Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2} \text{ when } abc=1$ So I have a possible proof but I'm not certain it's right:
$\text{As } \frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}\text{, we obtain } \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{a^3(b+c) + b^3(a+c) + c^3(a+b)}$
$\begin{align}
a^3(b+c) + b^3(a+c) + c^3(a+b) & = a^2(\frac{1}{b} + \frac{1}{c}) + b^2(\frac{1}{a} + \frac{1}{c}) c^2(\frac{1}{a} + \frac{1}{b}) \\
& \geq \frac{4a^2}{b+c} + \frac{4b^2}{a+c} + \frac{4c^2}{a+b} \text{ by the same inequality as above} \\
& \geq 4\frac{(a+b+c)^2}{2(a+b+c)} \text{ (same reason)} \\
& \geq 2(a+b+c)
\end{align}$
Now, we have:
$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{2(a+b+c)}$
I'm not sure if this is a logical progression to make though. Because if $a \geq b$ then $\frac{1}{a} \leq \frac{1}{b}$, so is this not true? I feel like it must be as I can get the desired result, but I'm confused about it.
However, if this is true it is then sufficient to prove that $a+b+c \geq 3$.
$a+b+c = a+b + \frac{1}{ab} \geq 2\sqrt{ab} + \frac{1}{ab}$
It can be proved with calculus that $\forall x\ge 0,f(x) = 2\sqrt{x} + \frac{1}{x} \geq 3$. Thus the inequality is true, and the proof is finished.
Can someone please explain either why my proof is true (namely what I am confused about) or why this false proof yields the right result?
| We have
$$\sum \frac{1}{a^3(b+c)} = \sum \frac{1}{a^2(ab+ac)} = \sum\frac{(bc)^2}{ab+ca}.$$
Replacing $ab=z,bc=x,ca=y$, we need to show
$$\sum\frac{x^2}{y+z}\ge\frac32,\tag1$$
for $xyz=1$.
Now (1) is trivial from C-S:
$$2(x+y+z) \sum\frac{x^2}{y+z}=\sum (y+z) sum\frac{x^2}{y+z}\ge (x+y+z)^2$$
and
$$x+y+z \ge 3\sqrt[3]{xyz}=3.$$
| {
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Binomial coefficients problem with modulo
Given a natural number $n \le 40$ and non-negative integer $k < 2^n$
Need to find all integer $c, 0 \le c \le 2^n-1$, for which $\binom {2^n-1} {c} \mod 2^n = k$
Where should I start to find out good solution for this problem?
| If it's feasible to solve the problem by checking $\binom{2^n-1}{c} \bmod {2^n}$ for $c = 0,1,2, \dots$ one at a time, then here's a reasonably efficient way to do the binomial computation.
We have $$\binom{2^n - 1}{c} = \frac{(2^n-1)(2^n-2)\dotsb(2^n-c)}{1\cdot 2\cdot \dotsb \cdot c} = \frac{2^n-c}{c} \binom{2^n-1}{c-1}.$$ If $c = 2^a \cdot b$, where $b$ is odd, then $$\frac{2^n - c}{c} = \frac{2^n - 2^a \cdot b}{2^a \cdot b} = \frac{2^{n-a}}{b} - 1.$$ But, because $b$ is odd, it has an inverse $b^*$ modulo $2^n$, so we also have
$$\binom{2^n - 1}{c} \equiv \left(2^{n-a}b^* - 1\right)\binom{2^n-1}{c-1} \pmod {2^n}.$$
In fact, to save effort, we don't need $bb^* \equiv 1 \pmod{2^n}$: because we'll be multiplying by $2^{n-a}$, we only need $bb^* \equiv 1 \pmod {2^a}$.
Anyway, this gives us a way to do the computation recursively:
*
*Suppose we've computed $x_{c-1} = \binom{2^n-1}{c-1} \bmod {2^n}$.
*Write $c$ as $2^a \cdot b$, where $b$ is odd.
*Use the extended Euclidean algorithm to find $b^*$ so that $b b^* \equiv 1 \pmod{2^a}$.
*Then $x_c = (2^{n-a} b^* - 1)x_{c-1} \bmod {2^n}$.
This still leaves us with $2^{39}$ (by symmetry) cases to check when $n=40$, but anything smarter than this would actually have to figure out a pattern to the results, which is tricky.
| {
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Discrete Math Questions? (checking my answers) 1.Assume $a$ and $b$ are irrational numbers. Find a counterexample showing that it is possible $a-b$ is rational.
*
*My answer: Suppose $a$ and $b$ are equal to $\sqrt{2}$. Therefore, $a-b= \sqrt{2}- \sqrt{2}=0$. This proof by counterexample that $a$ and $b$ are irrational but $a-b$ is rational?...(is this correct)?
2.If $x+1$ is odd, prove that $(x+1)^3$ is odd.
*
*My answer: let $x+1$ be odd therefore $x+1=2k+1$ for some integer $k$.
Suppose $x+1$ is not odd. Then $x=2k$ where
\begin{align}
(x+1)^3 & = x^3+3x^2+3x+1\\
& = (2k)^3+3(2k)^2+3(2k)+1\\
& = 2(4k^2+6k^2+3k)+1\\
& = 2(a)+1
\end{align}
for some integer $a$ where $a=4k^2+6k^2+3k$. Therefore, this proves that $(x+1)^3=2a+1$ is odd.
| 1. Done.
2. Suppose $x+1$ is odd. Then $x+1=2k+1$ for some integer $k$. Then
$$
\begin{aligned}[t]
(x+1)^3
= (2k+1)^3
&= (2k)^3+3(2k)^2+3(2k)+1\\
&= 8k^3+12k^2+6k+1\\
&= 2(4k^3+6k^2+3k)+1.
\end{aligned}
$$
Thus, $(x+1)^3=2m+1$, where $m$ is the integer $4k^3+6k^2+3k$, so $(x+1)^3$ is odd. Therefore, we have shown that if $x+1$ is odd, then $(x+1)^3$ is odd.
| {
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Show that $\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{r\cos\theta -r^2}{1-2r\cos\theta+r^2}$ whenever $0Note I have seen that this question has already been posted but I believe my concerns with the question have yet to be answered.
Question: Write $z=re^{i\theta}$, where $0<r<1$, in the summation formula $$\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}$$ whenever $|z|<1$. Then, with the aid of the following theorem,
Suppose that $z_n=x_n+iy_n$ ($n=1,2,\dots$) and $S=X+iY$. Then
$$\sum_{n=1}^\infty z_n=S \text{ if and only if }\sum_{n=1}^\infty x_n=X \text{ and } \sum_{n=1}^\infty y_n=Y$$
show that
$$\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{r\cos\theta -r^2}{1-2r\cos\theta+r^2} \text{ and } \sum_{n=1}^\infty r^n\sin(n\theta)=\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}$$ whenever $0<r<1$.
Proof: Let $z=re^{i\theta}$, where $0<r<1$. Recall $$\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}$$ whenever $|z|<1$. Replace $z$ by $re^{i\theta}$ in the summation.
$$\sum_{n=0}^\infty \left(re^{i\theta}\right)^n=\sum_{n=0}^\infty r^ne^{i\theta n}=\dfrac{1}{1-re^{i\theta}}$$ whenever $\left|re^{i\theta}\right|<1$. Replace $e^{i\theta}$ by $\cos\theta +i\sin\theta$.
\begin{equation*}
\sum_{n=0}^\infty r^ne^{i\theta n} =\dfrac{1}{1-r\cos\theta-ir\sin\theta} =\dfrac{1-r\cos\theta+ir\sin\theta}{((1-r\cos\theta)-ir\sin\theta)((1-r\cos\theta)+ir\sin\theta)}=\dfrac{1-r\cos\theta+ir\sin\theta}{(1-r\cos\theta)^2+(r\sin\theta)^2}=\dfrac{1-r\cos\theta+ir\sin\theta}{1-2r\cos\theta+r^2\cos^2\theta+r^2\sin^2\theta}=\dfrac{1-r\cos\theta+ir\sin\theta}{1-2r\cos\theta+r^2}
\end{equation*}
whenever $\left|re^{i\theta} \right|<1$.
Replace $e^{i\theta n}$ by $\cos(n\theta)+i\sin(n\theta)$.
$$\sum_{n=0}^\infty r^n(cos(\theta n)+i\sin(\theta n))=\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}+i \cdot \dfrac{r\sin\theta}{1-2r\cos\theta+r^2}$$
whenever $\left|re^{i\theta}\right|<1$.
By the theorem, we have the next two sums:
$$\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}$$ and $$\sum_{n=1}^\infty r^n\sin(n\theta)=\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}$$ whenever $\left|re^{i\theta}\right|<1$.
So my questions are:
*
*How can I get $1-r\cos\theta$ to become $r\cos\theta -r^2$?
*What do I do with the $n=0$ term of both sums?
| The answer to your questions can be answered in one fell swoop. Note that we have
$$\sum_{n=0}^\infty r^n\cos(n\theta)=\frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}\tag 1$$
The left-hand side of $(1)$ can be written
$$\sum_{n=0}^\infty r^n\cos(n\theta)=1+\sum_{n=1}^\infty r^n\cos(n\theta) \tag 2$$
Using $(2)$ in $(1)$ reveals
$$\sum_{n=1}^\infty r^n\cos(n\theta)=\frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}-1=\frac{r\cos(\theta)-r^2}{1-2r\cos(\theta)+r^2}$$
as was to be shown!
| {
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Asymptotic quality of rational approximations to $\pi$ Dalzell's integral
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$
is case $n=2$ of the generalization
$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx = \frac{p_n}{q_n}-\pi$$
Such an integral gives rational approximations to $\pi$ from above: $4$, $\dfrac{19}{6}$, $\dfrac{22}{7}$, $\dfrac{377}{120}$,...
The qualities M of the latter three fractions are $2.057, 3.429, 1.98669,$ according to the definition
$$\Bigg \|{\pi-\frac{p_n}{q_n}}\Bigg \|=\frac{1}{q^{M_n}}$$
How is the asymptotic quality of this approximation sequence $\displaystyle\lim_{n \to \infty} M_n$ computed?
| This answer combines @AntonioVargas and @GregMartin results.
Let us start from the approximation in Laplace's method (https://en.wikipedia.org/wiki/Laplace%27s_method)
$$\int_a^b h(x)e^{Mg(x)}dx \approx \sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)}$$
and rewrite the general integral in the question as
$$\int_0^1 \frac{4x^2}{1+x^2}e^{n\log\left(\dfrac{x(1-x)^2}{2}\right)} dx$$
so we may identify
$$h(x)=\dfrac{4x^2}{1+x^2}$$
$$g(x)=\log\left(\dfrac{x(1-x)^2}{2}\right)$$
$$g'(x)=\dfrac{1-3x}{x(1-x)}$$
$$g''(x)=-\dfrac{1-2x+3x^2}{x^2(1-x)^2}$$
The position of the unique global maximum of $g(x)$ is obtained from $g'(x_0)=0$
$$x_0=\frac{1}{3}$$
Substituting into $h(x)$, $g(x)$ and $g''(x)$, we obtain
$$h(x_0)=h\left(\dfrac{1}{3}\right)=\frac{4\left(\dfrac{1}{3}\right)^2}{1+\left(\dfrac{1}{3}\right)^2}=\frac{2}{5}$$
$$g(x_0)=log\left(\dfrac{x_0(1-x_0)^2}{2}\right) = log\left(\frac{2}{27}\right)$$
$$g''(x_0)=g''\left(\dfrac{1}{3}\right)=-\frac{27}{2}$$
Finally,
$$\sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)} = \sqrt{\dfrac{2\pi}{n\frac{27}{2}}}\frac{2}{5}e^{n\log\left(\dfrac{2}{27}\right)} = \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n,$$
so
$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx \sim \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n$$
for large $n$.
The asymptotic quality is therefore
$$\theta = \lim_{n \to \infty} -\frac{\log\left(\dfrac{p_n}{q_n}-\pi\right)}{\log(q_n)} = \lim_{n \to \infty} -\frac{\log\left( \dfrac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n \right)}{\log( (2e^3)^n)} = \frac{3log(3)-log(2)}{3+log(2)} \approx 0.7 < 1,$$
which is low.
| {
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Help differentiating $\frac{(x-1)^2(x+2)^2}{(x+1)^2}$ I need help differentiating this expression below. I know you can use a mix of the product rule and quotient rule, but that is tedious and long. Is there a shorter method?
$$\dfrac{(x-1)^2(x+2)^2}{(x+1)^2}$$
| \begin{align*}\displaystyle \frac{\mathrm d}{\mathrm dx} \frac{(x-1)^2(x+2)^2}{(x+1)^2}&=\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\frac{(x-1)(x+2)}{x+1}\right)^2\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \left( \frac{\mathrm d}{\mathrm dx} \frac{(x-1)(x+2)}{x+1} \right)\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \frac{\mathrm d}{\mathrm dx} \left( \frac{x^2+x-2}{x+1} \right)\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \frac{\mathrm d}{\mathrm dx} \left( \frac{(x+1)x-2}{x+1} \right)\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \frac{\mathrm d}{\mathrm dx} \left(x - \frac{2}{x+1} \right)\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \left(1 + \frac{2}{(x+1)^2} \right)\\
&=\displaystyle \frac{2(x-1)(x+2)}{x+1} \cdot \frac{x^2+2x+3}{(x+1)^2}\\
&=\displaystyle \frac{2(x-1)(x+2)(x^2+2x+3)}{(x+1)^3}.
\end{align*}
| {
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Are there any positive integers such that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1$?
Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.
I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.
| Rearrange the equality
$x^2z+y^2x+z^2y=xyz$
Since $x,y,z>0$, we have
$x^2z+y^2x+z^2y>x^2z$
Multiply both sides with $\frac{y}{x}$
$y^2z>xyz$
And $y>x$
On the other hand,
$x^2z+y^2x+z^2y>y^2x$
Multiply both sides with $\frac{z}{y}$ and you get
$xz^2>xyz$, and $z>y$
Lastly,
$x^2z+y^2x+z^2y>z^2y$ Multiply both sides with $\frac{x}{z}$ and get
$x^2y>xyz$ and $x>z$
So we got, $y>x, z>y, x>z$ which is clearly not possible when $x,y,z$ are positive.
| {
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minimum value of $3$ square root function of variables $x,y$ Finding minimum value of $$\sqrt{x^2+49-7\sqrt{2}x}+\sqrt{x^2+y^2-\sqrt{2}xy}+\sqrt{y^2+25-5\sqrt{2}y}, x,y \in \mathbb{R}$$
Attempt: $$\sqrt{(x+7)^2-14x-7\sqrt{2}x}+\sqrt{(x+y)^2-2xy-\sqrt{2}xy}+\sqrt{(y+5)^2-10y-5\sqrt{2}y}$$
could some help me how to solve it , thanks
| We can assume that $x$ and $y$ are non-negatives.
Consider three triangles $\Delta ABC$, $\Delta CBD$ and $\Delta DBE$,
where $\measuredangle ABC=\measuredangle CBD=\measuredangle DBC=45^{\circ}$, $AB=5$, $BE=7$, $CB=y$ and $DB=x$.
Thus, $AC+CD+DE\geq AE=\sqrt{25+35\sqrt2+49}=\sqrt{74+35\sqrt2}.$
The equality occurs, when $\{C,D\}\subset AE$, which is possible of course.
| {
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prove this by binomial theorem Prove that by binomial theorem :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}=\dbinom{6}{1}+\dbinom{6}{3}+\dbinom{6}{5}$
I started to prove it by moving the right side to the left :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}-\dbinom{6}{1}-\dbinom{6}{3}-\dbinom{6}{5}=0$
it should be done by the binomial coefficient ...?....
how ??
thanks in advance
| Hint: The number of subsets of even cardinality= The number of subsets with odd cardinality.
| {
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Evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ I need to evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ to solve a statistic problem. Mathematica gives an answer of $\frac{9}{2}+\frac{1}{2e^{10}}$. How can I evaluate explicitly without aid of computer?
| \begin{align*}
\sum\limits_{n=1}^{\infty}\dfrac{n-1}{2}\dfrac{e^{-10}10^n}{n!}&=\dfrac{1}{2}\sum_{n=1}^{\infty}n\dfrac{e^{-10}10^n}{n!}-\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{e^{-10}10^n}{n!}\\
&=\dfrac{10e^{-10}}{2}\sum_{n=1}^{\infty}\dfrac{10^{n-1}}{(n-1)!}-\dfrac{e^{-10}}{2}\sum_{n=1}^{\infty}\dfrac{10^{n}}{n!}\\
&=\dfrac{10e^{-10}}{2}\left(1+\dfrac{10}{1!}+\dfrac{10^2}{2!}+\cdots\infty\right)-\dfrac{e^{-10}}{2}\left(\dfrac{10}{1!}+\dfrac{10^2}{2!}+\dfrac{10^3}{3!}+\cdots\infty\right)\\
&=\dfrac{10e^{-10}}{2}\cdot e^{10}-\dfrac{e^{-10}}{2}\left[\left(1+\dfrac{10}{1!}+\dfrac{10^2}{2!}+\dfrac{10^3}{3!}+\cdots\infty\right)-1\right]\\
&=\dfrac{10e^{-10}}{2}\cdot e^{10}-\dfrac{e^{-10}}{2}\cdot\left(e^{10}-1\right)\\
&=\dfrac{10}{2}-\dfrac{1}{2}+\dfrac{e^{-10}}{2}\\
&=\dfrac{9}{2}+\dfrac{1}{2e^{10}}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$ Integrate $$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$$
What I first did is I tried to complete the square for the numerator
$\sqrt{4(x^2-2x+3/4)} = \sqrt{4(x^2-2x+1-1/4)} = \sqrt{4(x-1)^2 - 1}$
Now I did trig-substitution:
$(x-1) = \frac{1}{2}\sec(\theta)$
$dx = \frac{1}{2}\sec(\theta)\tan(\theta)d\theta$
$$=\int \frac{\tan^2(\theta)\sec(\theta)}{\frac{1}{2}\sec(\theta) }d\theta = 2\tan(\theta) - 2\theta + C = 2\frac{\sqrt{4x^2-8x+3}}{2} - 2\sec^{-1}\left(\frac{x-1}{2}\right) + C$$
| I think your last step is wrong.
