{d}y\mathrm{d}z$$
This is where I get stuck.
I am unsure about how to solve the triple integral, and how to further solve the surface integrals for the lid and the bottom surface.
| $$\iiint_{ \{ (x,y,z) \in \mathbb{R}^3: 4x^2 + (y-1)^2 = z^2, 1 \leq z \leq 2 \} } \mathrm{d}x \mathrm{d}y \mathrm{d}z =\\
= \int_{z=1}^{z=2} \left( \int_{ \{ (x,y) \in \mathbb{R}^2: 4x^2 + (y-1)^2 = z^2 \} } \mathrm{d}x \mathrm{d}y \right) \mathrm{d}z =\\ = \int_{z=1}^{z=2}\pi\cdot\frac{z}{2}\cdot z\ dz=\frac{7\pi}{6}$$
Top ($z=2$)
$$4x^2 + (y-1)^2= 2 \Leftrightarrow \frac{x^2}{\displaystyle \frac{1}{2}}+\frac{(y-1)^2}{2} = 1$$
so a parametrization is
$$\sigma: [0,1] \times [0,2\pi] \to \left( \frac{1}{\sqrt{2}}r\cos\theta,1+\sqrt{2}r\sin\theta,2 \right) \in \mathbb{R}^3$$
and a normal vector is
$$\mathbf{\nu} = \frac{ \partial \sigma }{\partial r} \times \frac{ \partial \sigma}{\partial \theta} = (0,0,r)$$
thus
$$\iint_{\mathrm{top}} \mathbf{F} \cdot \mathbf{\nu} \ \mathrm{d}x \mathrm{d}y = \iint_{ \{ (x,y) \in \mathbb{R}^2: 4x^2 + (y-1)^2 \leq 2 \} } \mathbf{F} \cdot \mathbf{\nu} \ \mathrm{d}x \mathrm{d}y =\\
= \iint_{ \{ (r,\theta): 0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi \} } F ( \sigma( r,\theta )) \cdot \nu(r,\theta) \cdot r \ \mathrm{d}r \mathrm{d} \theta =\\
= \int_{ \theta=0 }^{ \theta = 2\pi } \left[ \int_{r=0}^{r=1} \left( -\frac{3}{\sqrt{2}} r^3 \cos\theta + (\sqrt{2}-4) r^3 \sin \theta + r^2 \right) \mathrm{d}r \right] \mathrm{d}\theta = \frac{2\pi}{3}$$
Bottom ($z=1$)
$$4x^2+(y-1)^2 = 1 \Leftrightarrow \frac{x^2}{\displaystyle \frac{1}{4}} + (y-1)^2 = 1$$
so a parametrization is
$$\sigma:[0,1] \times [0,2\pi] \to \left( \displaystyle \frac{1}{2}r \cos \theta,1-r \sin \theta,1 \right) \in \mathbb{R}^3$$
and a normal vector is
$$\mathrm{\nu} = \frac{\partial\sigma}{\partial r} \times \frac{ \partial \sigma}{\partial \theta} = \left( 0,0,-\frac{r}{2} \right)$$
thus
$$\int_{\mathrm{bottom}} \mathbf{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y = \iint_{ \{ (x,y) \in \mathbb{R}^2:4x^2+(y-1)^2\leq 1 \} } \mathbf{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y=\\
= \iint_{ \{ (r,\theta): 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} } F( \sigma(r,\theta)) \cdot \nu (r,\theta) \cdot r \ \mathrm{d}r \mathrm{d} \theta =\\
= \int_{\theta=0}^{\theta=2\pi} \left[ \int_{r=0}^{r=1} \left( -\frac{r^2}{2} + \frac{r^3}{2} \sin \theta - \frac{r^4}{4}\sin\theta\cos\theta \right) \mathrm{d}r \right] \mathrm{d}\theta = -\frac{2\pi}{6}$$
Thus
$$\iint_{Y} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y = \iiint \nabla \cdot \mathrm{F} \ \mathrm{d}x \mathrm{d}y \mathrm{d}z - \iint_{\mathrm{top}} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y - \iint_{\mathrm{bottom}} \mathrm{F} \cdot \mathrm{\nu} \ \mathrm{d}x \mathrm{d}y=\\
= \frac{7\pi}{6} - \frac{2\pi}{3} + \frac{2\pi}{6} = \frac{5\pi}{6}$$
| {
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How do I show that a complex number $z$ is a root of $z^6-1$? This problem would be less confusing to me if one was not subtracted from $z^6$.
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:
$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$
$$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$
$$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$
$$r=1$$
$$\theta=Tan^1(\frac{b}{a})$$
$$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$
$$\theta=\frac{\pi}{6}$$
$$\therefore z=1\angle{\frac{\pi}{6}}$$
This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$
$$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$
Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$
The equation becomes...
$$(-\frac{13}{32})-1$$
Which equates to...
$$-\frac{45}{32}$$
Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$
$$z^6=(-\frac{45}{32}\angle{0rad})^6$$
$$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$
$$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$
$$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$
I get...
$$z^6=(-\frac{45}{32})^6\angle{\pi}$$
Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?
| Alt. hint (not using the polar form): $\;z=\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\,$, so $\,|z|=1\,$, and:
$$z^2 = \left(\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\right)^2= \dfrac{1}{4}+ 2i \cdot \dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2} + i^2 \dfrac{3}{4}=-\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2} = - \overline{z}$$
Multiplying with $\,z\,$ gives$\,z^3 = - z \cdot \overline z = -|z|^2 = -1\,$.
| {
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For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes. Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$
By induction: $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$. Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note: This argument was taken from Wikipedia.
(2) $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$
If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$
(3) For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$
From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$
(4) Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:
${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.
| We have the following Stirling approximations:
*
*$\sqrt{2\pi n}n^ne^{-n} \leq n! \leq e \sqrt{n} n^n e^{-n} $ for all $n \in \mathbb{N}$
$${{3n} \choose {n}} \geq \frac{\sqrt{2\pi 3n}(3n)^{3n}e^{-3n}}{e \sqrt{n} n^n e^{-n} \cdot e \sqrt{2n} (2n)^{2n} e^{-2n}} =\frac{\sqrt{3\pi}}{e^2\sqrt{n}} \frac{27^n}{4^n} > \frac{\sqrt{3\pi}}{e^2} \frac{\left(\frac{3}{2}\right)^n}{\sqrt{n}}4^n$$
So, if $n$ is large enough, the inequality holds.
| {
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The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c} $
The ellipse of the maximum area contained between the curves $\frac{\pm 1}{x^2+c}, c > 0 $
It seems to me that the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ of the maximum area when $ a = c; b = \frac{1}{c} $
That is, it looks like this:
The problem is that I'm not sure that an ellipse here can be a finite area
I will be glad to any hint or solution
| Given the function for the ellipse as: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}$$ It is clear that $b$ must be equal to $\frac1c$ so that the ellipse must be below the curve $\frac{\pm1}{x^2+c}$ Solving for $y$ in $(1)$ gives us: $$\left\{y\to\frac{b \sqrt{a^2-x^2}}{a}\right\},\left\{y\to -\frac{b \sqrt{a^2-x^2}}{a}\right\}$$ We know that $b=\frac1c$, therefore, half of the ellipse can be written as: $$y=\frac{\sqrt{a^2-x^2}}{a c}$$ Now, in order for the ellipse to be under the curve, we must ensure that: $$\frac{\sqrt{a^2-x^2}}{a c}<\frac{1}{c+x^2}$$ Which can be rewritten as: $$\frac{1}{c+x^2}-\frac{\sqrt{a^2-x^2}}{a c}>0$$ Now the equality is achieved when for the following solutions in terms of $x$: $$\left\{\{x\to 0\},\left\{x\to -\frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{-a \sqrt{a^2+4 c}+a^2-2 c}}{\sqrt{2}}\right\},\left\{x\to -\frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\},\left\{x\to \frac{\sqrt{a \left(\sqrt{a^2+4 c}+a\right)-2 c}}{\sqrt{2}}\right\}\right\}$$ Meaning, these are values of $x$ in the $xy$-plane where the ellipse and the curves "intersect". Now, we want this solutions equal to $0$, so that the graphs of the functions only ever intersect at $x=0$ (if that is meaningful). Now, we get the solutions as: $$a\to\pm\frac{\sqrt{c}}{\sqrt{2}}$$ Since $a$ must be greater than 0, we reject the negative answer. Now we have $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, now half of the ellipse is defined by: $$y=\frac{\sqrt{2} \sqrt{\frac{c}{2}-x^2}}{c^{3/2}}$$ Square both sides, and you get the function for the ellipse as: $$y^2=\frac{2 \left(\frac{c}{2}-x^2\right)}{c^3}$$
The area the ellipse is given by:
$$A=2\int_{-a}^{a}\frac{b \sqrt{a^2-x^2}}{a}dx$$
Given $a=\frac{\sqrt{c}}{\sqrt{2}}$ and $b=\frac1c$, the maximum area of the ellipse should be:
$$A=2\int_{-\sqrt{\frac{c}{2}}}^{\sqrt{\frac{c}{2}}}\frac{\sqrt2\sqrt{c/2-x^2}}{c^{3/2}}dx=2\frac\pi{2\sqrt{2c}}=\frac\pi{\sqrt{2c}}$$
| {
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What is the probability of getting at least two different numbers from first roll when three six-sided dice are rolled a second time? I roll three six sided fair dice. Three different numbers show. What is the probability that at least two different numbers show in the second roll that occurred in the first roll? For example, first roll (1,2,3), second roll (3,1,4)?
Attempt
\begin{align*}
\Pr(\text{of at least two the same}) & = 1 - \Pr(\text{exactly $0$ the same}) - \Pr(\text{exactly $1$ the same})\\
& = 1 - \frac{3}{6} \cdot \frac{3}{6} \cdot \frac{3}{6} - \frac{1}{3} \cdot \frac{3}{6} \cdot \frac{3}{6} \cdot 3\\
& = 0.625
\end{align*}
In the second term, I am multiplying by $3$ to account for the number of placements the number can be in. Is this approach correct? Thanks!
| Your answer is wrong.
The probability of "exactly $0$ the same" is indeed $\frac36\frac36\frac36=\frac{27}{216}$ as you suggest.
But the probability of "exactly $1$ the same" equals:$$\left[\frac16\frac16\frac16+3\frac16\frac16\frac36+3\frac16\frac36\frac36\right]3=\frac{111}{216}$$
If e.g. the first roll was $(1,2,3)$ then between the brackets $[$ and $]$ you find the probability that number $1$ appears in the second roll and the numbers $2$ and $3$ do not. The first term of the three gives the probability on roll $(1,1,1)$, the second term gives the probability of a roll with exactly twice a $1$ in it and the third term gives the probability of a roll with once a $1$ in it.
The final result is: $$1-\frac{27}{216}-\frac{111}{216}=\frac{78}{216}\approx0.3611$$
| {
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Computing the coefficient of the term of a certain degree in a polynomial Given the polynomial
${1\over8}((1+z)^9 + 3(1-z)^4(1+z)^5 + (1-z)^6(1+z)^3)$
(which is the weight enumerator of a code)
how do I find out the coefficient of $z^2$?
The solution given is ${1 \over 8}(36-12+0) = 3$.
I got $36$ for the $z^2$ coefficient of $(1+z)^9$ using the Binomial Theorem, but I don't know how to get $-12$ for the $z^2$ coefficient of $3(1-z)^4(1+z)^5$. By using the Binomial Theorem separately on $(1-z)^4$ and $(1+z)^5$ I get the following two polynomials, repsectively:
$z^4-4z^3+6z^2-4z+1$
$z^5+5z^4+10z^3+10z^2+5z+1$
I am unsure what to do next, or even if this is going in the right direction.
| We can work this out by establishing a general formula for
$$(1-z)^m(1+z)^n=(1-mz+\frac{m(m-1)}2-\cdots)(1+nz+\frac{n(n-1)}2z^2+\cdots).$$
The $z^2$ term has the coefficient
$$\frac{m(m-1)}2-mn+\frac{n(n-1)}2=\frac{(m-n)^2-(m+n)}2.$$
Hence,
$$9,0\to 36\\4,5\to-4,\\6,3\to0$$ and the solution is
$$\frac{36-3\cdot4}8.$$
| {
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If $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$ , then $\triangle ABC$ is a right triangle
Show that the triangle whose angles satisfy the equality
$$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$
is right-angled.
I've tried many times, but was unsuccessful.
| Hint: Show that
\begin{eqnarray*}
1-\cos^2 A-\cos^2 B-\cos^2C+ 2\cos A \cos B \cos C =0
\end{eqnarray*}
for any triangle.
| {
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Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$
i got this problem in olympiad mathematics .
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{66}\right)$
$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{33}\right)$
$\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$
Then what will I do?
| $\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{1}{i}+\frac{1}{101-i}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{101-i+i}{(i)(101-i)}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{101}{(i)(101-i)}$
$\frac{a}{b}=101.\left(\sum_{i=34}^{50} \frac{1}{(i)(101-i)}\right)$
therefore $101\mid a$.
thanks,Carl Schildkraut
| {
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Evaulating the trigonometric integral $\int \frac{1}{(x^2+1)^2} \, dx$ Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2} \, dx \\
\end{eqnarray*}
Answer:
To do this, I let $x = \tan u$. Now we have $dx = \sec^2 u du$.
\begin{eqnarray*} \int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\sec^2{u} \, du}{(\tan^2{u} + 1)^2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{1}{\sec^2{u}} \, du
= \int \cos^2{u} \, du \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \int \frac{\cos{(2u)} + 1}{2} \, du
= \frac{\sin(u)}{4} + \frac{u}{2} \\
\int \frac{1}{(x^2+1)^2} \, dx &=& \frac{\sqrt{1 - \cos^2{u}}}{4} + \frac{u}{2} \\
\end{eqnarray*}
Now, I think I am right so far but I do not know have to get rid of the $u$ in
the $\cos^2(u)$ term. Please help.
