Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know upon graphing that the limit is $5\over4$. What should I do here?
Note: To be done without the use of L'Hospital Rule
| Here are the steps
$$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
$$= \lim_{x \to 1}\frac{(x-1)(x^3- 2x^2-2x-2)}{(x-1)(x^2-4x-1)} $$
$$= \lim_{x \to 1}\frac{x^3- 2x^2-2x-2}{x^2-4x-1} $$
$$= \frac{1- 2-2-2}{1-4-1} = \frac{1- 6}{1-5} = \frac{5}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$y''+y'^{2}+y=0$ equation solution How would you solve this differential equation $y''+y'^{2}+y=0$? I can't apply the ansatz method (or more formally apply the characteristic polynomial method). Thanks
| $$
y'' + y'^2 + y = 0
$$
we can rewrite as this (prove it)
$$
\frac{1}{2}\frac{d}{dy}p^2 + p^2 + y = \frac{d}{dy}p^2 + 2p^2 + 2y = 0
$$
where $p = y'$.
$$
p^2 = \mathrm{e}^{-2y}\left(-\int 2y\mathrm{e}^{2y}dy + C \right)
$$
thus solution in terms of $p$
$$
p = \sqrt{\frac{1}{2} -y + C\mathrm{e}^{-2y}} = y'
$$
or
$$
\int \frac{dy}{\sqrt{\frac{1}{2} -y + C\mathrm{e}^{-2y}}} = \int dx = x + \lambda
$$
now the lhs probably has no closed form but if choose the special case $C=0$
$$
\int \frac{dy}{\sqrt{\frac{1}{2}-y}} = -2\left(\frac{1}{2}-y\right)^{1/2} = x + \lambda
$$
thus
$$
\frac{1}{2} - y = \left(-\frac{x}{2} + \lambda_1\right)^2 = \frac{x^2}{4} -\lambda_1 x + \lambda_1^2
$$
or
$$
y = -\frac{x^2}{4} +\lambda_1 x - \lambda_1^2 +\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Permutations with odd-length cycles I need to find - as a homework problem - the exponential generating function for the number of permutations of $n$ consisting of an even number of odd-length cycles. I can retrieve the exponential generating function for the number of permutations with just odd-length cycles, by
$$
E(z)=\exp(\sum_{k\geq 0}\frac{z^{2k+1}}{2k+1})=\exp(\sum_{k\geq 1}\frac{z^{k}}{k})/\exp(\sum_{k\geq 1}\frac{z^{2k}}{2k})=\frac{1}{1-z}(\sqrt{1-z^2})=\sqrt{\frac{1+z}{1-z}}.
$$
Since $n$ has permutations of an even number of odd-length cycles if and only if it is even, in which case all permutations with odd-length cycles have an even number of cycles, I believe that I have to find the power series obtained by $E(z)$ by just selecting the terms in $z^n$ with $n$ even. Is this correct? In any case, I don't know how to proceed.
| There are two possible interpretations here, the first, permutations
consisting of an even number of odd cycles and some even cycles and
second, permutations consisting of an even number of odd cycles only.
First interpretation.
Observe that the generating function of permutations with odd cycles
marked is
$$G(z, u) =
\exp\left(\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left((1-u)\sum_{k\ge 1} \frac{z^{2k}}{2k}
+ u \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\exp\left(\sum_{k\ge 1}\frac{z^{k}}{k}\right)
+ \frac{1}{2} \exp\left(2\sum_{k\ge 1} \frac{z^{2k}}{2k}
- \sum_{k\ge 1}\frac{z^{k}}{k}\right).$$
This simplifies to
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} (1-z) \frac{1}{1-z^2}$$
which is
$$\frac{1}{2} \frac{1}{1-z}
+ \frac{1}{2} \frac{1}{1+z}.$$
This simply says that when $n$ is even then there must be an even
number of odd cycles and when $n$ is odd there cannot be an even
number of odd cycles, which follows by inspection (parity).
Second interpretation.
Here we have
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right).$$
This is
$$G(z, u) =
\exp\left(u \sum_{k\ge 0}\frac{z^{k}}{k}
- u \sum_{k\ge 1}\frac{z^{2k}}{2k}\right).$$
To get the permutations with an even number of odd cycles use
$$\frac{1}{2} G(z,1)+\frac{1}{2} G(z, -1)$$
which yields
$$\frac{1}{2}\frac{1}{1-z}
\left(\frac{1}{1-z^2}\right)^{-1/2}
+ \frac{1}{2} (1-z)
\left(\frac{1}{1-z^2}\right)^{1/2}.$$
This gives the sequence
$$0, 1, 0, 9, 0, 225, 0, 11025, 0, 893025, 0,
108056025, 0, 18261468225,\ldots$$
which points us to
OEIS A177145,
where we find this computation confirmed.
This generating function maybe written as
$$\frac{1}{2}\frac{1}{1-z}
\sqrt{1-z^2}
+ \frac{1}{2} (1-z)
\frac{1}{\sqrt{1-z^2}}$$
or
$$\frac{1}{2}\sqrt{\frac{1+z}{1-z}}
+ \frac{1}{2}\sqrt{\frac{1-z}{1+z}}.$$
| {
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"url": "https://math.stackexchange.com/questions/1163681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying help: If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$. If ${n \choose 3} + {n+3-1 \choose 3} = (n)_3$, compute $n$.
So far, I have:
$\frac{n!}{3! (n-3)!} + \frac{(n+2)!}{3! (n-1)!}$
Then I simplified to:
$\frac{1}{6} (n-2)(n-1)(n) + \frac{1}{6} (n)(n+1)(n+2)$
Is this correct, and if so, what is the next move? It needs to equal $(n)_3$.
Thanks!
| You are nearly ready.
Substitute $(n)_3 = n \cdot (n-1) \cdot (n-2)$, and you will see both sides are divisable by n. Thus, converting the result to a polinomial form, you will get an at most 2-grade equation, which will be soon very easy to solve.
$$\frac{1}{6}(n-2)(n-1)n + \frac{1}{6}n(n+1)(n+2) = n(n-1)(n-2)$$
Multiply with 6:
$$(n-2)(n-1)n + n(n+1)(n+2) = 6(n-2)(n-1)n$$
Minus $(n-2)(n-1)n$:
$$n(n+1)(n+2) = 5n(n-1)(n-2)$$
Divide both side with $n$:
$$(n+1)(n+2) = 5(n-1)(n-2)$$
$$n^2+3n+2 = 5(n^2-3n+2)$$
$$n^2+3n+2 = 5n^2-15n+10$$
$$4n^2-18n+8 = 0$$
$$2n^2-9n+4 = 0$$
$$n_{1,2}=\frac{9\pm \sqrt{81-4*2*4}}{4}=\frac{9\pm 7}{4} \in \{ \frac{1}{2}, 4\}$$
Only $\underline{\underline{4}}$ is integer, thus this is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve: $-3(6t^3-1)^6 -3t[6(6t^3-1)^5(18t^2)]$ Solve: $-3(6t^3-1)^6 + -3t[6(6t^3-1)^5(18t^2)]$
I don't know how to multiply the two equations and then add them.
| Let $A = 6t^3-1$.
\begin{align} -3(6t^3-1)^6 + -3t[6(6t^3-1)^5(18t^2)] &= -3A^6-3t(108A^5t^2) \\
&= -3A^6-324A^5t^3\\
&=-3A^5(A+108t^3)\\
&=-3(6t^3-1)^5[(6t^3-1)+108t^3]\\
&=-3(6t^3-1)^5(114t^3-1)\\
\end{align}
So $t=\sqrt[3]{\frac{1}{6}}, \sqrt[3]{\frac{1}{114}}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by mathematical induction that $\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $ Prove by mathematical induction that:
$$\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $$
Step 1: Show that the statement is true for $n = 1$:
LHS = $$\frac{(-1)^{1+1}}{1} = 1$$
RHS = $$\frac{1}{1} = 1$$
Step 2: Show that "if true for n = p, then true for n = p + 1":
Starting with the LHS of equality for $n = p + 1$ and try to get to the RHS by using the equality for $n = p$. Simplifying:
$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k}$$
Breaking out the first two term:
$$ \sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} $$
The last term can now be exchanged for the RHS in the original equality:
$$\frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k}$$
Since $2p+3$ is odd and $2p+2$ is even, we get:
$$\frac{(-1)}{2p+2} + \frac{1}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k} = \frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k}$$
...because we can easily absorb the second term in the third term sum. However, I am stuck here because the first term is negative. I probably made a trivial error somewhere, but I am unable to find it. Any suggestions?
| When you go from $n=p$ to $n=p+1$, what you want to show is
$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k={\color{red}{(p+1)}}+1}^{2(p+1)} \frac{1}{k}. $$
The crucial part I think you're overlooking is in red. To get the desired result, you have to pull the first term out of your sum $\sum_{k=p+1}^{2p+1} \frac{1}{k}$.
$$\begin{align}\\
\frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k} &= \frac{-1}{2p+1} + \frac{1}{p+1} + \sum_{k=p+2}^{2p+1} \frac{1}{k},\\
&= \frac{-1}{2p+2} +\frac{2}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\
&= \frac{1}{2p+2} + \sum_{k=p+2}^{2p+1} \frac{1}{k}, \\
&= \sum_{k=p+2}^{2p+2} \frac{1}{k}.\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding square root of $-5-12i$ by formula and by De Moivre's Theorem I was trying to obtain the square root of $-5-12i$ by the formula for square root (given below) and also by De Moivre's theorem and verify that both give the same result. But the two results are somehow not matching for this complex number. I am writing my solution below in two cases for each method:
Case - I:
As given on pg - 3 of Complex Analysis - Newman and Bak, the equation $(x+iy)^2 = a+ib$ has the solution: $x=\pm\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$ and $y=\pm\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}.($sign $b)$
Putting $a=-5, b=-12$ & $($sign $b) = -ve$ in the above formula for $x$ & $y$, we get that the square roots of $-5-12i$ are $2-3i$ & $-2+3i$
Case - II:
By De Moivre's theorem, we know that given $z = r(cos \theta + i sin \theta)$; its $n$th root $z_k$ is given by $z_k = r^{1/n}(cos (\frac{\theta + 2k \pi}{n}) + i sin (\frac{\theta + 2k \pi}{n}))$, where $k=0,1,...,n-1$
Here, $z=-5-12 i = r(cos \theta + i sin \theta)$. Thus, $r=13$ and $\theta = atan(\frac{-12}{-5}) = 1.176005207$ (in radian)
Hence, $z_k = \sqrt{13}(cos (\frac{\theta + 2k \pi}{2}) + i sin (\frac{\theta + 2k \pi}{2}))$, where $k=0,1$
For $k=0$, $z_0 = \sqrt{13} (cos (\frac{\theta}{2}) + i sin (\frac{\theta}{2})) = \sqrt{13} (0.8320502943 + i 0.5547001962) = 3 + 2i$
For $k=1$, $z_1=\sqrt{13} (cos (\frac{\theta + 2 \pi}{2}) + i sin (\frac{\theta + 2 \pi}{2})) = \sqrt{13} (cos (\pi + \frac{\theta}{2}) + i sin (\pi + \frac{\theta}{2})) = - \sqrt{13} (cos (\frac{\theta}{2}) + i sin (\frac{\theta}{2})) = - \sqrt{13} (0.8320502943 + i 0.5547001962) = -3 - 2i$
Here, the square roots of $-5-12i$ are $3+2i$ and $-3-2i$
I think there must be some error in the solution because the square roots are coming out to be different. Thanks...
| There is a way without using $\arctan$.
Let the square root be $a+bi$. Then, $a^2+2abi-b^2 = -5-12i$, so $a^2-b^2 = -5 \text{(eq. 1)}$ and $2abi = -12i \ \text{(eq. 2)}; ab = -6 \ \text{(eq. 3)}$.
From equation $3$, $a = -\frac{6}{b}$. Substituting into equation $1$, $(-\frac{6}{b})^2-b^2+5=0$, and so:
$$\frac{36}{b^2} - b^2 + 5 = 0$$
$$\Rightarrow b^4 - 5b^2 - 36 = 0$$
$$\Rightarrow (b^2-9)(b^2+4) = 0$$
$$b^2 = 9; b^2 = -4$$
$$b = ±3$$
Substiuting $b = 3$ into equation $1$, we have $a = 2$; and substituting $b = -3$ we also get $a = 2$. Therefore, the two solutions are $2 + 3i$ and $2 - 3i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| You could do the enclidian division of $X^3 - Y^3$ by $X-Y$ in the ring $A[X]$, where $A = \mathbf{Z}[Y]$, has $X-Y$ has unit leading coefficient.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 8
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Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$
How prove this inequality. Thanks
a suggestion please to prove this inequality
| Let's transform $a^ab^bc^c\geq(abc)^{(a+b+c)/3}$ by taking $\log$:
$$
a^ab^bc^c\geq(abc)^{(a+b+c)/3}\iff a\log a+b\log b+c\log c\geq\frac{a+b+c}{3}\log(abc)\\
\iff\frac{a}{a+b+c}\log a+\frac{b}{a+b+c}\log b+\frac{c}{a+b+c}\log c\geq\log(\sqrt[3]{abc}).
$$
Let's prove the last inequality above. With $f(x)=-\log x=\log(1/x)$ being a convex function on $(0,\infty)$, we can use Jensen's inequality to infer that
$$
\frac{a}{a+b+c}f(1/a)+\frac{b}{a+b+c}\log f(1/b)+\frac{c}{a+b+c}\log f(1/c)\\
\geq f\left(\frac{3}{a+b+c}\right)=\log\left(\frac{a+b+c}{3}\right)\geq \log(\sqrt[3]{abc}).
$$
The last inequality above follows from the AM-GM inequality and the monotonicity of $\log$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which of following inequalities hold in interval 0 to pi/2
i tried using calculator and i got 1,2,4 correct .But i am not sure about how to prove them
| For C,
I recognize the right side
as the first 3 terms
of the binomial theorem expansion
of
$\sqrt{1+x}$.
If it is squared,
we get
$\begin{array}\\
(1+\frac{x}{2}-\frac{x^2}{8})^2
&=1+\frac{x^2}{4}+\frac{x^4}{64}
+2\frac{x}{2}-2\frac{x^2}{8}-2\frac{x^3}{16}\\
&=1+x-\frac{x^3}{8}+\frac{x^4}{64}\\
&=1+x-\frac{x^3}{8}(1-\frac{x}{8})\\
&< 1+x\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174120",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 0
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Solve $16x^{-3}=-2$
Solve $16x^{-3}=-2$.
My working:
\begin{align}
16x^{-3}&=-2\\
\frac{1}{16x^{3}}&=-2\\
\frac{16x^3}{16x^3}&=-32x^3\\
1&=-32x^{3}\\
-32x^{3}&=1\\
-32x&=\sqrt[3]{1}\\
-32x&=1\\
x&=\frac{-1}{32}
\end{align}
Is this right? What have I done wrong?
| $$16x^{-3} = -2$$
$$\frac{16}{x^3} = -2$$
$$-8 = x^3$$
$$x = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\frac{-x}{2}}{\sqrt{5-x}}$$
$f'(x) = 0$, when $x = 0, $ and is undefined when x= 5
Plugging in the roots of the derivative function and the end points of the interval into the original function:
\begin{align*}
f(0) & = 0\\
f(5) & = 0\\
f(-6) & = -6\sqrt{11}\\
f(4) & = 4 \cdot 1 = 4
\end{align*}
So why is the 4 not the absolute maximum value?
p.s. I assumed the first term goes to zero when taking a derivative by the product rule. I confused d/dx x = 1, with any number d/dd = 0
| $\dfrac{d(x\sqrt{5-x})}{dx}=\dfrac{10-3x}{2\sqrt{5-x}}$
$10-3x=0$ imply the critical number is $x=\dfrac{10}{3}$.
| {
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chain rule derivative This is an derivatives Problem.
find the tangent to $y=\sqrt{x^2-x+5\;}\;$ at $x=5$.
$y = \boxed{\;?\;}$
What I did was first find y by plugin x into the equation. The answer is $y = 5$. Then I found the derivative of both $x^2-x+5$ and $\sqrt{x^2-x+5\;}\;$ the answers for those are $2x-1$ and $\tfrac 1 2 (x)^{-1/2}$.
From here I know I have to find the answers by plugin in $5$ to $\tfrac 1 2 (x^2-x+5)^{-1/2} \cdot (2(x)-1)$ to find $m$.
After $y= mx +b$ to find $b$.
and help anyone Im on my last chance and really need the help.
| You are on a right track. You have found the derivative and now just plug $x=5$ in it and you have:
$$f'(x) = \frac{2x-1}{2\sqrt{x^2-x+5}}=\frac{10-1}{2\sqrt{25-5+2}} = \frac{9}{10}$$
This gives you the slope of the tangent at the point $(5,5)$ so you have $y=\frac 9{10}x + b$. But note that $(5,5)$ is a point that lies on both the tangent line and the function graph so you have:
$$5 = \frac 9{10}\cdot 5 + b \implies b = 5 - \frac 92 = \frac 12$$
So the final result is:$y = \frac 9{10}x + \frac 12$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the interval after substitution Given this problem
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1}\geq 9^{\sqrt{x}}$$
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+3^{2\sqrt[4]{x}+2}\geq 3^{2\sqrt{x}}\\8\cdot 3^{\sqrt{x}+\sqrt[4]{x}-2\sqrt{x}}+3^{2\sqrt[4]{x}+2-2\sqrt{x}}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+3^{2\sqrt[4]{x}-2\sqrt{x}+2}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$$
After simplifying I get $8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$ now putting $t=3^{\sqrt[4]{x}-\sqrt{x}}$ we get $8t+9t^2\geq 1$ and solving we get $t\in (-\infty,-1)\cup(\frac{1}{9},+\infty)$ since $t$ is exponential function with positive base then $t>0$ hence we're only looking at the interval $(\frac{1}{9},\infty)$.Now I have no idea how to find the interval for $x$,I've tried substituting $\sqrt[4]{x}-\sqrt{x}=-2$ and I get $x=16$ and substituting $\sqrt[4]{x}-\sqrt{x}\to \infty$ which is impossible so I have no idea what to do now.
| Your work seems correct: the only place equality takes place is at $x=16$. Substituting that into your original equation checks, with $6561$ on both sides.
Since both sides of your inequality are continuous for $x\ge 0$, the solution set for your inequality is either $[0,16]$ or $[16,\infty)$. A quick calculator check shows that the inequality is true for $x=15$ and false for $x=17$. Therefore the solution is
$x\in[0,16]$
This is confirmed with a graph:
I do not understand your last sentence: $\sqrt[4] x-\sqrt x\to-\infty$ as $x\to\infty$, but what does that have to do with the solution to this problem?
| {
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"timestamp": "2023-03-29T00:00:00",
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A fair 6-sided die is rolled 6 times, what's the probability the outcome has exactly 2 or 3 elements? A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements?