$2\tan\theta - 2\theta + c$
$= 2 \sqrt {\sec^2 \theta - 1} - 2 \theta + c$
As we have,
$\sec \theta = 2(x - 1)$
$\theta = \sec^{-1}2(x - 1)$
Put these values above,
$= 2 \sqrt {4x^2-8x+3} - 2 \sec^{-1}2(x - 1)+ c$
| {
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How is the factorial of $(2k)!=2^k k! 1.3.5 .....(2k-1)$ I want to figure out how $(2k)!=2^k \cdot k! \cdot 1\cdot 3\cdot 5\cdot \ldots\cdot(2k-1)$ I know that $(2k)!=2k(2k-1)(2k-2)\cdot \ldots \cdot1$. I tried to figure it but I cant. How does one get from the left side of the equation to the right side? Is there an intuitive explanation as for why this makes sense?
| Intuitive explanation
$$\begin{array}{rcl}
(2k)! &=& 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2k) \\
&=& (2 \cdot 4 \cdot 6 \cdots (2k)) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
&=& 2^k (1 \cdot 2 \cdot 3 \cdots k) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
&=& 2^k (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1)) \\
\end{array}$$
Proof
Use Mathematical induction on $k$.
Let $P(n)$ be the proposition that $(2n)! = 2^n (n!) \cdot (1 \cdot 3 \cdot 5 \cdots (2n-1))$, where $n$ is a natural number.
When $n=0$, $\text{LHS}= 0! = 1$ and $\text{RHS}=1$.
Assume $P(k)$ be true, where $k$ is a natural number.
Then, for $P(k+1)$:
$$\begin{array}{rcl}
\text{LHS} &=& (2(k+1))! \\
&=& (2k)!(2k+1)(2k+2) \\
&=& 2^k (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1))(2k+1)(2k+2) \\
&=& 2^{k+1} (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1))(2k+1)(k+1) \\
&=& 2^{k+1} (k!) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1))(k+1) \\
&=& 2^{k+1} (k!(k+1)) \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1)) \\
&=& 2^{k+1} (k+1)! \cdot (1 \cdot 3 \cdot 5 \cdots (2k-1) \cdot (2k+1)) \\
&=& \text{RHS}
\end{array}$$
Thus, by the principle of Mathematical Induction, $P(n)$ is true for all natural number $n$.
| {
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"url": "https://math.stackexchange.com/questions/2238069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital's Rule. The questions is: Evaluate $$\lim_{x\to 0} \frac{a^x -1}{x}$$ without applying L'Hopital's Rule.
Does this question fundamentally same as asking if the $\lim_{x\to 0} \frac{a^x -1}{x}$ exists? rather than straightway asking to find the limit. That means are questions (1) proving if the limit of a function exists and (2) asking what is the limit of that function, essentially same question?
| Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}$ without applying the L'Hospital's Rule.
I think that one of best way for evaluating of this limit is using of the power series expansion for the function $a^x, (a > 0)$ about the point $x = 0$. Namely, we can write that
$$
\begin{align*}
a^x &= \sum_{k = 0}^\infty \left( x \ln(a) \right)^k \\
&= 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 + \cdots \\
&= 1 + x \ln(a) \left( 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 + \cdots \right) \\
&= 1 + x \ln(a) \sum_{k = 0}^\infty \left( x \ln(a) \right)^k \\
&= 1 + x \ln(a) \lim_{n \to \infty} \sum_{k = 0}^n \left( x \ln(a) \right)^k \\
&= 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left(x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}
\end{align*}
$$
by the sum of the geometric sequence.
So, we have got
$$a^x = 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}.$$
Substitution of this expression in $\lim_{x \to 0} \frac{a^x - 1}{x}$ gives us the following:
$$
\begin{align*}
\lim_{x \to 0} \frac{a^x - 1}{x} &= \lim_{x \to 0} \frac{1 + x \ln(a) \lim_{n \to \infty} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} - 1}{x} \\
&= \lim_{x \to 0} \frac{x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}}{x} \\
&= \ln(a) \lim_{x \to 0} \left( \lim_{n \to \infty } \frac{1 - \left( { \ln(a)} \right)^{n + 1}}{1 - x \ln(a)} \right) \\
&= \ln(a) \lim_{n \to \infty } \left( \lim_{x \to 0} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} \right) \\
&= \ln(a) \frac{1 - 0}{1 - 0} \\
&= \ln(a).
\end{align*}
$$
So, $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}= \ln(a)$.
| {
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"url": "https://math.stackexchange.com/questions/2238417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $x^2+y^2+z^2+xy+yz+zx = 2w^2$
Solve in integers the equation $$x^2+y^2+z^2+xy+yz+zx = 2w^2.$$
A trivial solution to the equation is $x = y = z = w = 0$. We can rewrite the given equation as $$(x+y)^2+(x+z)^2+(y+z)^2 = 4w^2.$$ I then thought about doing a substitution, but didn't see how to do it without becoming computational. How can we continue?
| the primitive solutions to $$ a^2 + b^2 + c^2 = d^2, $$
that is
$$ \gcd(a,b,c,d) = 1, $$
are given by the quaternion norm; We must have $d$ odd and one of the others, say $a.$ as well. Then
$$ a = p^2 + q^2 - r^2 - s^2, $$
$$ b = 2(-ps +qr), $$
$$ c = 2(pr+qs), $$
$$ d = p^2 + q^2 + r^2 + s^2. $$
This is with $p+q+r+s$ odd, along with $\gcd(p,q,r,s)=1.$
This formula was surely known to Euler. However, the first acceptable proof that all primitive solutions occur this way was by L. E. Dickson, about 1920.
There is a way to get the formulas above. I have not used letter $t$ yet, take
$$ t = p + qi + rj + sk, $$
then
$$ \bar{t}i t = ai+bj+ck. $$
Note that, for your $e^2 + f^2 + g^2 = 4 w^2,$
we must have $e,f,g$ all even. So, take
$$ x+y = 2a, y+z = 2b, z+x = 2c, w = d. $$
Apparently
$$ x = a-b+c, \; \; y = a+b - c, \; \; z = -a +b+c. $$
Very Heronian, and all odd.
For the curious, the integer values of
$$ x^2 + y^2 + z^2 + yz + zx + xy $$
are exactly the same as the integer values of
$$ u^2 + v^2 + 2 w^2, $$
that is, all positive integers except
$$ 4^k (16 n + 14). $$
See ME
| {
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"url": "https://math.stackexchange.com/questions/2239261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find maximum value of $|M|$ Let z1 and z2 be two complex numbers such that $z_1 ^2 -4z_2 =16+i20$ . If a and b are roots of $x^2+z_1 x+z_2+M=0$(Where M is a complex number) and $|(a-b)^2|=28$ . then we have to find the minimum and maximum value of $|M|$
My try
$(a+b)^2 -4ab =28$
$z_1 ^2 -4z_2 -4M =28$
from this I got $|M|=5$
But I could not understand how to find minimum and maximum value .
| $$\begin{array}{rcl}
|(a-b)^2| &=& 28 \\
(a-b)^2 &=& 28e^{i\theta} \\
(a+b)^2 - 4ab &=& 28e^{i\theta} \\
z_1^2 - 4z_2 - 4M &=& 28e^{i\theta} \\
16+20i - 4M &=& 28e^{i\theta} \\
4M &=& 16 + 20i - 28e^{i\theta} \\
M &=& 4 + 5i - 7e^{i\theta} \\
\end{array}$$
Thus, $M$ is a circle of radius $7$ with center $4+5i$.
$$\begin{array}{rcccl}
||4+5i| - |7e^{i\theta}|| &\le& |M| &\le& |4+5i| + |7e^{i\theta}| \\
|\sqrt{41} - 7| &\le& |M| &\le& \sqrt{41}+7 \\
7-\sqrt{41} &\le& |M| &\le& \sqrt{41}+7 \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem when row reducing matrix with complex entries I have to find a row-reduced matrix which is row-equivalent to
$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right]$$
On one hand
$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right] \\ \xrightarrow{R_1 \leftrightarrow R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ i & -(1+i) & 0 \\ 1 & 2i & -1\\ \end{array}\right]$$
$$\xrightarrow{R_2 \gets -iR_1 + R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 1 & 2i & -1\\ \end{array}\right] $$
$$\xrightarrow{R_3 \gets -R_1 + R_3} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ \frac{1}{(-1+i)}R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ R_1\gets 2R_2+R_1} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$
$$\xrightarrow{\ R_3\gets -(2+2i)R_2+R_3} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 0 & -4\\ \end{array}\right] $$
and this last matrix is equivalent to the identity matrix $I_{3x3}$.
On the other hand I reduced it with other row operations and concluded the same as WOLFRAM ALPHA. Which one is wrong? Any help is appreciated.
| Your mistake is in 4th step. When you are taking $(-1+i)$ common from $R_2$. Then element at 2nd row and 3rd column is wrong I think.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem finding roots of a depressed quartic using the Ferrari Method. I attempted to follow the Ferrari method for a depressed quartic as described at wikipedia for the function $x^4+5/8x^2-5/8x-51/256$
The roots are $-1/4, 3/4, -1/4-i$, and $-1/4+i$. However, when I follow the method, I get a $1/4, -3/4, 1/4-i$, and $1/4+i$, or each root times $-1$.
I followed the process with other depressed quartics, and sometimes they were correct, and sometimes they were, again, the negative of the correct value.
For the quartic in question, I calculated the resolvent cubic of $x^3+5/8x^2-5/8x-51/256$. Which has one real root, m = 1/8. When I use this m, I get the negative of each of the roots.
I actually went through the derivation, and it seems correct. Does anyone have any insight as to why it is sometimes incorrect?
Edit: Here is the process as laid out on Wikipedia:
$$y^4+\frac{5}{8}y^2-\frac{5}{8}y-\frac{51}{256} = 0$$
Add $\frac{p^2}{4}$ to each side: $(y^2 + 5/16)^2 = 5/8y + 9/64
$
Add $2y^2m + pm+m^2$ to each side: $(y^2 + 5/16 + m)^2 = 2my^2 + 5/8y + m^2 + 5/8m + 9/64$
Set discriminant to zero and solve for m: $8m^3+5m^2+\frac{19}{8}m-\frac{25}{64}=0$
Use real root: $m=\frac{1}{8}$
Final equation is: $\left(y^2+\frac{5}{16}+\frac{1}{8}+\frac{1}{2}y+\frac{5}{8}\right)\left(y^2+\frac{5}{16}+\frac{1}{8}-\frac{1}{2}y-\frac{5}{8}\right)$
or: $\left(y^2+\frac{27}{16}+\frac{1}{2}y\right)\left(y^2-
\frac{3}{16}-\frac{1}{2}y\right)= 0$
The roots of the right side are $-\frac{1}{4}$ and $\frac{3}{4}$
I'll simplify the wikipedia final equations for comparison.
| I solve equations by Ferrari's method in the following stile.
We need to solve
$$256x^4+160x^2-160x-51=0$$ or
$$(16x^2+k)^2-((32k-160)x^2+160x+x^2+51)=0.$$
Now, we'll find a value of $k$, for which
$(32k-160)x^2+160x+x^2+51$ is a square.
For which we need $$80^2-(x^2+51)(32k-160)=0$$ or
$$k^3-5k^2+51k-455=0$$
and since $7$ is a divisor of $455$, we get a multiplier $k-7$, which gives
$$(k-7)(k^2+2k+65)=0$$
or $k=7$ only.
Thus, we need to solve
$$(16x^2+7)^2-(64x^2+160x+100)=0$$ or
$$(16x^2+7+8x+10)(16x^2+7-8x-10)=0$$ or
$$(16x^2+8x+17)(16x^2-8x-3)=0$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving an Olympiad type inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\le\sqrt{3(\frac{a}b+\frac{b}c+\frac{c}a)}$ Let $a,b,c>0$ be real numbers. Prove that:
$$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\leq\sqrt{3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
| By C-S
$$\left(\sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}.$$
Thus, it remains to prove that
$$\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}$$ or
$$\sum_{cyc}(3a^6c^3+3a^5b^4+6a^5c^4+a^6b^2c+2a^6c^2b+4a^5b^3c+4a^5c^3b+7a^4b^4c+$$
$$+a^5b^2c^2-11a^4b^3c^2-12a^4c^3b^2-8a^3b^3c^3)\geq0,$$
which is obviously true.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$ Let $$a=-\sqrt{99}+\sqrt{999}+\sqrt{9999}$$
$$b = \sqrt{99}-\sqrt{999}+\sqrt{9999}$$
$$c = \sqrt{99}+\sqrt{999}-\sqrt{9999}$$
Evaluate $$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$$
Edited work :
$$\frac{a^4(b-c) + b^4(c-a)+c^4(a-b)}{(a-b)(a-c)(b-c)}$$
$$\frac{(a-b)(a-c)(b-c)(a^2+b^2+c^2+ab+ac+bc)}{(a-b)(a-c)(b-c)}$$
$$a^2+b^2+c^2+ab+ac+bc$$
$$a(a+b) +b(b+c) +c(c+a)$$
Ans : $$2(99+999+9999)=22194$$
| HINT: simplifying at first and you will get $${a}^{2}+ba+ac+{b}^{2}+bc+{c}^{2} $$ and now you can plug in the given values
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding n terms of a series by multiplying two known series The exact question:
Find the first 4 non-zero terms of the series for $$e^{-x}\cos(x)$$ by
multiplying two known series. Hint: use pointers on the 2nd series.
I know $$e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... $$
and
$$\cos(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{2k!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + ... $$
so,
for $$e^{-x}\cos(x)$$ we just plug in the -x for x and get
$$e^{-x} = \sum_{k=0}^{\infty} \frac{(-1)^kx^{k}}{k!} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + ... $$
cos(x) stays the same, but...what do we do from there? He gave us the answer, which I think is (not sure because the prof just wrote it on the board and didn't really say if that was the "final answer"):
$$1 - x + \frac{x^{3}}{6} - \frac{x^{4}}{6} + ... $$
Can anyone explain to me what's going on here? What are pointers even? I went to the tutors at school and no one knows what that is and they told me I would just have to FOIL out the first four terms each series, but...is that even right? They didn't want to do it because it would take too long, but is that how this is supposed to be done?
| $(1 - x + \frac 12 x^2 - \frac 16 x^3 + \frac 1{24} x^4)(1 - \frac 12x^2 + \frac 1{24} x^4)$
We can ignore the higher powered terms. When we multiply two terms together the power of $x$ equals sum of the powers of $x$ in the two factors.
To get the $x^0$ coefficient then, it will be the product of the two constant terms. $a_0 = 1$
To get the $x^1$ coefficient, we multiply the constant term from each series by the $x^1$ terms from the other.
$a_1 = 1\cdot -1 + 1\cdot 0 = -1$
The $x^2$ terms, now it starts to get a little bit more complicated.
We multiply the each constant term by the others $x^2$ term and then multiply the two $x^1$ terms (if we had two $x^1$ terms)
$a_2 = \frac 12 - \frac 12 = 0$
$a_3 = -\frac 16 + \frac 12 = \frac 13$
$a_4 = \frac 1{24} + \frac {1}{24} - \frac 14 = -\frac 16$
or perhaps this is easier.
$\sum_\limits{n=0}^{\infty} a_n x^n \sum_\limits{m=0}^{\infty} b_m x^m = \sum_\limits{n=0}^{\infty} \sum_\limits{i=0}^n a_ib_{n-i} x^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the exact value of $\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha)$ Given that $\tan(\alpha) = -3$ find the exact value.