Thanks
Bob
| I suggest that you compute your primitive like this:\begin{align}\int\frac1{(x^2+1)^2}\,\mathrm dx&=\int\frac{x^2+1}{(x^2+1)^2}\,\mathrm dx-\int\frac{x^2}{(x^2+1)^2}\,\mathrm dx\\&=\arctan(x)+\frac12\int x\frac{-2x}{(x^2+1)^2}\mathrm dx\\&=\arctan(x)+\frac12\cdot\frac{x}{x^2+1}-\frac12\int\frac1{x^2+1}\,\mathrm dx.\end{align}
| {
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Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$
Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
Solution I tried:
$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$
$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$
I did not find how do multiply terms such that i get my result. Help me
| $$S_{(10,2)} = \sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
You can generalize the question as:
\begin{align}
S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{(n+r)-k}{r}\\
\end{align}
Set $(n+r) = m\,$ and we get
\begin{align}
S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{m-k}{r}= \binom{m+1}{2r+1}\\
\end{align}
(See Another form of the Chu–Vandermonde identity and this proof )
So
$$S_{(n,\,r)} = \binom{n+r+1}{2r+1}$$
In your case
$$S_{(10,2)} = \binom{10+2+1}{2\cdot 2+1} = \binom{13}{5} = 1287$$
| {
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Is this way of evaluating $\lim_{x\to 0}\frac{x-\sin x}{x^3}$ rigorous? Assume that you want to evaluate $$\lim_{x \to 0} \frac{x-\sin x}{x^3}.$$ Of course, you may apply $\textbf{de l'Hôpital's Rule}$, that's natural and simple. But what if you are required to avoid using the more advanced knowledge such as “de l'Hôpital”，“Taylor” and so on？Now,someone gives a solution for the problem as follows.
Since$$\begin{align*}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}&=\lim\limits_{x \to 0} \dfrac{2x-\sin 2x}{(2x)^3}\\[4pt]&=\lim\limits_{x \to 0} \dfrac{2x-2\sin x\cos x}{(2x)^3}\\[4pt]&=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x+\sin x-\sin x\cos x}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x(1-\cos x)}{x^3}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\lim\limits_{x \to 0} \dfrac{\sin x}{x}\cdot\lim_{x \to 0}\frac{1-\cos x}{x^2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{4}\cdot 1\cdot \frac{1}{2}\\[4pt] &=\frac{1}{4}\lim\limits_{x \to 0} \dfrac{x-\sin x}{x^3}+\frac{1}{8},\end{align*}$$ which sets up an equation on the limit, hence, by solving the equation, we obtain that$$\lim_{x \to 0} \frac{x-\sin x}{x^3}=\frac{1}{6}.$$
What about the solution? Is it rigorous? Hope to read your comments. Thank you!
| There's a typo: when you wrote (after the third equality) $\sin\cos x$, you meant $\sin x\cos x$.
This proves correctly that if the limit exists then it must be equal to $\frac16$, but that's all.
| {
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Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$
| I am continuing from where you left -
$$\int \dfrac {dx}{1+\sin x}$$
$$=\int \dfrac {dx}{1+\dfrac {2\tan (\dfrac {x}{2})}{1+\tan^2 (\dfrac {x}{2})}}$$
$$=\int \dfrac {1+\tan^2 (\dfrac {x}{2})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$
$$=\int \dfrac {1+(\dfrac {\sin^2 (\dfrac {x}{2})}{\cos^2 (\dfrac {x}{2})})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$
$$=\int \dfrac {\dfrac{\sin^2(\dfrac{x}{2})+\cos^2(\dfrac{x}{2})}{\cos^2(\dfrac{x}{2})}}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$
$$=\int \dfrac {\dfrac{1}{\cos^2(\dfrac{x}{2})}}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$
$$=\int \dfrac {\sec^2(\dfrac{x}{2})}{(1+\tan (\dfrac {x}{2}))^2}\,dx$$
$$ Let,$$
$$\quad \quad \quad \quad \quad \quad \quad \quad 1+\tan(\dfrac{x}{2})=z$$
$$\quad \quad \quad \quad \quad \quad \quad \quad \implies(\sec^2(\dfrac{x}{2}))dx=2dz$$
$$=\int 2\dfrac {dz}{z^2}$$
$$=\dfrac{-2}{z}+c$$
$$=-\dfrac {2}{(1+\tan (\dfrac {x}{2}))}+c$$
| {
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} |
Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| Hint. Notice that $\int\frac{1}{\sqrt{1-y^2}}\mathrm{d}y=\sin^{-1}(y)$, so you could try writing $y^2$ as $y^2-1+1$ and splitting into two integrals:
$$\int\frac{y^2}{\sqrt{1-y^2}}\mathrm{d}y=\int\left(\frac{y^2-1}{\sqrt{1-y^2}}+\frac{1}{\sqrt{1-y^2}}\right)\mathrm{d}y\\=\int\left (\frac{1}{\sqrt{1-y^2}}-\sqrt{1-y^2}\right )\mathrm{d}y=\int\frac{1}{\sqrt{1-y^2}} \mathrm{d}y-\int\sqrt{1-y^2} \mathrm{d}y$$
Solving $\int\sqrt{1-y^2} \mathrm{d}y$ can be made via trigonometric substitution, setting $y=\sin u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Evaluating $\lim_{(x,y) \to (\pi,0)} {\cos(x) + 1 + {y^2/2} \over (x - \pi)^2 + y^2}$ I'm struggling to evaluate this limit problem:
$$\lim\limits_{(x,y) \to (\pi,0)} {\cos x + 1 + {y^2/2} \over (x - \pi)^2 + y^2}$$
I've tried this with paths $x=\pi$, $y=0$, $y=x-\pi$, ... and so far all of them have resulted in $1 \over 2$. However when I try to evaluate this in WolframAlpha, it says that the limit does not exist.
I haven't found any paths that result in a value other than $1/2$. How can I prove that this limit does not exist?
| You should try $x' = x - \pi$ and then $$\lim\limits_{(x,y) \to (\pi,0)} {\cos x + 1 + {y^2/2} \over (x - \pi)^2 + y^2} = \lim\limits_{(x',y) \to (0,0)} {1-\cos x' + {y^2/2} \over (x')^2 + y^2}$$
$$= \lim\limits_{(x,y) \to (0,0)} {1-\cos x \over x^2 + y^2} + \frac{1}{2} {{y^2} \over x^2 + y^2}$$
$$= \lim\limits_{(x,y) \to (0,0)} { 2\sin^2(x/2) \over x^2 + y^2} + \frac{1}{2} { y^2 \over x^2 + y^2}$$
$$= \lim\limits_{(x,y) \to (0,0)} \frac{\frac{1}{2}x^2}{x^2+y^2} \left ({\sin^2(x/2)\over (x/2)^2} \right) + {\frac{1}{2}y^2\over x^2 + y^2}$$
$$= \lim\limits_{(x,y) \to (0,0)} \frac{\frac{1}{2}x^2}{x^2+y^2} \left (1+{\sin^2(x/2)\over (x/2)^2} - 1 \right) + {\frac{1}{2}y^2\over x^2 + y^2}$$
$$= \lim\limits_{(x,y) \to (0,0)} \frac{1}{2}\frac{x^2 + y^2}{x^2+y^2} + \lim\limits_{(x,y) \to (0,0)} \frac{1}{2}\frac{x^2}{x^2+y^2} \left ( {\sin^2(x/2)\over (x/2)^2} - 1\right ) $$
$$= \frac{1}{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which one of the following are correct?
Let, $A= \left[ {\begin{array}{cc} -1 & 2 \\ 0 & -1 \\
\end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$
$(C) B^2 = A$
$(D) B^2 = B$
Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So, options (A) and (C) can be eliminated. How do I eliminate further?. By Cayley-Hamilton Theorem $A^2+2A+I=0$. I found $A^3,A^4,...$ using Cayley-Hamilton Theorem. It was time-consuming. Given matrix is not diagonalizable. Can you please help me?
| The Jordan normal form of $A$ is:
$$A = PJP^{-1}$$
where $J = \begin{bmatrix}-1 & 1 \\ 0 & -1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 0 \\ 0 & \frac12\end{bmatrix}$.
For any entire function $f$ we have
$$f(A) = f(PJP^{-1}) = Pf(J)P^{-1} = P\begin{bmatrix}f(-1) & f'(-1) \\ 0 & f(-1)\end{bmatrix}P^{-1} $$
In particular, since $B = A + A^2 + \cdots + A^{50}$ we have
$$f(z) = z + z^2 + \cdots + z^{50} = z \cdot \frac{x^{50}-1}{z-1}$$
$$f'(z) = \frac{51z^{50} - 1}{z-1} - \frac{z(x^{50}-1)}{(z-1)^2}$$
hence $f(-1) = 0$ and $f'(-1) = -25$.
$$B = f(A) = P\begin{bmatrix}0 & -25 \\ 0 & 0\end{bmatrix}P^{-1} = \begin{bmatrix}0 & -50 \\ 0 & 0\end{bmatrix}$$
Therefore $B^2 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x} - \int\bigg[-2\cos x\sin x\cdot e^{-3x}-3\cos^2x\cdot e^{-3x}\bigg](-\cos x)\,dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x}+2\int \cos^2x\sin x \cdot e^{-3x} \, dx-3\int\cos^3x\cdot e^{-3x}\,dx
\end{align}
How do I solve this integral ?
| $$\int e^{-3x}\cos^{3}x dx$$
For this type of integrals use the reduction formula
$$\int \exp(\alpha \ x)cos^{n}(\beta\ x)dx=\frac{\beta\ n\ \sin(\beta\ x)+\alpha\ \cos(\beta \ x)}{\alpha^2+\beta^2\ n^2}+\frac{\beta^2\ (n-1)n\ \int\ \exp(\alpha \ x)\cos^{n-2} (\beta\ x)dx}{{\alpha^2+\beta^2\ n^2}}$$ where $\alpha=-3, \beta=1, n=3$
$$=-\frac16 e^{-3x}\cos^{3}(x)+\frac16 e^{-3x}\sin(x)\cos^{2}(x)+\frac13\int e^{-3x}\cos(x)dx$$
Then solve $\int e^{-3x}\cos(x)dx$ using Integration By Parts
Then we get $$\int e^{-3x}\cos(x)dx=-\frac{1}{10}e^{-3x}(3\cos(x)-\sin(x))$$
Finally,
$$\int e^{-3x}\cos^{3}x dx=\frac{1}{120}e^{-3x}(9\sin(x)+5\sin(3x)-27\cos(x)-5\cos(3x))+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $\frac{\sqrt{r+1}-\sqrt{r-1}}{\sqrt{r+1}+\sqrt{r-1}}=\log_2\left(|x-2|+|x+2|\right)-\frac{11}{9}$, where $r=\frac{1+x^2}{2x}$ The problems in my sophomore workbook are marked with three colors that signify how hard is the marked problem: green for D and C, yellow for B and A and red for advanced students. All problems are also ordered from the easiest to the hardest.
This problem is marked red and appears last in the section of logarithmic problems, which means that it's the hardest logarithmic problem in the whole workbook. I know what its solution is (there are solutions at the end of the workbook), but I'd like to see how do we arrive at it.
I wrote the solution here, but it's hidden until it's hovered over. I don't want to spoil fun to those who want to find it by themselves.
Find the value(s) of $x$ when the left-hand side is equal to the right-hand side.
$$\frac{\sqrt{\dfrac{1+x^2}{2x}+1}\;-\;\sqrt{\dfrac{1+x^2}{2x}-1}}{\sqrt{\dfrac{1+x^2}{2x}+1}\;+\;\sqrt{\dfrac{1+x^2}{2x}-1}}=\log_2(|x-2|+|x+2|)-\frac{11}{9}$$
The solution is
$$x_1=\frac{7}{9},x_2=\frac{9}{7}$$
| Simplify the LHS:
$$\frac{\sqrt{\frac{1+x^2}{2x}+1}-\sqrt{\frac{1+x^2}{2x}-1}}{\sqrt{\frac{1+x^2}{2x}+1}+\sqrt{\frac{1+x^2}{2x}-1}}=\frac{\sqrt{\frac{(x+1)^2}{2x}}-\sqrt{\frac{(x-1)^2}{2x}}}{\sqrt{\frac{(x+1)^2}{2x}}+\sqrt{\frac{(x-1)^2}{2x}}}\overbrace{=}^{x>0}{|x+1|-|x-1|\over |x+1|+|x-1|}.$$
Hence:
$${|x+1|-|x-1|\over |x+1|+|x-1|}=\log_2(|x-2|+|x+2|)-\frac{11}{9}.$$
Consider the intervals:
$$x\in (0,1]: \qquad {2x\over 2}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac79;\\
x\in (1,2]: \qquad {2\over 2x}=\log_2 4-\frac{11}{9} \Rightarrow x=\frac97;\\
x\in (2,+\infty): \qquad {2\over 2x}=\log_2 (2x)-\frac{11}{9} \Rightarrow \log_2x=\frac 1x+\frac29 \Rightarrow \emptyset, \text{because}:\\
\log_2x\ge 1>\frac{13}{18}\ge \frac 1x+\frac29, x\ge 2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integers and Fractions. Each of the numbers $2, 3, 4, 5, 6, 7, 8$ and $9$ is used once to fill in one of the
variables in the equation below to make it correct. Of the three fractions being added, what is the value of the largest one?
$\dfrac{1}{ab}+\dfrac{c}{de}+\dfrac{f}{gh}=1$
I have tried putting in different combinations of the numbers $2$ to $9$ in the blanks, but it's not leading me anywhere. Is there a smarter way to answer this question?
| Since no other numbers share a common divisor with $5$ and $7$, they must be in the numerator. Therefore, let $c = 5$ and $f = 7$:
$$\frac{1}{ab} + \frac{5}{de} + \frac{7}{gh} = 1$$
Let us start with the fraction with the biggest numerator: $\displaystyle \frac{7}{gh}$. However, $ \displaystyle \frac{7}{2 \cdot 3}$ already is greater than $1$.