I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)
| Exactly two numbers appear:
There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the same number is rolled six times. Thus, the probability of exactly two numbers appearing in six rolls of the die is
$$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$
Exactly three numbers appear:
There are $\binom{6}{3}$ ways for three of the six numbers to appear. On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers. However, we have counted the sequences in which not all three of the numbers appear. There are $\binom{3}{2}$ ways for two of the numbers to appear. For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times. Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear. The number of sequences in which exactly one of the three numbers appears is $3$. Hence, the probability that exactly three of the numbers appear is
$$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$
My thanks to @bof for clarifying my thinking about the first problem. Any errors that remain are entirely my responsibility.
| {
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Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$.
I tried many methods but I failed.
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}} $.
The denominator of RHS of the equation must have, minimum value. So it boils down to
$\frac{x^2y^2+y^2z^2+z^2x^2}{xyz} \leq 3 $. Don't know how to proceed after that.
Taking partial derivative in range $(0,3)$ for the function $(x^2+y^2+z^2)$ also didn't give me the required insight.
Thankyou.
| use AM-GM inequality
$$3=x+y+z\ge 3\sqrt[3]{xyz}\Longrightarrow xyz\le 1$$
and
$$x^2+y^2+z^2\ge 3\sqrt[3]{x^2y^2z^2}=3\sqrt[3]{(xyz)^2\cdot 1}\ge\sqrt[3]{(xyz)^2\cdot(xyz)}= 3\sqrt[3]{(xyz)^3}= 3xyz$$
| {
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Proving infinity limit of a multivariable function I have a function $f(x,y,z): \mathbb{R}^3\rightarrow \mathbb{R}$ which is $f(x,y,z)=3x^2+z^2+y^2-2xy+14$.
I'm trying to show that $f(x,y,z)\rightarrow \infty$ when $||(x,y,z)|| \rightarrow \infty$.
(Or formally: $\forall_{M>0}\exists_{R>0}\forall_{(x,y,z) \in \mathbb{R}^3}:||(x,y,z)||>R \rightarrow f(x,y,z)>M $)
So in the process of building the proof I'm trying to find a suitable $R$, and for that I'm trying to express $f(x,y,z)$ in terms of $||(x,y,z)||$ or $||(x,y,z)||^2$.
Try 1:
Let $(x,y,z)$ be such that $||(x,y,z)||>R$, then $f(x,y,z) = 3x^2+z^2+y^2-2xy+14$
$= x^2+z^2+y^2 +2x^2-2xy+14 \geq R^2+2(x^2-xy)+14 $
$\geq R^2-2(\sqrt{2x^2} \sqrt{x^2+y^2})+14 $
$\geq R^2-2(\sqrt{2x^2} \sqrt{x^2+y^2+z^2})+14 $
$\geq R^2-2(\sqrt{2x^2} R)+14 = R^2-2\sqrt{2}|x|R+14$.
From which I can't continue since if I try to bound $x$ by $R$ I get a negative expression.
Try 2:
Using $xy \leq \frac{x^2+y^2}{2}$, we get that
$f(x,y,z) = 3x^2+z^2+y^2-2xy+14 \geq 2x^2+z^2+14$,
from which I can't proceed since I can't express $R$ without $y$.
Try 3:
Let $R=\sqrt{3}M$, and since $||(x,y,z)||>R$ then $x^2+y^2+x^2>R^2=3M^2$.
Meaning that at least one of $x^2$, $y^2$, $z^2$ has to be greater or equal $M^2$.
$f(x,y,z) = 3x^2+z^2+y^2-2xy+14 = 2x^2+z^2+(y-x)^2+14$.
If $x^2>M^2$ or $z^2>M^2$ then obviously $f(x,y,z)>M$ and we're done. But I don't know how to handle the case of $y^2>M^2$.
Any help would be appreciated!
| Solved! Given $M>0$, setting $R=\sqrt{2M}$, then for all vectors $(x,y,z) \in \mathbb{R}^3$ for which $||(x,y,z)||>R$, it holds that:
$f(x,y,z)=3x2+z2+y2−2xy+14=$
$0.5(x^2+y^2+z^2)+0.5(x^2+z^2+(2x-y)^2+28)$
$\geq 0.5(x^2+y^2+z^2) > M$
| {
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Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+16}\,dx-\frac{1}{8}\int\frac{x^2-4}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{1+\frac{4}{x^2}}{\left(x-\frac{4}{x}\right)^2+\left(\sqrt{3}\right)^2}dx-\frac{1}{8}\int\frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2-\left(\sqrt{13}\right)^2}\,dx$$
Now Using $$\displaystyle \left(x-\frac{4}{x}\right)=t$$ and $$\displaystyle \left(1+\frac{4}{x^2}\right)dx = dt$$ in First Integral and
Using $$\displaystyle \left(x+\frac{4}{x}\right)=u$$ and $$\displaystyle \left(1-\frac{4}{x^2}\right)\,dx = du$$ in Second Integral.
So We Get $$\displaystyle = \frac{1}{8}\int\frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{8}\int\frac{1}{u^2-\left(\sqrt{13}\right)^2}\,du$$
So We Get $$\displaystyle = \frac{1}{8\sqrt{3}}\tan^{-1}\left(\frac{x^2-4}{\sqrt{3}x}\right)-\frac{1}{16\sqrt{13}}\ln \left|\frac{x^2-\sqrt{13}x+4}{x^2+\sqrt{13}x+4}\right|+{C}$$
My question is can we solve the above question any other Method, If yes then
please explain here. Thanks
| You could notice that $x^4-5x^2+16$ can be factored as $(x-a)(x-b)(x+a)(x+b)$ where $a$ and $b$ are complex numbers. Then partial fraction decomposition would lead to quite simple terms and the result for integral will just be the sum of four logarithms.
However, it is more likely that the recombination of the results in terms of reals could be quite tedious.
| {
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Quadratic Residues $\pmod {2^n}$ I'd imagine this is a duplicate question, but I can't find it:
How many quadratic residues are there $\pmod{2^n}$.
I tried small $n$: $n=1: 2, n=2:2, n=3: 3, n=4: 4, n=5: $not 5: 0, 1, 4, 9, 16, 25
No pattern :/
| There is a basic solution that only uses modular arithmetic.
Let $p(n)$ be the number of quadratic residues modulo $2^n$. We know that $p(1) = p(2) = 2$. Now suppose $n \geq 3$. We will proof 3 small lemmas.
Lemma 1. It suffices to consider the residues of the numbers $0,1,...,2^{n-2}$
Proof. Indeed, the following relations hold:
$$(2^n-x)^2 \equiv x^2 \mod 2^n, \quad (2^{n-1}-x)^2 \equiv x^2 \mod 2^n $$
So every square can be reduced to the square of one of the numbers $0,1,...2^{n-2}$
Lemma 2. If $x,y$ are distinct odd integers and $1 \leq x,y \leq 2^{n-2}-1$, then $x^2\not\equiv y^2 \mod 2^n$.
Proof. Suppose that $2^n \mid x^2-y^2 = (x-y)(x+y)$. Now $\text{gcd}(x+y,x-y)$ is divisible by $2$ but not by $4$ (because it divides $2x = (x+y)+(x-y)$). This implies that one of the two factors is divisible by $2$ and the other one is divisible by $2^{n-1}$. But $x+y \leq 2^{n-1}-2$, so $x-y$ should be divisible by $2^{n-1}$. We conclude that $x = y$.
These two lemmas imply that there are exactly $2^{n-3}$ odd quadratic residues modulo $2^n$. Now we look at the even residues:
Lemma 3. There are $p(n-2)$ even quadratic residues mod $2^n$.
Proof. Indeed, if $x = 2m$ and $y = 2n$, then $x^2 \equiv y^2 \mod 2^n$ if and only if $m^2\equiv n^2 \mod 2^{n-2}$. This gives the result immediately.
We conclude that $p(n) = p(n-2)+2^{n-3}$. From this recurrence relation it is quite easy to prove by induction that
*
*$p(n) = \frac{5+2^{n-1}}{3}$ if $n$ is odd
*$p(n) = \frac{1+2^{n-1}}{3}$ if $n$ is even
If you like to put everything together, we get the following result:
There are $$\frac{2^{n-1}+3+2.(-1)^{n-1}}{3}$$ quadratic residues modulo $2^n$
| {
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evaluate the double integral $\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$ evaluate the double integral
$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$
Hi all, could someone give me a hint on this question?
I've actually tried converting to polar coordinates but i cant seem to get the limits. But if polar coordinates are the way to go i'll just keep working on it. Thanks in advance.
edit: so the trick is to change the order of integration.
$\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy$
=$\int_0^2 \int_0^{x^2} \sqrt{x^2+y}\, dydx$
=$ \frac{2}{3}\int_0^2 (2x^2)^{3/2}-x^3\,dx$
=$ \frac{2}{3}(2\sqrt2-1) \int_0^2 x^3\,dx$
=$\frac{8}{3}(2\sqrt2-1)$
| Here is a polar treatment.
First get rid of the square root by $z=\sqrt{y}$ with $dy=2zdz$: $$I=\int_0^4 \int_{\sqrt{y}}^2 \sqrt{x^2+y}\, dxdy=2\int_0^2 \int_{z}^2 \sqrt{x^2+z^2}z\, dxdz$$
Now draw a graph to see the integration regions:
Therefore with $z=r\sin \theta$, $x=r \cos \theta$ and $dxdz=r dr d\theta$ we have
\begin{align}
I&=2\int_0^{\frac{\pi}{4}}\int_0^{\frac{2}{\cos \theta}}r^3 \sin \theta dr d\theta\\
&=8\int_0^{\frac{\pi}{4}}\frac{\sin \theta}{\cos^4 \theta} d\theta\\
&=8\times \Big(\frac13 (2\sqrt{2}-1)\Big)
\end{align}
Note: $$\int\frac{\sin \theta}{\cos^4 \theta} d\theta=\frac{1}{3\cos^3 \theta}$$
| {
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Show that $f(n) = -\sqrt{n} + n\log\left(1 + \frac{1}{\sqrt{n}}\right), f:\mathbb{N}\rightarrow \mathbb{R}$ is a decreasing function. The way I tried to approach this question was to consider it as a continuous $f: \mathbb{R} \rightarrow \mathbb{R}$. I then tried to show that $\frac{d}{dx}f < 0$. So
\begin{align*}
\frac{d}{dx}f(x) &= -\frac{1}{2\sqrt{x}} + \log\left(1 + \frac{1}{\sqrt{x}}\right) -\frac{1}{2} \frac{1}{1+ \sqrt{x}} \left(x^{-\frac{3}{2}}\right) \\
&= \log\left(1 + \frac{1}{\sqrt{x}}\right) - \frac{1}{2}\left(\frac{2 + \sqrt{x}}{x + \sqrt{x}}\right).
\end{align*}
Using the substitution $x = \frac{1}{u^2}$ gives
$$\log(1+u) - \frac{1}{2}\left(\frac{u^2}{u+1} + u\right).$$
At this point I'm stuck. How do I show this thing is negative for $u>0$?
| Hint: To complete what you had begun, compute the second derivative and study the function $f'$. Deduce its sign from its table of variations.
| {
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Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ?
Thanks
| hint:If $k^3 = 3k^2+9k+7 \to k(k^2-3k-9) = 7 \to k\mid 7 \to k = \pm 1, \pm 7$. Can you continue?
| {
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Locus of a midpoint Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of $M$. Then find the area of $S$. I am getting $81\pi$. See the figure
With one of my friends, I got this.Let C1 be centered at $(0,0)$ and let C2 be $(25, 0)$. The points that will lie on the boundary of S are
1. Midpoint of (-4, 0) and (11, 0) i.e. (3.5, 0)
2. Midpoint of (4, 0) and (39, 0) i.e. (21.5, 0)
3. Midpoint of the tangents joining C1 and C2
Let the angle that the point of intersection of the tanget to circle C1 and C2 makes with the x axis be $\theta $ then
The point of intersection on C1 and C2 are
$$(x_1, y_1) = (4\cos{\theta}, 4\sin{\theta})$$
$$(x_2, y_2) = (25+14\cos{\theta}, 14\sin{\theta})$$
The equation of the tangent is
$$ y = -\frac{x}{\tan{\theta}} + c$$
Putting the above two points in the line equation and eliminating c gives
$$ 10\sin{\theta} = -\frac{25 + 10\cos{\theta}}{\tan{\theta}}$$
$$ 10\sin{\theta}\tan{\theta} = -25 - 10\cos{\theta}
\frac{2}{\cos{\theta}} = -5
\cos{\theta} = -\frac{2}{5}$$
This gives
$$ \sin{\theta} = \pm \frac{\sqrt{21}}{5}$$
The two pairs of points of intersection in C1 and C2 are
$$ (x_1, y_1) = (-\frac{8}{5}, \frac{4\sqrt{21}}{5})$$
$$ (x_2, y_2) = (\frac{97}{5}, \frac{14\sqrt{21}}{5})$$
and
$$(x_1, y_1) = (-\frac{8}{5}, -\frac{4\sqrt{21}}{5})$$
$$ (x_2, y_2) = (\frac{97}{5}, -\frac{14\sqrt{21}}{5})$$
This gives the other two mid points as $(\frac{89}{10}, \frac{9\sqrt{21}}{5})$, $(\frac{89}{10}, -\frac{9\sqrt{21}}{5})$
Using all the mid points obtained and putting them into the ellipse equation
$$ \frac{(x - x_1)^2}{a^2} + \frac{(y - y_1)^2}{b^2}$$
following are obtained
$$ x_1 = \frac{25}{2}$$ and $$ y_1 = 0$$
$ a = 9$ and $b=9$.
Where is it wrong.($81\pi$ is wrong!) Please help, thanks.
| This is most likely analytic geometry, so I'd say we can put:
$$\Gamma_1\;:\;\; x^2+y^2=16\;\;;\;\;\;\Gamma_2\;:\;\; (x-25)^2+y^2=196$$
Take now for example
$$\begin{cases}A=\left(x\,,\,\sqrt{16-x^2}\right)\in\Gamma_1&,\;-4\le x\le4\\{}\\B'=\left(11\,,\,0\right)\in\Gamma_2\end{cases}$$
The midpoint of $\;AB'\;$ is given by
$$\left(\,\frac{x+11}2\;,\;\frac{\sqrt{16-x^2}}2\,\right)$$
We now have
$$\left(\frac{x+11}2-\frac{11}2\right)^2+\frac{16-x^2}4=\frac{x^2}{4}+4-\frac{x^2}4=4$$
and we thus get the circle $\;\left(x-\frac{11}2\right)^2+y^2=4\;$ upon taking both signs in the $\;y\;$-coordinate of $\;A\;$
Try to generalize the above to any point $\;B\in\Gamma_2\;$ .
| {
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Taylor series of $x/(x^2-4x+5)$ I'm supposed to find the Taylor series of this function (I can choose to center it at any A I want):
$$f(x)= x/(x^2-4x+5)$$
When I derivate, it only gets more and more confusing. How can I make any sense out of this?
| i think you can do a taylor series about a general center $x = a.$ of course there is more work to do. i am going to try to find a series about $x = a.$ make a change of variable $x = a + h,$ then we have
$$\frac{x}{x^2 - 4x + 5} = \frac{a + h}{(a+h)^2 - 4(a+h) + 5} = \frac{a + h}{\left(a^2 - 4a + 5 + 2h(a-2)+h^2\right)}=u_0+u_1h+\cdots+u_nh^n+\cdots$$
let $A = a^2 - 4a + 5, B = 2(a-2),$
then we have $$\left( u_0+u_1h+u_2h^2+\cdots+u_nh^n+\cdots \right)\left(A + Bh +h^2\right) = a + h$$ equating the coefficients of $h$ gives :
$$ Au_0 = a,\, Au_1+Bu_0 = 1,\, Au_2+Bu_1+u_0 =0, \,Au_n+Bu_{n-1}+u_{n-2}=0, n = 3, 4, \cdots, $$
we have the recurrence relation $$u_n = -\frac BA u_{n-1}- \frac1Au_{n-2}, \, u_0 =\frac aA,\, u_1 = - \frac {Ba}{A^2} + \frac 1A,\, u_2 = \frac {B^2a}{A^3} -\frac{B}{A^2} - \frac{a}{A^2}, \cdots $$
e.g.; $a = 2, A = 1, B = 0$ and the simpler recurrence relation is $$u_n = -u_{n-2}, u_0 = 2, u_1 = 1, u_2 = -2, u_3 = -1, \cdots, u_{2n}=(-1)^n2, u_{2n-1} = (-1)^{n-1}$$ and the series is $$ \frac{x}{x^2 - 4x + 5} = 2 +(x-2) -2(x-2)^2 -(x-2)^3 +2(x-2)^4+\cdots$$
| {
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
| Yet another way. Using $u=\tan(x)$, we get
$$
\begin{align}
\int\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x
&=\int\lower{2pt}{\frac{1-\frac{u}{\sqrt{1+u^2}}}{1+\frac{u}{\sqrt{1+u^2}}}\frac{\mathrm{d}u}{1+u^2}}\\
&=\int\left(1-\raise{2pt}{\frac{u}{\sqrt{1+u^2}}}\right)^{\!\!2}\,\mathrm{d}u\\
&=\int\left(2-\raise{2pt}{\frac{2u}{\sqrt{1+u^2}}}-\frac1{1+u^2}\right)\,\mathrm{d}u\\[6pt]
&=2u-2\sqrt{1+u^2}-\arctan(u)+C\\[14pt]
&=2\tan(x)-2\sec(x)-x+C
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\pi\frac{1-\sin(x)}{1+\sin(x)}\,\mathrm{d}x
&=(2-\pi)-(-2)\\[6pt]
&=4-\pi
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 5
} |
How many positive solutions are there (Positive 3 tuples)? I want to find how many positive solutions for the Diophantine equation $4x + 2y + 5z = 100$
I found a particular solution $(x,y,z) = (50,-50,0)$ then I found a general solution (basis) $s(-2,-1,2) + t(1,-2,0)$ for the homogeneous equation $4x + 2y +5z = 0$ and then I added that to the particular solution to get $(x,y,z) = (50-2s+t,-50-s-2t,2s)$ as the solution. However, Now I need to calculate how many positive solutions. So I have $$50-2s + t > 0$$ $$ -50 -s -2t > 0$$ $$2s > 0$$
I did some algebraic manipulation to get $t > -30$ and $s > 0 $ Now, I still don't understand how to count how many positive solutions are there ?