I tried:
$$\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha) = \\
(\sin(\pi)\cos(\alpha)+\cos(\pi)\sin(\alpha))+(\cos(\frac{3\pi}{2})\cos(\alpha)-\sin(\frac{3\pi}{2})\sin(\alpha))+\tan(-\frac{\pi}{2}+\alpha) = \\
-\sin(\alpha)+\sin(\alpha)+\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-(-\tan(\frac{\pi}{2})(-3)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-3\tan(\frac{\pi}{2})} = \\
???$$
What do I do next?
| *
*$\sin(\pi+\alpha)=-\sin\alpha$,
*$\cos\bigl(\frac{3\pi}2+\alpha\bigr)=\sin\alpha$,
*$\tan\bigl(-\frac\pi2+\alpha\bigr)=-\dfrac1{\tan\alpha}$, so
$$\sin(\pi+\alpha)+\cos\bigl(\tfrac{3\pi}2+\alpha\bigr)+\tan\bigl(-\tfrac\pi2+\alpha\bigr)=-\dfrac1{\tan\alpha}=\frac13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim_{z\to 0} \frac{zRe(z)}{\overline{z}}$
$$\lim_{z\to 0} \frac{z\cdot Re(z)}{\overline{z}}$$
$$\lim_{z\to 0} \frac{z\cdot Re(z)}{\overline{z}}=\lim_{(x,y)\to (0,0)} \frac{(x+yi)\cdot x}{x-yi}=\lim_{(x,y)\to (0,0)} \frac{x^2+yxi}{x-yi}\cdot\frac{x+yi}{x+yi}=\lim_{(x,y)\to (0,0)}\frac{x^3+yx^2i+y^2xi-y^2x}{x^2+y^2}=\lim_{(x,y)\to (0,0)}\frac{x^3-y^2x}{x^2+y^2}+i\frac{yx^2+y^2x}{x^2+y^2}$$
How should I continue?
| Bounding the numerators of real and imaginary parts:
$$|x^3-y^2x|\le|x|(x^2 + y^2),$$
$$|yx^2+y^2x|\le\max\{|x|,|y|\}(x^2 + y^2).$$
So the limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| Actually, there is a nice, generic solution to equations of this type, namely equations in the form
$$\sqrt[3]{f(x)}+\sqrt[3]{g(x)} = C,$$
especially if $f(x)$ and $g(x)$ sum to a constant.
Cubing and rearranging, we get
$$f(x) + g(x) + 3C\sqrt[3]{f(x)g(x)}=C^3,$$
or
$$\sqrt[3]{f(x)g(x)} = \frac{C^3-f(x)-g(x)}{3C}.$$
Cubing again, we get
$$f(x)g(x) = \frac{(C^3-f(x)-g(x))^3}{27C^3}.$$
If $f(x)$ and $g(x)$ are linear in $x$, this is a cubic equation. But if $f(x)+g(x)=A$ const., we get
$$f(x)[A-f(x)] = \frac{(C^3-A)^3}{27C^3},$$
or
$$f(x)^2 - Af(x) + \left[\frac{(C^3-A)}{3C}\right]^3 = 0,$$
which is quadratic in $f(x)$.
In the specific example, $f(x)=5x+7$, $g(x)=-5x+12$, $A=f(x)+g(x)=19$, $C=1$, and we get
$$f(x)^2 - 19f(x) - 216=0,$$
which yields $f(x)=5x+7=27$ or $f(x)=5x+7=-8$, i.e., $x=4$ or $x=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Can $\frac{a^4}{b^3}$ be arbitary close to $1$? I searched for positive integers $a$ and $b$ for which the difference of $a^4$ and $b^3$ is small compared to the powers $a^4$ and $b^3$. The most spectacular example I currently have is
$$9825757^4-2104527924^3=-137318688623$$ so $(a,b)=(9825757,2104527924)$ and $\frac{a^4}{b^3}=1-1.47\cdot 10^{-17}$
Questions :
*
*Is it true that $a^4-b^3=n$ has finite many integer solutions for every integer $n\ne 0$ ? I am not sure whether Falting's theorem can be applied here.
*Can $\frac{a^4}{b^3}$ be arbitary near to $1$ from above and below ? This seems to be the case, but I have no idea how to prove it.
*How can I find spectacular pairs $(a,b)$ in the above sense efficiently ? I found my example by brute force.
| Regarding dot number 2 and 3:
Find such integers $a$ and $b$ so $a^4=b^3$. It's rather obvious that there are infinitely many of them. Just set $a=c^3$ and $b=c^4$ for some integer $c$. Then the ratio $\frac{(a+1)^4}{b^3}$ is arbitrarily close to $1$. Analogously, $\frac{(a-1)^4}{b^3}$ is close from below. This also gives a method of generating examples.
Edit regarding dot number one:
If $n=1$ I have found a proof that no such solution exist.
$$
a^4 - b^3 = 1 \Leftrightarrow b^3 = a^4 - 1 = (a^2 + 1)(a^2 - 1) = (a^2 + 1)(a+1)(a-1)
$$
I will show that a common divisor of any two of three factors on the right side is at most $2$. Indeed, by Euclid's algorithm
$$gcd(a-1, a+1) = gcd(a-1, 2) \in \{ 1, 2\}$$
$$gcd(a-1, a^2+1) = gcd(a-1, a^2 + 1 - 2(a-1)) =$$$$= gcd(a-1, (a-1)^2+2) = gcd(a-1, 2) \in \{ 1, 2\}$$
The third proof is analogous.
Moreover, if $a$ is even, all three of these equal $1$ while for odd $a$, all three equal $2$. If $a$ is odd, then consider $b' = \frac b 2$. Else just set $b' = b$. This yields that integral $b'^3$ is a product of coprime integers, thus all three of them are a cube. However $a+1$ and $a-1$ differ only by $2$, while $\frac{a+1}2$ and $\frac{a-1}2$ differ by $1$, therefore they can't be both cubes, since the least difference between consecutive cubes is $2^3 - 1^3 = 7$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Let $a, b, c ∈ R$, $a^{2}+b^{2}+c^{2}=1$and $A = ab + bc + ca$. Then $A$:
Let $a, b, c ∈ R$, $a^{2}+b^{2}+c^{2}=1$and $A = ab + bc + ca$. Then
$A$:
(A) $-\tfrac{1}{2}< A < 1 $
(B).$ −1 < A < 1$
(C) $-\tfrac{1}{2}< A \leq 1 $
(D) $-\tfrac{1}{2}\leq A \leq 1 $
$$\left ( a+b+c \right )^{2}= \left ( \left ( a+b \right )+c \right )^{2}$$
$$\Rightarrow \left ( a^{2}+b^{2}+2ab \right )+c^{2}+ 2\left ( a+b \right )c$$
$$\Rightarrow \left ( a^{2}+b^{2}+c^{2} \right ) +2\left ( ab + bc + ca \right )$$
$$\Rightarrow 1+2A\geq 0$$
$$\Rightarrow A\geq -\frac{1}{2}$$
I got the left hand constraint of A but i don't know how to proceed and find the upper limit.
Is this correct? and so do i automatically select option (D)
Any help will be appreciated.
| It is obvious that $A=1$ can be reached, simply set $a=b=c$. This already gets rid of option (A) and (B). Since (D) includes (C) (that is, if (C) is true then (D) must also be true), one of the answers must be (D). The question now is: can $A=\frac12$ occur?
Well, in the reasoning
\begin{align}
(a+b+c)^2&\geq 0\\
a^2+b^2+c^2+2(ab+bc+ca)&\geq 0\\
1+2A&\geq 0\\
A&\geq -\tfrac12\\
\end{align}
what do we need for equality? We merely need to have the first expression to have an equality, that is, $a+b+c=0$. Now we need to find $a,b,c$ with $a+b+c=0$ and $a^2+b^2+c^2=1$. Substitution yields $a^2+b^2+(-a-b)^2=1$ or
$$a^2+ab+(b^2-\tfrac12)=0$$
which is simply a quadratic in $a$ and thus solvable, making $A=\frac12$ possible. The only answer can be (D).
To give explicit $a,b,c$ with $A=-\frac12$: solving the quadratic yields
\begin{align}
a&=\tfrac12-\tfrac 16\sqrt{3}\\
b&=\tfrac13\sqrt{3}\\
c&=-\tfrac12-\tfrac 16\sqrt{3}
\end{align}
so that $a^2+b^2+c^2=1$ and $A=ab+bc+ca=-\frac12$.
| {
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What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt{x}}}$ What is the first derivative and nth derivative of the following function $ y = \sqrt {2 +\sqrt {3 + \sqrt {x}}}$
I think taking the ln for both sides will remove the first square root only?
Could anyone give me a hint ?
| Alternative approach (same answer) ...
$y^2 = 2+ \sqrt{3+\sqrt{x}}$
$\Rightarrow y^2-2=\sqrt{3 + \sqrt{x}}$
$\Rightarrow y^4-4y^2+4=3+\sqrt{x}$
$\Rightarrow 4y^3\frac{dy}{dx} - 8y\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{8y(y^2-2)\sqrt{x}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Double integral - transformation I'm trying to calculate $$\iint_{\Omega } e^{(x+y^2)^{3/2}} \,\mathrm{d}A,$$ where $$\Omega =\{x,y>0 : x+y\leq 2\}. $$
Not sure where to go with it. I need to find a transformation and then calculate the integral. Any pointers?
| With the substitution
\begin{align*}
x &= (r\cos\theta)^2 \\
y &= r\sin\theta
\end{align*}
we find
\begin{align*}
\iint_{\Omega } e^{(x+y^2)^{3/2}} \,d A
&= \int_0^{\pi/2}\int_0^{R(\theta)} 2 e^{r^3}r^2\cos\theta \,dr\,d\theta \\
&= \frac{2}{3} \int_0^{\pi/2}(e^{R(\theta)^3}-1)\cos\theta\,d\theta,
\end{align*}
where
$$R(\theta) = \frac{-\sin\theta+\sqrt{8-7\sin^2\theta}}{2\cos^2\theta}.$$
($R(\theta)$ is found by solving $x(r,\theta)+y(r,\theta) = 2$ for $r$.)
Letting $\sin\theta = t$ we find
\begin{align*}
\iint_{\Omega } e^{(x+y^2)^{3/2}} \,\mathrm{d}A
&= \frac{2}{3} \int_0^1 \left(\exp\left[\left(\frac{-t+\sqrt{8-7t^2}}{2(1-t^2)}\right)^3\right] -1\right)dt \\
&= -\frac 2 3 + \frac{2}{3} \int_0^1 \exp\left(\frac{64}{(t+\sqrt{8-7t^2})^3}\right) dt.
\end{align*}
The final one-dimensional integral may be found numerically.
| {
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If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true? If one root of the equation $ax^2+bx+c=0$ be the square of the other then which is true?
$1$. $a^3+b^3+c^3-3abc=0$
$2$. $a^3+b^3+bc^2=3abc$
$3$. $b^3+a^2c+ac^2=3abc$
$4$. none.
My Attempt:
Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
$$(x-\alpha)(x-\alpha^2)=0$$
$$x^2-x(\alpha^2+\alpha)+\alpha^3=0$$ Comparing with $ax^2+bx+c=0$ we get,
$$a=1$$
$$b=-(\alpha^2+\alpha)$$
$$c=\alpha^3$$
|
Let one root be $\alpha $ then the other root will be $\alpha^2$. Then,
$$(x-\alpha)(x-\alpha^2)=0$$
Note that you need a factorization of the form $\color{red}{a}(x-\alpha)(x-\alpha^2)=0$ to be able to compare with $ax^2+bx+c=0$. Alternatively, you can divide by $a$ and continue with a quadratic where $a=1$.
Other than that, the approach is fine. Insert the expressions you find for $a$, $b$ and $c$ into the three options. But carefully looking at the powers, you can already see 1. (because of $c^3$, leaving a term in $\alpha^9$) and 2. (because of $b^2c$, leaving a term in $\alpha^7$) won't simplify to $0$. That leaves calculating expression 3 and concluding it's either 3. or 4.
Alternatively, still calling the roots $\alpha$ and $\alpha^2$, we know that the sum and product of the roots of $ax^2+bx+c=0$ are equal to $-\tfrac{b}{a}$ and $\tfrac{c}{a}$ respectively, so:
$$-\frac{b}{a}=\alpha+\alpha^2 \quad \mbox{and} \quad \frac{c}{a}=\alpha^3$$
And continue from there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Please verify my power series solution to the differential equation $y''-xy'+y=0$. Please verify my answer to the following differential equation:
$$y''-xy'+y=0$$
Let $y = {\sum_{n=0}^\infty}C_nx^n$, then $y' = {\sum_{n=1}^\infty}nC_nx^{n-1}$ and $y''={\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}$
Substituting this to the equation we get
$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-x{\sum_{n=1}^\infty}nC_nx^{n-1}+{\sum_{n=0}^\infty}C_nx^n = 0$$
$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $x^n$ term on all the terms
$${\sum_{n=0}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $0$th term from the first and the third summations we get
$$2C_2+C_0 + {\sum_{n=1}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=1}^\infty}C_nx^n = 0$$
Factoring $x^n$ we get
$$2C_2+C_0 + {\sum_{n=0}^\infty}[(n+2)(n+1)C_{n+2}-nC_n+C_n]x^n= 0$$
i.$$2C_2+C_0 = 0 => C_2 = \frac{-C_0}{2}$$
ii.$$(n+2)(n+1)C_{n+2}-nC_n+C_n = 0$$
Therefore solving ii. for $C_{n+2}$
$$C_{n+2}=\frac{(n-1)C_n}{(n+2)(n+1)}, n=0,1,2,3,...$$
If $n = 0$,
$$C_2 = \frac{-C_0}{2!}$$
If $n=1$,
$$C_3 = 0$$
If $n=2$,
$$C_4 = \frac{C_2}{3*4} = \frac{-C_0}{4!}$$
If $n=3$,
$$C_5 = \frac{2C_3}{4*5}=0$$
If $n=4$,
$$C_6 = \frac{3C_4}{5*6} = \frac{-C_0}{6!}$$
Upon seeing the pattern we realize that if $n=2m$ then
$$C_{2m} = \frac{-C_0}{2m!}$$
And if $n=2m+1$ then
$$C_{2m+1} = 0$$
So the final answer would be
$$y = {\sum_{n=0}^\infty}C_nx^n => {\sum_{m=0}^\infty}\frac{-C_0*x^{2m}}{2m!}$$
| You should be able to do a change of variable to transform this equation into Hermite's equation then compare the resulting coefficients to Hermite's.
Alternatively, you can determine the Rodrigue's Operator formulation of solutions of the equation and verify the operators have the desired effect on your series.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find minimum value of the function If $g(x) = \max|y^2 - xy|$ for $0\leq y\leq 1$. Then the minimum value of $g(x)$ is?