After the next largest choice, $\displaystyle \frac{7}{2 \cdot 4}$, the numbers left over are $3, 6, 8, 9$. There are $\displaystyle {4 \choose 2} = 6$ possibilities, but since addition is communitative, there are only $3$ possibilities. These are:
$$\frac{1}{3 \cdot 6} + \frac{5}{8 \cdot 9} = \frac{1}{8}$$
$$\frac{1}{3 \cdot 8} + \frac{5}{6 \cdot 9} \ne \frac{1}{8}$$
$$\frac{1}{3 \cdot 9} + \frac{5}{6 \cdot 8} \ne \frac{1}{8}$$
Therefore, $$\frac{1}{3 \cdot 9} + \frac{5}{6 \cdot 8} + \frac{7}{2 \cdot 4} = 1, $$
and the value of the largest fraction is $\frac{7}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Triangles geometric inequalities Is there any technique for geometric inequalities for triangles? For example, I tried to use Ravi’s substitution
$a=x+y, b=y+z, c=z+x, s=x+y+z, s-a=z, s-b=x, s-c=y,
A= \sqrt { xyz(x+y+z) }$
and
$r= \frac {A}{s} = \sqrt{\frac {xyz}{x+y+z}}),$
$R = \frac {abc} {4A}=\frac {(x+y)(y+z)(z+x) } {4 \sqrt { xyz(x+y+z) }}$
$\frac{r}{2R} = \frac {2xyz}{(x+y)(y+z)(z+x)} $
$r_a= \sqrt { \frac {xy}{z} (x+y+z)},$
to prove the inequality
$\sum_{cyc}^{} \frac {r_a}{a} \ge \sqrt {3(2+ \frac{r}{2R})}$
which becomes
$\sum_{cyc}^{} \sqrt {\frac {xy}{z} \frac{x+y+z}{x+y}} \ge \sqrt {3 (2+ \frac {2xyz}{(x+y)(y+z)(z+x) })}$
Then I tried to prove that
$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \sqrt {2 + \frac{r}{2R}} = \sqrt {2+ \frac {2xyz}{(x+y)(y+z)(z+x) }}$
Since
$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \frac {3}{2}$
then we have to prove
$\frac {3}{2} \le \sqrt {2+ \frac {2xyz}{(x+y)(y+z)(z+x) }}$
but I stuck. Thank you
| A proof of $$\sin \frac{A}{2} + \sin \frac{B}{2}+ \sin \frac{C}{2} \le \sqrt {2 + \frac{r}{2R}}.$$
We need to prove that
$$\sum_{cyc}\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}\leq\sqrt{2+\frac{\frac{2S}{a+b+c}}{\frac{abc}{2S}}}$$ or
$$\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{4bc}}\leq\sqrt{2+\frac{16S^2}{4(a+b+c)abc}}$$ or
$$\sum_{cyc}\sqrt{\frac{(a+b-c)(a+c-b)}{bc}}\leq\sqrt{\frac{8abc+(a+b-c)(a+c-b)(b+c-a)}{abc}}$$ or
$$\sum_{cyc}\sqrt{a(a+b-c)(a+c-b)}\leq\sqrt{8abc+(a+b-c)(a+c-b)(b+c-a)}$$ or
$$\sum_{cyc}\sqrt{(y+z)4yz}\leq\sqrt{8(x+y)(x+z)(y+z)+8xyz}$$ or
$$\sqrt{2(x+y+z)(xy+xz+yz)}\geq\sum_{cyc}\sqrt{xy(x+y)}$$ or
$$2\sum_{cyc}(x^2y+x^2z+xyz)\geq\sum_{cyc}(x^2y+x^2z+2\sqrt{xyxz(x+y)(x+z)})$$ or
$$\sum_{cyc}(x^2y+x^2z+2xyz)\geq2\sum_{cyc}x\sqrt{yz(x+y)(x+z)},$$
which is true by AM-GM:
$$2\sum_{cyc}x\sqrt{yz(x+y)(x+z)}\leq\sum_{cyc}x(y(x+z)+z(x+y))=\sum_{cyc}(x^2y+x^2z+2xyz).$$
Done!
A proof of
$$\sum_{cyc}^{} \frac {r_a}{a} \ge \sqrt {3(2+ \frac{r}{2R})}.$$
We need to prove that
$$\sum_{cyc}\frac{2S}{a(b+c-a)}\geq\sqrt{\frac{3(8abc+(a+b-c)(a+c-b)(b+c-a))}{4abc}}$$ or
$$\sum_{cyc}\frac{2\sqrt{(x+y+z)xyz}}{2x(y+z)}\geq\sqrt{\frac{3(8(x+y)(x+z)(y+z)+8xyz)}{4(x+y)(x+z)(y+z)}}$$ or
$$\sum_{cyc}yz(x+y)(x+z)\geq\sqrt{6(xy+xz+yz)xyz(x+y)(x+z)(y+z)}.$$
Now, let $xy=r$, $xz=q$ and $yz=p$.
Thus, we need to prove that
$$\sum_{cyc}(p+q)(p+r)\geq\sqrt{6(p+q+r)(p+q)(p+r)(q+r)}$$ or
$$\sum_{sum}(p^4-p^2q^2)\geq0$$ or
$$\sum_{cyc}(p^2-q^2)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
My Attempt
Using Partial fractions
$$
\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\
I_1=\bigg[\frac{1}{b^2}.\frac{1}{ab}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0=\bigg[\frac{1}{ab^3}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0
$$
$$
I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=\frac{b^2-a^2}{b^2}\bigg[\frac{x}{(a^2+b^2x^2)^2}-\int\frac{-2.2x.x}{(a^2+b^2x^2)^3}dx\bigg]
$$
How do I evaluate the integral $I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$ ?
| To evaluate $\int\limits_0^{+\infty}\frac{x^2\,dx}{(a^2+b^2x^2)^2}$
just take $\int\limits_0^{+\infty}\frac{dx}{a^2+b^2x^2}$ by parts.
And $\int\limits_0^{+\infty}\frac{dx}{(a^2+b^2x^2)^2}$ reduces to the previous integral by substitution $t=\frac{a}{bx}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving the DE: $y''+1/(2y^3)=0$. \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2}}{2}+C_1\\
p^2=\frac{1}{2y^2}+C_1\\
p=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
$p$ was equal to $y'$
\begin{align}
y'=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
I've tried to continue with trigonometric substitution but that didn't work for me.
Can you please help me out?
Thank you in advance!
Edit: adding final solution
Because of the answers below, I found the solution:
\begin{align}
\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}=\pm{x}dx\\
\int{\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}}=\int{\pm{x}dx}\\
\frac{1}{2C_1}\int{\frac{d({1+C_1y^2})}{\sqrt{1+C_1y^2}}}=\pm\frac{1}{\sqrt2}x\\
2\sqrt{1+C_1y^2}=\pm{\frac{2C_1}{\sqrt{2}}}(x+C_2)\\
\sqrt{1+C_1y^2}=\pm{\frac{C_1}{\sqrt{2}}}(x+C_2)\\
2(1+C_1y^2)=C_1^2(x+C_2)\\
\end{align}
| Multiplying both sides of the ODE by $2y'$ concludes that:$$2y'y''+y'/(y^3)=0$$by integrating$$(y')^2-\dfrac{1}{2}\dfrac{1}{y^2}=C_1$$or$$y'=\pm \dfrac{\sqrt 2}{2}\dfrac{\sqrt{C_2y^2+1}}{y}$$which yields to$$\dfrac{yy'}{\sqrt{C_2y^2+1}}=\pm \dfrac{\sqrt 2}{2}$$which results in$$\dfrac{\sqrt{C_2y^2+1}}{C_2}=\pm \dfrac{\sqrt 2}{2}x +C_3$$
| {
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(Solved) Sum of the series $\sum_{k=2}^\infty (k-1)(1-p)^{k-2}$? Where did I go wrong here?
As part of my worksheet for uni, I had to calculate the sum of the series $\sum_{k=2}^\infty (k-1)(1-p)^{k-2}$. ($p \in (0,1)$ denotes a probablity.) According to the German Wiki page on Geometric Series (relevant formula anchored in link, language not relevant), the series can be solved by differentiating. The result turns out to be $\frac{1}{p^2}$, which I know to be correct.
My initial calculations as detailed below, however, led to the incorrect result of $\frac{1}{p(1-p)}$. I would like to know where exactly I went wrong.
Let $S$ denote the sum of the series.
\begin{eqnarray}
S &:=& \sum_{k=2}^\infty (k-1)(1-p)^{k-2} \\
&=& (2-1)(1-p)^{2-2} + (3-1)(1-p)^{3-2} + (4-1)(1-p)^{4-2} + \dots \\
&=& 1(1-p)^0 + 2(1-p)^1 + 3(1-p)^2 + \dots \\
\Leftrightarrow (1-p)S &=& 1(1-p)^1 + 2(1-p)^2 + 3(1-p)^3 + \dots \\\\
\Leftrightarrow (1-p)S &=& (1(1-p)^0 - (1-p)^0)
+ (2(1-p)^1 - (1-p)^1) \\
&&+ (3(1-p)^2 - (1-p)^2)
+ (4(1-p)^3 - (1-p)^3) + \dots \\\\
\Leftrightarrow (1-p)S &=& (1(1-p)^0 + 2(1-p)^1 + 3(1-p)^2 + 4(1-p)^3 + \dots) \\\\
&& -((1-p)^0 + (1-p)^1 + (1-p)^2 + (1-p)^3 + \dots) \\\\
\Leftrightarrow S(1-p) &=& S - \sum_{i=0}^\infty (1-p)^i \\
&=& S - \frac{1}{1-p} \\
\Leftrightarrow S((1-p)-1) &=& - \frac{1}{1-p} \\
\Leftrightarrow S &=& - \frac{1}{(1-p)(-p)} = \frac{1}{p(1-p)}
\end{eqnarray}
Any help is greatly appreciated! Thanks in advance!
| Your problem is in the equality
\begin{align*}
S(1-p)&=S-\sum_{i=0}^\infty (1-p)^i\\
&=S-\frac{1}{1-p}.
\end{align*}
Remember that
$$
\frac{1}{1-x}=\sum_{i=0}^\infty x^i;
$$
do you see the error?
| {
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Which body has a minumum volume I would really appretiate a clear explanation.
Solve using lagrange functio.Which body, bounded by the planes $x = 0, y = 0, z = 0$ and the plane tangent to the
ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
has the minimal volume?
Solution : We need to minimiz $$V= \frac{abc}{6xyz}$$ knowing that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$ and we will solve it using Lagrange function.
However i do not understand how do we know that $V= \frac{abc}{6xyz}$?
And finding tangent plane $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^1} = 1$, we know that tangent has an equation $$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)$$ and then how do we get that tangent plane is $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}$$? and why did we put point (1,1,1) to solve the problem?
| The body $B$ of which we want to evaluate the volume as for vertices the origin and the intersections of the plane $P$ tangent to the ellipsoid with the coordinates axes.
Knowing the equation of $P$ that you gave in your question, the $x$-axis of coordinates has for equations $y=z=0$. Plug that with the equation of $P$, you get
$$\frac{2x}{a^2}(X-x)-\frac{2y}{b^2}y-\frac{2 z}{c^2}z=0$$
As $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, you obtain
$$X=\frac{a ^2}{x}.$$ And similarly for the other vertices
$$Y=\frac{b ^2}{y}, \, Z=\frac{c ^2}{z}.$$
You then get the formulae
$$V =\frac{1}{3}\left[\left(\frac{a^2 b^2}{2xy}\right) \times \frac{c^2}{z}\right]=\frac{a^2 b^2 c^2}{6xyz}$$
as the volume of a tetrahedron is a third of the basis area multiplied by the height.
| {
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Square inscribed in semicircle. Find some areas. $ABCD$, $AEFG$, $GHIJ$ are squares; $ABCD$ is inscribed in the semicircle, and $E$ is the middle of $[AD]$. Attention! $JG+GA+AO \neq r$. We have to find the areas of the squares.
It's easy to find that $Area(ABCD)=\frac{4r^2}{5}$ and $Area(AEFG)=\frac{r^2}{5}$, but I can't figure out how to find the area of the last square. Can you help me? Thanks!
| Let's let the radius of the circle be $1$ for a while and account for $r$ at the end. I'm also flipping your circle so that all of the interesting stuff happens in the first quadrant.
Consider the vertical line $x= x_i$. To build a square to the left of that with sides of length $s_i$, the upper right corner of that square will have coordinates of $(x_i+s_i, s_i)$ and will be on the unit circle. Hence
\begin{align}
(x_i + s_i)^2 + s_i^2 &= 1 \\
2s_i^2 + 2 x_i s_i - (1-x_i^2) &= 0 \\
s_i &= \dfrac{-2x_i + \sqrt{4 x_i^2 + 8 - 8 x_i^2}}{4} \\
s_i &= \dfrac{\sqrt{2 - x_i^2} - x_i}{2} \\
x_{i+1} &= \dfrac{\sqrt{2 - x_i^2} + x_i}{2} \\
\end{align}
If we start with $x_1 = \dfrac{1}{\sqrt 5}$
Then $s_2= \dfrac{1}{\sqrt 5}$ and $x_2 = \dfrac{2}{\sqrt 5}$
$s_3= \dfrac{\sqrt 6 - 2}{2\sqrt 5}$ and $x_3 = \dfrac{\sqrt 6 + 2}{2\sqrt 5}$
$s_4= \dfrac{2+\sqrt 6 - \sqrt{30-4 \sqrt 6}}{4\sqrt 5}$ and
$x_4 = \dfrac{2+\sqrt 6 + \sqrt{30-4 \sqrt 6}}{4\sqrt 5}$
and so on
| {
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Flipping two fair coins probability function $4$ Consider flipping two fair coins. Let $X = 1$ if the first coin is heads, and $X = 0$
if the first coin is tails. Let $Y = 1$ if the two coins show the same thing (i.e., both heads
or both tails), with $Y = 0$ otherwise. Let $Z = X + Y$, and $W = XY$.
(a) What is the probability function of $Z$?
Attempt:
$P(X = 1) = 1/2$
$P(X = 0) = 1/2$
$P(Y = 1) = 1/2$ since {HH, TT, TH, HT}, 2/4 times.
$P(Y = 0) = 1/2$ (not sure bout this)
(a)
Given $X = 1$, $Z = 2,1$
Given $X = 0$, $Z = 1, 0$
$P(Z = 2) = P(X = 1) + P(Y = 1) = 1/2 + 1/2 = 1$ [Book's answer is 1/4]
$P(Z = 1) = P(X = 1) + P(Y = 0) + P(X = 0) + P(Y = 1) = 2$ [1/2 book]
$P(Z = 0) = P(X = 0) + P(Y = 0) = 1$ [1/4 book]
Could someone explain to me why they multiply instead of adding them? Thanks
| You're on the right track. You can case out the whole function since there are only four cases.
\begin{array}{|c|c|c|c|c|c|}
\hline
Coin 1& Coin 2 & X & Y & Z & W \\ \hline
H & H & 1 & 1 & 2 & 1\\ \hline
H & T & 1 & 0 & 1 & 0\\ \hline
T & H & 0 & 0 & 0 & 0\\ \hline
T & T & 0 & 1 & 1 & 0\\ \hline
\end{array}
Note that $P(Z) \not= P(X) + P(Y)$
| {
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Another way to find the sum $S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}$ I want to know the sum
$$S=1-\dfrac{3}{2}+\dfrac{5}{4}-\dfrac{7}{8}+\cdots+(-1)^{n-1}\cdot\dfrac{2n-1}{2^{n-1}}.$$ This is my way.