I went to wolfram alpha and it said $106$ but I have no clue how ? Any Suggestions
http://www.wolframalpha.com/input/?i=50-2s+%2B+t+%3E+0+%2C+-50+-+s+-+2t+%3E+0+%2C+2s+%3E+0+
| Since $5z = 100 - 4x - 2y$, $z$ must be even. Since $x, y, z > 0$, there are only nine possible values of $z$, namely, $z = 2, 4, 6, 8, 10, 12, 14, 16, 18$. By substituting these values for $z$, we can reduce the problem to determining the number of solutions of the equations
\begin{align*}
4x + 2y & = 10\\
4x + 2y & = 20\\
4x + 2y & = 30\\
4x + 2y & = 40\\
4x + 2y & = 50\\
4x + 2y & = 60\\
4x + 2y & = 70\\
4x + 2y & = 80\\
4x + 2y & = 90
\end{align*}
in the positive integers.
If we divide each equation by $2$, we obtain
\begin{align*}
2x + y & = 5\\
2x + y & = 10\\
2x + y & = 15\\
2x + y & = 20\\
2x + y & = 25\\
2x + y & = 30\\
2x + y & = 35\\
2x + y & = 40\\
2x + y & = 45
\end{align*}
Since the value of $y$ is determined by the value of $x$, the number of solutions of each equation is the number of positive even integers less than the number on the right hand side of the equation. Thus, the number of solutions of the linear Diophantine equation $4x + 2y + 5z = 100$ in the positive integers is
$$2 + 4 + 7 + 9 + 12 + 14 + 17 + 19 + 22 = 106$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.
Any other approach using Fourier Series, or contour integration around a square, if that is possible?
| I do not know how much this could help you; so forgive me if I am out off topic.
Rewriting a little the expression $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}x^n=\frac{1}{8} \,\Phi \left(-x,3,\frac{1}{2}\right)$$ where appears the Lerch transcendent function. Now, using $x=1$, we can get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find solution of $(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$ The following differential equation is given:
$$(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$$
*
*Find the general solution of the differential equation at the interval $(-1,1)$ (with the method of power series).
*Are there solutions of the differential equation that are polynomials?
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence $R>0$.
$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$
$$y''(x)= \sum_{n=1}^{\infty} (n+1)n a_{n+1}x^{n-1}= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n$$
$$-x^2y''(x)= \sum_{n=0}^{\infty} -n(n-1)a_n x^n$$
We have:
$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1) a_n-na_n+p^2a_n\right]x^n=0, \forall x \in (-R,R)$$
It has to hold:
$$(n+2)(n+1) a_{n+2}+ \left[ -n(n-1)-n+p^2\right] a_n=0 \\ \Rightarrow (n+2)(n+1)a_{n+2}+ \left[ -n(n-1+1)+p^2 \right]a_n=0 \\ \Rightarrow a_{n+2}=\frac{n^2-p^2}{(n+2)(n+1)} a_n \forall n=0,1,2, \dots$$
For $n=0: a_2=-\frac{p^2}{2}a_0$
For $n=1: a_3= \frac{1-p^2}{2 \cdot 3} a_1$
For $n=2: a_4= \frac{(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot 4 } a_0$
For $n=3: a_5=\frac{(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5}a_1$
For $n=4: a_6=\frac{(4^2-p^2)(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}a_0$
For $n=5: a_7=\frac{(5^2-p^2)(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$
So:
$$a_{2n}=\frac{\prod_{i=0}^{n-1} (2i)^2-p^2}{(2n)!} a_0$$
$$a_{2n+1}=\frac{\prod_{i=0}^{n-1} (2i+1)^2-p^2}{(2n+1)!} a_1$$
So the solution $y$ can be written as follows:
$$y(x)=\sum_{n=0}^{\infty} a_{2n}x^{2n}+ \sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$$
Is it right so far? If so, then do we have to find for the power series $\sum_{n=0}^{\infty} a_{2n}x^{2n}$ and $\sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$ the radius of convergence and show that they define functions that are infinitely many times differentiable?
If so, then do we have to take the ratio
$$\left| \frac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}}\right|$$
? If yes, could we deduce something from the latter about the radius of convergence?
| The Equation can be rewritten as( D^2-(x/(1-x^2))×D+x^2/1-x^2)y=0.
This equation can easily be solved using one integral method .
As P+XQ=0 then solution will assume type of y=xv.
substitute dy/dx by xdv/dx+v ,d^2y/dx^2 by xd^2/dx^2+2dv/dx and y by xv
This equation will be converted to linear diffrential equation with constant coefficient or other forms which r very easy to evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y} \mid} \,\, = \,\, {\mid x^2+x-y^2-y \mid} \,\, = \,\, {\mid (x+y)(x-y)+(x-y) \mid} \,\, = \,\, {\mid (x-y)(x+y+1) \mid } < {\epsilon}^2 \Rightarrow \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid} < \epsilon$
So now we know that if we take $\delta = {\epsilon}^2$ the condition for uniform continuity of this function will be met because
${\mid x - y \mid} < (\delta = {\epsilon}^2) \Rightarrow {\mid f(x)-f(y) \mid} < \epsilon$
Is this proof valid? Or I miss something?
| For $\;x\in [1,\infty)\;$:
$$\left(\sqrt{x^2+x}\right)'=\frac{2x+1}{2\sqrt{x^2+x}}\le\frac{2x+1}{2x}=1+\frac1{2x}\stackrel{\text{Why?}}\le\frac32$$
Thus, having a bounded derivative makes $\;\sqrt{x^2+x}\;$ uniformly continuous in $\;[1,\infty]\;$, and being continuous in the bounded, closed interval $\;[0,1]\;$ it is unif. continuous there as well. Thus, it is unif. continuous in the whole non-negative interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the equation of tangent line to the curve $x=\cos(t) + \cos(2t)$, $y= \sin(t) + \sin(2t)$ at the point $(-1,1)$. I get the equation for the slope as $\frac{\cos(t) + 2\cos(2t)}{-\sin(t) -2\sin(2t)}$ but I'm unsure how to solve for the value of $t$. I know I need to sub in $-1$ and $1$ for $x$ and $y$ but the equations are hard to solve.
| Given is curve:
$$\begin{gathered}
x(t) = \cos (t) + \cos (2t) \hfill \\
y(t) = \sin (t) + \sin (2t) \hfill \\
\end{gathered}$$
$$\begin{gathered}
\varphi (t) = \left( {\begin{array}{*{20}{c}}
{x(t)} \\
{y(t)}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{\cos (t) + \cos (2t)} \\
{\sin (t) + \sin (2t)}
\end{array}} \right) \hfill \\
\frac{{d\varphi }}{{dt}}(t) = \left( {\begin{array}{*{20}{c}}
{ - \sin (t) - 2\sin (2t)} \\
{\cos (t) + 2\cos (2t)}
\end{array}} \right) \hfill \\
\end{gathered}$$
From graphs for sin and cos we find:
$$\begin{gathered}
\cos (\frac{\pi }{2}) = 0 \wedge \cos (\pi ) = - 1 \hfill \\
\sin (\frac{\pi }{2}) = 1 \wedge \sin (\pi ) = 0 \hfill \\
\end{gathered} $$
$$\begin{gathered}
\varphi (\frac{\pi }{2}) = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
1
\end{array}} \right) \hfill \\
\frac{{d\varphi }}{{dt}}(\frac{\pi }{2}) = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
{ - 2}
\end{array}} \right) \hfill \\
\end{gathered}$$
and calculate slope:
$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{1}{{\frac{{dx}}{{dt}}}} = - 2 \cdot \frac{1}{{ - 1}} = 2$$
Remark: Our tangent vector is normed and therfore has lenght equal to one. So it's
$$\frac{{d\varphi }}{{dt}} = \frac{1}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}}
{ - 1} \\
{ - 2}
\end{array}} \right)$$
Slope doesn't change, because of $\frac{1}{{\sqrt 5 }}$ cancels by division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How come, in this problem, the maximum product is always achieved using only $2$s and $3$s? Consider the following problem.
Given a number $N$, write it as a sum $n = n_1 + n_2 + \cdots + n_k$, such that the product $p = n_1 \times n_2 \times \cdots \times n_k$ is maximized.
For example, $11$ can be written as $2 + 3 + 3 + 3$ or as $7 + 4$, yielding the products $2 \times 3 \times 3 \times 3 = 54$ and $7 \times 4 = 28$.
So in this case, the answer is is $2\times 3\times 3\times 3$.
For $17$ we have the same situation: $2 \times 3 \times 3 \times 3 \times 3 \times 3 = 486$.
So my question is, why is the maximum product always achieved by using only $2$s and $3$s?
| For every sequence of natural numbers $\{a_1,\dots,a_n\}$ such that $N=\sum\limits_{k=1}^{n}a_k$:
*
*If $a_k=4$, then you can get an equivalent value by decomposing it into $2+2$ since $2\cdot2=4$
*If $a_k=5$, then you can get a better value by decomposing it into $2+3$ since $2\cdot3=6>5$
*If $a_k=6$, then you can get a better value by decomposing it into $3+3$ since $3\cdot3=9>6$
*If $a_k>6$, then you can get a better value by decomposing it into one of the previous values
In addition to that, you can get a better value by replacing every triplet $\{2,2,2\}$ with a pair $\{3,3\}$:
*
*$2+2+2=6=3+3$
*$2\cdot2\cdot2=8<9=3\cdot3$
Hence the best decomposition is into a list of $2$s and $3$s, with at most two $2$s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Combinatorial graph theory proof Why does $\binom{1/2}{n+1} * (-1)^n * 2^{2*n+1}$ equal $1/(n+1) * \binom{2n}{n}$? I came across this in an exercise in Graph Theory and Its Applications by Gross and Yellen. I haven't been able to solve it, can anyone give any insight?
| We know that,
$$\binom{x}k=\frac{x^{\underline k}}{k!}$$
where $x^{\underline k}$ is the falling factorial function defined as $x^{\underline k}=x(x-1)(x-2)\cdots (x-k+1)$
Also, $x!!$ is the double factorial function defined as,
$$x!!=\begin{cases}x(x-2)(x-4)\cdots 4\cdot 2\cdot 1~,~\textrm{when }x\textrm{ is even}\\x(x-2)(x-4)\cdots 5\cdot 3\cdot 1~,~\textrm{when }x\textrm{ is odd}\end{cases}$$
Now,
$$\binom{1/2}{n+1}=\frac{\frac{1}{2}\cdot \left(-\frac{1}{2}\right)\cdot \left(-\frac{3}{2}\right)\cdots \left(-\frac{2n-1}{2}\right)}{(n+1)!}=\frac{(-1)^n\cdot (2n-1)!!}{2^{n+1}\cdot (n+1)!}\\ \implies \binom{1/2}{n+1}\cdot (-1)^n=\frac{(1)^n\cdot (2n-1)!!}{2^{n+1}\cdot (n+1)!}=\frac{(2n-1)!!}{2^{n+1}\cdot (n+1)!}\\ \implies \binom{1/2}{n+1}\cdot (-1)^n\cdot 2^{2n+1}=\frac{2^n(2n-1)!!}{(n+1)!}$$
Now, multiply numerator and denominator of RHS with $n!$ to get,
$$\binom{1/2}{n+1}\cdot (-1)^n\cdot 2^{2n+1}=\frac{\{2^n\cdot n!\}\cdot (2n-1)!!}{(n+1)\cdot (n!)^2}=\frac{(2n-1)!!\cdot (2n)!!}{(n+1)\cdot (n!)^2}=\frac{(2n)!}{(n+1)\cdot (n!)^2}$$
Since $\dbinom{2n}{n}=\dfrac{(2n)!}{n!\cdot n!}=\dfrac{(2n)!}{(n!)^2}$, we can write that,
$$\binom{1/2}{n+1}\cdot (-1)^n\cdot 2^{2n+1}=\frac{(2n)!}{(n+1)\cdot (n!)^2}=\frac{1}{n+1}\cdot \binom{2n}{n}\\ \implies \boxed{\dbinom{1/2}{n+1}\cdot (-1)^n\cdot 2^{2n+1}=\dfrac{1}{n+1}\cdot \dbinom{2n}{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof : Limit of a sequence
Prove from the definition of the limit of a sequence that $$\lim_{n\to\infty} \frac{2n^2+\cos(n)} {n^2+1} = 2 $$
(that is, for a given $\epsilon > 0$, find an explicit $N_\epsilon$)
Please explain.
I choose $n_0$ such that ${n_0} > \sqrt {\frac{3}{\varepsilon }} $. Then how can I say that for all $n\ge n_0$, we have $\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} - 2} \right| <\varepsilon$. ? How to proceed further ?
| Let $\epsilon > 0$ be given. Choose an $n_0 \in \mathbb{N}$ such that $n_0 > \sqrt{\frac 1{\epsilon}}$. Then for all $n \in \mathbb{N}$ such that $n \ge n_0$, we have
\begin{align}
\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} -2 } \right|&=\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} - \frac{2(n^2+1)}{n^2+1}} \right|\\ &=\left|\frac{\cos n -2}{n^2+1} \right| \\
&=\left|\frac{\cos n}{n^2+1}-\frac 2{n^2+1} \right| \\
&\le\left|\frac{\cos n}{n^2+1} \right|+\left|\frac 2{n^2+1} \right| \\
&=\frac{|\cos n|}{n^2+1}+\frac 2{n^2+1} \\
&\le \frac 1{n^2+1}+\frac 2{n^2+1} \\
&=\frac 3{n^2+1} \\
&< \frac 3{n^2} \\
&\le \frac 3{n_0^2} \\
&<\epsilon
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Combinatorial sum inequality Prove the following inequality:
$$ \forall k\in\left\{4n+5:n\in\mathbb{N}\right\},\qquad\sum_{m=0}^{\frac{k-1}{2}}{\left( -1 \right) }^{m}\binom{k}{2m}2^{2m}\neq 1. $$
I'm particularly interested in a combinatorial or number-theoretical solution.
| We just have to compute:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{1}{2}\left(\sum_{k=0}^{4n+5}\binom{4n+5}{k}(2i)^k+\sum_{k=0}^{4n+5}\binom{4n+5}{k}(-2i)^k\right) $$
that is:
$$\sum_{m=0}^{2n+2}\binom{4n+5}{2m}(2i)^{2m}=\frac{(1+2i)^{4n+5}+(1-2i)^{4n+5}}{2}.$$
If we set:
$$ A_n = \frac{(1+2i)^{n}+(1-2i)^{n}}{2},\qquad B_n=A_{4n+5} $$
we have $A_0=1,A_1=1,B_0=-7,B_1=-527$ and:
$$ A_{n+2} = 2 A_{n+1} - 5 A_n,\qquad B_{n+2} = -14 B_{n+1}-625 B_n\equiv B_{n+1}\pmod{5},$$
so, by induction, the only possible residue class of $B_n\pmod{5}$ is $\color{red}{3}$ and $B_n$ cannot be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Differential equation : $y' = (x+1)/(xy+x)$ So, I have the following differential equation to solve : $$y' = \frac{x+1}{xy+x}$$
After several steps, I get here :
$t^2 + 2t = 2x + 2ln(x) + c$
How do I isolate $t ?$
thank you!
By the way, $ t=f(x)=y$
| Start by dividing through by $x$:
$$
y'=\frac{1+\frac{1}{x}}{y+1} \\
y'(y+1)=1+\frac{1}{x} \\
\int y'(y+1)dx=\int 1+\frac{1}{x}dx \\
\frac{y^2}{2}+2y=2x+2\ln(x)+c~\text{for constant $c$} \\
y^2+2y+1=2x+2\ln(x)+2c+1 \\
(y+1)^2=2x+2\ln(x)+2c+1 \\
y+1=\pm \sqrt{2x+2\ln(x)+2c+1} \\
y=\pm \sqrt{2x+2\ln(x)+2c+1}-1.~_{\square}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Lowering powers of $\cos^2x \sin^4x$ First, I will be straight up, this is a homework question.
I need to write $\cos^2 x \sin^4 x$ in terms of cosine to the first power. I know that $\sin^4x$ =
$$ \frac{3-4\cos 2x+\cos 4x}{8}$$
from there I go:
$$ \frac{1+\cos 2x}{2} \cdot \frac{3-4\cos 2x+\cos 4x}{8}$$
$$ \frac{3(1 + \cos 2x) - 4\cos 2x(1 + \cos 2x) + \cos 4x (1 + \cos 2x)} {16} $$
$$ \frac{3 + 3\cos 2x - 4\cos 2x \color{red}{+} 4 \cos^2 2x + \cos 4x + \cos 4x \cos 2x}{16}$$
$$ \frac{3 - \cos 2x + \cos 4x + 4 \frac{1 + \cos 4x}{2} \cos 4x \cos 2x}{16}$$
Then $$ 4 \frac{1 + \cos 4x}{2}$$ simplifies to $$ 2 + 2\cos 4x$$
adding this into the rest of the fraction gives me
$$ \frac{5 - \cos 2x + 3\cos 4x + \cos 4x \cos 2x}{16}$$
However, the answer sheet that I have says the answer is
$$ \frac{1 - \cos 2x - \cos 4x + \cos 2x \cos 4x }{16}$$
This is problem $13$ from section $7.3$ of Stewart, Redlin and Watson precalculus $6$th edition
Where is my mistake?
| I found:
$$cos^2(x)sin^4(x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
Verify the indenty:
$$32cos(x)^2sin(x)^4=2-cos(2x)-2cos(4x)+cos(6x)$$
$$6-2cos(2x)-4-4cos(4x)+2cos(4x)+cos(2x)+cos(6x)=2-cos(2x)-2cos(4x)+cos(6x)$$
$$\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)=\frac{1}{32}(-cos(2x)-2cos(4x)+cos(6x)+x)$$
The left hand side and right hand side are equal!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $|\sin^2 x+ 17 - x^2|=|16-x^2|+2 \sin ^2 x + \cos ^2x$ then $x$ lies in what interval If $|\sin^2 x+ 17 - x^2|=|16-x^2|+2 \sin ^2 x + \cos ^2x~$ then $x$ lies in what interval?