I am not being able to proceed. Tried drawing the graph.
| \begin{align*}
&\text{First suppose $x < 0$.}\\[6pt]
&\text{Then}\;\;x < 0\\[4pt]
&\implies\, y > x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(y-x) \ge 0\\[4pt]
&\implies\, y^2-xy \ge 0\\[4pt]
&\implies\, g(x) = \max(y^2-xy)\\[4pt]
&\qquad\qquad\qquad\;\;\text{(for $0 \le y \le 1$)}\\[4pt]
&\implies\, g(x) = \max(0,1-x)&&\text{[no local max, so}\\[2pt]
&&&\,\text{max must occur at $y=0$ or $y=1$]}\\[4pt]
&\implies\, g(x) = 1-x\\[12pt]
&\text{Next suppose $1 < x \le 2$.}\\[6pt]
&\text{Then}\;\;1 < x \le 2\\[4pt]
&\implies\, y < x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(x-y) \ge 0\\[4pt]
&\implies\, xy-y^2\ge 0\\[4pt]
&\implies\, g(x) = \max(xy-y^2)\\[2pt]
&\qquad\qquad\qquad\;\;\text{(for $0 \le y \le 1$)}\\[4pt]
&\implies\, g(x) = \max(0,x-1,x^2/4)&&\text{[max occurs at}\\[2pt]
&&&\,\text{$y=0,\;y=1$}\\[2pt]
&&&\,\text{or$\;y=x/2$ (critical point)]}\\[6pt]
&\implies\, g(x) = x^2/4&&\text{[since $x^2/4 \ge x-1$]}\\[12pt]
&\text{Next suppose $x > 2$.}\\[6pt]
&\text{Then}\;\;x>2\\[4pt]
&\implies\, y < x&&\text{[since $0 \le y \le 1$]}\\[4pt]
&\implies\, y(x-y) \ge 0\\[4pt]
&\implies\, xy-y^2\ge 0\\[4pt]
&\implies\, g(x) = \max(xy-y^2)\\[4pt]
&\implies\, g(x) = \max(0,x-1)&&\text{[max occurs at}\\[4pt]
&&&\,\text{$y=0,\;y=1$}\\[4pt]
&&&\,\text{($y = x/2 \notin [0,1]$)]}\\[4pt]
&\implies\, g(x) = x-1\\[12pt]
&\text{Next suppose $0 \le x \le 1$.}\\[6pt]
&\text{Then by previous logic}\\[4pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;1-x \ge x^2/4\\[3pt]
x^2/4 &\text{if}\;\;x^2/4 > 1-x
\end{cases}
\\[6pt]
&\text{Equivalently, for $0 \le x \le 1$}\\[4pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;0 \le x \le 2\sqrt{2}-2\\[3pt]
x^2/4 &\text{if}\;\;2\sqrt{2}-2 < x \le 1
\end{cases}
\\[12pt]
&\text{Collecting the results obtained so far}\\[6pt]
&\phantom{\implies\,}
g(x)
=
\begin{cases}
1-x &\text{if}\;\;x \le 2\sqrt{2}-2\\[3pt]
x^2/4 &\text{if}\;\;2\sqrt{2}-2 < x \le 2\\[3pt]
x-1 &\text{if}\;\;x >2
\end{cases}
\\[12pt]
&\text{Looking at each of the $3$ pieces in turn},\\[2pt]
&\text{it follows that $g$ is}\\[2pt]
&\phantom{\implies\,}{\small{\bullet}}\;\;\text{decreasing on the interval $(-\infty,2\sqrt{2}-2]$},\\[2pt]
&\phantom{\implies\,}{\small{\bullet}}\;\;\text{increasing on the interval $(2\sqrt{2}-2,\infty)$}\\[4pt]
&\text{hence the minimum value of $g$ is}\\[4pt]
&\phantom{\implies\,}\;\,g(2\sqrt{2}-2)\\[4pt]
&\phantom{\implies\,}=1 - \left(2\sqrt{2}-2\right)\\[4pt]
&\phantom{\implies\,}=3 - 2\sqrt{2}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find $2^2 + 5^2 + 8^2 + ... + (3n-1)^2 $
First prove that $1^2 + 2^2 + 3^2 +...+ n^2 = \frac{n(n+1)(n+2)}{6}$, then find $$2^2 + 5^2 + 8^2 + ... + (3n-1)^2.$$
I can prove the first part but I have no idea about the second part.
| Hint
$$\sum(3k-1)^2=9\sum k^2-6\sum k+\sum 1.$$
| {
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Simplifying expression $\sqrt{6+2\sqrt{5}} - \sqrt{6-2\sqrt{5}}.$ I'm stuck. Squaring it will change it's value. Is there any general method of simplifying expressions of the form $$\sqrt{a+b}-\sqrt{a-b} = c \ \ ?$$
| If I were you I would tackle the individual square roots to find$$\sqrt{6+2\sqrt{5}} = \sqrt{5}+1$$ $$\sqrt{6-2\sqrt{5}} = \sqrt{5}-1$$ and see that the difference is $2$.
Note: It is not always possible to denest a square root like this, but if our goal is to try to find integers $a$, $b$ such that $\sqrt{6+2\sqrt{5}}=\sqrt{a}+\sqrt{b}$, we can try to solve $a$, $b$ by $$6+2\sqrt{5} = a+b+2\sqrt{ab}\implies a+b=6, \ \ ab=5\implies (a,b)=(1,5) \ \text{or} \ (5,1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Condition for $f(x)=2x^3+ax^2+bx+4$ to have three real roots If $2x^3+ax^2+bx+4=0$ ( $a$ and $b$ are positive real numbers). Find minimum value of $a^3$ and $b^3$ so that the equation has $3$ distinct real roots.
Now $f(x)=2x^3+ax^2+bx+4$
For three distinct roots,
$f'(x)=6x^2+2ax+b$ must have two distinct real roots. Let us say roots are $p$ and $q$. Now for three real roots, should the condition $f(p) \times f(q)<0$ also be satisfied because if $f(p)$ and $f(q)$ have same sign, then we will get only one real root.
Is my approach correct? And is there any better way to approach this?
| Your method is fine so far, but I think it's worth pointing out a way to make the subsequent calculation easier. Since $p$ is a root of $6x^2+2ax+b=0$, we have
$$ p^2 = \frac{-2ap-b}{6}, $$
which simplifies the calculation of $f(p)$ to a linear one:
\begin{align}
f(p) &= 2p^3 + ap^2 + bp+4 \\
&= (2p+a)\frac{-2ap-b}{6} +bp+4 \\
&= -\frac{2a}{3}p^2-\frac{a^2+b}{3}p -\frac{ab}{6}+bp+4 \\
&= \frac{2a}{3} \frac{2ap+b}{6} +\frac{1}{3}(2b-a^2) p + 4-\frac{ab}{6} \\
&= \frac{6b-a^2}{9}p + 4-\frac{ab}{18}
\end{align}
Of course, exactly the same thing holds for $q$. But now we can expand the product and use the formulae
$$ p+q = -\frac{a}{3}, \qquad pq = \frac{b}{6} $$
to remove $p$ and $q$ entirely:
\begin{align}
f(p)f(q) &= \left(\frac{6b-a^2}{9}p + 4-\frac{ab}{18} \right)\left(\frac{6b-a^2}{9}q + 4-\frac{ab}{18} \right) \\
&= \frac{(6b-a^2)^2}{81}pq + \frac{12b-a^2}{9} \left(4-\frac{ab}{18}\right)(p+q) + \left( 4-\frac{ab}{18} \right)^2 \\
&= \dotsb = \frac{1}{108}(1728+16a^3+8b^3-a^2b^2-144ab)
\end{align}
We want this to be nonpositive. Also, from the discriminant of the quadratic, we have $a^2-6b \geq 0$. If we suppose that $a^2=6b$, the equation we obtained from the cubic becomes
$$ 1728 -8a^3 + \frac{1}{108}a^6, $$
a quadratic in $a^3$ with one real root, a double root at $a^3=432$. Hence this is the only possible answer, and plotting the regions where the inequalities are satisfied confirms this. $b^3>864$ is the other inequality.
| {
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"source": "stackexchange",
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If $a_n = \frac{{(n + 1)}^n}{n^{n - 1}} $ and $c > 0$ then $\lim_{n \to \infty} n^{1-c}(a^c_n-a^c_{n-1}) =ce^c $ If
$a_n
= \frac{{(n + 1)}^n}{n^{n - 1}}
$
and $c > 0$
then
$\lim_{n \to \infty}
n^{1-c}(a^c_n-a^c_{n-1})
=ce^c
$.
This is a generalization of
How do you calculate this limit?,
which is the case
$c = \frac12$.
Here is my solution.
As if often the case,
the algebra is
somewhat messy
and I am interested in seeing
if there is a
more elegant solution.
$a_n
=n\frac{{(n + 1)}^n}{n^{n }}
=n(1+1/n)^n
$.
We need the asymptotics of
$(1+1/n)^{cn}$.
$\begin{array}\\
(1+1/n)^{cn}
&=e^{cn\ln(1+1/n)}\\
&=e^{cn(1/n-1/(2n^2)+O(1/n^3))}\\
&=e^{c-c/(2n)+O(1/n^2))}\\
&=e^c\cdot e^{-c/(2n)+O(1/n^2))}\\
&=e^c\cdot (1-\frac{c}{2n}+O(1/n^2))\\
&=e^c-ce^c\frac{1}{2n}+O(1/n^2)\\
\end{array}
$
Note:
Wolfy says that
$(1+1/n)^{cn}
=e^c
- ce^c\left(\frac{1}{2 n}
- \frac{ 3 c + 8}{24 n^2}
+ \frac{c^2 + 8 c + 12}{48 n^3}
\right) + O(1/n^4)
$,
which is comforting.
Therefore
$\begin{array}\\
a_n^c
&=n^c(1+1/n)^{cn}\\
&=n^c(e^c-e^cc\frac1{2n}+O(1/n^2))\\
\end{array}
$
and
$\begin{array}\\
a_{n-1}^c
&=(n-1)^c(e^c-e^cc\frac1{2(n-1)}+O(1/n^2))\\
&=(n-1)^ce^c-e^cc\frac1{2(n-1)}+O(1/n^2))\\
\end{array}
$
so that
$\begin{array}\\
a_n^c-a_{n-1}^c
&=e^c(n^c-(n-1)^c)-\frac12 ce^c(\frac1{n}-\frac1{n-1})+O(1/n^2)\\
&=e^cn^c(1-(1-1/n)^c)-\frac12 ce^c(\frac1{n(n-1)})+O(1/n^2)\\
&=e^cn^c(1-(1-\frac{c}{n}+\frac{c(c-1)}{2n^2}))+O(1/n^2)\\
&=e^cn^c(\frac{c}{n}-\frac{c(c-1)}{2n^2})+O(1/n^2)\\
&=ce^cn^{c-1}-\frac{c(c-1)}{2n^{2-c}}+O(1/n^2)\\
\end{array}
$
so that
$n^{1-c}(a_n^c-a_{n-1}^c)
=ce^c-\frac{c(c-1)}{2n}+O(1/n^{1+c})
\to ce^c
$.
If $c = \frac12$,
this is
$\frac12 \sqrt{e}$
which agrees with
the original problem.
| I think the approach in my answer to the question which you have linked should work without much change.
To simplify typing I will use $A = a_{n}, B = a_{n - 1}$. Note that $A/n, B/n \to e$.
I first analyze $A^{c} - B^{c}$ and to that end we have
\begin{align}
A^{c} &= \exp(c\log A) = \exp\{c(n\log(n + 1) - (n - 1)\log n)\}\notag\\
&= \exp\{c(\log n + n\log(1 + 1/n))\}\notag\\
&= \exp(a)\text{ (say)}\notag
\end{align}
Similarly $$B^{c} = \exp\{c(\log n - (n - 2)\log(1 - 1/n))\} = \exp(b)$$ Note that $a - b \to 0$. And thus $$n^{1 - c}(A^{c} - B^{c}) = n^{1 - c}B^{c}\cdot\frac{\exp(a - b) - 1}{a - b}\cdot(a - b)$$ and thus it is sufficient to evaluate the limit of $$n^{1 - c}B^{c}(a - b) = (B/n)^{c}n(a - b)$$ Now $(B/n)^{c} \to e^{c}$ and $$n(a - b) = cn\{n\log(1 + 1/n) + (n - 2)\log(1 - 1/n)\}$$ which is same as $$c\{n^{2}\log(1 - 1/n^{2}) - 2n\log(1 - 1/n)\}$$ which tends to $c(-1 + 2) = c$. The final limit is thus $ce^{c}$. There is no need to use Taylor expansions. Just the limits $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ suffice.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove that powers of 5 are the sum of two squares using mathematical induction Using mathematical induction I need to prove $$5^n=a_n^2 +b_n^2$$
What I've tried
P(1): $$5^1 = 1^2 +2^2$$
Which is true
P(2): $$5^2 = 3^2 +4^2$$
Which is also true
Now for the induction hypothesis
P(k): $$5^k=a_k^2+b_k^2$$
Then the question suggests to prove for (k+2) which I guess still makes sense.
P(k+2): $$5^{k+2} = 5^2 . 5^k$$
$$=(3^2 +4^2)(a_k^2+b_k^2)$$
This is where I get stuck.
Apologies for formatting I'm fairly new to the site
| Continuing with your induction, assuming that $5^n=a_n^2+b_n^2$ then multiplying both sides by $5^2$ gives $5^{n+2}=(5a_n)^2+(5b_n)^2$. So $5^{n+2}$ is a sum of two squares if $5^{n}$ is. We know it is for $n=2$, so it must be for $n=2+2=4$, etc. Hence by induction, for all even $n \geq 2$ we have that $5^n$ is a sum of two squares. We also know $5^n$ is expressible as the sum of two squares for $n=1$, so it must be for $n=1+2=3$ and by induction for all odd $n \geq 1$. Because a positive integer is either even or odd, we have accounted for all positive integers.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
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Prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdot...\cdot\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$ We want to prove by induction that
$$\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$$ for all $n,k \in \mathbb{Z^+}$
For k=1 : $\frac1{2}\le\frac{1}{2}$ which it is true.
Assume the induction hypothesis $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2k-1}{2k}\le \frac{1}{\sqrt{3k+1}}$
We need to prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
We can take the induction hypothesis and multiply by $\frac{2(k+1)-1}{2(k+1)}$ on both sides.
Then, we need to prove that $\frac{1}{\sqrt{3k+1}}\cdot\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
$\frac{(2k+1)^2}{4(3k+1)(k+1)^2}\le \frac{1}{(3k+4)} \Leftrightarrow$
$(3k+4)(2k+1)^2\le 4(3k+1)(k+1)^2 \Leftrightarrow$
$12k^3+28k^2+19k+4\le 12k^3+16k^2+20k+4 \Leftrightarrow$
$12k^2-k\le 0$ which is not true!
Where is the mistake??
| Just for fun, here's an alternate method
$$2\cdot 4\cdot ...\cdot 2n=n!2^n$$
$$1\cdot 3\cdot ...\cdot (2n-1) =\frac{(2n)!}{2^nn!}$$
Then $$\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdot...\cdot\frac{2n-1}{2n} = \frac{(2n)!}{2^nn!}\cdot\frac{1}{n!2^n}=\frac{(2n)!}{(2^nn!)^2}$$
Now the problem is equivalent to proving $$\frac{(2n)!}{(2^nn!)^2} \leq \frac{1}{\sqrt{3n+1}},\quad n\in\mathbb{Z^+}$$
$n=1: LHS = \frac{2}{2^2}=\frac{1}{2}$
$RHS = \frac{1}{2}=LHS\quad $ so it holds for $n=1$. Then assume true for $n$; now we must prove true for $n+1$.
$$\frac{(2n+2)!}{(2^{n+1}(n+1)!)^2} =\frac{(2n+2)(2n+1)(2n)!}{(2(n+1)\cdot 2^nn!)^2} =\frac{(2n+2)(2n+1)(2n)!}{4(n+1)^2(2^nn!)^2} =\frac{(2n+1)(2n)!}{2(n+1)(2^nn!)^2}$$
Then we have $$\frac{(2n)!}{(2^nn!)^2} \leq \frac{1}{\sqrt{3n+1}} \implies \frac{(2n+1)(2n)!}{2(n+1)(2^nn!)^2} \leq \frac{(2n+1)}{2(n+1)\sqrt{3n+1}}$$
Consider $$\frac{1}{\sqrt{3n+4}}-\frac{(2n+1)}{2(n+1)\sqrt{3n+1}} = \frac{2(n+1)\sqrt{3n+1}-(2n+1)\sqrt{3n+4}}{2(n+1)\sqrt{3n+1}\sqrt{3n+4}}$$
If we can show $2(n+1)\sqrt{3n+1}-(2n+1)\sqrt{3n+4}\geq 0$, then we are done.
$$2(n+1)\sqrt{3n+1}\geq (2n+1)\sqrt{3n+4} \iff 4(n+1)^2(3n+1)\geq (2n+1)^2(3n+4)$$ which expands into $n\geq 0$ (which we are given from the question since $n\in\mathbb{Z^+}$.
Therefore we have $$\frac{1}{\sqrt{3n+4}}-\frac{(2n+1)}{2(n+1)\sqrt{3n+1}}\geq 0\implies \frac{1}{\sqrt{3n+4}}\geq \frac{(2n+2)!}{(2^{n+1}(n+1)!)^2}$$ as required.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Consider $f(x) = ax +\frac{1}{x+1}$. Let $ $L = the max of $ S$ = the min of $f(x)$ for $x ∈[0, 1]$. Show that $L-S >\frac{1}{12}$ for any $a > 0$. Consider the function $f(x) = ax +\frac{1}{x+1}$, where $a$ is a positive constant. Let $L$ = the largest value of $f(x)$ and $S$ = the smallest value of $f(x)$ for $x ∈[0, 1]$. Show that $L-S >\frac{1}{12}$ for any $a > 0$.
| $f'(x)= a - \dfrac{1}{(x+1)^2}= 0 \iff x = \dfrac{1}{\sqrt{a}} - 1$. So if $a = 1$, then $f'(x) = 0 \iff x = 0$, Thus $L = f(1) = \dfrac{3}{2}, S = f(0) = 1 \implies L-S = \dfrac{3}{2} > \dfrac{1}{2}$. If $a > 1 \implies L = f(1) = a+\dfrac{1}{2}, S = f(0) = 1\implies L-S = a-\dfrac{1}{2} > \dfrac{1}{2} > \dfrac{1}{12}$. If $a < 1$. We have two cases: $a) 0 < a < \dfrac{1}{4} \implies \dfrac{1}{\sqrt{a}} - 1 > 1\implies S = f(1) = a+ \dfrac{1}{2}, L = f(0) = 1\implies L-S = 1- (a - \dfrac{1}{2}) > \dfrac{3}{2} - a > \dfrac{3}{2} - \dfrac{1}{4} = \dfrac{5}{4} > \dfrac{1}{12}$.