First, we find the sum
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}=\sum _{k=1}^n (2 k-1) x^{k-1}.$$
We have
$$\sum _{k=1}^n (2 k-1) x^{k-1} = 2\sum _{k=1}^n k\cdot x^{k-1}-\sum _{k=1}^n x^{k-1}. $$
Note that
$$\sum _{k=1}^n k\cdot x^{k-1} = \left (\sum _{k=1}^n x^k\right)' = \left (\dfrac{x\left(x^n-1\right)}{x-1}\right)'=\dfrac{n x^{n+1}-(n+1)x^n+1}{(x-1)^2}.$$
Another way
$$\sum _{k=1}^n x^{k-1} = \dfrac{x^n-1}{x-1}.$$
From the above results, we have
$$1 + 3 x + 5x^2 + \cdots + (2n-1) \cdot x^{n-1}= \dfrac{(2 n-1) x^{n+1}-(2 n+1)x^n+x+1}{(x-1)^2}.$$
With $x=-\dfrac{1}{2}$,we get
$$S=\dfrac{2^n + (-1)^{n+1} \cdot(6n+1)}{9 \cdot 2^{n-1}}.$$
How to find the result with another way?
| The general term of your series is on the form $(-1)^{n-1}\frac{2n-1}{2^{n-1}} = -4nx^n + 2x^n$ where $x = -\frac{1}{2}$. When summed the latter term gives rise to a geometrical series and for the former note that $$(1-x)\left[x + 2x^2 + 3x^3 + 4x^4 + \ldots + nx^n\right] = x+x^2+x^3+\ldots+x^{n} - nx^{n+1}$$ which is just another geometrical series. This gives you
$$\sum_{k=1}^n-4kx^k + 2x^k = -4\frac{\left(g_n(x) - nx^{n+1}\right)}{1-x} + 2g_n(x)$$
where $g_n(x) = x+x^2+\ldots+x^n =\frac{x-x^{n+1}}{1-x}$. Taking $x=-1/2$ gives the same result as you have found already.
| {
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Prove that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1}I tried using mathematical induction to prove this, but the problem I faced was that there are a lot of numbers between $\frac{1}{2^k-1}$ and $\frac{1}{2^{k+1}-1}$. Is it possible to prove this with induction or is there a better method?
| Given the hints above, we have during the induction process that
$$\frac{1}{2^n}+\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^{n+1}-1}<1$$
If we multiply both sides by $2^n$ and note that there are $2^n$ terms being added, we have that
\begin{eqnarray*}
\frac{2^n}{2^n}+\frac{2^n}{2^n+1}+\frac{2^n}{2^n+2}+...+\frac{2^n}{2^{n+1}-1}&<&2^n\\1+\frac{2^n+1-1}{2^n+1}+\frac{2^n+2-2}{2^n+2}+...+\frac{2^n+(2^n-1)-(2^n-1)}{2^{n}+2^n-1}&<&2^n
\\1+1-\frac{1}{2^n+1}+1-\frac{2}{2^n+2}+...+1-\frac{2^n-1}{2^{n+1}-1}&<&2^n\\2^n-\frac{1}{2^n+1}-\frac{2}{2^n+2}-...-\frac{2^n-1}{2^{n+1}-1}&<&2^n\\-\left(\frac{1}{2^n+1}+\frac{2}{2^n+2}+...+\frac{2^n-1}{2^{n+1}-1}\right)&<&0
\end{eqnarray*}
which is true for all $n\ge2$
| {
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Solving an equation for positive integral values of variables. Question
Solve
$$y^3+3y^2+3y=x^3+5x^2-19x+20$$
for positive integers $x$ and $y$.
My approach:
I factorised the equation as
$(y-x)(x^2+y^2+3x+3y+3)=2[(x-1)(x-10)]$
and got two ordered pairs of $(x,y)$ i.e. $(1,1);(10,10)$
I can't go further.
Thanks for your help.
PS: Please consider this question without my approach also, because it may mislead you.
| Write $$(y+1)^3= y^3+3y^2+3y+1=x^3+5x^2-19x+21$$
now, since a discriminant of $5x^2-19x+21$ is $19^2-20\cdot 21<0$ we have
$$x^3<x^3+5x^2-19x+21$$
Also
$$x^3+5x^2-19x+21<x^3+6x^3+12x+8 =(x+2)^3$$
since $x^2+31x-13>0$ for positive $x$. So $$x^3<(y+1)^3<(x+2)^3$$
which is only possible iff $y+1=x+1$ so $x=y$...
| {
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How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.
| HINT
By induction we have
*
*Base case: $n=1\implies 0<\frac14<1$
*Induction step, we assume true
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
and we need to prove
$$P(n+1) := 1^3 + 2^3+\cdots+(n-1)^3+n^3 < \frac{(n+1)^4}{4} < 1^3 + 2^3 + \cdots + n^3+(n+1)^3$$
then we have
$$1^3 + 2^3+\cdots+(n-1)^3+n^3 < n^3 +\frac{n^4}{4}$$
$$\frac{n^4}{4}+(n+1)^3< 1^3 + 2^3 + \cdots + n^3+(n+1)^3$$
then it suffices to prove that
$$n^3 +\frac{n^4}{4}< \frac{(n+1)^4}{4}<\frac{n^4}{4}+(n+1)^3$$
that is
$$n^3 < \frac{(n+1)^4}{4}-\frac{n^4}{4}<(n+1)^3$$
| {
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Same integration with 2 different answers? $$\int x(x^2+2)^4\,dx $$
When we do this integration with u substitution we get
$$\frac{(x^2+2)^5}{10}$$
as $u=x^2+2$
$du=2x\,dx$
$$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$
Although when we expand the fraction and then integrate the answer we get is different:
$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$
$$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$
we get
$$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$
For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1
Now,
$x=0$
$$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$
$$\therefore C= 1$$
AND
$$\frac {(0+2)^5}{10} + C= 1$$
$$\therefore \frac {32}{10} + C = 1$$
$$\therefore C = 1 - 3.2 = -2.2$$
| You can check an antiderivative by differentiating.
$$\left(\frac{(x^2+2)^5}{10}\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$
and
$$\left(\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\right)'=x^9+8x^7+24+32x^3+16x$$
and the two expressions are indeed equivalent.
Now the long explanation.
Consider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$.
$$2x(x^2+a)^m$$
which integrates as
$$\frac{(x^2+a)^{m+1}}{m+1}.$$
By the binomial theorem, the terms in the development of this antiderivative are
$$\frac1{m+1}\binom{m+1}kx^{2(m+1-k)}a^k.$$
On the other hand, the development of the initial integrand gives terms
$$2\binom mkx^{2(m-k)+1}a^k,$$ and after integration
$$\frac1{m-k+1}\binom mkx^{2(m-k)+2}a^k.$$
It is easy to see that all terms coincide, because
$$\frac1{m+1}\frac{(m+1)!}{k!(m+1-k)!}=\frac1{m-k+1}\frac{m!}{k!(m-k)!}=\frac{(m-1)!}{k!(m-k+1)!}.$$
Anyway, the first development holds for $0\le k\le m+1$, giving a constant term $\dfrac{a^m}{m+1}$, but the second for $0\le k\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant.
| {
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If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime.
Answer:
We have to prove $\gcd(a^2+b^2,a^2b^2)=1$, $\gcd(a,b)=1$. Then $\gcd(a^2,b^2)=1$. Now we have $\gcd(a^2,a^2+b^2)=1$ and $\gcd(b^2,a^2+b^2)=1$ by $\gcd(a,b)=\gcd(p\pm q)$.
Then I am lost.
| Suppose $(a,b)=1$. Bezout's Identity says there are $x,y\in\mathbb{Z}$ so that
$$
ax+by=1\tag1
$$
Then
$$
1=(ax+by)^3=\underbrace{\left(ax^3+3bx^2y\right)}_ua^2+\underbrace{\left(3axy^2+by^3\right)}_vb^2\tag2
$$
Rearranging $(2)$, we get
$$
u\left(a^2+b^2\right)+(v-u)b^2=1\tag3
$$
and
$$
(u-v)a^2+v\left(a^2+b^2\right)=1\tag4
$$
Multiplying $(3)$ and $(4)$ yields
$$
\left(\left(a^2+b^2\right)uv+\left(a^2u-b^2v\right)(u-v)\right)\left(a^2+b^2\right)-(u-v)^2a^2b^2=1\tag5
$$
Thus, $\left(a^2+b^2,a^2b^2\right)=1$.
| {
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Evaluating $\cos\frac{k\pi}{p}+\cos\frac{2k\pi}{p}+\cos\frac{3k\pi}{p}+\dots +\cos\frac{2\frac{(p - 1)}{2}k\pi}{p}$
If $p$ is a prime then what is the value of the series
$$\cos\frac{2\pi}{p}+\cos\frac{4\pi}{p}+\cos\frac{6\pi}{p}+\dots +\cos\frac{(p - 1)\pi}{p}$$ In general what is the value of the following series
$$\cos\frac{k\pi}{p}+\cos\frac{2k\pi}{p}+\cos\frac{3k\pi}{p}+\dots +\cos\frac{2\frac{(p - 1)}{2}k\pi}{p}$$
| In general, if $n$ is an odd positive integer,
$$\sum_{k=0}^{(n-1)/2}\cos(2\pi k/n)=\Re
\sum_{k=0}^{(n-1)/2}\exp(2\pi i k/n)
=\Re\frac{e^{2\pi i (n-1)/2n}-1}{e^{2\pi i/n}-1}
=\Re\frac{-e^{\pi i/n}-1}{e^{2\pi i/n}-1}
=\frac12$$
so
$$\sum_{k=1}^{n-1}\cos(2\pi k/n)=\frac12-1=\frac12.$$
The second sum can be handled similarly.
| {
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Proof for a modular arithmetic identity. How can I prove the following? Or is there a reference to the proof for this?
$$d \mod abc = d \mod a + a\left(\left\lfloor\frac{d}{a}\right\rfloor \mod b\right)+ab\left(\left\lfloor\frac{d}{ab}\right\rfloor \mod c\right)$$
| Numerical example:
$$\begin{align}95 \mod \color{red}2\cdot \color{green}3\cdot \color{blue}7 = \color{purple}{\mathbf{11}} \Rightarrow \frac{95}{\color{red}2\cdot \color{green}3\cdot \color{blue}7}&=\frac{(1+94)}{\color{red}2\cdot 3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 47}{2\cdot 3\cdot 7}\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot (2+45)}{2\cdot \color{green}3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 2}{2\cdot 3\cdot 7}+\frac{2\cdot 3\cdot 15}{2\cdot 3\cdot 7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{2\cdot 2}{2\cdot 3\cdot 7}+\frac{2\cdot 3\cdot (1+14)}{2\cdot 3\cdot \color{blue}7}=\\
&=\frac{1}{2\cdot 3\cdot 7}+\frac{\color{red}2\cdot 2}{2\cdot 3\cdot 7}+\frac{\color{red}2\cdot \color{green}3\cdot 1}{2\cdot 3\cdot 7}+2=\\
&=\frac{\color{purple}{\mathbf{11}}}{\color{red}2\cdot \color{green}3\cdot \color{blue}7}+2.\end{align}$$
| {
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How to show a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$?
How to show that a rational polynomial is irreducible in $\mathbb{Q}[a,b,c]$? For example, I try to show this polynomial $$p(a,b,c)=a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b)-4(a+b)(a+c)(b+c)(*)$$ is irreducible, where $a,b,c\in \mathbb{Q}$.
The related problem is Ask for the rational roots of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4.$. Could I consider the points $(*)$ intersect with $L_{\infty}$ are three? $L_{\infty}$ is the infinity line in a projective space $\mathbb{C}P^2$.
| To ask about complete (three linear factor) reducibility over the complexes, we take the Hessian matrix of second partials. The entries are linear in the named variables. Next, let $\Delta$ be the determinant of the Hessian. This $\Delta$ is once again a cubic form. The original cubic (homogeneous) ternary form factors completely if and only if $\Delta $ is a constant multiple of it. I have finished this first test, your cubic does not factor completely.
More difficult if the cubic might be a linear times an irreducible quadratic. In that case, there is still a conclusive test:
For your problem, the coefficients in order from $0$ to $9$ are
$$ 1,-3,-3,-3,-5,-3,1,-3,-3,1 $$
where the best looking diagram is a triangle as in bowling. The $-5$ refers to $-5abc \; ,$ the $1$s refer to $a^3,b^3,c^3 \; .$
Alright, I wote a little program to correctly type the 8 by ten matrix, then put that into gp-Pari. The original polynomial is irreducible over the complex numbers.
?
? sch = [ 0, 3, 0, -6, -6, 0, -3, -5, -3, 0; 0, 0, 3, 0, -6, -6, 0, -3, -5, -3; -3, -6, -5, 3, -6, -3, 0, 0, 0, 0; 0, 0, -3, 0, -6, -5, 0, 3, -6, -3; -3, -5, -6, -3, -6, 3, 0, 0, 0, 0; 0, -3, 0, -5, -6, 0, -3, -6, 3, 0; 1, 0, -3, 3, 0, -3, -2, 3, 0, 1; 1, -3, 0, -3, 0, 3, 1, 0, 3, -2]
%25 =
[ 0 3 0 -6 -6 0 -3 -5 -3 0]
[ 0 0 3 0 -6 -6 0 -3 -5 -3]
[-3 -6 -5 3 -6 -3 0 0 0 0]
[ 0 0 -3 0 -6 -5 0 3 -6 -3]
[-3 -5 -6 -3 -6 3 0 0 0 0]
[ 0 -3 0 -5 -6 0 -3 -6 3 0]
[ 1 0 -3 3 0 -3 -2 3 0 1]
[ 1 -3 0 -3 0 3 1 0 3 -2]
? matrank(sch)
%26 = 8
?
?