Hints Please? I dont know how to approach this problem.
| Observe that $|\sin^2x+1+16-x^2|=|16-x^2|+\sin^2x+1$ $(*)$
Set $y=\sin^2x+1$ and $z=16-x^2$. Then $(*)$ reduce to
$|z+y|=|z|+y$. Taking square of both sides we get $y(z-|z|)=0$
Hence, since $\sin^2x+1\neq0$ we have $16-x^2=|16-x^2|$ and this is true only for $ -4\leq x\leq 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving $\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$ How can I prove
$$\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$$
I wanted to prove it by using the Taylor series of $\sqrt{2}$, but I couldnt do.
| We have to prove:
$$\frac{1}{\sqrt{2}}=\frac{3}{4}-\sum_{n\geq 1}\frac{(2n-1)!!}{(n+1)!2^{2n+2}}=\frac{3}{4}-\frac{1}{4}\sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!8^{n}}$$
that follows from:
$$ \sum_{n\geq 1}\frac{(2n)!}{n!(n+1)!}z^n = \frac{1-\sqrt{1-4z}-2z}{2z}.\tag{1}$$
$(1)$ is just a minor variation of the generating function for Catalan numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$
| Since the Taylor series of $g(x)=\frac{x^4}{(1+2x)^3(1-2x)^2}$ in a neighbourhood of the origin is given by $x^4+o(x^4)$, it is enough to compute the Taylor series of:
$$\begin{eqnarray*} h(x)&=&\frac{1}{(1+2x)^3(1-2x)^2}\\&=&\left(1-6 x+24 x^2-80 x^3+240 x^4+o(x^4)\right)\left(1+4 x+12 x^2+32 x^3+80 x^4+o(x^4)\right)\\&=&1 - 2 x + 12 x^2 - 24 x^3 + 96 x^4+o(x^4)\tag{1}\end{eqnarray*}$$
giving:
$$ \frac{1+x^4}{(1+2x)^3(1-2x)^2}=1 - 2 x + 12 x^2 - 24 x^3 + 9\color{red}{7} x^4+o(x^4).\tag{2}$$
In $(1)$ we just used the Taylor series for $\frac{1}{(1-z)^m}$, given by:
$$\frac{1}{(1-z)^m}=\sum_{n\geq 0}\binom{n+m-1}{n}\, z^n\tag{3}$$
and took the Cauchy product of two such series. A different approach is given by:
$$ \frac{1}{(1-4x^2)^2}=1+8 x^2+48 x^4+o(x^4),\tag{4} $$
that just leads to $(1)$ since:
$$ \frac{1}{1+2x}=1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4+o(x^4).\tag{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Find the best polynomial approximation of the piecewise function Find the best approximation of the function
$$ f(x)= \begin{cases}
1 - x \quad\text{ for } 0 \le x \le 1 \\
1 + x \quad\text{ for } -1 \le x \le 0 \\
\end{cases}$$
in the interval $[-1,~ 1]$ by a polynomial $p(x)$ of degree $\le 4$, meaning that the integral
$$\int_{-1}^{1} (|f(x) - p(x)|)^2\,{\rm d}x $$ is minimal possible.
We know that the best approximation is unque and coinced with the orthogonal projection. The minimal distance is equal to to length of orthogonal projection of $x$ into ${\rm ort}_{U}x$.
| $\newcommand{\Er}{{\rm Error}}$
Define the problem and solve the differential calculus:
$$\begin{align}
%
\Er(A, B, C, D, E)
%
&= \int_{-1}^{1} (Ax^4 + Bx^3 + Cx^2 + Dx + E - f(x))^2 ~dx
%
\\ \\ &= \int_{-1}^{0} (Ax^4 + Bx^3 + Cx^2 + Dx + E - 1 - x)^2 ~dx \\
\\ &+ \int_{0}^{1} (Ax^4 + Bx^3 + Cx^2 + Dx + E - 1 + x)^2 ~dx
%
\\ \\ &= \frac{2\,{A}^{2}}{9} + \frac{2\,{B}^{2}}{7} + \frac{2\,{C}^{2}}{5} + \frac{2\,{D}^{2}}{3} + \frac{2\,{E}^{2}}{1}
\\ &- \frac{2\,A}{15} - \frac{C}{3} - 2\,E
\\ &+ \frac{4\,A\,E}{5} + \frac{4\,C\,E}{3} + \frac{4\,B\,D}{5} + \frac{4\,A\,C}{7} + \frac{2}{3}
%
\end{align}$$
Solve the system of 5 linear equations :
$$\frac{\partial~\Er }{\partial~ A} =
\frac{\partial~\Er }{\partial~ B} =
\frac{\partial~\Er }{\partial~ C} =
\frac{\partial~\Er }{\partial~ D} =
\frac{\partial~\Er }{\partial~ E} = 0
$$
To get:
$$A=\frac{105}{128} ,~ B=0 ,~ C=-\frac{105}{64} ,~ D=0 ,~ E=\frac{113}{128} $$
$$\Er\left(\frac{105}{128} ,~ 0 ,~ -\frac{105}{64} ,~ 0 ,~ \frac{113}{128} \right) = \frac{1}{384}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
| This is only a solution for even values of $x$:
For $x$ even, $x^2$ is even and $x+1$ is odd. So $\gcd(16, x+1) = 1$. So $16 \mid x^2$. So $4 \mid x$. So $x = 4k$ for some $k$.
$16k^2(4k+1)=16(2^z+1)$
$k^2(4k+1)=2^z+1$
If $k$ is even then the RHS is even and we have a contradiction. So $k$ must be odd. So $k = 2t+1$ for some $t$.
$(2t+1)^2 (8t+5) = 2^z + 1$
$(4t^2 + 4t + 1)(8t + 5) = 2^z + 1$
$32t^3 + 52t^2 + 28t + 4 = 2^z$
We see $z \geq 2$. Let $w = z - 2$.
$8t^3 + 13t^2 + 7t + 1 = 2^w$
For $w \geq 1$, $t^2 + t + 1 \equiv 0 \pmod{2}$ which has no solution in $t$.
So $w = 0$.
$8t^3 + 13t^2 + 7t = 0$
$t(8t^2 + 13t + 7) = 0$
Take the discriminant of the quadratic, so we get $169 - 224 < 0$.
So $t = 0$ and $w = 0$.
So for $x$ even, $x = 4k = 4(2t + 1) = 4$ and $y = z + 4 = w + 6 = 6$.
You can also bound (bind?) $x$, giving it in terms of $y$. Go back to $x^3+x^2−16=2^y$. Consider $f(x,y) = x^3 + x^2 - 16 - 2^y$. Evaluate $f(2^{y/3},y) = 2^y + 2^{2y/3} - 16 - 2^y = 2^{2y/3} - 16$ which equals $0$ when $y = 6$ and is bigger than $0$ when $y > 6$. Evaluate $f(2^{y/3} - 1, y)$ which you can check here is less than $0$ for all $y$. This shows that if $x$ is an integer and $f(x,y) = 0$ then $x = \left \lfloor 2^{y/3} \right \rfloor$.
As far as working with $x = \left \lfloor 2^{y/3} \right \rfloor$, you can write it as $x = \left \lfloor 2^q2^{r/3} \right \rfloor$ where $0 \leq r < 3$ which can be expressed as $\left \lfloor 10_2^q 10_2^{r/3} \right \rfloor$ in binary. As $q$ increases (which increases $y$ by 3), the value of $x$ either gets doubled, or is double plus one. I'm not sure if that's useful.
How to finish this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
The greatest common divisor Problem : Confirm the following properties of the greatest common divisor:
If $\gcd(a,b) = 1$, and $c \mid (a+b)$, then $\gcd(a,c) = \gcd(b,c) = 1$.
Is this right? This is my answer:
*
*$\gcd(a,b) = 1$ => There exist integers $x$ and $y$ such that: $ax + by = 1$.
*$c \mid (a+b)$ => There exist an integer k such that: $ck = a + b$ => $a = ck - b$ and $b = ck - a$.
*Because $ax + by = 1$ and $a = ck - b$ => $(ck - b)x + by = 1$ <=> $c(xk) + b(y - x) = 1$ ( $xk$ and $(y - x)$ are integers ) => $\gcd(c, b) = 1$.
*Because $ax + by = 1$ and $b = ck - a$ => $ax + (ck - a)y = 1$ <=> $a(x - y) + c(ky) = 1$ ( $(x - y)$ and $ky$ are integers ) => $\gcd(a, c) = 1$.
=> $\gcd(a,c) = \gcd(b,c) = 1$.
| I think your reasoning is right, and I can provide a shorter proof: $\gcd(a,a+b)=\gcd(a,b)=1$ and similarly $\gcd(b,a+b)=\gcd(b,a)=1$. Then $c\mid a+b$ implies $\gcd(a,c) = \gcd(b,c) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that a subset is normal Let $H$ be the subset of $GL(2, \mathbb R)$ consisting of all matrices of the form
$$ \left[
\begin{array}{cc}
x & 0\\
0 & x\\
\end{array}
\right] $$
where $x \neq 0$
Prove that H is normal.
UPDATE
So if I show $\left[
\begin{array}{cc}
h & 0\\
0 & h\\
\end{array}
\right]$
commutes with everything in $ GL(2, \mathbb R)$, I can do something like $$\left[
\begin{array}{cc}
a & b\\
c & d\\
\end{array}
\right]\left[
\begin{array}{cc}
h & 0\\
0 & h\\
\end{array}
\right]\left[
\begin{array}{cc}
a & b\\
c & d\\
\end{array}
\right]^{-1}$$ $$= \left[
\begin{array}{cc}
h & 0\\
0 & h\\
\end{array}
\right]\left[
\begin{array}{cc}
a & b\\
c & d\\
\end{array}
\right]\left[
\begin{array}{cc}
a & b\\
c & d\\
\end{array}
\right]^{-1}$$
$$ = \left[
\begin{array}{cc}
h & 0\\
0 & h\\
\end{array}
\right] \in H$$
Yes?
| You are indeed on the wrong track. Sounds like you have shown $H$ to be a subgroup. As for proving $H$ is a normal subgroup, we pick any arbitrary element $\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right] \in \text{GL}_2(\mathbb{R})$ and prove that $\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right]\left[\begin{array}{cc}
x & 0\\
0 & x\\
\end{array}\right]\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right]^{-1} \in H$. It will probably be helpful to recall $$\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right]^{-1} = \frac{1}{ad-bc}\left[\begin{array}{cc}
d & -b\\
-c & a\\
\end{array}\right]$$ So simply (although a bit tiresome) calculate by hand that the product $$\left[\begin{array}{cc}
a & b\\
c & d\\
\end{array}\right]\left[\begin{array}{cc}
x & 0\\
0 & x\\
\end{array}\right]\left(\frac{1}{ad-bc}\left[\begin{array}{cc}
d & -b\\
-c & a\\
\end{array}\right]\right)$$ and show it will equal some matrix of the form $\left[\begin{array}{cc}
z & 0\\
0 & z\\
\end{array}\right]$ with $z \neq 0 $. This will complete the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...+n\cdot n) = \frac{1}{24}n(n+1)(n+2)(3n+1).$$
Could somebody shed some light on me?
| Base case:
Prove true for n=1
$ 1 * 1 = \frac{1(1 + 1)(1 + 2)(3(1) + 1)}{24} $
Inductive step:
Assume for n = k, prove for n = k+1. If we let the property be equal to P(n)
$ P(k+1) = P(k) + ((k+1) * 1 + ... +(k+1)*(k+1)) $
$ P(k+1) = \frac{k(k + 1)(k + 2)(3k + 1)}{24} + ((k+1) * 1 + ... +(k+1)*(k+1)) $
From here you should be able to do the rest, ask if you have any more issues
| {
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"url": "https://math.stackexchange.com/questions/1225311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solving second order differential equation in a hurry During last test I had to solve second order ODE: $y'' + 3y = e^x + 1$. I managed to write the general solution of homogenous equation (with RHS replaced by zero):
$$y = c_1 \cos \left(\sqrt{3} x\right) + c_2 \sin \left(\sqrt{3} x\right).$$
That wasn't hard. But how to find (at least one) the general solution? Using sosmath (Method of Variation of Parameters) I found out that:
\begin{align}
u_1 & = \sqrt{3} \int \sin \left(\sqrt{3} x\right) (e^x + 1) \,\textrm{d}x = \frac{\sqrt{3}}{12} \cdot \left(3e^x \sin \left(\sqrt 3 x\right) - \sqrt 3 (4+3e^x) \cos \left(\sqrt 3 x\right) \right) + C_1 \\
u_2 & = \sqrt{3} \int \cos \left(\sqrt{3} x\right) (e^x + 1) \,\textrm{d}x = \frac{\sqrt{3}}{12} \cdot \left(3e^x \cos \left(\sqrt 3 x\right) + \sqrt 3 (4+3e^x) \sin \left(\sqrt 3 x\right) \right) + C_2,
\end{align}
after what I don't know how to proceed. There must be a simpler way, because one can guess the solution to be $$y(x) = \frac{4 + 3e^x}{12}.$$
| try the method undetermined coefficients; try a particular solution of the form $$ y = ae^x + b, y' = ae^x, y'' = ae^x. $$ sub this in $$y'' + 3y = e^x + 1 \to ae^x + 3(ae^x + b)=e^x + 1 \to a = \frac 14, b = \frac 13 $$
therefore, a particular solution is $$y_p = \frac 14 e^x + \frac 13, y_g = Ce^{-3x}+y_p $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$
Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?
| Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$. Now,
$$
|x^3-c^3| = |x-c||x^2+xc+c^2|.
$$
If $|x-c|<1$, then we have that $-1<x-c<1$ or simply $c-1<x<c+1$ so that
$$
x^2+xc+c^2<(c+1)^2+(c+1)(c)+c^2=(c^2+2c+1)+(c^2+c)+c^2=3c^2+3c+1,
$$
and so
$$
|x^3-c^3| = |x-c||x^2+xc+c^2|<(3c^2+3c+1)|x-c|.
$$
So if we take $\delta=\min\left\{1,\frac{\epsilon}{3c^2+3c+1}\right\}$, then $0<|x-c|<\delta$ implies that
$$
|x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon}{3c^2+3c+1}\cdot(3c^2+3c+1)=\epsilon.
$$
Thus, by the definition of a limit, we have that
$$
\lim_{x\to c}x^3=c^3. \blacksquare
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bound for the sum of a finite sequence Consider ${\bf c} = (a,b) \in \mathbb{R}^2$ with $0< \|{\bf c}\| < 1.$ Let $n \in \mathbb{N} $ and define
\begin{align*}
F_{n}(k) & := \frac{ [a + x_{n}(k)]^2}{ [a + x_{n}(k)]^2 + [b + y_{n}(k)]^2}
\end{align*}
where $\omega := 2\pi/n,$ $x_{n}(k) := \cos(2 k \pi/n)$ and $y_{n}(k) := \sin(2 k \pi/n).$
The question is to find an upper bound $h = h(n,{\bf c})$ on the sum $S_n$
such that
$$ S_n \leq h(n,{\bf c}),$$
where
$$S_n =\left( \frac{2 \pi}{n}\right) \sum_{k=0}^{n-1} F_{n}(k).$$
Is there any result that can help to find such a bound ?
| One idea is the following:
Let $z_k=x_n(k)+iy_n(k)=e^{ik\omega}$ and $c=a+ib=re^{i\theta}$ with $0<r<1$. Then, $F_n(k)=\cos^2\theta_k,\ 0\le k\le n-1$ where $\theta_k=\angle (z_k+c)$ so that $F_n(k)=\pi+\frac{\pi}{n}\sum_{k=0}^{n-1}\cos 2\theta_k$.
I think this can be helpful in finding some bounds.
Edit: One can note that $\tan\theta_k=\frac{r\sin \theta+\sin k\omega}{r\cos \theta+\cos k\omega}$. Then, $$\cos2\theta_k=\frac{1-\tan^2{\theta_k}}{1+\tan^2\theta_k}=\frac{r^2\cos 2\theta+2r\cos(\theta+k\omega)+\cos2k\omega}{r^2+1+2r\cos(\theta-k\omega)}$$ So, $$S_n=\pi+\frac{\pi}{n}\sum_{k=0}^{n-1}\frac{r^2\cos2\theta+2r\cos(\theta+k\omega)+\cos2k\omega}{r^2+1+2r\cos(\theta-k\omega)}=\pi+\frac{\pi}{n}(A\cdot r^2\cos2\theta+B+C)$$
Now, we can write $$A=\frac{1}{r^2+1+2r\cos(\theta-k\omega)}=\frac{1}{r^2+1}\left(1-a\cos(\theta-k\omega)+a^2\cos^2(\theta-k\omega)+\cdots\right),\ a=\frac{2r}{r^2+1}\\=\frac{1}{r^2+1}\sum_{l=0}^{\infty}(-1)^la^l\cos^l(\theta-k\omega)\\=\frac{1}{r^2+1}\sum_{l=0}^{\infty}\sum_{m=0}^{\lfloor l/2\rfloor}\frac{a^l}{2^l}\binom{l}{2m}\cos((l-2m)(\theta-k\omega))$$ So that $$A=\frac{1}{r^2+1}\sum_{l\ge 0}\left(\frac{a^2}{4}\right)^{l}\binom{2l}{l}\\=\frac{1}{r^2+1}\frac{d}{dx}\left(xC(x)\right)\mid_{x=a^2/4}$$ where $C(x)$ is the generating function of Catalan numbers. Hence $\displaystyle xC(x)=(1-\sqrt{1-4x})/2\implies A=\frac{1}{(r^2+1)\sqrt{1-a^2}}=\frac{1}{1-r^2}$ With similar manipulation one can attempt to find exact expressions for $B,C$ and can find a suitable upper bound of $S_n$, though I guess those exact expressions will not be easy to handle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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multivariable limit of $\frac{x^2-y^2}{\sqrt{x^2+y^2}}$ Calculate multivariable limit of $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
How to do that? I was trying to figure out any transformations e.g. multiplying by denominator but I do not think it gives me any solution.
| $$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$$
Transforming $x=r\cos\theta, y=r\sin\theta$.
$$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}=\lim_{r \rightarrow 0}\frac{r^2\cos^2\theta-r^2\sin^2\theta}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}}=\lim_{r \rightarrow 0}\frac{r^2(\cos^2\theta-\sin^2\theta)}{r}=\lim_{r \rightarrow 0}{r(\cos^2\theta-\sin^2\theta)}=\lim_{r \rightarrow 0}{0\cdot(\cos^2\theta-\sin^2\theta)}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The Galois closure of $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ Consider the field extension $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ where $y$ is a root of the polynomial $g(T) = f(x,T) \in \mathbb{F}_2(x)[T]$, with
$$
\begin{array}{l}
f(x,y) = x^{12} + x^{10}y^2 + x^{10}y + x^{10} + x^9y^2 + x^9y + x^8y^4 + x^8y^2 + x^8 \\
{}+ x^6y^6 + x^6y^5 + x^6y^4 + x^6y^3 + x^6y^2 + x^6y + x^6 + \\
x^5y^6 + x^5y^5 + x^5y^4 + x^5y^3 + x^5y^2 + x^5y + x^4y^8 + \\
x^4y^6 + x^4y^5 + x^4y^4 + x^4y^3 + x^4y^2 + x^4 + x^3y^6 + \\
x^3y^5 + x^3y^4 + x^3y^3 + x^2y^{10} + x^2y^9 + x^2y^8 + x^2y^6 + \\
x^2y^5 + x^2y^4 + x^2y^2 + x^2y + x^2 + xy^{10} + xy^9 + xy^6 + \\
xy^5 + xy^2 + xy + y^{12} + y^{10} + y^8 + y^6 + y^4 + y^2 + 1
\end{array}
$$
I want to show that $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$ is not a Galois extension.