$b) \dfrac{1}{4} \le a < 1$. Thus $0 < x_c = \dfrac{1}{\sqrt{a}} - 1 \le 1$. For if $a = \dfrac{1}{4} \implies x_c = 1\implies L = f(0) = 1, S = f(1) = a+\dfrac{1}{2} = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}\implies L - S = 1-\dfrac{3}{4} = \dfrac{1}{4} > \dfrac{1}{12}$. Thus for the final case that $\dfrac{1}{4} < a < 1 $. We have: $f(0) = 1, f(1) = a+\dfrac{1}{2}, f\left(\dfrac{1}{\sqrt{a}}-1\right)= a\left(\dfrac{1}{\sqrt{a}} - 1\right) + \sqrt{a}=2\sqrt{a} - a$. We have: $2\sqrt{a}-a -\left(a+\dfrac{1}{2}\right)= 2\sqrt{a} - 2a-\dfrac{1}{2}= -\dfrac{\left(2\sqrt{a}-1\right)^2}{2} < 0$, and $2\sqrt{a} - a - 1 = -\left(\sqrt{a}-1\right)^2 < 0$. This implies that $L = 1, S = 2\sqrt{a} - a$ if $\dfrac{1}{4} < a \le \dfrac{1}{2}\implies L-S = \left(1-\sqrt{a}\right)^2 \ge \left(1-\dfrac{1}{\sqrt{2}}\right)^2= \dfrac{3}{2} - \sqrt{2} > \dfrac{1}{12}$ as you can check it yourself. Lastly, $L = a+\dfrac{1}{2}, S = 2\sqrt{a} - a$ if $\dfrac{1}{2} \le a < 1$. For this case, $L - S = 2a - 2\sqrt{a} + \dfrac{1}{2}= \dfrac{(2\sqrt{a}-1)^2}{2} \ge \dfrac{(\sqrt{2}-1)^2}{2} > \dfrac{1}{12}$ ( verify thisfact ! ). Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of rolling 2 and 3 numbers in a sequence when rolling 3, 6 sided dice I am wondering what the probability is to roll numbers in a sequence with 3, 6 sided dice.
" If 3 fair six-sided dice are rolled. What is the probability that at least two of the numbers will be in a sequence (ex: 1,2 or 5,6). Also, what is the probability that all 3 of them will be in a sequence. These dice are all rolled at one time and can be put in any order."
I've tried solving it but I couldn't get it. What is the probability to get these outcomes.
| An approach to this problem, a bit lengthy but having the advantage to provide a clear picture,
might be the following.
Start from considering the dice marked.
The set of possible, equi-probable, outcomes is represented by $6^3=216$ triples.
Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 5} & {k + 1} & \forall \\
{prob} & & {5/6} & {1/6} & 1 \\
\end{array}
$$
the probability of getting such a scheme is $5/36$.
Now, since in our problem order does not matter, we shall swap (permute) the above configuration.
But we cannot perform that without considering the value of die 3 (call it $j$) compared with the result of die 1 and 2.
In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways.
Moreover, we shall exclude the permuted triples that fall within the range of those already considered.
So, the prospect of the possible ordered configurations and number of ways to swap them is the following
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60} \\
{\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0 \\
{\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36} \\
\hline
{{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm = } & {126} \\
\end{array}
} $$
We see that the fourth configuration is cancelled as being already included in the first.
The prospect for the complementary case (no consecutive outcomes) will give
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6 \\
{\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18} \\
{\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6 \\
\hline
{{\rm none}\,{\rm consecutive}} & {} & {\rm = } & {90} \\
\end{array}
} $$
In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 4} & {k + 1} & {k + 2} \\
{prob} & & {4/6} & {1/6} & {1/6} \\
\end{array}
$$
and since each possible triple has distinct values, we can permute them to obtain:
$$ \bbox[lightyellow] { p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$.
You can verify by direct counting that the values above are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
| Well, $(1)+(2+3)$ is the same as $1+2+3$, so if you have $n$ terms, then you can find the sum by $$\sum_{i=1}^n i = \frac{(n)(n+1)}{2}$$
Let us try for example, 5:
$\frac{(5)(6)}{2}$
And you get 15, then $1+2+3+4+5=15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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Proving the trig identity $\sin(20^\circ)\cos(65^\circ)-\cos( 20^\circ)\sin(65^\circ)$ Using the trigonometric identities I have to prove that:
$$
\sin(20^\circ)\cos(65^\circ)-\cos(20^\circ)\sin(65^\circ)=-\frac{1}{\sqrt{2}}
$$
I solved
$$
\sin(20^\circ)\cos(65^\circ) = \frac{\sin(-45^\circ)+\sin(85^\circ)}{2}$$
and
$$
-\cos(20^\circ)\sin(65^\circ)=\frac{\sin(45^\circ)+\sin(85^\circ)}{2}
$$
How do I continue from here to prove that it is equal $-\dfrac{1}{\sqrt{2}}$
| Use fact that
$$
\sin(a - b) = \sin a\cos b - \sin b \cos a
$$
$$\text{Therefore, }\,
-\sin (65 - 20) = -\sin 65 \cos 20 + \sin 20 \cos 65 = -\left(\frac{1}{\sqrt{2}}\right)
$$
$$\text{That is },\,
\sin 45 = \frac{1}{\sqrt{2}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $
Let $n$ be postive integer,show that
$$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$
I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?
| Let
$$P(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64$$
Then $P(n+11)\equiv P(n)$ mod $11$, so if $11\mid P(n)$ for some integer $n$, then $11\mid P(n)$ for some even integer $n=2m$. But $P(2m)=64(m^6-m^5+m^4-m^3+m^2-m+1)$. If $11\mid P(2m)$, then, since $64\not\equiv0$ mod $11$, we must have
$$m^6-m^5+m^4-m^3+m^2-m+1\equiv0\mod11$$
After checking that $m=-1$ is not a solution, since
$$(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1=7\not\equiv0\mod11$$
we can multiply both sides by $m+1$ to get the equivalent condition
$$m^7+1\equiv0\mod11$$
or $m^7\equiv-1$ mod $11$. Cubing both sides implies $m^{21}\equiv-1$ mod $11$. After checking that $m=0$ is not a solution, we note that $m^{10}\equiv1$, which leads to $m\equiv-1$ mod $11$, which has already been ruled out.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $\lim_{x \to \sqrt{3}^+}\sqrt{x^{2}-3}=0$ I understand that given any $\epsilon>0$, I need to find a $\delta > 0$ such that $|f(x) - \sqrt{3}|<\epsilon$ whenever $0 < x - \sqrt{3} < \delta$.
Starting with $|\sqrt{x^{2}-3}|$, how do I simplify this to show this is $< \epsilon$ for an appropriately choosen $\delta$ using $0 < x - \sqrt{3} < \delta$?
| We have that $0<x-\sqrt{3}< \delta$.
Take $\delta<1$
Thus $|\sqrt{x^2-3}|=|\sqrt{x+\sqrt{3}}||\sqrt{x-\sqrt{3}}|<|\sqrt{2 \sqrt{3}+ \delta}||\sqrt{x-\sqrt{3}}|< \sqrt{2 \sqrt{3}+ \delta}\sqrt{\delta}<(\sqrt{2 \sqrt{3}+ 1}) \sqrt{\delta}$
Finaly you can take $\delta=\min\{\frac{\epsilon^2}{\sqrt{2 \sqrt{3}+1}},1\}$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\frac{1}{x-1}\geq\frac{a}{x+1}$, provided $a>1$ Moving everything to the left,
$$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{ax-x-a-1}{x^2-1}\leq 0.$$
Finding roots to nominator and denominator:
$$\begin{array}{lcl}
ax-x-a-1 & = & 0 \Leftrightarrow \frac{a+1}{a-1}\\
x^2-1 & = & 0 \Leftrightarrow x_1 =1 \ \text{och} \ x_2=-1.\\
\end{array}$$
And that's it. How to proceed?
| $$\frac{1}{x-1}-\frac{a}{x+1}\geq0\Longleftrightarrow\frac{x-ax+a+1}{x^2-1}\geq 0$$
Last inequality is the same that
$$\frac{(a-1)\left(-x+\frac{a+1}{a-1}\right)}{(x+1)(x-1)}\ge 0\quad\iff\quad \frac{-x+\frac{a+1}{a-1}}{(x+1)(x-1)}\ge 0$$
Now, we can test for $x$ in each interval $(-\infty,-1)$, $(-1,1)$, $\left(1,\frac{a+1}{a-1}\right]$ and $\left[\frac{a+1}{a-1},\infty\right)$. Which give us that the solution set is $(-\infty, -1)\cup \left(1,\frac{a+1}{a-1}\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the trick to get the limit of a sequence with roots like $a_n = -3n + \sqrt{9n^2+11n+16}$? I know of some "tricks" for limits of sequences like
$$b_n = \frac{n^2+7}{3n+1} = \frac{n^2+7}{3n+1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} = \frac{\frac{n^2}{n^2}+\frac{7}{n^2}}{\frac{3n}{n^2}+\frac{1}{n^2}}$$
Surely, there has to be a trick like this for the above example $a_n = -3n + \sqrt{9n^2+11n+16}$? Thanks. :-)
| An alternative to "rationalising the denominator" is
$$
\begin{align}
-3n+\sqrt{9n^2+11n+16}
&=-3n+3n\sqrt{1+\frac{11}{9n}+\frac{16}{9n^2}}\\
&=-3n+3n\left(1+\frac{11}{9n}+O(n^{-2})\right)^{1/2}\\
&=-3n+3n\left(1+\frac{11}{18n}+O(n^{-2})\right)\\
&=\frac{11}{6}+O(n^{-1}).
\end{align}\
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Coordinates of the point of an ellipsoid closest to the origin. Given the ellipse $\frac{(x-r)^2}{a^2} + \frac{(y-s)^2}{b^2} =1$
$(a^2 > b^2)$. Find the coordinates of the point of the ellipsoid closest to the origin in terms of a,b,r and s.
I tried three different method so far but all lead to highly complicated and lengthy computations. Anyone has a shortcut?
Thanks
| The reason this gets so messy is that the solution depends on solving a quartic equation whose coeffiicents are functions of $a,b,r,s$.
Start by writing the ellipse in parametric form, as
$$
x= a\cos t + r \\
y= b \sin t + s
$$
Then the derivative of the square of the distance is
$$
\cos t \left( 2bs + 2(b^2-a^2) \sin t -2ar \tan t\right)
$$
and unless the solution is at the special point $(x=r,y=s\pm b)$ we can divide by $\cos t$. Letting $u=\sin t$ and writing $\tan t = \frac{u}{\sqrt{1-u^2}}$
we find that the derivative is zero when
$$
2bs + 2(b^2-a^2) u = 2ar \frac{u}{\sqrt{1-u^2}}\\
\left( 2bs + 2(b^2-a^2) u\right)^2 = \frac{4a^2r^2u^2}{1-u^2} \\
(b^2-a^2)^2 u^4 + 2(b^2-a^2) bsu^3-\left((b^2-a^2)^2 -a^2r^2 -b^2s^2 \right)u^2 - 2(b^2-a^2)bs u +b^2s^2 = 0
$$
The solution to this quartic can be obtained in closed form but it is spectacularly messy.
So if you work the problem in any other way (including Lagrange multipliers) that spectacular messiness has to come in to the calculations at some point, becasue it remains at the end of the solution.
| {
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How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 21$, Question
How many solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$,
such that
all of $x_{i}$ where $i=1,2,3,4,5$ are non negative
and
$0\leq x_1 \leq 3$
$1\leq x_2 \lt 4$
and
$x_3 \geq 15$
Attempt
first used $0\leq x_1 \leq 3$ and
$1\leq x_2 \lt 4$ and then find the number of solutions
violating $0\leq x_1 \leq 3$ $\\,\,$and$\,\,$$1\leq x_2 \lt 4$
will give $x_1 \gt 3$$\\,\,$and$\,\, x_2 \gt 3$
Now number of solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$,
=$\binom{21+5-1}{21}=12,650$
Now number of solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$, such that $0\leq x_1 \leq 3$ and $1\leq x_2 \lt 4$
=$12650-$number of solutions are there to the equation
$x_1 \gt 3$$\\,\,$and$\,\, x_2 \gt 3$
$-----------------------------------------------$
solving number of equation for $x_1 \gt 3$$\\,\,$and$\,\, x_2 \gt 3$
let $x_1=x_1^{'}+3$
$x_2=x_2^{'}+3$
our equation becomes
$x_1^{'}+3+x_2^{'}+3+x_3+x_4+x_5=21$
$\Rightarrow x_1^{'}+x_2^{'}+x_3+x_4+x_5=15$
$\therefore $ number of equation=
$\binom{15+5-1}{15}=3876$
Now,
Now number of solutions are there to the equation
$x_1 + x_2 + x_3 + x_4 + x_5 = 21$, such that $0\leq x_1 \leq 3$ and $1\leq x_2 \lt 4$
=$12650-3876=8,774$
now among $8,774$ we have to find $x3 \geq 15$
for $x_3 \geq 15$,
let $x_3 =x_3 ^{'}+15$
our equation becomes
$x_1 + x_2 + x_3^{'} +15+ x_4 + x_5 = 21$,
$x_1 + x_2 + x_3 + x_4 + x_5 = 6$
$\therefore$ number of equation =$\binom{6+5-1}{6}=210$
so final answer$=8,774-210=8,564$
But the answer is $106$
Where am i wrong??
Please correct me or else give me the correct way
Thanks!
,
| Let $S_{a, b, c, d, e} $ be the number of solutions with $x_1\ge a$, $x_2 \ge b$, $x_3\ge c$, $x_4\ge d$ and $x_5\ge e$.
The equation can be written as
$$(x_1-a)+(x_2-b)+(x_3-c)+(x_4-d)+(x_5-e)=21-a-b-c-d-e$$
So we have
$$S_{a, b, c, d, e} =\binom{21-a-b-c-d-e+4}{4}$$
if $a+b+c+d +e\le21$ and is $0$ otherwise.
The answer to this question is
$$S_{0, 1,15,0,0}-S_{4,1,15,0,0}-S_{0,4,15,0,0}+S_{4,4,15,0,0}=\binom{9}{4}-\binom{5}{4}-\binom{6}{4}+0=106$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The maximum value of a trigonometric expression (without derivative) For how trivial it may appear, here I am with this question: find the value of $x$ for which this expression has its maximum.
$$f(x) = 1 + 4\sin^2(x) + 4\sin(x)\cos(x)$$
Without making use of derivatives, and only basing on trigonometric expressions / formulas.
What I have tried is to manipulate a bit the expression, but I couldn't be able to reach any interesting expression.
| Basics:
$$
\begin{align}
%
\sin^{2} x &= \frac{1}{2} \left( 1 - \cos 2 x \right) \\[3pt]
%
\cos x \sin x &= \frac{1}{2} \sin 2x
%
\end{align}
$$
The target function becomes
$$
f(x)
= 1 + 4 \sin^{2} x + 4 \sin x \cos x
= 3 + 2 \left( \sin 2 x - \cos 2 x \right)
\tag{1}
$$
When does $\sin \theta - \cos \theta$ achieve maximal value? When
$$
\theta = \frac{3}{4}\pi + 2\pi k, \quad k \in \mathbb{Z}
$$
Therefore, the function (1) achieves maximum value at
$$
\boxed{x = \frac{3}{8}\pi + \pi k, \quad k \in \mathbb{Z}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Conditional probability for bivariate random variables question: $$ f_{(X,Y)(x,y)}=
\begin{cases} 8xy, &0\le y\le x \le 1\\0.&\text{elsewhere} \end{cases} $$
Find $ P\left( X\le \frac{1}{2}|Y\ge \frac{1}{4}\right) $
What is wrong with my solution:
$$
\frac{\int_{\frac{1}{4}}^{\frac{1}{2}}\int_{\frac{1}{4}}^x8xydydx}{\int_{\frac{1}{4}}^{1}4y-4y^3dy}
$$
seeing that $ 4y-4y^3 $is the marginal pdf of Y
| Regards @matt . If I may contribute. Isn't
$$ P \left( \small{ X \le \frac{1}{2} | Y \ge \frac{1}{4} } \right) = \frac{P(X \le \frac{1}{2} \cap Y \ge \frac{1}{4} )}{ P (Y \ge \frac{1}{4})} \:\:\:\: ?$$
The numerator should be : $$ \int_{\frac{1}{4}}^{\frac{1}{2}} \int_{\frac{1}{4}}^{x} \: (8xy) \: dy \: dx $$
to avoid confusion, you could draw the $x$-$y$ diagram to make a clear view of the area. The area for the numerator is the area below the line $y(x) = x$, with $ \frac{1}{4} \le x \le \frac{1}{2} $, its a triangle.