?
| {
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Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution I am trying to solve this differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ using substitution, with the initial value $y(1)=0$ This is what I've done so far:
$$
\begin{align*}
y'(t) &= \frac{t^2+3y(t)^2}{2ty(t)} = \frac{t^2}{2ty(t)} + \frac{3y(t)^2}{2ty(t)} \\
&= \frac{t}{2y(t)} + \frac{3y(t)}{2t} \\
&= \frac{1}{\frac{2y(t)}{t}} + \frac{3y(t)}{2t} \\
\\
\text{Substitution:}\\
f\Big(\frac{y(t)}{t}\Big) &= y'(t) \ \ \text{with} \ \ f(z(t)) = \frac{1}{2z(t)} + \frac{3}{2}z(t) \\
z(t)&= y(t) \ \ \text{and} \ \ z_0 = z(t_0) = z(1) = 0 \\
\\
z'(t) &= \frac{1}{t} \Big( f(z(t) - z(t)\Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{3z(t)}{2} - \frac{2z(t)}{2} \Big) \\
&= \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)}{2} \Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)^2}{2z(t)} \Big) \\
&= \frac{1}{t} \frac{1+z(t)^2}{2z(t)} \\
\\
\text{Seperation of variables:}\\
g(t) &= \frac{1}{t} \ ,\ \ h(z(t)) = \frac{1+z(t)^2}{2z(t)} \\
\\
\int_{t_0=1}^t \frac{1}{t} d\tau &= \Big[\ln(\tau)\Big]_1^t = ln(t) - 0 = ln(t) \\
\\
\int_{z_0=0}^{z(t)} \frac{1+x}{2x} dx &= \int_{0}^{z(t)} \frac{1}{2x} dx + \frac{x}{2x} dx = \int_{0}^{z(t)} \frac{1}{2x} dx + \int_{0}^{z(t)} \frac{1}{2} dx \\
&= \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{1}{2} \int_{0}^{z(t)} 1 \ dx = \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} \\
&= \frac{1}{2} \int_{\alpha}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} = \frac{1}{2} \Big[ \ln(x) \Big]_\alpha^{z(t)} + \frac{z(t)}{2} \ \ \ \Big( \alpha \in \big(0,z(t)\big)\Big) \\
&= \frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} \\
\\
\lim_{\alpha \rightarrow 0} \ &\frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} = \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2} \\
\\
\text{Now solve for z(t):} \\
ln(t) &= \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2}
\end{align*}
$$
This is the point where I don't know how to continue.
| You make it more complicated than it is.
Once you substitute $y=tu$ the differential equation becomes $$2ttu(tu'+u)=t^2+3t^2u^2$$
Which after simplification and reordering gives $$\dfrac{2uu'}{1+u^2}=\dfrac 1t$$
And this integrates to $1+u^2=c|t|$ with $c\ge 0\qquad$ (i.e. $\ln|1+u^2|=\ln|t|+cst$).
Finally $$y=\pm t\sqrt{c|t|-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help me with $\int_{0}^{\pi/2} \frac{e^{a \cos^2x}}{b^2\cos^2x + c^2\sin^2x}dx$ I have this integral
$$ \int_{0}^{\pi/2} \frac{e^{a \cos^2x}}{b^2\cos^2x + c^2\sin^2x}dx.$$
I know that
$$\int_{0}^{\pi/2} \frac{dx}{b^2\cos^2x + c^2\sin^2x}=\frac{\pi}{2bc},$$
and
$$\int_{0}^{\pi/2} e^{a \cos^2x}dx=\frac{\pi}{2}e^{a/2} I_n\left(0,\frac{a}{2}\right).$$
Using these integrals is it possible to solve the first integral?
| Since $$ \int_{0}^{\pi/2} \cos(2mx) e^{a\cos^2(x)}\,dx =\frac{\pi}{2}e^{a/2} I_m(a/2)\tag{1}$$
it is enough to find the Fourier cosine series of $\frac{1}{b^2\cos^2 x+c^2 \sin^2 x}$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
$$\frac{1}{b^2\cos^2 x+c^2 \sin^2 x}=\frac{1}{bc}+\frac{2}{bc}\sum_{m\geq 1}\left(\frac{c-b}{c+b}\right)^m \cos(2mx) \tag{2}$$
immediately leads to
$$\int_{0}^{\pi/2}\frac{e^{a\cos^2(x)}}{b^2\cos^2 x+c^2 \sin^2 x}\,dx = \frac{\pi}{2bc}+\frac{\pi e^{a/2}}{bc}\sum_{m\geq 1}\left(\frac{c-b}{c+b}\right)^m I_m(a/2).\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Duality by computing $P+Q$ Given $C$ is symmetric and positive definite, consider
\begin{align}
\min: &Q(y) = \frac{1}{2} y^T C^{-1} y - b^Ty\\
\text{ subject to :}&A^Ty = f
\end{align}
and
$$\max: -P(x) = -\frac{1}{2}(Ax-b)^TC(Ax-b) + x^Tf.$$
I am asked to prove duality by computing $P(x)+Q(y)$. Here duality means that $-P(x)$ has a max and $Q(y)$ has a min, so there exists a saddle point $(x,y)$.
I am really confused about this problem...
We showed when we defined the Langrange multiplier for the constraind problem,
$$L(x,y) = Q(y) - x^T(A^Ty-f)$$
then $L(x,y)$ has a saddle point at where $Q(y)$ has a min (subject to the constrain) and $-P(x)$ has a max.
| We write out the sum and substitute $A^Ty= f$
\begin{align}
P+Q & = \left[\frac{1}{2}(Ax-b)^T C (Ax-b) - x^TA^T y\right] + \left[\frac{1}{2} y^T C^{-1} y - b^Ty\right]\\
& = \left[\frac{1}{2}(Ax-b)^T C (Ax-b) - x^TA^T y\right] + \left[\frac{1}{2} (C^{-1}y)^T C (C^{-1}y) - b^Ty\right]\\
& = \frac{1}{2}\bigg[(Ax-b)^T C (Ax-b) + (C^{-1}y)^T C (C^{-1}y) \bigg] +( x^TA^t - b^T)y \\
& = \frac{1}{2}\bigg[(Ax-b + C^{-1}y)^T C (Ax-b + C^{-1}y) - 2 (Ax-b)^T C C^{-1}y \bigg] +( x^TA^t - b^T)y \\
&= \frac{1}{2} (Ax-b + C^{-1}y)^T C (Ax-b + C^{-1}y)
\end{align}
Since $C$ is positive definite, we know $P+Q \geq 0$, and it can achieve zero when we have
$$Ax-b + C^{-1} y = 0 \text{ and }A^T y = f,$$
This condition implies that $Q(y)$ is a min and $-P(x)$ is a max. So we have
$$\min_{constrain} Q(y) = \max -P(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear Differential Equation $y'-3y-2x^2=0$ Power Series Method I'm trying to solve this first order linear differential equation using power series method. I have obtained a answer but I can't make it to the simplest form. Here's the problem:
$y'=2x^2 + 3y$
My work:
\begin{align*}
y' &= a_1 + 2a_2x + 3a_3x^2 + \ldots = \sum_{n=1}^{\infty}na_nx^{n-1} \\
y &=a_0 + a_1x + a_2x^2 + \ldots = \sum_{n=0}^{\infty}a_nx^{n}\\
\text{so} \\
&a_1 + 2a_2x + 3a_3x^2 + \ldots - 3(a_1x + a_2x^2 + \ldots )-2x^2= 0 \iff (a_1-3a_0)+x(2a_2-3a_1) + x^2(3a_3-3a_2 - 2) + x^3 (4a_4 -3a_3)+x^4(5a_5 - 3a_4) + \ldots = 0
\end{align*}
Then, I found that,
$a_1 = 3a_0, a_2 = \frac{9}{2}a_0, a_3 = \frac{9}{2}a_0 + \frac{2}{3}, a_4 = \frac{27}{8}a_0 + \frac{6}{12}, a_5 = \frac{81}{40}a_0 + \frac{18}{60}, \ldots$
By substituting each of coefficient to the original $y$, I have
\begin{align*}
y &= \sum_{n=0}^{\infty} a_nx^n\\
y &= a_0 + 3a_0 x + \frac{9}{2}a_0 x^2 + \left( \frac{9}{2} a_0 + \frac{2}{3} \right) x^3 + \left(\frac{27}{8}a_0 + \frac{1}{2} \right)x^4 + \left( \frac{81}{40}a_0 + \frac{18}{60} \right) x^5 + ...\\
y &= a_0 + 3a_0 x + \frac{9}{2}a_0 x^2 + \frac{9}{2} a_0 x^3 + \frac{27}{8}a_0 x^4 + \frac{81}{40}a_0 x^5 + \ldots + \frac{2}{3} x^3 + \frac{1}{2} x^4 + \frac{18}{60} x^5 + \ldots\\
y &= a_0 \underbrace{\left[ 1 + 3x + \frac{9}{2}x^2 + \frac{9}{2}x^3 + \frac{27}{8} x^4 + \frac{81}{40}x^5 + \ldots \right] }_{e^{3x}} + \frac{2}{3} x^3 + \frac{1}{2} x^4 + \frac{18}{60} x^5 + \ldots
\end{align*}
That's my final result, I can't figure out the rest. But I've tried by using integrating factor and I've got this : $y(x) = c_1 e^{3x} - \frac{2}{3}x^2 - \frac{4}{9}x - \frac{4}{27}$
Please help me out, your help means a lot to me. Thanks in advance.
| Note that in your last coefficients,
$$
\frac23=\frac{4}{3^3}\frac{3^3}{3!},~~\frac12=\frac{4}{3^3}\frac{3^4}{4!},~~ \frac{18}{60}=\frac{36}{5!}=\frac{4}{3^3}\frac{3^5}{5!}
$$
and so on, so that you get another exponential series for $\frac{4}{27}e^{3x}$ if you compensate for the missing first terms
$$
\frac{4}{3^3}(1+3x+\frac92x^2)=\frac4{27}+\frac49x+\frac23x^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to determine the intersection of a plane and a cone? I want to know what is the intersection of the plane $2x+y+z-2=0$ and $x^2+y^2=z^2$. My guess is that it is an ellipse or an hyperbola, but I can't understand how to determine which one.
Subsituting $z=2-2x-y$ in the cone equation, we obtain
$$
5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
and I can't undertsand whether it is an hyperbola or an ellipse.
Will you please help me figure it out?
Thanks
| Making the change of variables ( a rigid rotation does not alter the form)
$$
x = X\cos\theta-Y\sin\theta\\
y = X\sin\theta+Y\cos\theta
$$
in
$$
f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
we get at
$$
f(x,y)\equiv F(X,Y) = \frac{1}{2} (4 \sin (2 \theta)+3 \cos (2 \theta)+7)X^2+(4 \cos (2 \theta)-3 \sin (2 \theta))X Y+\frac{1}{2} (7-4 \sin (2 \theta)-3 \cos (2 \theta))Y^2-12\cos\theta X+12\sin\theta Y+4=0
$$
and now choosing $\theta$ such that $4\cos(2\theta)-3\sin(2\theta) = 0\to \theta = \frac 12 \arctan(\frac 43)$ we obtain finally
$$
F(X,Y) =6 X^2-\frac{24 X}{\sqrt{5}}+Y^2+\frac{12 Y}{\sqrt{5}}+4 = 0
$$
which is clearly an ellipse.
NOTE
As can be verified the intersection of the cone $x^2+y^2=z^2$ and the plane $2x+y+z-2=0$ gives
$$
f(x,y) = 4 - 8 x + 3 x^2 - 4 y + 4 x y = 0
$$
and not
$$
f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
as assumed. In this case with the same rotation the true intersection curve should be
$$
F(X,Y) = 4 X^2-4 \sqrt{5} X-Y^2+4
$$
which is a hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve the recurrence $T(n) = T(n-1) + 2T(n-2) + 2$ $T(0) = 1,
T(1) = 0$
I ain't able to get answer from any of the methods.
Substitution:
$T(n) = x^n $
\begin{align}
& x^n = x^{n-1} + 2x^{n-2} + 2 \\
& x^2 = x + 2 + 2x^2\\
& x^2 + x + 2 =0
\end{align}
solving this I will get a complex root.
\begin{align}
x & = \frac{-1 \pm \sqrt{1-8}}{2}
x & = \frac{-1 \pm 7i}{2}
\end{align}
Now how to go further.
General:
\begin{align}
T(2) & = 4\\
T(3) & = 6\\
T(4) & = 16\\
T(5) & = 30\\
T(6) & = 64\\
T(7) & = 126\\
T(8) & = 256
\end{align}
From this i can deduce if $n$ is even then $2^n$.
I cant deduce for if $n$ is odd.
| Another way to solve the problem: note that you have
$$
\begin{bmatrix}
a_{n+1}\\ a_n \\ 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 2 & 2 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
a_{n}\\ a_{n-1} \\ 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 2 & 2 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}^n
\begin{bmatrix}
0 \\ 1 \\ 1
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequalities with absolute values My question is:
Show that for all $|x-1|+|x-2|+\dots+|x-10| > 23$
I have solved above problem as below,
If $x-1 > 0$ and $x-2 > 0$ and ......$x-10 > 0$ then
LHS $= x-1+x-2+x-3+\dots+x-10 = 10x-55 > 23$ (because $x>10$)
If $x-1<0$ and $x-2 < 0$ and ......$x-10 < 0$ then
LHS $= -x+1-x+2-x+3.....-x+10 = -10x+55 > 23 $ (because $x<1$)
If the above solution is wrong, please give me the correct method
| Hint: by the triangle inequality $|x-1|+|x-10| = |x-1|+|10-x| \ge |x-1+10-x|=9$. The same way, $\,|x-2|+|x-9|\ge 7\,$, $\,|x-3|+|x-8|\ge 5\;\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Solving a system of algebraic equations with Maple I want to solve the following algebraic equation system using solve command. But, it gives only the trivial solution. I want to find a,b and c in terms of k. k is a constant here. Thanks in advance.
Input:
solve({-6*c+(3/2)*c^2-2*b-3*b*c+(3/2)*b^2-3*a*c+(k^2)*b-b+(3/2)*(a^2)+(k^2)*a-a=0,
-2*b-3*b*c+3*(b^2)-6*a*c+2*(k^2)*c-2*c-9*a*b+3*(k^2)*b-3*b+6*(a^2)+4*(k^2)*a-4*a=0,
(3/2)*(b^2)-3*a*c+(k^2)*c-c-9*a*b+3*(k^2)*b-3*b+9*(a^2)+6*(k^2)*a-a=0,
-3*a*b+(k^2)*b-b+6*(a^2)+4*(k^2)*a-4*a=0,
(3/2)*(a^2)+(k^2)*a-a=0},{a,b,c});
Output:
{a=0, b=0, c=0}
| On Maple 2018.1. Add option to solve: allsolutions = true
solve({3/2*(a^2)+k^2*a-a = 0, -6*c+3/2*(c^2)-2*b-3*b*c+3/2*(b^2)-3*a*c+k^2*b-
b+3/2*(a^2)+k^2*a-a = 0, 3*b*k^2+2*c*k^2-9*a*b-6*a*c+3*b^2-3*b*c-2*b-2*c-
3*b+6*a^2+4*k^2*a-4*a = 0, 3/2*(b^2)-3*a*c+k^2*c-c-9*a*b+3*k^2*b-
3*b+9*a^2+6*k^2*a-a = 0, 4*a*k^2+b*k^2+6*a^2-3*a*b-4*a-b = 0}, {a, b, c, k},
allsolutions = true)
#{a = 0, b = 0, c = 0, k = k}, {a = 0, b = 0, c = 4, k = 1}, {a = 0, b = 0, c = 4, k = -1}
Mathematica 11.3 give the same answer:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a nice way to evaluate $\iiint_{E}\, \frac{dx\,dy \,dz}{\sqrt{x^2+y^2+(z-b)^2}}$ where $E:x^2+y^2+z^2\leq a^2$
Is there a nice way to evaluate $$\displaystyle\iiint_{E}\, \dfrac{dx\,dy\,dz}{\sqrt{x^2+y^2+(z-b)^2}}$$ where $E:x^2+y^2+z^2\leq a^2$ with
$0<a<b$
If I use the standard spherical coordinates (Is there a better transformation?)
$$x=p\sin(\phi)\cos(\theta) $$
$$y=p\sin(\phi)\sin(\theta)$$
$$z=p\cos(\phi)$$
$$|J|=p^2\sin(\phi)$$
$$\iiint_{E}\, \frac{dx\,dy \,dz}{\sqrt{x^2+y^2+(z-b)^2}} $$
$$=\int_{\phi=0}^{\pi}\,\int_{\theta=0}^{2\pi}\,\int_{p=0}^{a}\, \frac{p^2\sin(\phi)\,dp\,d\theta \,d\phi}{\sqrt{p^2\sin^2(\phi) + (p\cos(\phi)-b)^2}}$$
$$=\int_{\phi=0}^{\pi}\,\int_{\theta=0}^{2\pi}\,\int_{p=0}^{a}\, \frac{p^2\sin(\phi)\,dp\,d\theta \,d\phi}{\sqrt{(p-b\cos(\phi))^2 + b^2\sin^2(\phi)}}$$
Which I'm finding bit difficult.
| Fubini split w.r.t. the $z$-component. Let $J$ be the integral to be calculated. I will use
$B(0,r)$ for a ball centered in $0$ with radius $r$ (in the dimension that
can be extracted from the context.)