We know that $f(x,y)$ is the unique non-linear factor of
$$
(x^{16}-x)(y^2-y)-(y^{16}-y)(x^2-x).
$$
We have
$$
(x^{16}-x)(T^2-T)-(T^{16}-T)(x^2-x) = T(T+1)(T+x)(T+x+1)g(T).
$$
In $\mathbb{F}_2(x,y)[T]$ we have
$$
g(T) = (T+y)(T+y+1)(T+y+x)(T+y+x+1)h(T),
$$
with $h(T)$ of degree $8$.
I want to find the Galois closure of $\mathbb{F}_2(x,y) / \mathbb{F}_2(x)$. Thus I can see that it is not Galois extension.
Any help is appreciated.
| Let $p(X,Y) = (X^{16}+X)(Y^2+Y)+(Y^{16}+Y)(X^2+X)$.
If $p(x,y) = p(x,z) = 0$ then $p(x,y+z) = 0$ so for a given value of $x$, the roots of the corresponding polynomial form a $\Bbb F_2$-subspace, usually $4$-dimensional.
Notice that for any $x \in \overline{\Bbb F_2}$,
$p(x,Y)$ has a double root $\iff$ $0$ is a double root of $p(x,Y)$ $\iff$ $x^{16}+x^2 = 0 \iff x^8+ x = 0 \iff x \in \Bbb F_8$ .
We already know that $p$ has $4$ roots $0,1,x,x+1 \in \overline {\Bbb F_2}(x)$.
Suppose there's another root $f \in \overline {\Bbb F_2}(X)$ such that $p(X,f(X)) = 0$.
By the rational root theorem, $f = a/b$ where $a$ divides $(X^8+X)^2$ and $b$ divides $(X^2+X)$. In particular, $f \in \Bbb F_8(X)$.
However there are lots of values of $x \in \Bbb F_{64}$ for which $p(x,Y)$ have only $4$ simple roots in $\Bbb F_{64}$, so this is impossible.
Next, let $y$ be a new root of $P$ and suppose that there's some $f \in \overline {\Bbb F_2}(X,y)$ such that the roots of $P(X,Y)$ are the subspace $\langle 1,X,y,f(X,y) \rangle$.
Then denoting $f^2 = Frob_2(f)$ (we square every coefficient of $f$), if $f \neq f^2$ then $f^2 + f \in \langle 1,X,y \rangle \subset \Bbb F_2(X,y)$. Hence $f^4 + f^2 = f^2 + f$ and $f^4 = f$ so that $f \in \Bbb F_4(X,y)$
However there are lots of values of $x \in \Bbb F_{2^{10}}$ for which $P(x,Y)$ have only $8$ simple roots in $\Bbb F_{2^{10}}$, so this is also impossible.
This shows that the splitting field of $P$ over$\Bbb F_2(X)$ has degree $12 \times 8 = 96$, and the Galois group is isomorphic to the matrix group
$G = \left\{ \begin{pmatrix} I_2 & A \\ 0 & B \end{pmatrix} \in GL_4(\Bbb F_2) \right\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to validate the basis of the space of solutions of a system of linear equations Problem
Solve the following nonhomogeneous system of linear equations:
$$
\begin{aligned}
x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 4,\\
2 x_{1} + 2 x_{2} + 3 x_{3} + x_{4} &= 5,\\
3 x_{1} + 3 x_{2} + 4 x_{3} + x_{4} &= 6
\end{aligned}$$
Please enter the specific solution first and then the basis in the space of solutions of the corresponding homogeneous system.
Comment of the teacher
Example. If $x_s=\begin{pmatrix} 0 \\ 1 \\ 0 \\1 \end{pmatrix}$ is a specific solution and $\left\{\begin{pmatrix} 1 \\ 2 \\ 3/2 \\ 4 \end{pmatrix}, \begin{pmatrix} 0 \\ -2 \\ 2 \\ 9 \end{pmatrix}\right\} $ is a basis of $\{x:Ax=0\}$ then please enter
[0,1,0,1],[1,2,3/2,4],[0,-2,2,9]
There is an automated system here, checking the solution.
Solution
Augmented matrix:
$$
\begin{pmatrix}
1 & 2 & 3 & 1 & 4\\
2 & 2 & 3 & 1 & 5\\
3 & 3 & 4 & 1 & 6\\
\end{pmatrix}
$$
I managed to transform the augmented matrix to the diagonal form:
$$
\begin{pmatrix}
1 & 0 & 0 & 0 & 1\\
0 & 1 & 0 & -1 & -3\\
0 & 0 & 1 & 1 & 3\\
\end{pmatrix}
$$
So, the specific solution is the last row + 0 at the bottom: $x_s=\begin{pmatrix} 1 \\ -3 \\ 3 \\ 0 \end{pmatrix}$. I checked, and it is really a solution of the original problem.
1-st equation:
$$
1 \cdot 1 + 2\cdot(-3) + 3 \cdot 3 + 1 \cdot 0 = 1 - 6 + 9 = 4
$$
2-nd equation:
$$
2 \cdot 1 + 2\cdot(-3) + 3 \cdot 3 + 1 \cdot 0 = 2 - 6 + 9 = 5
$$
3-rd equation:
$$
3 \cdot 1 + 3\cdot(-3) + 4 \cdot 3 + 1 \cdot 0 = 3 - 9 + 12 = 6
$$
But, why does the basis in the comment consist of 2 columns? I think, it must be only 1: $\begin{pmatrix} 0 \\ -1 \\ 1 \\ -1 \end{pmatrix}$ --- the 4-th column of the diagonal matrix and the last element (-1) is an element of negated identity matrix.
When I tried this solution: [1, -3, 3, 0],[0, -1, 1, -1] for the first time, the automated checker said: "incorrect". Then I wrote this question; but before posting it, I tried it again, and it said: "correct". Probably, I made a typo first time :).
How can I check, that $\begin{pmatrix} 0 \\ -1 \\ 1 \\ -1 \end{pmatrix}$ is really the basis of the space of solutions, like I did, validating the specific solution?
| I'm not sure if the teacher's comment was for specifically this question, but you're right. The solution to this system lies on a line and not a plane as the teacher's comment suggests.
To check that your answer is correct, substitute the general form back into the system. Here's an example with the first equation:
$$x_1 + 2x_2 + 3x_3 + x_4 = 4$$
$$\implies 1 + 2(-3 + x_4) + 3(3- x_4) + x_4 = 4$$
$$\implies 1 - 6 + 9 + 3x_4 - 3x_4 = 4$$
$$\implies 4 = 4$$
and so on for the equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $x\ln(x+1)$? I am trying to compute $\int x\ln (x+1)\, dx$. I tried integrating by parts and ended up with:
$$\int x\ln(x+1)\,dx = \frac{1}{2}x^2\ln(x+1) - \frac{1}{2}\int\frac{x^2}{x+1}\,dx$$ but I'm stuck here.
| $$\int \dfrac{x^2}{x+1}dx = \int \dfrac{x^2-1}{x+1}dx + \int \dfrac{dx}{x+1} = \int (x-1) dx + \int \dfrac{dx}{x+1} = \dfrac{x^2}2 - x + \ln(x+1)+ c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$
Find the function that equals to $1-x^3+x^6-x^9+ \cdots$ for all $|x| < 1$
I know that $\frac{1}{1+x} = 1-x+x^2-x^3+...$ But I couldn't find the pattern here
| Hint: if $|f(x)| < 1$ then $$\frac{1}{1+f(x)} = 1 - f(x) + f(x)^2 - f(x)^3 + \dotsb$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
| If you accept a proof without induction, here is one:
\begin{align*}
\frac 1n+\Bigl(\frac 1{n+1}+\dots+ \frac1{n^2}\Bigr)>\frac 1n +(n^2-n)\cdot\frac 1{n^2}=\frac 1n+1-\frac1n=1.
\end{align*}
This computation is valid if $n^2>n$, i.e. if n>1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How do you find the factorial of a decimal or negative number and what does it show us? I know that you can find the factorial of positive integers where n!= n(n-1)...2 x 1. However, what if you want to find the factorial of a negative integer or a decimal? I tried to do it on my calculator and it gave an answer however, I wasn't able to understand how they calculator got the answers.
I did some research and came across the gamma function which supposedly allowed you to solve such questions. However, I found it very hard to understand and still don't see the purpose of finding the factorial of a negative integers or decimals.
Help would be appreciated.
Thank you :)
| As already noted by @ncmathsadist, the Gamma Function $$\Gamma(z)=(z-1)!$$ can be extended to a meromorphic function defined on the complex plane without the non-positive integers.
Here's an instructive example how to work with negative factorials presented in section $3.6$ of $A=B$ by M. Petkovsek, H. Wilf and D. Zeilberger.
Challenge: Find a closed expression for
\begin{align*}
f(n)=\sum_{k=0}^{2n}t_k=\sum_{k}(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k}
\end{align*}
We start with checking if the ratio $\frac{t_{k+1}}{t_k}$ gives rise to a known hypergeometric series. Indeed, with
\begin{align*}
\frac{t_{k+1}}{t_k}&=\frac{(-1)^{k+1}\binom{2n}{k+1}\binom{2k+2}{k+1}\binom{4n-2k-2}{2n-k-1}}
{(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k}}
=\frac{(k+\frac{1}{2})(k-2n)^2}{(k+1)^2(k-2n+\frac{1}{2})}
\end{align*}
we derive
\begin{align*}
f(n)=\binom{4n}{2n} {}_{3}F_{2}\left(-2n,-2n,\frac{1}{2};1,-2n+\frac{1}{2};1\right)
\end{align*}
It turns out, that this hypergeometric series matches Dixon's identity and
we obtain
\begin{align*}
f(n)=\binom{4n}{2n}\frac{(-n)!(-2n-\frac{1}{2})!(n-\frac{1}{2})}{(-2n)!n!(-n-\frac{1}{2})!(-\frac{1}{2})!}\tag{1}
\end{align*}
At first glance this expression is rather distressing, since it contains factorials of negative integers which are precisely the values, where the gamma function is not defined!
The clou: We have a ratio of two factorials at negative integers and if we can take an appropriate limit, the singularities will cancel leaving a pleasant limiting ratio. As the authors point out, this situation happens fairly frequently when using this approach.
$$$$
We start analysing the ratio
\begin{align*}
\frac{(-n)!}{(-2n)!}\tag{2}
\end{align*}
Let's assume, that $n$ is near a positive integer, but is not equal to a positive integer. Then we can use the reflection formula for the $\Gamma$-function
\begin{align*}
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}
\end{align*}
once more in the equivalent form
\begin{align*}
(-z)!=\frac{\pi}{(z-1)!\sin \pi z}
\end{align*}
So, when $n$ is near a positive integer, the expression (2) becomes
\begin{align*}
\frac{(-n)!}{(-2n)!}=\frac{\pi}{(\sin n\pi)(n-1)!}\frac{(\sin 2n\pi)(2n-1)!}{\pi}=\frac{2(2n-1)!\cos n\pi}{(n-1)!}
\end{align*}
and we observe, if $n$ approaches a positive integer
\begin{align*}
\frac{(-n)!}{(-2n)!}\longrightarrow(-1)^n\frac{(2n)!}{n!}
\end{align*}
The expression (1) becomes
\begin{align*}
f(n)=(-1)^n\binom{4n}{2n}\binom{2n}{n}\frac{(-2n-\frac{1}{2})!(n-\frac{1}{2})}{(-n-\frac{1}{2})!(-\frac{1}{2})!}
\end{align*}
Similarly, we find
\begin{align*}
\frac{(-2n-\frac{1}{2})!}{(-n-\frac{1}{2})!}=\frac{(-1)^n(n-\frac{1}{2})!}{(2n-\frac{1}{2})!}
\end{align*}
and we obtain
\begin{align*}
f(n)=\binom{4n}{2n}\binom{2n}{n}\frac{(n-\frac{1}{2})!^2}{(2n-\frac{1}{2})!(-\frac{1}{2})!}\tag{3}
\end{align*}
For every positive integer $m$,
\begin{align*}
(m-\frac{1}{2})!&=(m-\frac{1}{2})(m-\frac{3}{2})\cdots(\frac{1}{2})(-\frac{1}{2})!\\
&=\frac{(2m-1)(2m-3)\cdots 1}{2^m}(-\frac{1}{2})!\\
&=\frac{(2m)!}{4^mm!}(-\frac{1}{2})!
\end{align*}
This way we can simplify the expression (3) to $f(n)=\binom{2n}{n}^2$ and we have shown the identity
\begin{align*}
\sum_{k}(-1)^k\binom{2n}{k}\binom{2k}{k}\binom{4n-2k}{2n-k}=\binom{2n}{n}^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
&=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
\end{align}$
$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\
&=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\
\end{align}$
$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\
&=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\
\end{align}$
is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$
I have studied maths upto $12$th grade.
| I do not know if you know multiple angle trig formulas.
Let $ A = 18 ^\circ. $ In a right angled triangle if acute angles are $ 2A= 36 ^\circ, \,3A= 54^\circ$,
$ \sin 2 A = \cos 3A $
$ 2 \sin A \cos A = 4 \cos^3 A -3 \cos A $
simplifying and solving for $ \sin A $ gives you
$$ \sin 18^\circ =\dfrac{\sqrt{5}-1} {4}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$ b)$\frac{\pi}{6}\quad$ c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$
| Use the identity $$1+\tan^2(x) = \sec^2(x)$$ Then multiply your equation through by $\cos^2(x)$ to get $$\begin{align*}1 = 8\sin^2(x)\cos^2(x) \\ = 2(2\sin(x)\cos(x))^2 \\ = 2(\sin(2x))^2 \\ \implies \frac{1}{2} = (\sin(2x))^2 \\ \implies \pm \frac{\sqrt{2}}{2} = \sin(2x)\end{align*}$$ Can you take it from here?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value of product $abc$.
| Since $a^2=2b^3$, this means $a = \sqrt{2b} \cdot b$. Hence, $b=2l^2$. This gives us $a=4l^3$.
We also need $a^2 = 3c^5 \implies 16l^6 = 3c^5 \implies c = \left(\dfrac{2^4l}3\right)^{1/5} \cdot l$. This means $l = 3\cdot 2 \cdot m^5$.
Hence, we have $(a,b,c) = (864m^{15},72m^{10},12m^6)$. The minimum is obtained when $m=1$. Hence, $(a,b,c) = (864,72,12)$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has the form $2^a \cdot 5^b$, and a decomposition of 1000000 into a product of five factors has the form
$$
1000000 = (2^{a_1} \cdot 5^{b_1})(2^{a_2} \cdot 5^{b_2})(2^{a_3} \cdot 5^{b_3})(2^{a_4} \cdot 5^{b_4})(2^{a_5} \cdot 5^{b_5})
$$
where $a_i$ and $b_i$ are nonnegative integers which satisfy the
conditions
$$
a_1 + a_2 + a_3 + a_4 + a_5 = 6, b_1 + b_2 + b_3 + b_4 + b_5 = 6
$$
The total number of systems of $a_i$ which satisfy the first equation is $210$ and the same number is for $b_i$. Thus the total number of decompositions is $210 \cdot 210 = 44100$.
However, in this enumeration, factorizations which differ only in the brder of the factors have been counted separately; that is, some factorizations are counted several times each.
To get the number of distinct unordered decompositions I must, at first, substract the number of ordered decompositions with at least two identical factors and, at second, divide resulting number by $5!$ to leave only unordered ones.
And I'm stuck at the step of counting the number of ordered decompositions with at least two identical factors.
The number of decompositions with $k$ identical factors is $(\lfloor\dfrac{a}{k}\rfloor + 1)(\lfloor\dfrac{b}{k}\rfloor+1){5 \choose k}$, that is, the number of decompositions with two identical factors is $16 \cdot {5 \choose 2} = 160$, three identical factors - $9 \cdot {5 \choose 3} = 90$, four ones - $4 \cdot {5 \choose 4} = 20$ and five ones - $4 \cdot {5 \choose 5} = 4$. Thus the number of distinct decompositions must be equal $\dfrac{44100-160-90-20-4}{5!}$, but this number is not integral. I suppose that the numbers of decompositions with $k$ identical factors overlap and I misuse inclusion-exclusion principle. But I have no idea how I can count that overlapped decompositions.
Please, help!
Thanks!
| The number in question is the coefficient of $x^6 y^6 z^5$ in the product
$$\prod_{0\le i,j\le 6}(1+x^i y^j z).$$
Here $z$ counts the number of factors (we want $5$); $x$ and $y$ tag the powers of $2$ and $5$ respectively; we want both of those to be $6$. Distinctness is ensured because, for each factor $2^i 5^j$, we either include it once (thereby multiplying by $x^i y^j z$) or we don't include it (and multiply by $1$).
Multiplying out the entire product is unwieldy, but you can compute it mod $(x^7,y^7)$, which eliminates a whole mess of terms you don't need. The coefficient of $x^6 y^6$ turns out to be
$$5 z^7+64 z^6+194 z^5+235 z^4+123 z^3+24 z^2+z,$$
which enumerates the number of ways to write $1000000$ as a product of $k$ distinct factors, for all values of $k$. Taking $k=5$ we confirm Christian's answer of $194$.
| {
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How to express sum as triple summation I am trying to express the following sequences as summations:
$$
1+2^2+3^2+4^4+5^4+6^4+7^4
$$
and
$$
1+(2+3)^2 + (4+5+6+7)^4
$$
as summations. I think they will likely be triple summations, so something like this:
$$
\sum\sum\sum n^k
$$
but I can't think of what to put on the top and bottom of each sum.
Any help would be appreciated. Thanks!
| To me,
it looks like
at each successive iteration,
the number of terms doubles
((1), (2, 3), (4, 5, 6, 7))
and the exponent
doubles
(1, 2, 4).
To get the elements in the terms,
since
$1+2+4+...+2^n
=2^{n+1}-1
$,
if stage 1 is just $(1)$,
the terms added at stage $n$
are from
$2^{n-1}$
to
$2^n-1$
and the exponent
is $2^{n-1}$.
As a check,
$n=1$ gives $(1)$ with exponent $1$,
$n=2$ gives $(2, 3)$ with exponent $2$,
and
$n=3$ gives $(4, 5, 6, 7)$ with exponent $4$.
So the next stage,
$n=4$,
would be
$(8, 9, 10,11, 12, 13, 14, 15)$
with exponent $8$.
The the $n$-th stage summation
for the first sum
would be
$\sum_{k=2^{n-1}}^{2^n-1} k^{2^{n-1}}$
and the sum for all stages
up to $N$
would be
$\sum_{n=1}^N\sum_{k=2^{n-1}}^{2^n-1} k^{2^{n-1}}$.
Since,
if $2^{n-1} \le k \le 2^n-1$,
$n
=\lfloor \log_2 k \rfloor +1
$,
this last sum can be written as
$\sum_{k=1}^{2^N-1} k^{2^{\lfloor \log_2 k \rfloor}}
$.