Regards, Arief.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $22\ldots 22$ ($1980$ twos) $\equiv 0 \pmod{1982}$
Prove that $$\underbrace{22\ldots 22}_{1980} \equiv 0 \pmod{1982}$$
I tried and found that we need to prove that $991$ divides $\underbrace{11\ldots 11}_{1980}$. I am stuck after this.
| $\underbrace{22\ldots 22}_{1980} = \frac {2}{9}\sum_\limits {i=0}^{1979} 10^i = \frac {2}{9} (10^{1980} -1)$
$10^{990}\equiv 1 \pmod {991}$ by Fermat's little theorem.
$10^{1980} = (10^{990})^2\equiv 1^2 \pmod {991}$
$10^{1980} - 1 \equiv 0 \pmod {991}\\
(10^{1980} - 1) \equiv 0 \pmod {991}$
$\frac 19(10^{1980} - 1)$ is whole number.
$991$ is prime and $991|9\cdot \frac 19(10^{1980} - 1)$ and implies $991|9$ or $991| \frac 19(10^{1980} - 1)$
If $991$ divides $\frac {1}{9}(10^{1980} - 1), 1982$ divides $\frac {2}{9}(10^{1980} - 1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the derivative of $y^3-xy^2+\cos xy=2$ Find the derivative of $y^3-xy^2+\cos xy=2$
My Attempt:
$$y^3-xy^2+\cos xy=2$$
$$\dfrac {d}{dx} [y^3-xy^2+\cos xy]=\dfrac {d}{dx} [2]$$
$$3y^2.\dfrac {dy}{dx} -[1.y^2+2xy\dfrac {dy}{dx}] + (-\sin xy) \dfrac {dy}{dx}(xy)=0$$
$$3y^2 \dfrac {dy}{dx} - y^2 - 2xy\dfrac {dy}{dx} - \sin xy (y+\dfrac {dy}{dx} x)=0$$
How do I proceed further?
| So your last line is:
\begin{align}
3y^2\dfrac{dy}{dx}-y^2-2xy\dfrac{dy}{dx}-\left(y+\dfrac {dy}{dx}x\right)\sin (xy)&=0\\
\implies\left(3y^2-2xy-x\sin(xy)\right)\dfrac{dy}{dx}&=y^2+y\sin(xy)\\
\implies\dfrac{dy}{dx}&=\dfrac{y^2+y\sin(xy)}{3y^2-2xy-x\sin(xy)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\lim_{n\to \infty}\{(1+\frac{2}{n})^n,(\sqrt{\frac{n+1}{4n-1}})\}$
$$\lim_{n\to \infty}\left\{\left(1+\frac{2}{n}\right)^n,\;\sqrt{\frac{n+1}{4n-1}}\right\}$$
$$\lim_{n\to \infty}\left(1+\frac{2}{n}\right)^n=\lim_{n\to \infty}\left(1+\frac{2}{n}\right)^{\frac{n}{2}\cdot 2}$$
$m=\frac{n}{2}$
$$\lim_{n\to \infty}(1+\frac{2}{n})^{\frac{n}{2}\cdot 2}=\lim_{n\to \infty}(1+\frac{1}{m})^{m\cdot 2}=\lim_{n\to \infty}(1+\frac{1}{m})^{{m}^{2}}=e^2$$
$$\lim_{n\to \infty}\sqrt{\frac{n+1}{4n-1}}=\lim_{n\to \infty}\sqrt{\frac{\frac{n}{n}+\frac{1}{n}}{\frac{4n}{n}-\frac{1}{n}}}$$
Because square root is a continuos function
$$\sqrt {lim_{n\to \infty}{\frac{\frac{n}{n}+\frac{1}{n}}{\frac{4n}{n}-\frac{1}{n}}}}=\sqrt {{\frac{1+0}{4-0}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$$
So
$$\lim_{n\to \infty}\{(1+\frac{2}{n})^n,(\sqrt{\frac{n+1}{4n-1}})\}=(e^{2},\frac{1}{2})$$
Are all the moves correct?
| Almost everything is correct, just a slight slip in the $e^2$.
$$\lim_{n\to\infty} \left(1+ \frac2n\right)^{\frac n2 \cdot 2} = \left(\lim_{n\to\infty} \left(1+ \frac2n\right)^{\frac n2}\right)^2 \\
=\left(\lim_{m\to\infty} \left(1+ \frac1m\right)^m\right)^2 \\ = e^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Let $w_1 = 3\cdot cis(\frac{4\pi}{3})$ and $w_2=1/3 \cdot cis(\alpha)$. Determine $\alpha$ so that $w_1\cdot w_2$ is a real number I tried:
$$3 \operatorname*{cis}(\frac{4\pi}{3})\cdot \frac{1}{3} \operatorname*{cis}(\alpha) = \operatorname*{cis}(\frac{4\pi}{3}+\alpha) = \cos(\frac{4\pi}{3}+\alpha) + \sin(\frac{4\pi}{3}+\alpha)i$$
Since this has to be a real number I did:
$$\sin(\frac{4\pi}{3}+\alpha) = 0 \Leftrightarrow \frac{4\pi}{3}+\alpha = 2k\pi \lor \frac{4\pi}{3}+\alpha = \pi+2k\pi \Leftrightarrow \\
\alpha = 2k\pi-\frac{4\pi}{3} \lor \alpha = \pi+2k\pi-\frac{4\pi}{3}$$
But my book states the solution is $\alpha = \frac{2\pi}{3}+k\pi$. What went wrong?
By the way I know theres a shorter way of representing the solution and it goes something like $\alpha = \text{something }+(-1)^k\arcsin(\text{number})$, anyone knows it?
| $$\sin (\alpha +\frac {4\pi}{3})=0$$
$$\implies \alpha+\frac {4\pi}{3}=k'\pi\;\;,k'\in\mathbb Z $$
$$\implies \alpha=k'\pi-\frac {4\pi}{3} $$
$$=(k'-2)\pi+\frac {2\pi}{3} $$
$$=k\pi+\frac {2\pi}{3} \;\;,k\in\mathbb Z$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rewrite complex number using de Moivre We have the following number $$z = (\dfrac{1}{2}-i)(1+i)^n + (\dfrac{1}{2}+i)(1-i)^n$$ and we want to rewrite it using de Moivre, in the neatest way I guess.
What I did is $$(1+i)^n = (\cos(\pi) + i \sin(\dfrac{1\pi}{2}))^n =(\cos(n\pi) + i \sin(\dfrac{n\pi}{2})) $$
Doing it for $(1-i)^n$ we obtain $$cos(n\pi) + i \sin (\dfrac{3n\pi}{2})$$
If we then multiply by the remaining factors, I get the form $$\cos(\pi n) + i \sin (\dfrac{\pi n}{2}) + \sin (\dfrac{\pi n}{2}) - \sin (\dfrac{3\pi n}{2})$$
However I'm wondering if this is the best way to use de Moivre here to clean the expression up. Wolfram alpha seems to give a different result.
| Let $ \tan \phi = 2$ then
\begin{eqnarray*}
1+i = \sqrt{2} e^{ \frac{i\pi}{4}} \\
1-i = \sqrt{2} e^{- \frac{i\pi}{4}} \\
\frac{1}{2}+i =\frac{\sqrt{5}}{2} e^{i \phi} \\
\frac{1}{2}-i =\frac{\sqrt{5}}{2} e^{-i \phi}
\end{eqnarray*}
So the quantitiy can be written as
\begin{eqnarray*}
z = (\dfrac{1}{2}-i)(1+i)^n + (\dfrac{1}{2}+i)(1-i)^n = \dfrac{\sqrt{5} \sqrt{2}^n}{2} \left( e^{i (- \phi+ \frac{n\pi}{4})} +e^{i ( \phi- \frac{n\pi}{4})} \right)
\end{eqnarray*}
This can be rewritten as
\begin{eqnarray*}
z = \sqrt{5} 2^{\frac{n}{2} }\cos ( \phi- \frac{n\pi}{4}).
\end{eqnarray*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Estimate arc length of an ellipse; proving the inequality $\pi(a+b) \le L \le 2\pi \sqrt{a^2+b^2}$ By $\gamma(t) := \space(a\cos(t), b\sin(t))^T$ with $a,b > 0$ and $0 \le t \le 2\pi$ we define an ellipse.
Now I want to prove that the arc length is bound by $a$ and $b$ so that the following inequality hold:
$$\pi(a+b) \le L \le 2\pi \sqrt{a^2+b^2}$$
I tried using the definition and got:
$$L := \int_0^{2\pi} \sqrt{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
which is equal to:
$$a\int_0^{2\pi} \sqrt{1-\frac{a^2-b^2}{a^2}\cos^2(t)}dt$$
Since the integral can't be solved by integrating I tried generating the Taylor Series with $c := \frac{\sqrt{a^2+b^2}}{a}$:
$$1-\frac{1}{2}c^2\cos^2(t)-\frac{1}{8}c^4\cos^4(t)... $$
However I don't know how this will help me.. Has anyone an idea or another way to prove this inequality?
Kind Regards
EDIT:
Upper Bound:
$$L = \int_0^{2\pi} \sqrt{a^2\sin^2(t)+b^2\cos^2(t)}dt \space \le \int_0^{2\pi} \sqrt{a^2*1+b^2*1}dt = 2\pi \sqrt{a^2+b^2}$$
Lower Bound:
See Macavity for a full answer.
| So effectively we need to show
$$\pi (a+b) \leqslant L = 4\int_0^{\frac{\pi}2}\sqrt{a^2\sin^2t + b^2\cos^2t } \: dt \leqslant 2\pi \sqrt{a^2+b^2}$$
For the right inequality, note for $x, y \geqslant 0$, we have $\sqrt{x^2+y^2} \leqslant x+y$, so $\sqrt{a^2\sin^2t + b^2\cos^2t } \leqslant a \sin t + b \cos t$, which integrated in $[0, \frac{\pi}2]$ gives the tighter bound $L \leqslant 4(a+b)$.
Using QM-AM, to conclude if needed.
For the left inequality, note from Cauchy-Schwarz we get:
$$(a^2\sin^2t + b^2\cos^2t)(\sin^2t + \cos^2t) \geqslant (a\sin^2t + b \cos^2t)^2$$
$$\implies L = 4\int_0^{\frac{\pi}2}\sqrt{a^2\sin^2t + b^2\cos^2t } \geqslant 4\int_0^\frac{\pi}2 (a \sin^2t + b \cos^2t) dt = \pi(a+b) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of reciprocals of Fibonacci numbers convergence I am trying to prove the convergence/divergence of the series´
$$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$
${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by:
$${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2} $$
I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.
Can anyone help me? Thank you
| For $n \ge 3$, we have
$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}}
= \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n}
= \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\
= \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n}
$$
The partial sums is monotonic increasing and bounded from above.
$$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n}
\le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$
As a result, the series converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
| By definition, this is $f'(8)$ for $f(x) = (x+1)^{-1/2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?
| simplify the term $$\frac{2+\sqrt{2}+\sqrt{3}}{4+3 \sqrt{2}+\sqrt{3}+\sqrt{6}}+\frac{4+3
\sqrt{2}+\sqrt{3}+\sqrt{6}}{2+\sqrt{2}+\sqrt{3}}$$
$$\sqrt{16}=4,\sqrt{18}=3\sqrt{2}$$
the final result is $2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Find $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$ $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$
I feel that this question will be solved by telescoping series but i cannot express it as telescoping series.
$\sin^4x=(\frac{1-\cos2x}{2})^2=\frac{1+\cos^22x-2\cos2x}{4}$
I am not able to solve it further.
The options are $(A)\sin^4x,(B)\sin^2x,(C)\cos^2x$(D)does not exist
| Given $$\sin^4 x = \sin^2 x(1-\cos^2 x) = \sin^2 x-\frac{1}{4}\sin^2(2x)$$
So $$\sin^4(2x) = \sin^2(2x)-\frac{1}{4}\sin^2(4x)$$
So $$\sin^4(4x) = \sin^2(4x)-\frac{1}{4}\sin^2(8x)$$
.................................
.................................
.................................
So $$\sin^4(2^n x) = \sin^2(2^n x)-\frac{1}{4}\sin^2(2^{n+1}x)$$
$$\lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{1}{2^k}\sin^4(2^k x) = \lim_{n\rightarrow \infty}\bigg[\sin^2 x-\frac{1}{4^{n+1}}\sin^2 (2^{n+1})\bigg]=\sin^2 x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Finding the square root $\sqrt{4-2\sqrt{3}}$ How do you solve this:
$$X^2=4-2\sqrt{3}$$
I don't really know how to solve for $X$. I initially got $X=\sqrt{4-2\sqrt{3}}$ but I don't know how to simplify it.
| $\surd$ simplifies if what's inside is a square.
Since $(x+y)^2=x^2+2xy+y^2$ we will search for $\begin{cases} x^2+y^2=4 \\ 2xy=-2\sqrt{3}\end{cases}\iff\begin{cases} x^2y^2+y^4=4y^2 \\ xy=-\sqrt{3}\end{cases}$
The quadratic equation $\big[y^4-4y^2+3=0\big]$ gives $y^2=1$ or $3$ and this leads to $x+y=\pm(1-\sqrt{3})$
If you want to explore further some methodology, I invite you to read this interesting paper :
Jeffrey-Rich : Simplifying square roots of square roots by denesting
Applying theorem $\mathbf{(4.12)}$ : $\sqrt{\vphantom{\big(}X\pm Y}=\sqrt{\frac 12X+\frac 12\sqrt{X^2-Y^2}}\pm\sqrt{\frac 12X-\frac 12\sqrt{X^2-Y^2}}$
With $X=4$ and $Y=2\sqrt{3}$
Gives $\sqrt{\vphantom{\big)}4-2\sqrt{3}}=\sqrt{2+\frac 12\sqrt{4}}-\sqrt{2-\frac 12\sqrt{4}}=\sqrt{3}-\sqrt{1}=\sqrt{3}-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Triangular distribution of a+b sums for rationals a/b of bounded height Let $a/b$ be a rational in $(0,1)$ expressed in lowest terms.
The height of $a/b$ is $\max \{a,b\}$.
For each rational in $(0,1)$ of height $\le h$, form the sum $a+b$.
So, for example, here are the rationals of height $\le 6$
and their sums:
\begin{array}{ccccccccccc}
\frac{1}{6} & \frac{1}{5} &
\frac{1}{4} & \frac{1}{3} &
\frac{2}{5} & \frac{1}{2} &
\frac{3}{5} & \frac{2}{3} &
\frac{3}{4} & \frac{4}{5} &
\frac{5}{6} \\
7 & 6 & 5 & 4 & 7 & 3 & 8 & 5
& 7 & 9 & 11 \\
\end{array}
Now form a histogram of these sums. So above, the bin for $3$ gets $1$ count
because only $\frac{1}{2}$ leads to $3$, but the bin for $7$ gets a count
of $3$ because each of $\{ \frac{1}{6}, \frac{2}{5}, \frac{3}{4} \}$ sum to $7$.
Here is the histogram for $h \le 24$:
You can see the bin for $3$ still has a count of $1$, the
bin for $47$ has a count of $1$ for $\frac{23}{24}$, and the bin
for $46$ is empty. The tallest bin is for $23$, whose sum is
achieved by $11$ fractions: $\{ \frac{1}{22}, \frac{2}{21}, \ldots, \frac{11}{12} \}$.
Here is the histogram for $h \le 256$:
My question is: What explains the features of this plot.
I would appreciate explanations of some of the structure that seems to be
emerging in this plot,
from the left-right symmetry to more subtle features.
Perhaps I am hallucinating structure where there
is none, but it seems one can discern a series of nested triangles that roughly
demarcate regions of differing density, something like this:
| Effectively, you have a histogram of $a + b$ for all $a < b \leq n$ such that $\gcd(a, b) = 1$.
For $a = 1$, we get all $b > 1$.
For $a = 2$, we get roughly half of all $b > 2$.
What does that look like when we add them together?
We continue like this, for $a = 3$, roughly one third, etc.
For $a \leq 5$, we get this:
Does this illuminate the pattern?
| {
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Complex: $z^2 = 1+2i$ (find real- and imaginary part of $z$) these are my toughs:
$$z^2 = 1 + 2i \Longrightarrow (x+yi)(x+yi) = 1 + 2i$$
so: $x^2-y^2 = 1$ and $2xy = 2$
then i got that $x = 1/y$ but i cant continue to find the real- and imaginary part of z anymore. Appriciated any help
| An alternative approach would be to use DeMoivre's Theorem.