Then:
$$
\begin{aligned}
J &=
\iiint_{B(0,a)}\frac 1{\sqrt{x^2+y^2+(b-z)^2}}\; dx\; dy\; dz
\\
&=
\int_{z\in B(0,a)}dz
\iint_{(x,y)\in B(0,\sqrt{a^2-z^2})}\frac 1{\sqrt{x^2+y^2+(b-z)^2}}\; dx\; dy
\\
&=
\int_{-a}^a dz
\int_{r\in[0,\sqrt{a^2-z^2}]}
\int_{t\in[0,2\pi]}
\frac 1{\sqrt{r^2+(b-z)^2}}\; r\; dr\; dt
\\
&=
\int_{-a}^a dz
\int_0^{\sqrt{a^2-z^2}}
2\pi\Big(\ (r^2+(b-z)^2)^{1/2}\ \Big)'\; dr
\\
&=
2\pi
\int_{-a}^a dz\;
\Big[\ (r^2+(b-z)^2)^{1/2}\ \Big]_0^{\sqrt{a^2-z^2}}
\\
&=
2\pi
\int_{-a}^a dz\;
\Big[\ (a^2-z^2+(b-z)^2)^{1/2}-(0^2+(b-z)^2)^{1/2}\ \Big]
\\
&=
2\pi
\int_{-a}^a dz\;
\Big[\ (a^2-2bz+b^2)^{1/2}-(b-z)\ \Big]
\\
&=
2\pi
\left[\ -\frac 1{2b}\cdot \frac 1{3/2}(a^2-2bz+b^2)^{3/2}
+\frac 12(b-z)^2\ \right]_{-a}^a
\\
&=
\frac {2\pi}{3b}\Big[\ (b+a)^3-(b-a)^3\ \Big]
-
\pi\Big[\ (b+a)^2-(b-a)^2\ \Big]
\\
&
=
\frac {2\pi}{3b}\cdot 2a^3
=
\frac {4\pi\, a^3}{3b}\ .
\end{aligned}
$$
Check using sage:
sage: var( 'r,t,z,a,b' );
sage: assume(a>0)
sage: assume(b>a)
sage: J = integral( integral( integral( r/sqrt(r^2+(z-b)^2), t,0,2*pi ), r,0,sqrt(a^2-z^2) ), z,-a,a )
sage: J.canonicalize_radical()
4/3*pi*a^3/b
Here is the computation of the integral for its version "without the square root in the denominator"...
$$
\begin{aligned}
J &=
\iiint_{B(0,a)}\frac 1{x^2+y^2+(b-z)^2}\; dx\; dy\; dz
\\
&=
\int_{z\in B(0,a)}dz
\iint_{(x,y)\in B(0,\sqrt{a^2-z^2})}\frac 1{x^2+y^2+(b-z)^2}\; dx\; dy
\\
&=
\int_{-a}^a dz
\int_{r\in[0,\sqrt{a^2-z^2}]}
\int_{t\in[0,2\pi]}
\frac 1{r^2+(b-z)^2}\; r\; dr\; dt
\\
&=
\int_{-a}^a dz
\int_0^{\sqrt{a^2-z^2}}
2\pi\cdot\frac 12\Big(\ \ln(r^2+(b-z)^2)\ \Big)'\; dr
\\
&=
\pi
\int_{-a}^a dz\;
\Big[\ \ln(r^2+(b-z)^2)\ \Big]_0^{\sqrt{a^2-z^2}}
\\
&=
\pi
\int_{-a}^a dz\;
\Big[\ \ln(a^2-z^2+(b-z)^2)-\ln(0^2+(b-z)^2)\ \Big]
\\
&=
\pi
\int_{-a}^a dz\;
\Big[\ \ln(a^2-2bz+b^2)-2\ln(b-z)\ \Big]
\\
&=
\pi
\left[\ -\frac 1{2b}(a^2-2bz+b^2)(\ln(a^2-2bz+b^2)-1)+2(b-z)(\ln(b-z)-1)\ \right]
_{-a}^a
\\
&=
\pi
\left[\
-\frac 1{2b}(b-a)^2(2\ln(b-a)-1)
+\frac 1{2b}(b+a)^2(2\ln(b+a)-1)
\right.
\\
&\qquad\qquad
\left.
+2(b-a)(\ln(b-a)-1)
-2(b+a)(\ln(b+a)-1)
\ \right]
\\
&=
2\pi\ln(b+a)\cdot\left[\frac 1b(b+a)^2-2(b+a)\right]
-
2\pi\ln(b-a)\cdot\left[\frac 1b(b-a)^2-2(b-a)\right]
+2\pi a
\\
&=
2\pi\frac 1b(b^2-a^2)\ln\frac {b-a}{b+a}+2a\pi\ .
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
| For those amongst you folks, like egreg and myself, who enjoy completing the square, we can also have at it like this:
Set
$J = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}; \tag 1$
then our equation reads
$X^2 + X = J, \tag 2$
from which we may write, completing the square,
$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4} = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 3$
our task is now to find the matrices $C$ such that
$C^2 = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 4$
to this end we observe that, setting
$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \tag 5$
whence
$P^2 = I, \tag 6$
we have
$\left (\dfrac{1}{2}I + P \right )^2 = \dfrac{1}{4} I^2 + 2\left ( \dfrac{1}{2}I \right ) P + P^2 = \dfrac{1}{4} I + I + P = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}, \tag 7$
and also
$\left (I + \dfrac{1}{2} P \right )^2 = I^2 + 2I \left (\dfrac{1}{2}P \right ) + \dfrac{1}{4} P^2 = I + \dfrac{1}{4}I + P = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 8$
we are thus motivated to generalize by seeking solutions of the form $aI + bP$:
$(aI + bP)^2 = a^2 I^2 + 2abIP + b^2 P^2 = a^2 I + b^2 I + 2ab P$
$= (a^2 + b^2)I + 2ab P = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix} = \dfrac{5}{4}I + P; \tag 9$
thus,
$a^2 + b^2 = \dfrac{5}{4}; \; 2ab = 1; \tag{10}$
it follows from (10) that the ordered pairs $(a, b)$ lie on both the circle of radius $5/4$ centered at the origin in the $(a, b)$ plane, and also on the hypberbola $ab = 1/2$; since the points $\pm (1/\sqrt 2, 1/\sqrt 2)$ lie in the interior of the circle, the branches of this hyperbola cross the circle at exactly four points, and we have seen in (8)-(9) that two of them are $(1, 1/2)$ and $(1/2, 1)$; it is then evident from the symmetry $(a, b) \leftrightarrow (-a, -b)$ of the solutions to (10) that there are a total of precisely four pairs $(a, b)$ which solve (10) and hence (9), viz.
$(a, b) = \pm \left (1, \dfrac{1}{2} \right ), \; \pm \left ( \dfrac{1}{2}, 1 \right ); \tag{11}$
since
$X + \dfrac{1}{2}I = aI + bP = \pm \begin{bmatrix} 1 & \dfrac{1}{2} \\ \dfrac{1}{2} & 1 \end{bmatrix}, \pm \begin{bmatrix} \dfrac{1}{2} & 1 \\ 1 & \dfrac{1}{2} \end{bmatrix}, \tag{12}$
we have
$X = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{1}{2} \end{bmatrix}, \; \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix}, \; \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -\dfrac{3}{2} & -\dfrac{1}{2} \\ -\dfrac{1}{2} & -\dfrac{3}{2} \end{bmatrix}, \tag{13}$
as the solutions to (2).
One question does, however, remain: how do we know every $C$ as in (4) may be cast in the form $aI + bP$? Well, as pointed out by Lord Shark the Unknown in his answer, $XJ = JX$ implies
$X = aI + bJ; \tag{14}$
but
$J = I + P, \tag{15}$
so in fact
$a I + b J = aI + b(I + P) = (a + b)I + bP, \tag{16}$
which shows that $X$ may be represented in the form $aI + bP$ as well; thus so may
$C = X + \dfrac{1}{2}I = (a + \dfrac{1}{2})I + bJ = (a + b + \dfrac{1}{2})I + bP. \tag{17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
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Find the volume common to sphere $x^2+y^2+z^2$ and the cylinder $x^2+y^2Find the volume common to sphere $x^2+y^2+z^2<1$ and the cylinder $x^2+y^2<ax$
I set up the following integral :
$$I=2\cdot \iiint_{z=0}^{\sqrt{a^2-x^2-y^2}} dz\,dy\,dz = 2\cdot\iint_E\sqrt{a^2-x^2-y^2} \, dy\,dx$$ where $E:x^2+y^2=ax$
Now under polar coordinates
$E:r=a\cos(\theta)$
and so $0\leq r \leq a\cos(\theta)$ and $-\pi/2\leq \theta \leq \pi/2$
$$I= 2\cdot\int_{\theta=-\pi/2}^{\pi/2}\,\int_{r=0}^{a\cos(\theta)} \sqrt{a^2-r^2}\,r\,dr\,d\theta=2\cdot (1/2)\cdot(2/3)\int_{\theta=-\pi/2}^{\pi/2}a^3\cdot(1-\sin^3(\theta))d\theta = \frac{2a^3}{3}\cdot\{\int_{\theta=-\pi/2}^{\pi/2} d\theta - \int_{\theta=-\pi/2}^{\pi/2}\sin^3(\theta)d\theta\} = \frac{2\pi a^3}{3}$$
However answer given to me is $$\frac{2a^3}{3}\cdot(\pi - \frac43)$$
Where am I making the mistake? Or is it the case that the answer given to me is incorrect?
| $$I= 2\cdot\int_{\theta=-\pi/2}^{\pi/2}\,\int_{r=0}^{a\cos(\theta)} \sqrt{a^2-r^2}\,r\,dr\,d\theta=2\cdot (1/2)\cdot(2/3)\int_{\theta=-\pi/2}^{\pi/2}a^3\cdot(1-\color{red}{|\sin^3(\theta)|})d\theta$$
Specifically,
$$\int_0^{a\cos t} \sqrt{a^2 - r^2}rdr = \frac{1}{2}\int_{a^2\sin^2t}^{a^2} u^{1/2}dr = \frac{a^3}{3}\left(1-(\sin^2t)^{3/2}\right)= \frac{a^3}{3}\left(1-|\sin^3t|\right)$$
The rest follows as you expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A strange thing when evaluating a basic limit I have attempted to evaluate the limit $\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)$. I understand, intuitively, that the limit is equal to zero, but when trying to show it, something strange happened. I've tried to evaluate the limit based on a technique that uses the fact that $\lim _{x\to \infty }\left(\frac{1}{x^n}\right)\:=\:0,\:n>0$, and divided both the numerator and the denominator of the fraction by $x$:
\begin{align}
& \lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{x}}\right) =\lim_{x\to \:\infty} \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{\sqrt{x^2}}}\right) =\lim_{x\to \:\:\infty \:\:}\left(\frac{\frac{1}{x}}{\sqrt{\frac{x-1}{x^2}}}\right) \\[10pt]
= {} & \lim_{x\to \:\:\:\infty \:\:\:} \left(\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}\right)=\frac{\frac 1 \infty}{\sqrt{\frac 1 \infty -\frac{1}{\infty ^2}}}=\frac{0}{\sqrt{0-0}}=\frac{0}{0}= \text{indeterminate}
\end{align}
Moreover, when I divided the numerator and the denominator by $\sqrt{x}$, I've actually gotten the correct result:
$$\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{\sqrt{x}}}{\frac{\sqrt{x-1}}{\sqrt{x}}}\right)=\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{\frac{x-1}{x}}}\right)=\lim_{x\to \:\infty \:}\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{1-\frac{1}{x}}} \right) = \frac{\frac{1}{\sqrt{\infty }}}{\sqrt{1-\frac{1}{\infty }}}=\frac{0}{\sqrt{1-0}} = \frac{0}{1}=0$$
Why is there a difference in the results? Why does dividing by $\sqrt{x}$ is allowed, but not by dividing by $x$, or have I done an arithmetic/algebraic mistake that caused the result to be wrong?
Help will be much appreciated :)
| Yes for that limit we have the indeterminate form
$$\lim_{x\to\infty} \frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\frac{0}{0}$$
but we can eliminate the indetermination by
$$\lim_{x\to\infty}\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\lim_{x\to\infty}\frac{\sqrt{\frac1x} }{\sqrt{\frac1x}}\frac{\sqrt{\frac1x}}{\sqrt{1-\frac{1}{x}}}=\frac{0}1=0$$
therefore it seems there is nothing special to that.
The latter manipuilation explain also the reason for a different result in the second case when we divide numerator and denominator by $\sqrt x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trig function bounded on interval (without calculus), prove that $x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$.
If $0 \lt x \lt \dfrac{\pi}{2}$, prove that
$$x^{3/2}\sin x + \sqrt{9-x^3}\cos x \leq 3$$
This question must be done without calculus. First, I tried splitting it into the intervals $(0,\pi/4)$ and $(\pi/4, \pi/2)$, hoping that, $\sin x$ was bound tightly enough on the interval that it'd be less than 3 even if $\cos x = 1$ (which doesn't work -- letting $\sin x = \dfrac{1}{\sqrt{2}}$ and $\cos x = 1$ produces a result greater than 3).