For the second sum,
arguing similarly,
the total up to stage $N$
would be
$\sum_{n=1}^N\left(\sum_{k=2^{n-1}}^{2^n-1} k \right)^{2^{n-1}}$.
Because
$\begin{array}\\
\sum_{k=2^{n-1}}^{2^n-1} k
&=\frac{(2^n-1)2^n}{2}-\frac{(2^{n-1}-1)2^{n-1}}{2}\\
&=\frac{2^{2n}-2^n-2^{2n-2}+2^{n-1}}{2}\\
&=\frac{3\cdot 2^{2n-2}-2^{n-1}}{2}\\
&=3\cdot 2^{2n-3}-2^{n-2}\\
&=2^{n-2}(3\cdot 2^{n-1}-1)\\
\end{array}
$
the last sum simplifies to
$\sum_{n=1}^N\left(2^{n-2}(3\cdot 2^{n-1}-1)\right)^{2^{n-1}}
=\sum_{n=1}^N2^{(n-2)2^{n-1}}\left(3\cdot 2^{n-1}-1\right)^{2^{n-1}}
$.
| {
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Factorize Trigonometric Equation: $ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $ I have a problem with the following trigonometric equation:
$$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$
It's from the book Engineering Mathematics 7th edition by Stroud.
The book is giving the answer, but I can't seem to be able to find out how to factorize it. I can't figure it out.
Consider the following solution:
This equation can be written as $3\sin ^2x-\cos ^2x = 2\sin x \cos x.$
That is: $3\sin ^2x-2\sin x \cos x-\cos ^2 x = 0.$
That is: $(3\sin x + \cos x)(\sin x - \cos x) = 0.$
So that $3 \sin x \cos x = 0$ or $\sin x - \cos x = 0.$
If $3 \sin x + \cos x = 0,$ then $\tan x = \frac{-1}{3},$ and so $x = -0.3218 ± n \pi,$ and if $\sin x - \cos x = 0,$ then $\tan x = 1,$ and so $x = \frac{\pi}{4}.$
If anyone could help me understand how to factorize this equation to get the one shown in the image it would help me very much.
Thank you in advance.
| Think of it like this:
Take $a=\sin x$ and $b=\cos x$. Then you have,
$$3a^2-2ab-b^2=3a^2-3ab+ab-b^2=3a(a-b)+b(a-b)=(a-b)(3a+b)$$
Substitute everything back and you have,
$$3\sin^2 x-2\sin x\cos x-\cos^2 x=(\sin x-\cos x)(3\sin x+\cos x)$$
| {
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Solving $a_n=5a(n/3)-6a(n/9)+2log_3n$ using domain transformation $a_n=5a(n/3)-6a(n/9)+2log_3n$, For $n\ge9$ and n is a power of 3.
$a_3=1$, and $a_1=0$
Transforming the first two terms is straightforward, but I'm not sure what to do with the log term. Should I rewrite it somehow? The fact that it isn't attached to the function, but just to n, is also somewhat confusing.
| Here is a closely related recurrence that has the same complexity as
the one in the OP and admits an exact solution for all $n.$ This
computation resembles the following MSE
link, the difference
being that this one does not depend on the digits of $n.$
Suppose we start by solving the following recurrence for $n\ge 3$:
$$T(n) = 5 T(\lfloor n/3 \rfloor) - 6 T(\lfloor n/9 \rfloor) +
\lfloor \log_3 n \rfloor$$
where $T(0) = 0$ and $T(1) = T(2) = 1.$
We unroll the recursion to obtain an exact formula for $n\ge 3$
$$T(n) = [z^{\lfloor \log_3 n \rfloor}] \frac{1}{1- 5 z+ 6 z^2} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
[z^j] \frac{1}{1- 5 z+ 6 z^2}
(\lfloor \log_3 n \rfloor - j).$$
where the first term represents the base cases with $n\lt 3$ and the
second the contribution from the logarithmic term.
Observe that the roots of $$1- 5 z + 6 z^2
\quad\text{are}\quad\rho_0=\frac{1}{2}
\quad\text{and}\quad\rho_1=\frac{1}{3}$$
and $$\frac{1}{1- 5 z+ 6 z^2}
= \frac{3}{1-3z} - \frac{2}{1-2z}.$$
It follows that the coefficients of the rational term have the form
$$[z^j] \frac{1}{1-5z+6z^2}
= 3^{j+1} - 2^{j+1}.$$
This gives the exact formula for $T(n):$
$$T(n) = 3^{\lfloor \log_3 n \rfloor+1}
- 2^{\lfloor \log_3 n \rfloor+1}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
(3^{j+1}-2^{j+1})
(\lfloor \log_3 n \rfloor - j)
\\ = 3\times 3^{\lfloor \log_3 n \rfloor}
- 2\times 2^{\lfloor \log_3 n \rfloor}
\\ + \frac{9}{4} 3^{\lfloor \log_3 n \rfloor}
- \frac{3}{2} \lfloor \log_3 n \rfloor
- \frac{9}{4}
\\ - 4 \times 2^{\lfloor \log_3 n \rfloor}
+ 2\times \lfloor \log_3 n \rfloor
+ 4$$
which simplifies to
$$\frac{21}{4} 3^{\lfloor \log_3 n \rfloor}
- 6 \times 2^{\lfloor \log_3 n \rfloor}
+ \frac{1}{2} \lfloor \log_3 n \rfloor
+ \frac{7}{4}.$$
It follows that $T(n)$ has the dominant asymptotic
$$T(n)\in\Theta\left(3^{\lfloor \log_3 n \rfloor}\right)
= \Theta\left(3^{\log_3 n}\right)
= \Theta(n).$$
Observe that $5/3-6/9= 5/3-2/3=3/3=1,$ which checks.
Here is another Master Theorem computation at MSE.
| {
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} |
How can I simplify $\sqrt{3^2 + 3^2\tan^2\theta}$? $$\sqrt{3^2 + 3^2\tan^2\theta}$$
$$ = (3)(3\tan\theta) = 9\tan\theta $$
I've simplified it like this but I'm not sure if that's correct.
| $$\begin{align} \sqrt{3^2 + 3^2\tan^2\theta} &= \sqrt{9(1+\tan^2 \theta)} \\
& = \sqrt 9 \cdot \sqrt {1 + \tan^2 \theta} \\
& = 3\sqrt{\sec^2\theta} \\
&= 3|\sec \theta| \end{align}$$
| {
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Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$
Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$
Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.
| Another approach
we see that $n(n-1)(n+1)$ divides both $n^7-n^3=n^3(n-1)(n+1)(n^2+1)$ and $n^{21}-n^{13}=n^{13}(n-1)(n+1)(n^2+1)(n^4+1)$
and we know that $(n-1)n(n+1)$ is a multiple of $3$ because the product of three consecutive integers is divisible by $3$.
| {
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A way to split this determinant as a product of two? The question asks if the det of the following 3*3 matrix is divisible by $$ x^4,x^3,x^2,x $$
$$
\begin{bmatrix}
a^2+x^2 & ab & ac \\
ab & b^2+x^2 & bc \\
ac & bc & c^2+x^2 \\
\end{bmatrix}
$$
I just shovelled through using the standard method and got $$ x^4(b^2+c^2+a^2+x^2) $$ So the answer is "Yes" but is there a simpler way like splitting the det as a product of two dets? If not that is there anything else?
| The matrix can be written as
$$A = x^2I_3 + \begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}$$
Now by Sylvester determinant theorem, we have
\begin{align}
\det(A) & = x^6 \det\left(I_3 + \dfrac1{x^2}\begin{bmatrix}a\\b\\c\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix}\right) = x^6 \left(1+\dfrac1{x^2}\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}\right)\\
& = x^4(x^2+a^2+b^2+c^2)
\end{align}
| {
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} |
Help with solving mathematical induction problem I need help with the following:
Use mathematical induction to prove that for every $n\in N$,
$$
\sum_{k=1}^n\frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(n+1)x-\tan x}{\sin x}
$$
For $n=1$, the statement is true.
Suppose that the statement is true for $n=m\in N$, and prove that it is true for $n=m+1$.
$$\sum_{k=1}^{m+1} \frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(m+2)x-\tan x}{\sin x}(*)$$
Proof:
$$\frac{\tan(m+2)x-\tan x}{\sin x}=\frac{\tan(m+1)x-\tan x}{\sin x}+\frac{1}{\cos(k+1)x \cos(k+2)x}$$
Using trigonometry rule $\cos x \cos y$, the right side of equation is
$$\frac{(\cos x+\cos(2k+3)x)(\tan(m+1)x-\tan x)+2\sin x}{\sin x(\cos x+\cos(2k+3)x)}$$
How to further transform this expression to become $(*)$
Thanks for replies.
| this is a telescoping series. you have $$\frac{\sin (k+1)x}{cos(k+1)x} - \frac{\sin kx}{\cos kx}=\frac{\sin( k+1)x\cos kx - \sin kx\cos(k+1)x}{cos(k+1)\cos kx} = \frac{\sin x}{cos(k+1)\cos kx} \tag 1$$
adding the equations for $k = 1, 2, \cdots, n$ gives you $$\frac{\sin (n+1)x}{cos(n+1)x} - \frac{\sin x}{\cos x} = \sin x \sum_{k=1}^n\frac{1}{coskxcos(k+1)x}$$
| {
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How to evaluate the $\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series? How to evaluate the $\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series?
It made sense to first try and build the numerator using power series that are commonly used:
$\displaystyle2\sin(x)=\sum_{k=0}^\infty \dfrac{2(-1)^kx^{2k+1}}{2k+1!} = 2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb$
$\displaystyle-\arctan(x)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{2k+1}}{2k+1} = -x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb$
$\displaystyle-x\cos(x^2)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{4k+1}}{2k!} = -x +\frac{x^5}{2}+ \dotsb$
Hence,
$\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5} =
\lim\limits_{x\to 0} \dfrac{[2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb] + [-x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb] + [x +\frac{x^5}{2}+ \dotsb]} {x^5}$
In similar problems, the there is an easy way to take out a common factor that would cancel out with the denominator, resulting in an easy-to-calculate limit. Here, however, if we were to take a common factor from the numerator, say, $x^6$, then we would end up with an extra $x$
What possible strategies are there to solve this question?
| Since the denominator has degree $5$, you just need to stop at $x^5$ in the power series expansions. So
\begin{align}
2\sin x&=2x-\frac{x^3}{3}+\frac{x^5}{60}+o(x^5)\\
-\arctan x&=-x+\frac{x^3}{3}-\frac{x^5}{5}+o(x^5)\\
-x\cos(x^2)&=-x+\frac{x^5}{2}+o(x^5)
\end{align}
Summing up we get
$$
\left(\frac{1}{60}-\frac{1}{5}+\frac{1}{2}\right)x^5+o(x^5)=
\frac{19}{60}x^5+o(x^5)
$$
Note that $o(x^5)$ means some unspecified function (but computable doing the math), with the property that $\lim\limits_{x\to0}\dfrac{o(x^5)}{x^5}=0$, so the sum of two of them is still $o(x^5)$.
Thus the limit is $19/60$.
You can be sure that going beyond $x^5$ would not really matter, because when dividing by $x^5$ the numerator would have summands with a factor $x$ that goes to $0$. If all terms of degree at most $5$ would cancel out, the limit would be $0$.
Note also the small error you have in the expansion of the arctangent.
| {
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Solve the following recurrence relation: $S(1) = 2$; $S(n) = 2S(n-1)+n2^n, n \ge 2$ Solve the following recurrence relation:
$$\begin{align}
S(1) &= 2 \\
S(n) &= 2S(n-1) + n 2^n, n \ge 2
\end{align}$$
I tried expanding the relation, but could not figure out what the closed relation is:
+-------+------------------------------------------------+
| n | S(n) |
+-------+------------------------------------------------+
| 1 | 2 |
| 2 | 2S(1)+2*2^2 = 2*2 + 2*2^2 = 2^2 + 2*2^2 = |
| | = 4 + 8 = 12 |
| 3 | 2S(2)+3*2^3 = 2*(2^2 + 2*2^2) + 3*2^3 = |
| | = 8 + 16 + 24 = 48 |
| 4 | 2S(3)+4*2^4 = 2*(2*2^2 + 2*2*2^2 + 3*2^3) + |
| | + 4*2^4 = |
| | = 16 + 32 + 48 + 64 = 160 |
| 5 | 2S(4)+5*2^5 = 2*(2*2*2^2 + 2*2*2*2^2 + |
| | + 2*3*2^3 + 4*2^4) + 5*2^5 = |
| | = 32 + 64 + 96 + 128 + 160 |
| | = 384 + 96 = 480 |
+-------+------------------------------------------------+
Edit: is it $S(n)=2^n\frac{n(n+1)}{2}$?
| Given
$$
S(n)=2S(n-1)+n2^n\tag{1}
$$
Consider $T(n)=2^{-n}S(n)$. $T(1)=1$ and, multiplying $(1)$ by $2^{-n}$, we have the recurrence
$$
T(n)=T(n-1)+n\tag{2}
$$
Therefore, $T(n)=\frac{n^2+n}2$ and thus,
$$
S(n)=(n^2+n)\,2^{n-1}\tag{3}
$$
| {
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Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$ finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$
My attempt is:
let $x=\sqrt{u}$
$dx=\frac{1}{2\sqrt{u}}$
$$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$
here I stopped because I don't know how to complete this solution. Any help please.
| $$\int_{0}^{\infty }\frac{1}{1+x^4}dx=$$
$$\int_{0}^{\infty }\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right)dx$$
The antiderivative of $\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right) $ is:
$$\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))$$
So we got the limit of b goes to infinity:
$$\left[\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))\right]^b_0$$
So we get:
$$\int_{0}^{\infty }\frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}$$
| {
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Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Attempt:
$$ \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy
= \int_0^1 \int_\sqrt{y}^1 x^2 + y^2 dx dy
= 1/3 + 1/3 - 2/15 - 2/7 = \frac{26}{105}.$$
However, the solution should be $26/35$ so it appears that I'm off by a factor of $3$ but I don't see where this fits in. Where did I go wrong?
| $$ \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy
= \int_0^1 \int_\sqrt{y}^1 ( x^2 + y^2) dx dy
$$
$$\int_\sqrt{y}^1 ( x^2 + y^2) dx=\left(\frac{1}{3}x^3+y^2x\right)_{x=\sqrt{y}}^{x=1} =\frac{1}{3}+y^2-\frac{1}{3}y^{3/2}-y^{5/2}$$
$$
\int_0^1(\frac{1}{3}+y^2-\frac{1}{3}y^{3/2}-y^{5/2})dy=\left( \frac{1}{3}y +\frac{1}{3}y^3 -\frac{2}{15}y^{5/2} -\frac{2}{7}y^{7/2}\right)_{y=0}^{y=1}=$$ $$=\frac{1}{3}+\frac{1}{3}-\frac{2}{15}-\frac{2}{7}=\frac{26}{105}$$
| {
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Continued product in $\sin$ series Find the value of the product $$(\sin 1°)(\sin 3°)(\sin 5°)\ldots(\sin 89°)$$ I tried multiplying and dividing by $2$ and then combining and then converting into cosine, but doesn't work out.
| $$\prod_{i=1}^{45}\sin(2i-1)^\circ=\prod_{j=1}^{45}\cos(2j-1)^\circ$$
Let $\cos45x=\cos45^\circ$
$\implies45x=360^\circ n\pm45^\circ\iff x=8^\circ n\pm1^\circ$ where $n$ is any integer
Considering the '+' sign,
$1\le8n+1\le90\iff0\le n\le11 \ \ \ \ (1)$
$91\le8n+1\le180\iff12\le n\le22, \ \ \ \ (2)$
$n=12\implies\cos(97^\circ)=\cos(180^\circ-83^\circ)=-\cos83^\circ$
$n=22\implies\cos(177^\circ)=\cos(180^\circ-3^\circ)=-\cos3^\circ$
$181\le8n+1\le270\iff23\le n\le33, \ \ \ \ (3)$
$n=23\implies\cos(185^\circ)=\cos(180^\circ+5^\circ)=-\cos5^\circ$
$n=33\implies\cos(265^\circ)=\cos(180^\circ+85^\circ)=-\cos85^\circ$
$271\le8n+1<360\iff34\le n<45, \ \ \ \ (4)$
$n=34\implies\cos(273^\circ)=\cos(360^\circ-87^\circ)=+\cos87^\circ$
$n=44\implies\cos(353^\circ)=\cos(360^\circ-7^\circ)=+\cos7^\circ$
There are $11+11$ negative values
Using this, the coefficient of $\cos^nx$ in $\cos(nx),$ is $1+\binom n2+\binom n4+\cdots=(1+1)^{n-1}$ which is validated by this
$\implies\cos45x=2^{45-1}\cos^{45}x+\cdots$
So, the roots $2^{n-1}\cos^{45}x+\cdots=\cos45x=\cos45^\circ$ are $(1),(2),(3),(4)$
$\implies(-1)^{22}\prod_{j=1}^{45}\cos(2j-1)^\circ=\dfrac{\dfrac1{\sqrt2}}{2^{44}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$f(x)$ is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2)$), given $f(2)=5.$
If f(x) is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2))$, given $f(2)=5.$
ATTEMPT:- $f(f(2))=f(5)$, We can find $f(0)$,$f(1)$ and $f(1/2)$ to be $1,2$ and $5/4$ respectively.
we can change the function to the form $g(x)*g(y)=g(xy)$ by basic transformation.
but how to get $f(5)$ without using transformations.
| Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$.
Now, suppose that $x_0 \neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$ is a zero of $f(x)-1$ for all non-negative integers $k$. But $f(x)-1$ is a non-zero polynomial, and thus, can only have finitely many zeros, a contradiction.
Therefore, the only zero of $f(x) - 1$ is $0$, i.e. $f(x)-1 = Cx^n$ for some constant $C$ and some non-negative integer $n$.
Now, use the fact that $f(2x)-1 = 4(f(x)-1)$ to determine $n$, and then use the fact that $f(2) = 5$ to determine $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
The solution doesn't need the use of the rules of modular-arithmetic. How can I demonstrate it?
| $11\mid 3(2a+4b)(2a-4b)-11a^2+11\cdot 2b^2=a^2-5b^2$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
A probability question - Dice Throw Sum of three dice (six faced) throws is $15$. What is the probability that first throw was $4$?
The way I thought of solving this was...
- Given - sum of second and third throw is $11$
- Probability of getting first throw = $4$ is $1$ out of $6$, that is $\frac{1}{6}$
Is this correct?
| Joffan pretty much pointed how to find the answer.
$P(D_{1}=4\mid Sum=15) = \frac {ways\ to\ get\ 11\ with\ 2\ dices}{ways\ to\ get\ 15\ with\ 3\ dices}$ (Note that getting 11 with 2 dices is like getting 4 with the first one).