$1+2i=\sqrt{5}e^{i\arctan{2}}$ so $z=\sqrt[4]{5}e^{i\arctan 2/2}$ or $z=\sqrt[4]{5} e^{(i\arctan 2 +2\pi i)/2}$ If $\tan \theta =2, \tan \frac{\theta}{2} = \frac{2}{1+\sqrt{5}}$ giving $$z=\sqrt[4]{5}\frac{1+\sqrt{5}}{\sqrt{10+2\sqrt{5}}}+i\sqrt[4]{5}\frac{2}{\sqrt{10+2\sqrt{5}}}$$ The other root being $-z$.
| {
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Cannot understand trig manipultion $a$ is a radius in a origin centered circle.
Any point on a circle circumference has coordinates $(a\ \cos\theta, a \ \sin\theta)$.
I do not understand the last transformation in this expression $x = \sqrt{a^2 - y^2} = \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $
How do we get $ \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $ ???
Note: this is just a part of calc lecture in MIT ocw
| Important note:
$$\sqrt{a^2-a^2\sin^2\theta} = |a\cos\theta |$$
as $\sqrt{\Delta^2} = |\Delta|$ and not $\Delta$.
Typically for indefinite integrals it's assumed the substitution is done over an appropriate interval (sometimes this is hand-waved away for whatever reason), as in your case it is required that $-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$, or similar, so that $\cos\theta\ge 0$. For definite integrals, this changes things.$$\\$$
For example, in integrating
$$\int_{-\pi}^{-3\pi/2} \dfrac{1}{\sqrt{2^2-2^2\sin^2(\theta)}} d\theta$$
One would obtain $\sqrt{2^2-2^2\sin^2\theta} = |2\cos\theta| = -2\cos\theta$ on this interval.
| {
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Find all integer solutions of the equation $x^2+15^a=2^b$ I found an integer solution for $$x^2+15^a=2^b$$ Which is $x=7$, when $a=1$ and $b=6$
In general: $x$ must be odd, $a$ must be odd and $b$ must be even
The problem is that I don't know wheater there are more solutions. Otherwise, how can I prove there is no other solution?
| I'll do the case where $x,a$ and $b$ are positive, the other one is easier but a bit intricate.
Working mod $3$ we have that $b$ must be even.
Let $b=2v$
We now have: $15^a=2^b-x^2=(2^v-x)(2^v+x)$
Clearly $x$ must be coprime to $15$, and so $(2^v-x)=3^a$ and $(2^v+x)=5^a$. or $2^v-x=1$ and $2^v+x=15^a$.
We deal with the first case first ($(2^v-x)=3^a$ and $(2^v+x)=5^a$):
So $3^a+5^a=2(2^v)$.
So in fact finding solutions to your equation is equivalent to $3^a+5^a$ being a power of $2$. When $a$ is even it is not possible (work mod $4$). When $a$ is odd it only works with $a=1$, this is easy to observe with the lifting the exponent lemma.
Since we have $2^v-x=3$ and $2^v+x=5$ we have $x=1,v=2,a=1$ or $x=1,b=4,a=1$.
We now deal with the second case: ($2^v-x=1$ and $2^v+x=15^a$):
We get $15^a+1=2(2^v)$.
So by Catalan's theorem $a=1$, and so $v=3$. Hence we get $x=7,b=6,a=1$. And this gives all solutions with $x,a,b$ positive.
Clearly the sign of $x$ is irrelevant, and simple arguments show there are no solutions with $a$ or $b$ negative.
However, there remains one other possibility, $x=0$ (neither positive not negative). For this case $15^a=2^b $, and as $15$ is not any rational power of $2$ there can be no nonzero solutions for $a $ and $b $. Only $a=b=0$ works for the case $x=0$.
The complete solution set is then
$(x,a,b) \in \{(0,0,0),(1,1,4),(7,1,6)\} $
| {
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Expressing $A\sin x + B\cos x$ as a sine -- $A,B \in \mathbb{C}$ Suppose we have the expression $P=A\sin x + B\cos x$ , where $x$ is real and $A,B$ are complex numbers. Can we then write $P$ in the form $C \sin y$ , where obviously $C=C(A,B)$ and $y=y(A,B,x)$, with $y$ preferably being real ?
In the case where $A,B$ are real positive numbers, it's fairly easy to do. In fact it holds that $P=R\sin(x-a)$ where $R=\sqrt{A^2 +B^2}$ and $a=\arctan\frac{B}{A}$. We can find that by demanding $$A\sin x+B\cos x = R\sin(x+a) = R(\sin x \cos a + \cos x \sin a) $$
From which we can see that we need $R\cos a= A $ $(1)$ and $R\sin a=B$ $(2)$.
Thus, by doing the following two steps:
$$(2)/(1) \implies \tan a= \frac{B}{A} $$
$$ (1)^2 + (2)^2 \implies R^2 = A^2 + B^2$$
However when they are complex, dividing (1) and (2) gives us again $\tan a= \frac{B}{A} $ but now $a$ and thus $\tan a$ is real, but $ \frac{B}{A}$ is generally complex. Is there another (any?) way of doing this?
| There is no way to work with $y$ real.
Indeed, when $x$ varies $A\sin x+B\cos x$ describes an ellipse in the complex plane, while $C\sin(y)$ is a line segment.
| {
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evaluating $\int\sqrt{81-x^2} \, dx$ So I know this problem needs trig sub to solve and I put it on the side opposite $\theta$ and I got a different answer since from the key since they put it on the length and not the height of the triangle.
I got a very similar answer, the only difference is that everything except C is - and instead of arcsin I had arccos.
Integral:
$$\int\sqrt{81-x^2} \, dx$$
Answer key:
$$\frac{1}{2}x\sqrt{81-x^2}+\frac{81}{2}\arcsin\frac{x}{9}+C$$
My answer:
$$-\frac{1}{2}x\sqrt{81-x^2}-\frac{81}{2}\arccos\frac{x}{9}+C$$
To explain what I mean by opposite of $\theta$. In my triangle, $\cos\theta=\frac{x}{9}$. With the answer key's set up, $\cos\theta=\frac{\sqrt{81-x^2}}{9}$.
My question is how do you know where to put the radical on your triangle? It seems like it shouldn't matter since the triangle works.
| $\int \sqrt {81-x^2}\ dx$ don't forget the index of integration.
Either one of these substitutions is acceptable.
$x = 9\cos t$ or $x = 9\sin t$
Lets demonstrate both ways
$\int \sqrt {81-x^2}\ dx\\
x = 9\cos t\\
dx = -9\sin t\\
\int \sqrt {81-81\cos^2 t}(-9\sin t)\ dt\\
\int - 81 \sin^2 t dt\\
\int - \frac {81}{2} + \frac {81}2\cos 2t\ dt\\
- \frac {81}{2}t + \frac {81}4\sin 2t\\
- \frac {81}{2}t + \frac {81}2 \sin t\cos t+C\\
t = \arccos \frac {x}{9}\\
- \frac {81}{2}\arccos \frac {x}{9} + \frac {81}2 (\sqrt {1-\frac {x^2}{81}})( \frac {x}{9})+C\\
- \frac {81}{2}\arccos \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + C
$
$\int \sqrt {81-x^2}\ dx\\
x = 9\sin t\\
dx = 9\cos t\\
\int \sqrt {81-81\sin^2 t}(9\cos t)\ dt\\
\int 81 \cos^2 t dt\\
\int \frac {81}{2} + \frac {81}2\cos 2t\ dt\\
\frac {81}{2}t + \frac {81}4\sin 2t\\
\frac {81}{2}t + \frac {81}2 \sin t\cos t+C\\
t = \arcsin \frac {x}{9}\\
\frac {81}{2}\arcsin \frac {x}{9} + \frac {81}2 \frac {x}{9}\sqrt {1-\frac {x^2}{81}} +C\\
\frac {81}{2}\arcsin \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + C
$
Both are fine ways to go.
$- \frac {81}{2}\arccos \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + C = \frac {81}{2}\arcsin \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + D$
as
$ -\frac {81}{2}\arccos \frac {x}{9}$ differs from $\frac {81}{2}\arcsin \frac {x}{9}$ by a constant.
Generally, you will see the $x = a \sin t$ substitution because it has fewer minus signs.
| {
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Prove that $\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$
Prove that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My attempt,
Let an equation $x^{2n+1}-1=0$, which has roots $$\cos \frac{2\pi}{2n+1}+i \sin \frac{2\pi}{2n+1}, \cos \frac{4\pi}{2n+1}+i \sin \frac{4\pi}{2n+1},...,\cos \frac{4n\pi}{2n+1}+i \sin \frac{4n\pi}{2n+1}$$
$$(1+\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+\cos \frac{4n\pi}{2n+1})+(\sin \frac{2\pi}{2n+1}+\sin \frac{4\pi}{2n+1}+...+\sin \frac{4n\pi}{2n+1})i=0$$
$$\cos \frac{2\pi}{2n+1}+ \cos \frac{4\pi}{2n+1}+...+ \cos \frac{4n\pi}{2n+1}=-1$$
Since $$\cos \frac{2\pi}{2n+1}=\cos \frac{4n\pi}{2n+1}$$
$$\cos \frac{4\pi}{2n+1}=\cos \frac{(4n-2)\pi}{2n+1}$$
So, $$2(\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+ \cos \frac{2n\pi}{2n+1})=-1$$
which proves that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My question:
Is my attempt with using complex number too tedious and long? Is there another way to solve this question?
| Let $\theta=\dfrac {2\pi}{2n+1}$.
Consider the summation
$$\sum_{r=1}^{2n+1}e^{ir\theta}$$
as plotted out on the Argand diagram for each step of the summation.
The summation traces out, starting from $(0,0)$, a regular polygon with $(2n+1)$ sides and unit side length. The last segment is from $(-1,0)$ to $(0,0)$. The polygon is symmetrical about $x=-\frac 12$, and has $n$ non-horizontal segments on either side of this line.
The topmost point of the polygon has coordinates
$$\left(\sum_{r=1}^n \cos r\theta, \sum_{r=1}^n \sin r\theta\right)=\left(-\frac12, H\right)$$
by symmetry.
Hence $$\sum_{r=1}^n \cos r\theta=\sum_{r=1}^n \cos \frac {2\pi r}{2n+1}=\color{red}{-\frac 12}$$
An example is shown below for the case where $n=3$ (i.e. $2n+1=7$).
Note
This method can also be used to find $\displaystyle\sum_{r=1}^n\sin\frac {2\pi r}{2n+1}$.
Assume that the polygon has radius $r$ and interior angle $2\alpha$.
Given that the exterior angle is $\theta$, we have
$\alpha=\frac {\pi}2-\frac \theta2$. Hence, $\sin\alpha=\cos\frac\theta 2$ and $\cos\alpha=\sin\frac\theta 2$. Also, we know that $r\cos\alpha=\frac 12$.
It can be shown by simple geometry that
$$\small H=r(1+\sin\alpha)=\frac {1+\sin\alpha}{2\cos\alpha}=\frac {1+\cos\frac\theta 2}{2\sin\frac \theta 2}=\frac {2\cos^2\frac \theta4}{4\sin\frac\theta 4\cos\frac\theta 4}=\frac{\cos\frac\theta 4}{\sin\frac\theta 4}=\frac 12\cot\frac\theta 4=\frac 12 \cot\frac{\pi}{2(2n+1)}$$
Hence
$$\sum_{r=1}^n \sin r\theta=\sum_{r=1}^n\sin\frac {2\pi r}{2n+1}=\frac 12\cot\frac {\pi}{2(2n+1)}$$
| {
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Prove inequality $a^3b^3+2b^3c^3+3a^3c^3\le 0$ For $a^3+b^3+c^3=0$. Prove that $a^3b^3+2b^3c^3+3a^3c^3\le 0$
i think from $a^3+b^3+c^3=0$ we have one of the three numbers is zero, the other two are opposites.Suppose $a;b$ opposites and $c=0$
| You can make:
$$a^3b^3+2b^3c^3+3a^3c^3=b^3(a^3+2c^3)+3a^3c^3$$
but, $a^3+2c^3=c^3-b^3$, so
$$b^3(c^3-b^3)+3a^3c^3=-b^6+c^3(b^3+3a^3)$$
but, $b^3+3a^3=-2b^3-3c^3$, so
$$-b^6+c^3(-2b^3-3c^3)=-[(b^3+c^3)^2+2c^6]\le 0$$
| {
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Generating functions (Expansion) Problem: Give an interpretation in partition terms for the coefficient $x^{12}$ in the expansion $(1+x^2+x^4+x^6+x^8+x^{10}+x^{12})(1+x^4+x^8+x^{12})(1+x^6+x^{12})(1+x^8)(1+x^{10})(1+x^{12})$
I'm not sure if I have to find a generating function or only determine when $x^{12}$ appears or only try to describe his behavior.
Thanks for your help.
| That's the same as the coefficient of $x^6$ in
$$(1+x+x^2+x^3+x^4+x^5+x^6)(1+x^2+x^4+x^6)(1+x^3+x^6)(1+x^4)(1+x^5)(1+x^6)$$
which is the same as the coefficient of $x^6$ in
$$(1+x+x^2+\cdots)(1+x^2+x^4+
\cdots)(1+x^3+x^6+\cdots)\cdots=
\prod_{n=1}^\infty\frac1{1-x^n}$$
which must be the number of partitions of some or other number....
| {
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Writing 616 as a sum of four squares I am using the fact that $S_4$ is closed under multiplication and Euler's four square identity:
$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2) =\\
\quad(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4)^2 + (a_1b_2+a_2b_1+a_3b_4-a_4b_3)^2
+(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2)^2 + (a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1)^2$$
So $616= 2^3\cdot 7\cdot 11$
$= 8 \times (2^2+1^2+1^2+1^2)(3^2+1^2+1^2+0^2)$
$=8 \times(6 − 1 − 1)^2 + (2 + 3 + 0 − 1)^2 + (2 − 0 + 3 + 1)^2 + (0 + 1 − 1 + 3)^2$
$=8(4^2+4^2+6^2+3^2)$
But then from here I am stuck, I can't see how to get the four integers.
| If the sum of four squares is divisible by $8,$ then each of the four numbers is even and we may divide through by $2.$ That is, we write $616/4 = 154$ as the sum of four squares (this allows $0$ and $1$ as squares if convenient). Then just double those.
$$ 154 = 144 + 9 + 1 + 0 $$
$$ 154 = 121 + 25 + 4 + 4 $$
$$ 154 = 121 + 16 + 16 + 1 $$
and so on
| {
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How to solve this limit without using L'Hopital's Rule: $\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}$?
Find:
$$\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}\cdot$$
I tried to simplify the expression, but I kept getting stuck. I also tried to make a substitution $u=\frac{1}{x}$, but I got stuck again. Please give me hint or explain what to do. Thanks.
| Just added as a too long comment.
Many ways to get the answer have been given in answers. Using Taylor expansions with one more term, you could even know how is approached the limit since $$x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)=9 x^4-\frac{35 x^6}{3}+O\left(x^8\right)$$ making $$x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)-x^4=8 x^4-\frac{35 x^6}{3}+O\left(x^8\right)$$
$$\sqrt{x^4 \cos ^2(x)+2 x^2 \sin ^2(2 x)-x^4}=2 \sqrt{2}\, x^2-\frac{35 }{12 \sqrt{2}}x^4+O\left(x^6\right)$$ $$\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}=2 \sqrt{2}-\frac{35 x^2}{12 \sqrt{2}}+O\left(x^4\right)$$ which shows the limit and how it is approached.
Moreover, this provides an easy estimation of the function "close" to $x=0$. For example $$x=\frac \pi 6 \implies \text{exact} \approx 2.28503 \qquad \text{approximation} \approx 2.26301$$
$$x=\frac \pi 4 \implies \text{exact} \approx 1.65598 \qquad \text{approximation} \approx 1.55624$$
| {
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Complex series sum Show that for $|z|<1$,
$$\sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z}$$
and
$$\sum_{n=0}^\infty \frac{2^n z^{2^n}}{1+z^{2^n}}=\frac{z}{1-z}$$
As a hint is given to use the dyadic expansion of an integer.
I have no idea how to proceed.Please help.
| The LHS of the first is
\begin{align}
&(z+z^3+z^5+z^7+\cdots)\\
&+(z^2+z^6+z^{10}+z^{14}+\cdots)\\
&+(z^4+z^{12}+z^{20}+z^{28}+\cdots)+\cdots.
\end{align}
The LHS of the second is
\begin{align}
&(z-z^2+z^3-z^4+z^5-\cdots)\\
&+(2z^2-2z^4+2z^6-2z^8+2z^{10}-\cdots)\\
&+(4z^4-4z^8+4z^{12}-4z^{16}+4z^{20}-\cdots)+\cdots.
\end{align}
Do these give a clue?
| {
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What is the remainder when dividing $11^{(345^{678})}$ by 13? So basically I figured out that this problem comes down to the following:
We want to find the answer to $11^{(345^{678})} \ \text{mod} \ 13$.