The other thing I noticed was that inside the square root sign, we have $\sqrt{9-x^3} = \sqrt{(3-x^{3/2})(3+x^{3/2})}$, and an $x^{3/2}$ appears in the first term, but I'm not sure how useful the similarity there is.
Advice on how to proceed?
| For all $0\le x\le 3^{2/3}$ (which is $\ge\frac\pi2$) \begin{align}&x^{3/2}\sin x+(9-x^3)^{1/2}\cos x=\\&=\sqrt{(x^{3/2})^2+9-x^3}\left(\frac{x^{3/2}}{\sqrt{(x^{3/2})^2+9-x^3}}\sin x+\frac{(9-x^3)^{1/2}}{\sqrt{(x^{3/2})^2+9-x^3}}\cos x\right)=\\&=3\left(\frac{x^{3/2}}{3}\sin x+\frac{(9-x^3)^{1/2}}{3}\cos x\right)=3\sin\left(x+\arccos\frac{x^{3/2}}3\right)\le 3\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| $\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}$The integrand is periodic with period $\frac \pi 2$ hence:
\begin{align}
I:=\int^{2\pi}_0 \frac{1}{\sin^4 (x)+\cos^4(x)}\,dx = 4 \int^{\pi/2}_0 \frac{1}{\sin^4(x)+\cos^4(x)}\,dx
\end{align}
Applying Weierstrass substitution now leads to high order polynomial in the denominator, you can reduce the powers by noticing that:
\begin{align}
I &= 4\int^{\pi/2}_0 \Re\frac{1}{\sin^2(x)+i\cos^2(x)}-\Im \frac{1}{\sin^2(x)+i\cos^2(x)} \,dx \\
&= 4(\Re J- \Im J)
\end{align}
where $$J:= \int^{\pi/2}_0 \frac{1}{\sin^2(x)+i\cos^2(x)} \,dx$$ now we set $t=\tan(x)$ to get $$J = \int^\infty_0 \frac{1}{t^2+i}\,dt = \frac 1 2 \int^\infty_{-\infty}\frac{1}{t^2+i}\,dt$$
This integral is just a standard application of the Residue Theorem:
$$J=\frac \pi 2 e^{-i\pi/4}$$
So:
$$I = 2\pi \left[\cos\left(-\frac \pi 4 \right) -\sin\left(-\frac \pi 4 \right)\right] = 2\sqrt[]{2}\ \pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
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Finding expected value of number of kings drawn before the first Ace and its lower bound? Consider a standard deck of cards. We draw cards from the top until we get either a King or an Ace.
1.Write a lower bound for the expected value of the number of Kings drawn before the first Ace.
The solution given in this problem is 1/2.
I'm not sure how they arrive at this solution. I assumed that both Kings and Aces are equally likely to be drawn so shouldn't the # of kings being drawn before first Ace appear be $n* \frac {1}{52}$ ?
2. Find the expected value of the number of Kings drawn before the first Ace.
Here since the number of kings in a deck is 4 and there are 52 cards in total shouldn't the expected number of kings drawn before the first Ace be $n* \frac {4}{52}$ ?
| Part 2. For the purpose of analysis, we may pretend the deck consists only of four kings and four aces. Let $X$ be the number of kings before the first ace.
Then $P(X>0)$ is the probability that the first card is a king, so $P(X>0) = \frac{4}{8}$.
$P(X > 1)$ is the probability that the first two cards are kings, so $P(X>1) = \frac{4}{8} \cdot \frac{3}{7}$.
Similarly, we see that $P(X>2) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}$, and $P(X>3) = \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5}$.
So $$E(X) = P(X>0) + P(X>1)+P(X>2)+P(X>3) = 0.8$$
Here we have made use of the theorem that
$$E(X) = \sum_{n=0}^{\infty} P(X>n)$$
which holds for any random variable $X$ which takes on only values in $\{0, 1, 2, 3, \dots \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction $\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$ In this problem I have to prove the following equation using mathematical induction:
$$\sum_{i=1}^n 3i(i+4) = \frac{(n)(n+1)(2n+13)}{2}$$
So far I've proved that $P_1$ is true, and written $P_k$ as $$15+36+63+...+3k(k+4)= \frac{(k)(k+1)(2k+13)}{2}$$
I then attempt to solve $P_{k+1}$. I start by using $P_k$ and substituting it into the equation to get $$\frac{(k)(k+1)(2k+13)}{2}+(3k+1)(k+5)=\frac{(k+1)(k+2)(2k+14)}{2}$$
I then multiply $(3k+1)(k+5)$ by $\frac{2}{2}$ to get $$\frac{(k)(k+1)(2k+13)(3k+1)(2k+10)}{2}=\frac{(k+1)(k+2)(2k+14)}{2}$$
Here its obvious that the LHS isnt going to be equal to the RHS, seeing as the LHS would have $k^5$ amd the RHS would have $k^3$. Any ideas?
| Alternatively, prove a very useful identity: $$\sum_{k=0}^n\,\binom{k}{r}=\binom{n+1}{r+1}$$ for all integers $n,r\geq 0$. Then, note that
$$3k(k+4)=6\binom{k}{2}+15\binom{k}{1}\,.$$
You can prove it by induction on $n$, or using some kind of combinatorial arguments.
Then, we have
$$\begin{align}
\sum_{k=1}^n\,3k(k+4)&=\sum_{k=0}^n\,3k(k+4)=6\,\sum_{k=0}^n\,\binom{k}{2}+15\,\sum_{k=0}^n\,\binom{k}{1}=6\binom{n+1}{2+1}+15\binom{n+1}{1+1}
\\&=6\binom{n+1}{3}+15\binom{n+1}{2}=(n+1)n(n-1)+\frac{15(n+1)n}{2}\\&=\frac{n(n+1)\big(2(n-1)+15\big)}{2}=\frac{n(n+1)(2n+13)}{2}\,.\end{align}$$
Here is another proof, using the well known sum
$$\sum_{k=0}^n\,k^2=\frac{n(n+1)(2n+1)}{6}\text{ for all integer }n\geq 0\,.$$
Note that $$\sum_{k=1}^n\,3k(k+4)=3\,\sum_{k=1}^n\,\big((k+2)^2-4\big)=3\,\left(\sum_{k=1}^n\,(k+2)^2-\sum_{k=1}^n\,4\right)\,.$$
That is,
$$\sum_{k=1}^n\,3k(k+4)=3\,\left(\sum_{k=3}^{n+2}\,k^2-4n\right)=3\,\left(\sum_{k=0}^{n+2}\,k^2-5-4n\right)\,.$$
Now, plugging in $\sum_{k=0}^{n+2}\,k^2=\frac{(n+2)(n+3)(2n+5)}{6}$ yields
$$\sum_{k=1}^n\,3k(k+4)=3\,\left(\frac{(n+2)(n+3)(2n+5)}{6}-5-4n\right)\,.$$
The rest is just some work with simple algebraic manipulations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Domain of definition of :$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$ let $f$ be a function of the variable $x$ defined as : $$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$$ , Really i didn't succeed to determine the domain of definition of that function since i'm not able to get when does that converges , Then my question here is :What is the domain of definition of :$f(x)=\int_{0}^{\infty}\frac{t^x}{1+t^4 \sin ^2 t}dt$ ?
| Since the integrand is positive, we can write
\begin{align}
f(x) &= \int \limits_0^\infty \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t \\
&= \int \limits_0^{\pi/4} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t + \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t + \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{1}{4}\right)\pi}^{\left(k+\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4 \sin^2 (t)} \, \mathrm{d} t \\
&\equiv f_1 (x) + f_2 (x) + f_3 (x) \,
\end{align}
and estimate the three terms separately.
We have
$$ \frac{1}{1+2^{-9} \pi^4} \int \limits_0^{\pi/4} t^x \, \mathrm{d} t \leq f_1 (x) \leq \int \limits_0^{\pi/4} t^x \, \mathrm{d} t \, , $$
so $f_1(x) < \infty$ holds if and only if $x > -1$ .
For $k \in \mathbb{N}$ and $ \left(k-\frac{3}{4}\right)\pi \leq t \leq \left(k-\frac{1}{4}\right)\pi$ we have $\frac{1}{2} \leq \sin^2 (t) \leq 1$ , so
$$ \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+t^4} \, \mathrm{d} t \leq f_2(x) \leq \sum \limits_{k=1}^\infty \int \limits_{\left(k-\frac{3}{4}\right)\pi}^{\left(k-\frac{1}{4}\right)\pi} \frac{t^x}{1+ \frac{1}{2} t^4} \, \mathrm{d} t \, . $$
We conclude that $f_2(x)$ is finite if and only if $x<3$ .
For the last term we can use
\begin{align}
f_3 (x) &\leq \pi^x \sum \limits_{k=1}^\infty \left(k+\frac{1}{4}\right)^x \int \limits_{\left(k-\frac{1}{4}\right)\pi}^{\left(k+\frac{1}{4}\right)\pi} \frac{1}{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4 \sin^2 (t)} \, \mathrm{d} t \\
&= \pi^x \sum \limits_{k=1}^\infty \left(k+\frac{1}{4}\right)^x \left[\frac{\arctan\left\{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4} \tan(x) \right\}}{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}} \right]_{x= \left(k-\frac{1}{4}\right) \pi}^{x= \left(k+\frac{1}{4}\right) \pi} \\
&= 2 \pi^x \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}} \arctan\left\{\sqrt{1+\left[\left(k - \frac{1}{4}\right)\pi\right]^4}\right\} \\
&\leq \pi^{x-1} \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\left(k - \frac{1}{4}\right)^2}
\end{align}
and a similar estimate from below to find
$$ \frac{\pi^{x-1}}{2\sqrt{2}} \sum \limits_{k=1}^\infty \frac{\left(k-\frac{1}{4}\right)^x}{\left(k + \frac{1}{4}\right)^2} \leq f_3(x) \leq \pi^{x-1} \sum \limits_{k=1}^\infty \frac{\left(k+\frac{1}{4}\right)^x}{\left(k - \frac{1}{4}\right)^2} \, . $$
This implies $f_3(x) < \infty ~ \Leftrightarrow ~ x < 1$ .
Combining these results we see that $f(x) < \infty$ holds if and only if $-1 < x < 1$ , so $f$ is well-defined on the interval $(-1,1)$ .
| {
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"url": "https://math.stackexchange.com/questions/2847170",
"timestamp": "2023-03-29T00:00:00",
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How can I prove $\tan15^\circ=2-\sqrt3$ using the triple-angle formulas?
Prove $\tan15^\circ=2-\sqrt3\,\,$ using following results $$\sin3x=3\sin{x}-4\sin^3x$$
$$\cos3x=4\cos^3x-3\cos{x}$$
I know that if I substitute $x=15^\circ\,$I can write$$\sin45=3\sin15-4\sin^315$$
$$\cos45=4\cos^315-3\cos15$$ What can I do next? any hints?
| Let $s=\sin 15^\circ$, $c=\cos 15^\circ$ and $t=s/c=\tan 15^\circ\in (\tan 0^\circ,\tan 45^\circ)=(0,1)$ then, from your equations, since $\sin 45^\circ=\cos 45^\circ$, it follows that
$$s(3-4s^2)=c(4c^2-3)\implies s(3(s^2+c^2)-4s^2)=c(4c^2-3(s^2+c^2))$$
and, after dividing both sides by $c^3$, we obtain
$$t(3-t^2)=(1-3t^2)\implies (t+1)(t^2-4t+1)=0.$$
(the factorization can be obtained by noticing that the equation on the left is satisfied by $t=-1$).
Can you take it from here and find the root $t\in (0,1)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumption is correct then $x^2+1=(x-i)(x+i)$
Now for $x=\pm i$ using synthetic division it leaves remainder 0
Am I at the right direction, please tell
| HINT:
Note that $$\begin{align}x^{20}+x^{15}+x^{10}+x^5&=x^5((x^5)^3+(x^5)^2+x^5+1)\\&=x^5(x^5+1)((x^5)^2+1)\\&=x^5(x^5+1)(x^{10}+1)\end{align}$$ and $$(-i)^{10}=i^{10}=i^2=-1$$ so $x-i$ and $x+i$ are factors of the polynomial $x^{10}+1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
System of Diophantine equations $a^2+b^2=2x^2+1,c^2+d^2=2y^2+1,ac-bd=1$ has no natural solutions. How to prove that system
$$
a^2+b^2=2x^2+1, \\
c^2+d^2=2y^2+1, \\
a\cdot c-b \cdot d=1
$$
has no natural solutions?
It can be proved that system equal to the equation
$$(2х^2+1)(2у^2+1)=4z^2+1$$
In one direction
from Fibonacci identity
$$(2x^2+1)(2у^2+1)=(а^2+b^2)(c^2+d^2)=(аd+bc)^2+(аc-bd)^2=(аd+bc)^2+1$$
and using Pythagorean quadruple $$(1; 2z; х^2-у^2; 1+х^2+у^2)$$
$$1+(2z)^2+(x^2-у^2)^2=(1+х^2+у^2)^2$$
$$1=m^2+n^2-p^2-q^2; 1+x^2+у^2=m^2+n^2+p^2+q^2$$
$$х^2-у^2$$ equal $$2(mq+np)$$ or $$2(nq-mp)$$
Let
$$х^2-у^2=2(mq+np)$$
We get
$$2х^2+1=(m+q)^2+(n+p)^2; 2у^2+1=(m-q)^2+(n-p)^2$$
and
$$2х^2+1=а^2+b^2; 2у^2+1=c^2+d^2$$. Besides
$$1=m^2+n^2-p^2-q^2=(m-q)(m+q)-(p-n)(p+n)=аc-bd$$
and we get system again.
For equation $$(2х^2+1)(2у^2+1)=z^2+1$$
I have tried rewrite it
$$4x^2y^2+2x^2+2y^2=z^2$$
Let $$z = 2xy+k$$
$$4z^2=4x^2y^2+4xyk+k^2 $$
$$4x^2y^2+2x^2+2y^2 = 4x^2y^2+4xyk+k^2$$
$$2x^2+2y^2 =4xyk+k^2$$
We may assume $$k=2m$$
$$2x^2+2y^2 =8xym+4m^2$$
$$x^2+y^2= 4xym +2m^2$$
$$x^2+y^2-4xym =2m^2$$
$$(x-2my)^2+y^2-4m^2y^2 =2m^2$$
$$(x-2my)^2-y^2(4m^2-1) =2m^2$$
And have tried standard techniques like Legendre theorem,but it didn't help. .