$ways\ to\ get\ 11\ with\ 2\ dices = 2$ (both $(5,6)$ and $(6,5)$)
$ways\ to\ get\ 15\ with\ 3\ dices =10$
You can have 15 with $(3,6,6), (4,5,6),(5,5,5)\ and\ its\ permutations: (6,3,6), (6,6,3), (4,6,5), (5,4,6), (5,6,4), (6,4,5)\ and\ (6,5,4)$
All together gives us that $P(D_{1}=4\mid Sum=15) =\frac {2}{10}=0.2$
$0.2$, is the probability you were looking for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$ for a 4-th degree monic polynomial If $f(x)=x^4+ax^3+bx^2+cx+d$.
Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$.
This problem has troubled me a lot.The more I try to solve it,it becomes lengthier.
My problem is that there are four unknowns and only three equations.
Please help me.
| Let $g(x) = f(x) - 10x$, we have
$$g(1) = g(2) = g(3) = 0\quad\implies\quad (x-1)(x-2)(x-3) \;\;\text{ divides }\;\;g(x)$$
As a result, $f(x)$ has the form
$$f(x) = 10x + (x-t)(x-1)(x-2)(x-3)$$
for suitably chosen constant $t$. Notice
$$\begin{align}
(12-1)(12-2)(12-3) &= 990\\
(-8-2)(-8-1)(-8-3) &= -990
\end{align}$$
we have
$$\begin{align}
f(12) + f(-8)
&= 10(12 - 8) + 990(12-t) - 990(-8-t)\\
&= 40 + 990 \times 20 = 19840
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274133",
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"question_score": "5",
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How prove $\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc} $? Let $a \ge b \ge c >0$ . How can I prove
$$\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc}. $$
Maybe a simple way?
| I think a simplest way here is a full expanding:
$$\sum\limits_{cyc}(5a^7b^2+a^7c^2+13a^6b^3+9a^6c^3+18a^5b^4+18a^5c^4+2a^7bc+15a^6b^2c+11a^6c^2b+$$
$$+11a^5b^3c+15a^5c^3b+16a^4b^4c-6a^5b^2c^2-50a^4b^3c^2-50a^4c^3b^2-28a^2b^2c^2)\geq0,$$
which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is $\frac{a^2+b^2}{2}=c^2$ possible? I am looking for an integer solution to the equation:
$$\frac{a^2+b^2}{2}=c^2(a\neq b\neq c)$$
That is a square number that is the mean of two other square numbers, is this possible? And if so please can you give me an example?
| We know that $(x-y)^2+(x+y)^2=2(x^2+y^2).$
So, if $a=x+y$ and $b=x-y$, then $c^2=x^2+y^2.$
Now see Formulas for generating Pythagorean triples.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Proving $9^n - 8n - 1$ is divisible by $8^2$ for $n\ge 0$? My textbook provided the following proof:
*Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.
Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64k$. Therefore:
$$\begin{align}
9^{n+1}-8(n+1)-1&=9^{n+1}-8n-9\\
&=9^{n+1}-72n-9+64n\\
&=9\left(9^n-8n-1\right)+64n\\
&=9(64k)+64n\\
&=64(9k+n)
\end{align}$$
so $64\mid\left(9^{n+1}-8(n+1)-1\right)$.
What I don't understand is how the equation goes from
$$= 9^{n+1} - 8n - 9$$
to
$$= 9^{n+1} - 72n - 9 + 64n$$
| Use induction.
You know that the assertion is true for $n = 1.$
Assume that the assertion is true for $n = N.$
Then, the problem reduces to showing that
$$(64) ~\text{divides}~ \{[9^{N+2} - 8(N+1)] - [9^{N+1} - 8N]\}$$
$$ ~=~ [9^{N+1}(9 - 1)] - 8 = 8 \times [9^{N+1} - 1].$$
So, the problem reduces to showing that
$$(8) ~\text{divides}~ [9^{N+1} - 1].$$
However, this is immediate, since it is known that for any positive integer $n$, the polynomial $(x^n - 1)$ is divisible by $(x - 1).$
Alternatively, you could reason that since $9 \equiv 1\pmod{8}$, you must have that $9^{N+1} \equiv 1\pmod{8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$ Question:
Show that:
$$\sum_{k = 0}^{4} (1+x)^k = \sum_{k=1}^5 \binom{5}{k}x^{k-1}$$
then go on to prove the general case that:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k=1}^n \binom{n}{k}x^{k-1}$$
Attempted solution:
It might be doable to first prove the general case and then say that it is true for the specific case, but for the specific case I decided to write out the term and show that they were identical since they were so few.
For the LHS
$$\sum_{k = 0}^{4} (1+x)^k = (1+x)^0 + (1+x)^1 + (1+x)^2 + (1+x)^3 + (1+x)^4$$
$$ = (1) + (1 + x) + (x^2 +2x + 1) + (x^3 + 3x^2 + 3x + 1) + (x^4+4 x^3+6 x^2+4 x+1)$$
$$=5 + 10x + 10x^2 + 5x^3 + x^4$$
For the RHS:
$$\sum_{k=1}^5 \binom{5}{k}x^{k-1} = \binom{5}{1}x^{1-1} + \binom{5}{2}x^{2-1} + \binom{5}{3}x^{3-1} + \binom{5}{4}x^{4-1} + \binom{5}{5}x^{5-1}$$
$$= \frac{5!}{1!4!} x^0 + \frac{5!}{2!3!} x^1 + \frac{5!}{3!2!} x^2 + \frac{5!}{4!1!} x^3 + \frac{5!}{5!0!} x^4$$
$$ = 5 + 10x + 10x^2 + 5x^3 + x^4$$
That completes the first step of the question. So far so good.
For the general case, I started by using the binomial theorem for the binom and then writing out the inner-most sum:
$$\sum_{k = 0}^{n-1} (1+x)^k = \sum_{k = 0}^{n-1} \sum_{i = 0}^{k} \binom{k}{i}x^i = \sum_{k = 0}^{n-1} \left( \binom{k}{0}x^0 + \binom{k}{1}x^1 + ... + \binom{k}{k}x^{k}\right)$$
I can imagine that each step in the sum will decide the coefficients for the various powers of x and thus be identical to the RHS of the general case.
However, I run of out steam here and do not at the moment see any obvious way forward. What are some productive approaches for the general case? Am I doing things too complicated?
| Using $a^n-1=(a-1)(\sum_{k=0}^{n-1}a^k)\implies\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}$ for the first equality below (with $a=x+1$), we have
$$
\sum_{k=0}^{n-1}(x+1)^k=\frac{(x+1)^n-1}{x+1-1}=\frac{1}{x}(-1+(x+1)^n)=\frac{1}{x}\left(-1+\sum_{k=0}^n\binom{n}{k}x^k\right)=\frac{1}{x}\sum_{k=1}^n\binom{n}{k}x^k
$$
which simplifies to your RHS.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How would I make a graph for $\sqrt{x+1}-3$ I have to make a table for $\sqrt{x+1}-3$ and I can't figure out how to find the $x$ values. I know that I have to get the middle section to equal perfect squares, which are $0,\ 1,\ 4,\ 9$ but I don't know how. Please help!
Is it $2,\ 3,\ 6$ and $12$?
The problem is when I get the $x$ values, the $y$ values are in decimal form. Are they supposed to be?
$$\begin{array}{|c|c|c|}
\hline
x & \sqrt{x+1}-3 & y\\
\hline
? & & \\
\hline
? & & \\
\hline? & & \\
\hline? & & \\
\hline? & & \\
\hline
\end{array}$$
| To obtain integer values for $y = \sqrt{x + 1} - 3$, $x + 1$ must be a perfect square. The first five perfect squares are $0, 1, 4, 9, 16$. Thus, we must set $x + 1$ equal to $0, 1, 4, 9, 16$.
If $x + 1 = 0$, then $x = -1$. If $x = -1$, then $y = \sqrt{-1 + 1} - 3 = \sqrt{0} - 3 = -3$. Hence, the point $(-1, -3)$ is on the graph.
If $x < -1$, then $x + 1 < 0$, so $\sqrt{x + 1}$ is not a real number. Hence, $(-1, 0)$ is the endpoint of the graph.
If $x + 1 = 1$, then $x = 0$. If $x = 0$, then $y = \sqrt{0 + 1} - 3 = \sqrt{1} - 3 = 1 - 3 = -2$. Hence, the point $(0, -2)$ is on the graph.
If $x + 1 = 4$, then $x = 3$. If $x = 3$, then $y = \sqrt{3 + 1} - 3 = \sqrt{4} - 3 = 2 - 3 = -1$. Hence, the point $(3, -1)$ is on the graph.
As you can check, setting $x + 1 = 9$ yields the point $(8, 0)$ and setting $x + 1 = 16$ yields the point $(15, 1)$.
You can plot these points on the coordinate plane and connect them with a smooth curve to obtain the graph.
Note that the graph of $y = \sqrt{x + 1} - 3$, which is shown in green, can be obtained from the graph of $y = \sqrt{x}$, which is shown in blue, by shifting the graph of $y = \sqrt{x}$ to the left by one unit and down by three units.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality for sides and height of right angle triangle
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2xy$
$x^2+y^2+2h^2=x^2+y^2+2xy$
so $2h^2=2xy$ and $$xy=h^2$$
also $Area={ab\over 2}={ch\over 2}$ so $$ab=ch$$
$(a+b)^2=a^2+b^2+2ab$
$(c+h)^2=c^2+h^2+2ch=a^2+b^2+xy+2ab$
therefore $$(a+b)^2+xy=(c+h)^2$$
so $$c+h>a+b$$
I feel like I have made a mistake somewhere, is this incorrectly generalised? And is there an easier way to show the inequality. Also, if this is correct can it be expanded to non right angle triangle, I tried to do this using trig but was pretty much going in circles, thanks!
| I think we get $$a+b<c+h$$
Squaring we obtain
$$a^2+b^2+2ab<c^2+h^2+2hc$$
thus we have
$$2ab<h^2+2hc$$
$$2hc<h^2+2hc$$
because $ab=ch$ (area formulas) and we get $$h^2>0,$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with an Inverse Trigonometry Integral 2 Evaluate
$$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$
The Solution: We try to eliminate $\cos^{-1}\frac{2x}{1+x^2}$ by using the relation $$\pi - cos^{-1}(a) = cos^{-1}(a)$$
Consider the first 3 steps:$$$$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x$$
Putting $x=-t$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{(-t)^4}{1-(-t)^4}\cos^{-1}\frac{2(-t)}{1+(-t)^2} (-dt)$$
$$I=\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t) $$ $$$$
Now, how do we add both of these:
$$\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{x^4}{1-x^4}\cos^{-1}\frac{2x}{1+x^2} \mathrm{d}x +\int^{1/{\sqrt{3}}}_{-1/{\sqrt{3}}} \frac{t^4}{1-t^4}(\pi -cos^{-1}\frac{2t}{1+t^2}) (-\mathrm{d}t)$$ Could somebody please show this in detail?
| Let $I$ be given by
$$I=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx$$
Upon substituting $x\to -x$ we find that
$$\begin{align}
I&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{-2x}{1+x^2}\right)dx\\\\
&=\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \left(\pi-\arccos\left(\frac{2x}{1+x^2}\right)\right)dx\\\\
&=\pi\,\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4}dx-\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} \arccos\left(\frac{2x}{1+x^2}\right)dx\\\\
&=\pi\,\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4}dx-I
\end{align}$$
Thus,
$$I=\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} dx$$
The integral $I=\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{x^4}{1-x^4} dx$ can be evaluated in closed form. First we note that
$$\frac{x^4}{1-x^4}=-1+\frac{1}{1-x^4}$$
Thus, $I$ becomes
$$I=-\pi\sqrt{3}/3+\frac{\pi}{2}\int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1-x^4} dx$$
Next, we use partial fraction expansion to write
$$\frac{1}{1-x^4}=\frac{1/4}{1-x}+\frac{1/4}{1+x}+\frac{1/2}{1+x^2}$$
whereby we find that
$$\begin{align}
I&=-\pi\sqrt{3}/3+\frac{\pi}{8} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1-x}dx+\frac{\pi}{8} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1+x}dx+\frac{\pi}{4} \int_{-\sqrt{3}/3}^{\sqrt{3}/3}\frac{1}{1+x^2}dx\\\\
&=-\pi\sqrt{3}/3+\frac{\pi}{4}\log\left(2+\sqrt{3}\right)+\pi^2/12\\\\
&=\frac{\pi}{4}\log(2+\sqrt{3})+\frac{\pi^2}{12}-\frac{\pi}{\sqrt{3}}
\end{align}$$
Finally, $abcdef+1=577$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving $a = \sqrt{b + x} + \sqrt{c + x}$ for $x$ I'm trying to solve a very simple looking square root equation but nothing seems to work. The equation has this form (solve for $x$):
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
Squaring both sides obviously doesn't help since it will still give me a square root. Rearranging and then squaring doesn't help either.
The problem looks very simple to me but I have no idea on how to approach this.
| To solve
$$
a = \sqrt{b + x} + \sqrt{c + x}
$$
you must square twice, since you have two square roots.
First, for $\sqrt{b + x}$ and $\sqrt{c + x}$ to both exist, it must be $x\ge-b$ and $x\ge-c$, so
$$
x\ge\max(-b,-c).
$$
Then:
$$
a - \sqrt{b + x} = \sqrt{c + x}.\tag{1}
$$
Squaring once we find
\begin{align}
& a^2 + b \color{red}{+ x} - 2a\sqrt{b + x} = c \color{red}{+ x}\\
& a^2 + b - c = 2a\sqrt{b + x}\\
\end{align}
and squaring the second time:
$$
(a^2 + b - c)^2 = 4a^2(b + x),
$$
so
$$
x=\frac{(a^2 + b - c)^2}{4a^2} - b
$$
or
$$
x=\left(\frac{a^2 + b - c}{2a}\right)^2 - b.
$$
If you rewrite $(1)$ as
$$
a - \sqrt{c + x} = \sqrt{b + x},
$$
you will find
$$
x=\frac{(a^2 + c - b)^2}{4a^2} - c
$$
or
$$
x=\left(\frac{a^2 + c - b}{2a}\right)^2 - c
$$
which is the same.
Here $a$ (not $x$) is a symmetric function of $b$ and $c$ so one may wonder if $b<c$ or $b>c$ but it doesn't really matter.
| {
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"timestamp": "2023-03-29T00:00:00",
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Formal power series coefficient problem Find the coefficient of:
$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$
I have figured out that I need to use this identity:
$(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $
But I have no clue how to proceed with this, I have been stuck with this for hours please help.
| $$
\begin{align}
[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}
&=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}\\
&+[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37}
\end{align}
$$
The binomial theorem gives
$$
\begin{align}
(1+5x^2)^{-13}
&=\sum_{j=0}^\infty\binom{-13}{j}\left(5x^2\right)^j\\
&=\sum_{j=0}^\infty\binom{j+12}{12}\left(-5x^2\right)^j\\
\end{align}
$$
and
$$
\begin{align}
(1-8x^3)^{-37}
&=\sum_{k=0}^\infty\binom{-37}{k}\left(-8x^3\right)^k\\
&=\sum_{k=0}^\infty\binom{k+36}{36}\left(8x^3\right)^k\\
\end{align}
$$
Therefore,
$$
\begin{align}
[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37}
&=\sum_{2j+3k=10}\binom{j+12}{12}\binom{k+36}{36}(-5)^j8^k\\
&=\overbrace{\binom{17}{12}\binom{36}{36}(-5)^58^0}^{j=5,\,k=0}+\overbrace{\binom{14}{12}\binom{38}{36}(-5)^28^2}^{j=2,\,k=2}\\[15pt]
&=83019300
\end{align}
$$
Given
Coefficient[Series[(x+x^3)(1+5x^6)^-13(1-8x^9)^-37,{x,0,33}],x,33]
Mathematica returns 83019300
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\\
\end{align}$
I also know a relation like $a+b>c\\b+c>a\\a+c>b$
I have studied maths up to $12th$ grade.
| As given $(a+b+c)(b+c-a)=\alpha bc$ then we can simply as follows
$$(b+c)^2-a^2=\alpha bc \implies \alpha=\frac{b^2+c^2+2bc-a^2}{bc}=\frac{b^2+c^2-a^2}{bc}+2=2+2\cos A$$ Now, for a triangle to exist, we have a condition for angle $A$ as $0<A<\pi$ thus we get $-1<\cos A<1$ hence, we get
$$(2-2)<\alpha<(2+2) \implies 0<\alpha<4$$ Thus, the value of $\alpha$ lies between 0 & 4.
But including two extreme values for a triangle being a line, we get $0\leq\alpha\leq 4$ which shows that the triangle is a line for two extreme values $\alpha=0$ & $\alpha=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x + (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4} (x-1)^4 + \ldots \\
g(x) &= 1+\cos πx = 1+\left[ 1+\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6 +\ldots \right] \\
\frac {f(x)}{g(x)} & = \frac{\frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} {2-\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6+\ldots}
\end{align*}$$
I have no idea where to go to solve for $-\frac{1}{π^2}$ now. Please help
| You do not need the Tailor expansion but only the first derivative, the rule of L'Hôpital gives us:
$$
\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=\lim\limits_{x \to 1} \frac{\frac{1}{x}-1}{-\pi\,\sin\pi x}=\lim\limits_{x \to 1} \dfrac{\dfrac{1}{x^2}}{-\pi^2\cos\pi x} = -\dfrac{1}{\pi^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding the equation of a curve where the gradient is $ax + b$ at all points. The gradient of a curve is $ax + b$ at all points, where $a$ and $b$ are constants. Find the equation of the curve given that it passes through the points $(0,4)$ and $(1,3)$ and that the tangent at $(1,3)$ is parallel to the $x$-axis.
My workings so far are:
$\int ax + b$ $dx$ $= \frac{ax^2}{2} + bx + c$
for $(0,4)$:
$ 4 = \frac{a * (0^2)}{2} + b * 0 + c $
$ 4 = 0+c $
therefore $ c = 4 $
making the equation $\frac{ax^2}{2} + bx + 4$
for $(1,3)$:
$3 = \frac{a*(1^2)}{2} + b*1 + 4 $
$3 = \frac{a}{2} + b + 4$
The answer given is $y = x^2 - 2x + 4$
So I feel like I've got most of the way there, I'm just stuck on where to go next e.g. finding the values for $a$ and $b$, so any help would be greatly appreciated.
Thanks
| Well, the tangent line at $(1,3)$ is parallel to the $x$-axis, meaning that the gradient when $x=3$ is $0.$
On the other hand, we know that the gradient is always of the form $ax+b$ at any given $x.$ In particular, the gradient when $x=3$ is $3a+b.$
Now you have two linear equations with variables $a$ and $b,$ which will let you finish the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all 4 digits numbers that $ABCD=(CD)^2$ Please help me to solve following problem:
Find all 4 digits numbers such that $ABCD=(CD)^2$.(any of $A,B,C,D$ is a digit!)