I have seen this problem in one social math community on the russian site like facebook.
| COMMENT.-It is clear that variables $a, b$ have distinct parity; let $a$ be even and $b$ odd.
It is known that the general solution of the equation
$$x^2+y^2+z^2=w^2$$ is given by the identity $$(2XZ)^2+(2YZ)^2+(Z^2-X^2-Y^2)^2=(X^2+Y^2+Z^2)^2$$ It follows because of the first equation is equivalent to $$a^2+b^2+x^4=(x^2+1)^2$$ we have
$$\begin{cases}a=2XZ\\b=Z^2-X^2-Y^2\\x^2=2YZ\\x^2+1=X^2+Y^2+Z^2\end{cases}$$
Then the three parameters $X,Y,Z$ are related by $$(Y-Z)^2+X^2=1$$
Similar reasoning with the second equation.
I have no time to try to get the end of the proof in case this remark is useful. Can you do it?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Examine convergence $\sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right) $ I need some help with following series:
$$\sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right)$$
I have no idea how I should modify or compare it with something else.
| Also, from
$$\frac{x}{1+x}\leq \ln{(1+x)}\leq x, \forall x>-1$$
we have
$$\sqrt{\frac{1}{n+1}}\leq \sqrt{\ln{\left(1+\frac{1}{n}\right)}}\leq \frac{1}{\sqrt{n}}$$
or
$$0\leq \frac{1}{\sqrt{n}}-\sqrt{\ln{\left(1+\frac{1}{n}\right)}} \leq \frac{1}{\sqrt{n}}-\sqrt{\frac{1}{n+1}}$$
where
$$0\leq \frac{1}{\sqrt{n}}-\sqrt{\frac{1}{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}< \frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n(\sqrt{n+1}+\sqrt{n})}< \frac{1}{2\sqrt{n^3}}$$
As a result
$$0\leq\sum\limits_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right)<
\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$
which converges, since $\frac{3}{2}>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Algebraic Inequality involving AM-GM-HM If $$a,b,c \;\epsilon \;R^+$$
Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$
This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$
Not sure what to to do next.
Any help would be appreciated.
| By C-S
$$\sum_{cyc}\frac{bc}{b+c}\leq\sum_{cyc}\frac{bc}{(1+1)^2}\left(\frac{1^2}{b}+\frac{1^2}{c}\right)=\frac{a+b+c}{2.}$$
| {
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"url": "https://math.stackexchange.com/questions/2861098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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what does it mean to reject a hypothesis? This is a simple problem I cannot understand. The task is to solve:
$$\frac{2}{3} \times |5-2x| - \frac{1}{2} = 5$$
*
*As a first step, I isolate the absolute value, like this:
\begin{align}
2/3 \times |5-2x| &= 5 + 1/2\\
2/3 \times |5-2x| &= 11/2\\
|5-2x| &= 11/2 ÷ 2/3 \\
|5-2x| &= 33/4\\
|5-2x| &= 8.25
\end{align}
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =\begin{cases} 5-2x & x \geq 2.5\\
5+2x & x < 2.5.
\end{cases}
$$
*Let's solve the equation $|5-2x| = 8.25$
if $x \geq 2.5$, then:
\begin{align}
5-2x &= 8.25\\
-2x &= 3.25\\
x &= -1.625
\end{align}
But we said that $x \geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
\begin{align}
-5+2x &= 8.25\\
2x &= 13.25\\
x &= 6.625
\end{align}
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis?
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)
| A good way to solve this kind of equation is consider two cases
1) For $5-2x\ge 0 \iff x\le \frac25$ we have
$$\frac23 |5-2x| - 1/2 = 5 \iff \frac23 (5-2x) - 1/2 = 5 \iff 20-8x-3=30 \\\iff 8x=-13 \iff x=-\frac{13}8$$
that solution is acceptable since it is consistent to the assumption $x\le \frac25$.
2) For $5-2x< 0 \iff x> \frac25$ we have
$$\frac23 |5-2x| - 1/2 = 5 \iff \frac23 (2x-5) - 1/2 = 5 \iff 8x-20-3=30 \\\iff 8x=53 \iff x=\frac{53}8$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Evaluating a common integral for $\arcsin(x)$ using a different method
The common method of evaluating the integral below is by using the substitution $x = a \sin u$ which gives:
$$\int\frac{dx}{\sqrt{a^2-x^2}}= \arcsin\Bigl(\frac xa\Bigr) + C.$$
But, using this substitution led me down a path that seems to give a different answer (or, at least, a different form). Did I make a mistake, or are the two forms somehow equivalent?
$$x = ai\tan u $$
$$x^2=-a^2\tan^2(u)du$$
$$dx = ai\sec^2(u)du$$
$$1+\tan^2(u)=\sec^2(u)$$
\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}} & = \int\frac{1}{\sqrt{a^2+a^2\tan^2(u)}}\cdot ai\sec^2(u)du \\
& = ai\int \frac{\sec^2(u)}{\sqrt{a^2(1+\tan^2(u))}}du \\
& = \frac{ai}{a} \int \frac {\sec^2(u)}{\sec(u)}du \\
& = i \int \sec(u)du \\
& =i \ln \lvert \sec(u)+\tan(u)\rvert + C \\
& = i \ln \bigl| \frac {\sqrt{x^2-a^2}}{ai} + \frac {x}{ai} \bigr| + C \\
& = i \ln \bigl| \frac {-i\sqrt{x^2-a^2}}{a} - \frac {ix}{a} \bigr| + C \\
& = i \ln \bigl| -i\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C\\
& = i \Big[ \ln \lvert -i \rvert + \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| \Big] + C\\
& = i \ln \bigl|\frac {\sqrt{x^2-a^2}+x}{a} \bigr| + C
\end{align}
| First, we have for complex $u$
$$\int \sec(u)\,du=\log(\sec(u)+\tan(u))+C$$
Next, we note that for $x=ia\tan(u)$ and $a\ge x$
$$\begin{align}
i\log(\sec(u)+\tan(u))+C&=i\log\left(\frac{\sqrt{x^2-a^2}}{ia}+\frac{x}{ia}\right)+C\\\\
&=i\log(x+i\sqrt{a^2-x^2})+C'\\\\
&=-\arctan\left(\frac{\sqrt{a^2-x^2}}{x}\right)+C''\\\\
&=\arctan\left(\frac{x}{\sqrt{a^2-x^2}}\right)+C'''\\\\
&=\arcsin(x/a)+C''''
\end{align}$$
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding a Moment Generating Function where the density function has $e$ in it Problem:
Consider a r.v. $X$ with pdf
\begin{eqnarray*}
f_x(x) &=& \frac{e^{ -\frac{(x+7)^2}{32} }}{ \sqrt{32\pi} }
\,\,\, -\infty < x < \infty \\
\end{eqnarray*}
Find the moment generating function of $X$.
Answer:
\begin{eqnarray*}
M_x(t) &=& E(e^{tX}) =
\int_{-\infty}^{\infty} e^{tx} \frac{e^{ -\frac{(x+7)^2}{32} }}{ \sqrt{32\pi} }
\,\, dx \\
M_x(t) &=&
\int_{-\infty}^{\infty}
e^{tx} \frac{e^{ -\frac{x^2+14x+49}{32} }}{ \sqrt{32\pi} } \,\, dx =
\int_{-\infty}^{\infty}
\frac{e^{ -\frac{x^2+32tx+14x+49}{32} }}{ \sqrt{32\pi} } \,\, dx
\\
\end{eqnarray*}
Is what I have done so far right? If it is, how do I perform the integration?
Thanks,
Bob
Here is my second attempt to solve the problem. This time, I come up with the answer, but I come up with the wrong answer. I am hoping somebody can tell me where I went wrong.
Answer:
\begin{eqnarray*}
M_x(t) &=& E(e^{tX}) =
\int_{-\infty}^{\infty} e^{tx} \frac{e^{ -\frac{(x+7)^2}{32} }}{ \sqrt{32\pi} }
\,\, dx \\
M_x(t) &=&
\int_{-\infty}^{\infty}
e^{tx} \frac{e^{ -\frac{x^2+14x+49}{32} }}{ \sqrt{32\pi} } \,\, dx =
\int_{-\infty}^{\infty}
\frac{e^{ -\frac{x^2-32tx+14x+49}{32} }}{ \sqrt{32\pi} } \,\, dx
\\
\end{eqnarray*}
Now to perform this integration, we complete the square:
\begin{eqnarray*}
x^2 - 32tx + 14x + 49 &=& (x + 7 - 16t)^2 - 256t^2 + 224t \\
M_x(t) &=&
\int_{-\infty}^{\infty}
\frac{e^{ -\frac{(x + 7 - 16t)^2 - 256t^2 + 224t}{32} }}{ \sqrt{32\pi} } \,\, dx
\\
M_x(t) &=&
e^{8t^2-7t} \int_{-\infty}^{\infty}
\frac{e^{ -\frac{(x + 7 - 16t)^2 }{32} }}{ \sqrt{32\pi} }
\,\, dx \\
\end{eqnarray*}
Now to preform the integration, let $u = x + 7 - 16t$ and that makes the integral
like the one for the normal distribution.
\begin{eqnarray*}
\int_{-\infty}^{\infty} \frac{e^{ -\frac{(x + 7 - 16t)^2 }{32} }} \,\, dx &=& 0 \\
M_x(t) &=& 0 \\
\end{eqnarray*}
Thanks,
Bob
| In the last step you forgot the minus sign in front of $32 t x$, everything else looks fine. The idea for the integration is to complete the square in the exponent:
$$ x^2 + 2 (7 - 16t) x + 49 = (x + 7 - 16t)^2 - 256 t^2 + 224 t \, .$$
Then you are left with a standard Gaussian integral:
$$ M_X (t) = \mathrm{e}^{8 t^2 - 7 t} \int \limits_{-\infty}^\infty \frac{\mathrm{e}^{-(x + 7 - 16 t)^2 /32}}{\sqrt{32 \pi}} \, \mathrm{d} x \, . $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$
Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.
I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.
| Hints:
*
*$(1-a) \cdot (1+a) = 1 - a^2\,$, $\,(1-a) \cdot (1+a)(1+a^2) = 1 - a^4\,$, $\;\ldots$
*$(1+i)^2 = 2i$
*$1 - \dfrac{1+i}{2}=\dfrac{1-i}{2} = \dfrac{1}{1+i}$
| {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solution to Putnam 2007 A-1
2007 A-1: Find the values of $\alpha$ for which the curves $y=\alpha x^2 + \alpha x + \frac{1}{24}$ and $x=\alpha y^2 + \alpha y + \frac{1}{24}$ are tangent to each other.
My solution: Notice that the second equation is obtained from the first by swapping $x$ and $y$. This happens when you reflect the graph of the first equation across the $x=y$ line. Hence, the two curves will intersect only once, and therefore be tangent to each other at that point of intersection, if each curve individually also intersects the $y=x$ line only once. Hence $x=\alpha x^2 + \alpha x + \frac{1}{24}$, on solving which we get $\alpha=\frac{43}{12},-\frac{17}{12}$.
This is not the answer or solution given on Kedlaya's putnam page. Where am I going wrong?
| $$(\alpha - 1)^2 - \frac{\alpha}6=0$$
$$\alpha^2-2\alpha-\frac{\alpha}6+1=0$$
$$6\alpha^2-12\alpha - \alpha + 6 = 0$$
$$6\alpha^2-13\alpha+6=0$$
$$(2\alpha-3)(3\alpha-2)=0$$
Also, here is the visualization for the other solutions.
| {
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"url": "https://math.stackexchange.com/questions/2866985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the natural number $a$ for which $\sum^n_{k=1}f(a+k)=16(2^n-1)$, given that $f(x+y)=f(x)\,f(y)$.
Find the natural number $a$ for which
$$\sum^n_{k=1}f(a+k)=16(2^n-1)\,,$$
where the function $f:\mathbb{N}\to\mathbb{N}$ satisfies the relation
$$f(x+y)=f(x)f(y)\text{ for all }x,y\in \mathbb{N}$$
and further $f(1) = 2$.
Please solve this.
| The first thing to find is what $f(x)$ is
$$f(1) = 2 = f(1+0) = f(1)*f(0)$$
Therefore, $f(0) = 1$
$$f(2) = f(1)f(1) = 2^2$$
$$f(3) = f(2)f(1) = f(1)f(1)f(1) = 2^3$$
Clearly, $f(x) = 2^x, x \in \mathbb{N}$
Now
$$\sum_{k=1}^n 2^{a+k} = 16(2^n-1)$$
$$2^a\sum_{k=1}^n 2^k = 16(2^n-1)$$
$$2^a \cdot \frac{2^{n+1} - 2}{2-1} = 16(2^n-1)$$
$$2^a(2 \cdot(2^n-1)) = 16(2^n-1)$$
Clearly $a = 3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?
| For $x=0$ we get a value $\frac{7}{3}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{x^4+5x^2+7}{x^2+3}\geq\frac{7}{3}$$ or
$$x^2(3x^2+8)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the least value of the expression $x^2+2xy+2y^2+4y+7$ Find the least value of the expression
$x^2+2xy+2y^2+4y+7$
I am not able to solve this equation though i am able to differentiate it.
| Given $x^2+2y^2+2xy+4y+7$$$x^2+2y^2+2xy+4y+7$$$$x^2+y^2+2xy+y^2+4y+7$$$$x^2+y^2+2xy+y^2+4y+4+3$$$$(x+y)^2+(y+2)^2+3$$Minimum of square of any number is $0$. So, minima will be at $x=2,y=-2$
Therefore, $(2+(-2))^2+(-2+2)^2+3=3$
Therefore, the least value of the expression $x^2+2y^2+2xy+4y+7$ is $3$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$
If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$.
I tried to solve this problem by C-S. But I can't sovle it.
Things I have done so far:
$\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq \frac{(\sum\limits_{cyc}a)^2}{2.\sum\limits_{cyc}a^2+\sum\limits_{cyc}ab}\geq \frac{(\sum\limits_{cyc}a)^2}{3.\sum\limits_{cyc}a^2}=\frac{\sum\limits_{cyc}a^2+2.\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}=\frac{1}{3}+\frac{2\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}\geq ?$
| By C-S and Muirhead
$$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2c^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2c^2+a^2bc)}=$$
$$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4)}=1.$$
Also, by Holder and AM-GM:
$$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^3}{b^2a+c^2a+abc}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(ab^2+ac^2+abc)}=$$
$$=\frac{(a+b+c)^3}{\sum\limits_{cyc}(abc+3a^2b+3a^2c+2abc)}\geq\frac{(a+b+c)^3}{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.