I know one of solutions is $5776=(76)^2$.
| We need $D^2\equiv D\pmod{10}$ hence $D(D-1)$ must be a multiple of $10$. This implies that $D\in\{0,1,5,6\}$.
Next, the tens digit of $(CD)^2=(10\cdot C+D)^2=100\cdot C^2+20\cdot C\cdot D+D^2$ is determined by the ones digit of $2\cdot C\cdot D$ and the tens digit of $D^2$.
*
*For $D=0$ we need $2\cdot 0\cdot C\equiv C\pmod {10}$, so $C=0$.
*For $D=1$ we need $2\cdot 1\cdot C\equiv C\pmod {10}$, so $C=0$.
*For $D=5$ we need $2\cdot 5\cdot C+2\equiv C\pmod {10}$, so $C=2$
*For $D=6$ we need $2\cdot 6\cdot C+3\equiv C\pmod {10}$, so $C=7$
Thus the full list of answers is
$$00^2=0000\quad 01^2=0001\quad 25^2=0625\quad 76^2=5776 $$
and possibly you won't count the first three as valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
| It is true that both $a/n$ and $b/n$ tend to $0$ and $n\to\infty$, however the factor of $n$ in that term is approaching $\infty$ at the same time. So, analyzing this way, the second term gives the form $\infty\cdot 1$ and so the entire expression gives the indeterminate form $\infty-\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
Hexagon Numbering Problem
So in the above hexagon figure, I have to arrange 1 to 7, inclusive, into the circles such that the three dark red triangles have the same sum. How many distinct arrangements can there be?
| First find which numbers will be common for all shaded triangles (center circle) so the remaining numbers can be ordered in three pairs with the same sum.
If you put $\boxed{1 \Rightarrow 2+7=3+6=4+5}$
If you put $2$ there is no possible arrangament
If you put $3$ there is no possible arrangament
If you put $\boxed{4 \Rightarrow 3+5=2+6=1+7}$
If you put $5$ there is no possible arrangament
If you put $6$ there is no possible arrangament
If you put $\boxed{7 \Rightarrow 1+6=2+5=3+4}$
So in the center you can put $3$ numbers, then you select one number for the first saded triangle $\binom{6}{1}$, then by the second $\binom{4}{1}$ and last by the third $\binom{2}{1}$.
Therefore our result is $$3\times\binom{6}{1}\binom{4}{1}\binom{2}{1}=72$$
Note that I only select one number by triangle because the other one is definite by the first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
how many possible acute triangles with perimeter given How many possible acute triangles exist with perimeter 18? All sides are positive integers. The triangle (7,7,4) is the same as (4,7,7). I need the work in a way that a geometry 9th grade student would be able to come up with.
| Let's assume that $c \geq a \land c \geq b$; this forms the first constraint:
$$
c \geq \frac{18}{3} \implies c \geq 6
$$
A valid triangle requires $a + b > c$; substituting the perimeter constraint gives us:
$$
a + b > c \implies (18 - c) > c \implies c < 9
$$
You can work out the values for $a$ and $b$ from the three possible values of $c$ and the following inequality between $a$ and $b$:
$$
a = (18 - c) - b\\
18 - 2 \cdot c \leq a \leq c
$$
*
*$(6, 6, 6)$
*$(4, 7, 7)$
*$(5, 6, 7)$
*$(2, 8, 8)$
*$(3, 7, 8)$
*$(4, 6, 8)$
*$(5, 5, 8)$
These findings must also satisfy the condition of an acute triangle; from the Pythagorean theorem's converse, we learn that:
If $a^2 + b^2 > c^2$, then the triangle is acute.
After substituting $c$ for $18 - a - b$:
$$
a^2 + b^2 > (18 - a - b)^2\\
\implies a + b - \frac{a\cdot b}{18} > \frac{18^2}{2\cdot18}\\
\implies a + b - \frac{a\cdot b}{18} > 9
$$
This eliminates $(4, 6, 8)$, $(3, 7, 8)$ and $(5, 5, 8)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of matrices without expanding Show that $$\begin{array}{|ccc|}
-2a & a + b & c + a \\
a + b & -2b & b + c \\
c + a & c + b & -2c
\end{array} = 4(a+b)(b+c)(c+a)\text{.}$$
I added the all rows but couldn't get it.
| Let $x=b+c,y=c+a,z=a+b$. We claim that
$$
\left|\begin{pmatrix}
x-y-z & z & y\\
z & y-z-x & x\\
y & x & z-x-y
\end{pmatrix}\right|=4xyz.
$$
When $x=0$, add column 1 to columns 2 and 3 to obtain
$$
\left|\begin{pmatrix}
-y-z & -y & -z\\
z & y & z\\
y & y & z
\end{pmatrix}\right|=0.
$$
Thus by symmetry, $xyz$ divides the determinant. Setting $x=y=z=1$ yields
$$
\left|\begin{pmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{pmatrix}\right|=
\left|\begin{pmatrix}
-1 & 0 & 0\\
1 & 0 & 2\\
1 & 2 & 0
\end{pmatrix}\right|=4.
$$
Therefore the determinant is $4xyz$, since it is of degree 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$$
| $$\lim\limits_{n\to \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }=
\lim\limits_{n\to \infty} {\frac{\frac{3^{n+1} + 4}{3^n}\cdot \frac{7^n-47}{7^n}}{\frac{7^{n+1}-47}{7^n}\cdot\frac{3^n +4}{3^n}} }=
\lim\limits_{n\to \infty} {\frac{\left(3+\frac{4}{3^n}\right)\left( 1-\frac{47}{7^n}\right)}{\left(7-\frac{47}{7^n}\right)\left(1+\frac{4}{3^n}\right)} }=
{\frac{\left(\lim\limits_{n\to \infty} 3+\frac{4}{3^n}\right)\left(\lim\limits_{n\to \infty} 1-\frac{47}{7^n}\right)}{\left(\lim\limits_{n\to \infty} 7-\frac{47}{7^n}\right)\left(\lim\limits_{n\to \infty} 1+\frac{4}{3^n}\right)} }=\frac{(3+0)(1+0)}{(7+0)(1+0)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\int dx \frac{1}{\cosh^2 x +a^2}$ How to integrate that function
$$\int \frac{1}{\cosh^2 x +a^2}dx$$,
What I did was rewrite $$\cosh^2 x = ({\frac {{e}^x+{e}^{-x}}{2}})^2 $$
then $$\int \frac{1}{({\frac {{e}^x+{e}^{-x}}{2}})^2 +a^2}dx=$$ $$\frac{1}{4}\int \frac{1}{e^{2x}+e^{-2x}+4{a}^2 }dx$$,
Multiplu by ${e}^x $
$$\frac{1}{4}\int \frac{e^{x}}{e^{x}(e^{2x}+e^{-2x}+4{a}^2) }dx$$,
Let's take ${e}^x=t $, it follows
$$\frac{1}{4}\int \frac{dt}{t(t^{2}+t^{-2}+4{a}^2) }dx$$,
this integral is,
$$\frac{1}{4}\ \frac{t}{t(4a^{2}+t^{2}+{t}^{-2}) }$$
Finally, we only need to rewrite this expression taking care ${e}^x=t $
| Expand of the comment from lab.
\begin{align}
&\int \frac{1}{\cosh^2 x +a^2} dx\\
=&\int \frac{\mathrm{sech} ^2x}{1 +a^2\mathrm{sech} ^2x} dx \\
=&\int \frac{1}{1 +a^2(1-\tanh^2x)} d\tanh x \qquad\text{let $y=\tanh x$}\\
=&\int \frac{d y}{(1+a^2)-a^2y^2}\\
=&\frac{1}{a^2}\int \frac{d y}{-y^2+\frac{1+a^2}{a^2}}\qquad \text{let $y=\sqrt{\frac{1+a^2}{a^2}}\mathrm{tanh}\,z$}\\
=&\frac{1}{a^2}\sqrt{\frac{a^2}{1+a^2}} z\\
=&\frac{1}{a^2}\sqrt{\frac{a^2}{1+a^2}}\mathrm{arctanh}\frac{y}{\frac{\sqrt{1+a^2}}{a}} \\
=&\frac{1}{a\sqrt{1+a^2}}\mathrm{arctanh} \frac{ay}{\sqrt{1+a^2}}\\
=&\frac{1}{a\sqrt{1+a^2}}\mathrm{arctanh}\frac{a \tanh x}{\sqrt{1+a^2}}
\end{align}
| {
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"source": "stackexchange",
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How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.
Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After that I am unable to solve the problem.
| My Solution:: Given $$f(\theta) = \cos \theta \left(\sin \theta + \sqrt{\sin^2 \theta + 3}\right)$$
Now let $$y=\sin \theta \cdot \cos \theta +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}$$
Now using the Cauchy-Schwarz inequality, we get
$$\left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \left(\sqrt{\sin^2 \theta + 3}\right)^2\right\}\geq \left\{\sin \theta \cdot \cos \theta +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}\right\}^2$$
So we get $$y^2 \leq \left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \sin^2 \theta + 3\right\}=2^2$$
So we get $$-2 \leq y\leq 2\Rightarrow y\in \left[-2,2\right]$$
| {
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"source": "stackexchange",
"question_score": "5",
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How to prove that this matrix is positive definite? Let $\mathbf{A}=\begin{pmatrix}a^2+b^2 & b^2 & b^2 & ... & b^2 \\ b^2 & a^2+b^2 & b^2 & ... & b^2\\ \vdots & b^2 & \ddots & & b^2 \\ b^2 & \dots & & & a^2+b^2 \end{pmatrix}$, where $a,b\ne 0$. How can I be sure that this matrix is positive definite?
Any help would be appreciated.
| Notice that
$$x^TAx=\sum_{j=1}^n(a^2+b^2)x_j^2+2\sum_{i\lt j\leqslant n}b^2x_ix_j=b^2\left(\sum_{j=1}^nx_j\right)^2 +a^2\sum_{j=1}^nx_j^2\geqslant a^2\sum_{j=1}^nx_j^2 $$
which is positive unless $x=0$. Note that we only have to require that $a\neq 0$, $b$ can be any real number.
| {
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"question_score": "3",
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Prove sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$ I am asked to verify that the sequence $\left(\frac{1}{6n^2+1}\right)$ converges to $0$:
$$\lim \frac{1}{6n^2+1}=0.$$
Here is my work:
$$\left|\frac{1}{6n^2+1}-0\right|<\epsilon$$
$\frac{1}{6n^2+1}<\epsilon$, since $\frac{1}{6n^2+1}$ is positive
$$\frac{1}{\epsilon}<6n^2+1$$
$$\frac{1}{\epsilon}-1<6n^2$$
$$\frac{1}{6\epsilon}-\frac{1}{6}<n^2$$
At this point, I am stuck. I'm not sure if I take the square root of both sides if I then have to deal with $\pm\sqrt{\frac{1}{6\epsilon}-\frac{1}{6}}$. That doesn't seem right.
The book provides the answer:
$$\sqrt{\frac{1}{6\epsilon}}<n$$
But I don't understand (1) what happened to the $\frac{1}{6}$, and (2) why there's not a +/- in front of the square root.
Any help is greatly appreciated.
| Let $\epsilon > 0$. Let's sketch it first. $$\frac{1}{6n^2+1} < \frac{1}{6n^2} < \epsilon.$$ Look: $$\frac{1}{6n^2} < \epsilon \iff 1 < 6n^2\epsilon \iff \frac{1}{6\epsilon} < n^2 \iff n > \frac{1}{\sqrt{6\epsilon}}.$$
Now we begin. Let $\epsilon > 0$. Then exists $n_0 \in \Bbb N$ such that $n_0 > 1/\sqrt{6\epsilon}$. If $n \geq n_0$, then: $$\left|\frac{1}{6n^2+1}\right| = \frac{1}{6n^2+1} < \frac{1}{6n^2} \leq \frac{1}{6n_0^2} < \epsilon.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Roots of the equation $x^2 + px + q = 0$ If $\tan A$ and $\tan B$ are the roots of the equation $x^2 + px + q = 0 $ , show that $$\sin^2(A+B) + p \sin(A+B)\cos(A+B) + q \cos^2(A+B) = q$$
I tried using the result that $\tan A + \tan B = -p $ and $(\tan A)(\tan B) = q$ and tried substituting in the original equation but was unable to make any headway.
| you have $$\tan A + \tan B = -p, \tan A \tan B = q \implies \tan(A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B}=\frac p{q-1} $$ and $$\sin(A+B) = \pm\frac p{\sqrt{p^2 + (q-1)^2}},\quad \cos(A+B) = \pm \frac {q-1}{\sqrt{p^2 + (q-1)^2}} $$
now, $$\begin{align}\sin^2(A+B) & + p \sin(A+B)\cos(A+B) + q \cos^2(A+B)\\
& = \frac1{p^2+(q-1)^2}\left(p^2+p^2(q-1)+q(q-1)^2\right)\\
&= \frac1{p^2+(q-1)^2}\left(p^2q+q(q-1)^2\right)\\
&= q\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}$ The question:
Let $\gamma$ be a contour such that $0 \in I(\gamma),$ where $I$ is the interior of the contour. Show that
$$\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}$$
By taking $\gamma$ as the ellipse
$$\{ (x,y) : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \},$$
show that $$ \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}.$$
The question's answer uses the Deformation theorem and the fact that the first integral has the given value if the contour is a unit circle. However, the two contours must not overlap, so it seems like this should only be true for a contour that either always has magnitude less than one or greater than one. In Mathematica, the final integral was true for the case that $a = 1.5$ and $b=0.4$. What am I missing?
Edit: The radius of the circle cancels in the integral if $n = -1$, so then it does hold irrespective of the radius.
| We assume that $a$ and $b$ are real-valued and positive-valued parameters.
Let $z=e^{it}$ so that $\cos t = \frac12(z+z^{-1})$, $\sin t=\frac1{2i}(z-z^{-1})$, $dt=dz/iz$ and $t$ goes from $0$ to $2\pi$. Then,
$$\begin{align}
\int_0^{2\pi} \frac{dt}{a^2\cos^2 t+b^2\sin^2t}&=\oint_C \frac{1}{a^2\frac14(z+z^{-1})^2-b^2\frac14(z-z^{-1})^2}\frac{dz}{iz}\\\\
&=-\frac{4i}{a^2-b^2}\oint_C\frac{zdz}{z^4+2\frac{a^2+b^2}{a^2-b^2}+1}
\end{align}$$
where $C$ is the unit circle. Roots of the denominator are at $\pm i \sqrt{\frac{a-b}{a+b}}$ and $\pm i \sqrt{\frac{a+b}{a-b}}$. For $a>b$, the roots inside the unit circle are $\pm i \sqrt{\frac{a-b}{a+b}}$. The residues of the integrand are easy to compute and both are given by
$$\frac12\frac{a^2-b^2}{4ab}.$$
Thus, we have that
$$\begin{align}
\int_0^{2\pi} \frac{dt}{a^2\cos^2 t+b^2\sin^2t}&=-\frac{4i}{a^2-b^2}2\pi i\sum \text{Res}\left(\frac{z}{z^4+2\frac{a^2+b^2}{a^2-b^2}+1}\right)\\\\
&=-\frac{4i}{a^2-b^2}2\pi i\left(2\frac12\frac{a^2-b^2}{4ab}\right)\\\\
&=\frac{2\pi}{ab}
\end{align}$$
as was to be shown. If $b>a$, symmetry considerations show that the value of the integral is unchanged.
NOTE:
This integral can be evaluated without appeal to contour integration. We see that
$$\begin{align}
\int_0^{2\pi} \frac{dt}{a^2\cos^2 t+b^2\sin^2t}&=\frac{4}{a^2}\int_0^{\pi/2}\frac{\sec^2 t}{1+\frac{b^2}{a^2}\tan^2 t}\,dt\\\\
&=\frac{4}{ab}\int_0^{\infty}\frac{du}{1+u^2}\\\\
&=\frac{4}{ab}\frac{\pi}{2}\\\\
&=\frac{2\pi}{ab}
\end{align}$$
as expected!!
NOTE 2:
If one chooses to parameterize with an ellipse, then we let $z=a\cos t+ib\sin t$ and $dz=(-a\sin t+ib \cos t)dt$. Note that the integral around the ellipse $C'$ is
$$\begin{align}
\int_0^{2\pi}\frac{dt}{a^2 \cos^2t+b^2\sin^2t}&=\oint_{C'}\frac{1}{ab}\text{Im}\left(\frac{\bar z\,dz}{|z|^2}\right)\\\\
&=\oint_{C'}\frac{1}{ab}\text{Im}\left(\frac{dz}{z}\right)\\\\
&=\frac{1}{ab}\text{Im}\left(2\pi i\right)\\\\
&=\frac{2\pi}{ab}
\end{align}$$
again as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim\limits_{x\to+\infty}\frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{x^n},n\in \mathbb{N}$ If we use the following
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)=u\times t$$
$$u=x+\sqrt{x^2-1}-x+\sqrt{x^2-1}=2\sqrt{x^2-1}=2x\sqrt{1-\frac{1}{x}}$$
Now, the limit is
$$2\lim\limits_{x\to+\infty}\frac{x\sqrt{1-\frac{1}{x}}\times t}{x^n}$$
What to do next, if this is a good approach?
| $$\lim\limits_{x\to+\infty}\frac{(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n}{x^n}
$$
$$= \lim\limits_{x\to+\infty}(1+\sqrt{1-x^{-2}})^n+ \lim\limits_{x\to+\infty}(1-\sqrt{1-x^{-2}})^n = 2^n + 0 = 2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding distance from point to line which is perpendicular to another line Find the distance of the point $(1,1,1)$ from $x+y+z=1$ measured perpendicular to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$
| The equation of the straight line perpendicular to the given line $\frac{x}{2}=\frac{y}{3}=\frac{z}{6}$ and passing through the point $P(1,1,1)$ is:
$$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$$
where, $2a+3b+6c=0$.
Clearly, $(a,b,c)=(3,2,-2)$ is a solution.
Let $$\frac{x-1}{3}=\frac{y-1}{2}=\frac{z-1}{-2}=t$$
Then, any point on this line can be written as $(1+3t,1+2t,1-2t)$. Let this line intersect the plane $x+y+z=1$ at $Q(1+3t,1+2t,1-2t)$.
Then,
$$(1+3t)+(1+2t)+(1-2t)=1 \Rightarrow t=-\frac{2}{3}$$
So, the point of intersection is $Q\left(-1,-\frac{1}{3},\frac{7}{3}\right)$, and the required distance is:
$$PQ=\sqrt{{(2)}^{2}+{\left(\frac{4}{3}\right)}^{2}+{\left(-\frac{4}{3}\right)}^{2}}=\sqrt{\frac{68}{9}}$$
$$\Rightarrow PQ=\frac{2\sqrt{17}}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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