Then, because $11$ and $13$ are coprime, we know that $11^{\phi(13)} = 1 \
\text{mod} \ 13$. Because 13 is prime, we know that $\phi(13) = 12$. So
$11^{12x} = 1 \ \text{mod} \ 13 $ for any integer x.
Then we need to find how many times 12 fits into $345^{678}$, so $345^{678} \ \text{mod} \ 12 $. However, $345$ and $12$ are not coprime, so we can't use Euler's theorem. This is the part where I don't actually know what to do anymore.
I tried the following: 345 mod 12 = 9 mod 12, so $345^{678} \ \text{mod} \ 12 = 9^{678} \ \text{mod} \ 12$.
We know that $9^2 \ \text{mod} \ 12 = 81 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12$. And we know that $9^4 \ \text{mod} \ 12 = 9^2 \cdot 9^2 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12$. So $9^2$, $9^4$, $9^8$, $9^{16}$, $9^{32}$, $9^{64}$,$9^{128}$, $9^{256}$ and $9^{512}$ are all 9 in mod 12.
So then, $9^{678} \ \text{mod} \ 12 = 9^{512} \cdot 9^{128} \cdot 9^{32} \cdot 9^{4} \cdot 9^2 \ \text{mod} \ 12 = 9^5 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12 $.
Then, going back to the original question, we know that we need to calculate $11^{9} \ \text{mod} \ 13 = 11^8 \cdot 11 \ \text{mod} \ 13$.
Using the same trick as with the 9, I get that $11^{9} \ \text{mod} \ 13 = 8 \ \text{mod} \ 13$.
Is this correct? If not, what am I doing wrong? Is there a better way to attack such problems?
Thanks for reading,
K.
| As $11\equiv-2\pmod{13},$
$11^{(345^{678})}\equiv(-2)^{(345^{678})}\pmod{13}$
Now $(-2)^6\equiv-1\pmod{13}\implies(-2)^{12}\equiv1,$
and $(-2)^{(345^{678})}\equiv(-2)^{(345^{678}\pmod{12})}\pmod{13}$
Now as $345\equiv0\pmod3\implies345^{678}\equiv0\ \ \ \ (1)$
and $345\equiv1\pmod4\implies345^{678}\equiv1\ \ \ \ (2)$
Using $(1),(2)$ by observation, $345^{678}\equiv9\pmod{12}$
$(-2)^{(345^{678}\pmod{12})}\equiv(-2)^9\pmod{13}\equiv?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given a $2 \times 2$ matrix $A$ with $A^3=O$, find the sum of all the entries of matrix $A^2$
Given a matrix $$A=\begin{bmatrix} a & b\\c &d \end{bmatrix}$$ satisfying
$$A^3=O$$ find the sum of all the elements of matrix $A^2$.
My try:
Since $A$ is $2 \times 2$ matrix it can be expressed as
$$A^2=pA+qI$$ where $p$ and $q$ are some scalars. Then we have
$$A^3=pA^2+qA=O$$
So
$$p(pA+qI)+qA=O$$ that is
$$A=\frac{-pq}{p^2+q} I$$
So
$$A^2=\frac{p^2q^2}{(p^2+2q)^2} I$$
so sum of elements of $A^2$ is $\frac{2p^2q^2}{(p^2+2q)^2}$. But the answer is zero. Can I know my mistake?
| Taking a path filled with work the matrix
$$A=\left(\begin{matrix} a & b\\c & d \end{matrix}\right)$$
with the condition $A^3 = O_{2}$ leads to the set of equations
\begin{align}
a^3 + 2 a b c + b c d &=0 \\
c( a^2 + a d + b c + d^2 ) &= 0 \\
b( a^2 + ad + bc + d^2 ) &= 0 \\
a b c + 2 b c d + d^3 &= 0.
\end{align}
If $b=c=0$ then it is evident that $a^3 = d^3 = 0$. If $b\neq 0$, or equivalently $c \neq 0$ then the equations reduce to
\begin{align}
a^3 + 2 a b c + b c d &=0 \\
a^2 + a d + b c + d^2 &= 0 \\
a b c + 2 b c d + d^3 &= 0.
\end{align}
Some possible solutions can be then found to be: $(a,b,c,d) \in \{(0,b,0,0), (0,0,c,0), (a,-a,a,-a)\}$.
This then leads to the set of matrices being:
\begin{align}
A \in \left\{ \left(\begin{matrix} 0 & 0\\0 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & b\\0 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 0\\c & 0 \end{matrix}\right), \left(\begin{matrix} a & -a\\a & -a \end{matrix}\right) \right\}.
\end{align}
In this set the sum of the elements is then $\{0,b,c\}$. These matrices demonstrate that $A^2 = O_{2}$ and $A^3 = O_{2}$ and $tr(A) = 0$.
Another example may be seen as:
$$A=\left(\begin{matrix} a & b\\c & a \end{matrix}\right)$$
for which
$$A^3 = \left(\begin{matrix} a(a^2 + 3 b c) & b(3a^2 + c)\\c(3 a^2 + b) & a(a^2 + 3 b c) \end{matrix}\right).$$
Taking $b = c= \frac{1}{9}$ leads to
$$A^3 = \frac{1+ 27 a^2}{81} \, \left(\begin{matrix} 3a & 1\\1 & 3a \end{matrix}\right).$$
Now it is left to solve $1 + 27 a^2 = 0$, or $a = \pm i/(3 \sqrt{3})$ for which
\begin{align}
A \in \left\{ \frac{1}{9} \, \left(\begin{matrix} i \sqrt{3} & 1\\1 & i \sqrt{3} \end{matrix}\right), \frac{1}{9} \, \left(\begin{matrix} - i \sqrt{3} & 1\\1 & - i \sqrt{3} \end{matrix}\right) \right\}.
\end{align}
The sum of the elements is $\{\frac{2(1+i\sqrt{3})}{9}, \frac{2(1-i \sqrt{3})}{9}\}$, $tr(A) \in \{\pm \frac{2i \sqrt{3}}{9}\}$, $Det(A) = - \frac{4}{81}$.
From Show that a matrix... the equation, with $t=tr(A)$ and $d = Det|A|$,
$A^2 - t \, A + d \, I_{2} = 0_{2}$ as been established. Multiplying this by $A$ and using the equation to reduce the resulting equation yields
$$A^{3} + (d - t^2) \, A + t d \, I_{2} = 0_{2}.$$
Since $A^3 = 0_{2}$ then $(d - t^2) \, A = - t d \, I_{2}$, or
$$A_{2} = - \frac{t \, d}{d - t^{2}} \, I_{2}.$$
The question asks about the sum of the elements of $A$ for which, by this last equation, it can be seen that
$$A_{\text{sum}} = - \frac{2 \, t \, d}{d - t^2}.$$
In essence the matrix has been diagonalized and it can be stated that
$$t = - \frac{2 \, t \, d}{d - t^2}$$
which reduces to $t^{2} = 3 d$, ie $Tr(A_{2}) = 3 \, Det(A_{2})$. The matrices presented in this solution all satisfy this requirement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do you solve $\sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x+im^k}{m^{k+1}} \right \rfloor$? Let $m > 1$ be an integer, x is a real number and $f(x)=\sum_{k=0}^{\infty} \sum_{i=1}^{m-1} \left \lfloor \frac{x+im^k}{m^{k+1}} \right \rfloor$
Do you have any idea how can I show this?
$f(x)=
\begin{cases}
\left \lfloor x \right \rfloor & \text{if } x\geq 0\\
\left \lfloor x+1 \right \rfloor & \text{if } x<0
\end{cases}$
| The key to solving the sum is realising that for any $k$ you can write $x$ uniquely as $x = m^{k+1} q_{k+1} + m^k r_{k}$ with $q_{k+1} \in \mathbb{Z}$ and $r_k \in [0, m) \subset \mathbb{R}$.
You can think of this as writing $x$ in base $m$ and splitting into two numbers corresponding to the first couple of digits and the remaining digits. For example if $x=1234.56$ and $m=10$ then $q_2 = 12$ and $r_{1} = 3.456$. We also have that
$$m^{k+1} q_{k+1} + m^k r_k = m^k q_k + m^{k-1} r_{k-1} = m^{k+1} q_{k+1} + m^k ((q_k - m q_{k+1}) + r_{k-1} /m)$$ so $r_k = (q_k - m q_{k+1}) + r_{k-1}/m$ which implies $\lfloor r_k \rfloor = (q_k - m q_{k+1})$. Which makes sense if you think of the integral part of $r_k$ as the 'next digit'.
Motivation aside, this notation makes it possible to simplify the sum as follows:
$$
\begin{align}
\sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor
\frac{x+tm^k}{m^{k+1}}
\right \rfloor &=
\sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor
\frac{m^{k+1} q_{k+1} + m^{k} r_{k} + tm^k}{m^{k+1}}
\right \rfloor \\&=
\sum_{k=0}^{\infty} \sum_{t=1}^{m-1} \left \lfloor
q_{k+1} + \frac{(r_k +t)m^k}{m^{k+1}}
\right \rfloor \\&=
\sum_{k=0}^{\infty} \sum_{t=1}^{m-1}
q_{k+1} + \left \lfloor
\frac{r_k + t}{m}
\right \rfloor \\
&= \sum_{k=0}^{\infty} (m-1) q_{k+1} + \lfloor r_k \rfloor\\
&= \sum_{k=0}^{\infty} (m-1) q_{k+1} + (q_k - mq_{k+1})\\
&= \sum_{k=0}^{\infty} q_k - q_{k+1}
\end{align}
$$
Now note that
$$
\lim_{N\to\infty} \sum_{k=0}^{N} q_k - q_{k+1} = \lim_{N\to\infty} q_0 - q_N = q_0 = \begin{cases}
\left \lfloor x \right \rfloor & \text{if } x\geq 0\\
\left \lfloor x+1 \right \rfloor & \text{if } x<0
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Verify if $\sum(\sqrt{n+1}-\sqrt{n})$ is convergent or divergent
Verify if the series $\sum a_n$ with $a_n=\sqrt{n+1}-\sqrt{n}$ is
convergent or divergent.
What I did is
$$a_n=(\sqrt{n+1}-\sqrt{n})\times\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt
{n+1}+\sqrt{n}}$$
$$=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
$$<\frac{1}{2\sqrt{n}}=b_n$$
Since $b_n$ is monotone decreasing and $b_n\rightarrow 0$ when $n\rightarrow \infty$ then $b_n$ is convergent.
Using the comparison test, we have that $0\leq a_n\leq b_n$. If $b_n$ is convergent then $a_n$ is convergent.
Is it right?
| The series diverges. This can be seen using the fact that it is a "telescoping series".
The $k^\text{th}$ partial sum can be written as follows:
\begin{align}
S_k &= \displaystyle\sum_{n=1}^k \Big(\sqrt{n+1} - \sqrt{n}\Big)\\\\
&= \big(\sqrt{2} - \sqrt{1}\,\big) + \big(\sqrt{3} - \sqrt{2}\,\big) + \big(\sqrt{4} - \sqrt{3}\,\big) + \cdots +\big(\sqrt{k+1} - \sqrt{k}\,\big)\\\\
&=-\sqrt{1} + \big(\sqrt{2} -\sqrt{2}\,\big) + \big(\sqrt{3} -\sqrt{3}\,\big) + \big(\sqrt{4} - \cdots -\sqrt{k}\,\big) + \sqrt{k+1}\\\\
&=\sqrt{k+1} - 1\\
\end{align}
As $k$ goes to infinity, this diverges, so the infinite sum does not converge.
$$\displaystyle\sum_{n=1}^\infty \Big(\sqrt{n+1} - \sqrt{n}\Big) = \lim_{k\to\infty} S_k = \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proof of a logarithmic equation If \begin{align}\log_{16}{15} &= a\\
\log_{12}{18} &= b\\
\log_{25}{24} &= c\end{align}
then prove that $$c=\frac{5-b}{2(8a - 4ab -2b +1)}$$
My attempt: I tried to prove it by applying the standard logarithmic formulas such as $$\log_{a}{b} = \log_{c}{b}\times \log_{a}{c} $$ but the denominator is becoming too complex to perform an LCM of all the terms. Can there be a simpler way to prove this?
Any help is appreciated.
| Without using logs, purely by manipulation of indices:
$$\begin{align}
12^b&=18\\
2^{2b}\cdot 3^b&=2\cdot 3^2\\
2^{2b-1}&=3^{2-b}\\
3&=2^{\frac{2b-1}{2-b}}\tag{1}\\
\text{Also,}\hspace{3cm}\\
16^a&=15\\
2^{4a}&=3\cdot 5=2^{\frac{2b-1}{2-b}}\cdot 5
&&\scriptsize(\text{using}\;\; (1))\\
2^{4a-\frac {2b-1}{2-b}}&=5\\
2^{\frac {8a-4ab-2b+1}{2-b}}&=5\\
2&=5^\frac{2-b}{8a-4ab-2b+1}\tag{2}\\
\text{And finally, }\hspace{3cm}\\
25^c&=24\\
5^{2c}&=3\cdot 2^3\\
&=2^{\frac{2b-1}{2-b}}\cdot 2^3
&&\scriptsize(\text{using}\;\; (1))\\
&=2^{\frac {2b-1}{2-b}+3}\\
&=2^\frac{5-b}{2-b}\\
&=\left(5^\frac{2-b}{8a-4ab-2b+1}\right)^\frac{5-b}{2-b}
&&\scriptsize(\text{using}\;\; (2))\\\\
&=5^\frac{5-b}{8a-4ab-2b+1}\\
\text{Equating indices and dividing by $2$,}\\
\color{red}c&\color{red}{=\frac{5-b}{2(8a-4ab-2b+1)}\;\;\blacksquare}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Tricky simplification problem $\frac{ \sqrt{x^2+a^2}\,\sqrt{(c-x)^2+b^2}\,\sin(180-\arctan(\frac ax)-\arctan(\frac {b}{c-x}))}2 $ So when solving an optimising the area of a triangle problem (maximising the area), I found a formula for the area listed below:
$$\text{Area}=\frac{\sqrt{x^2+a^2}\,\sqrt{(c-x)^2+b^2}\,\sin\Bigl(180-\arctan\bigl(\frac ax\bigl)-\arctan\bigl(\frac {b}{c-x}\bigl)\Bigl)}{2}$$
As you can see, it is rather ugly and a pain to optimize for $x$ (when $a$, $b$ and $c$ are constant). I later came upon a much better and easier formula for the area, which was very simple and easy to optimize.
Afterwards, it occurred to me out the monster above and the easier formula for area I worked out below, were equivalent - but I have no idea how to simplify the one above into the one below, so explanations and steps are very much appreciated.
For those (crazy) people who want to try simplifying the above without knowing the end result, the answer below is hidden.
$$\text{Area}=\frac{ac-x(a-b)}{2}$$
| let $\tfrac{a}{x}=u$ and $\tfrac{b}{c-x}=v$. In the case of $uv<1:$ $arctg(u)+arctg(v)=arctg\tfrac{u+v}{1-uv}$
$$sin(arctgx)=\tfrac{x}{\sqrt{1+x^2}}$$
$$sin(arctg\tfrac{u+v}{1-uv})=\tfrac{u+v}{1-uv}\cdot\tfrac{1}{\sqrt{1+(\tfrac{u+v}{1-uv})^2}}=\tfrac{u+v}{\sqrt{1+(uv)^2+u^2+v^2}}=\tfrac{u+v}{\sqrt{(u^2+1)(v^2+1)}}$$
using $sin(\pi-\alpha)=sin\alpha$ and back to our substitution:
$$Area=\tfrac{1}{2}\sqrt{x^2+a^2}\cdot\sqrt{(c-x)^2+b^2}\cdot\tfrac{\tfrac{a}{x}+\tfrac{b}{c-x}}{\sqrt{\tfrac{a^2+x^2}{x^2}\cdot\tfrac{b^2+(c-x)^2}{(c-x)^2}}}=$$
$$=\tfrac{1}{2}\cdot(c-x)x(\tfrac{a}{x}+\tfrac{b}{c-x})=\tfrac{ac-x(a-b)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.
| $$a+b+c=6\tag{1}$$
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{47}{60}\tag{2}$$
$$S = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$
$S = \left(\frac{a+b+c}{b+c}-1\right)+\left(\frac{a+b+c}{c+a}-1\right)+\left(\frac{a+b+c}{a+b}-1\right) =6\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3=6\cdot\frac{47}{60}-3$
$S =\frac{47}{10}-3 =\frac{17}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 4
} |
prove that $ x-\frac{1}{6}x^3<\sin(x)
prove that :
there exists a deleted neighborhood of $x=0$
such that :
$$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5
$$
MyTry:
let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$
And :
$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$
Now what ?
| From $$3^k\,\sin\frac{x}{3^k}-3^{k-1}\,\sin\frac{x}{3^{k-1}}=4\cdot3^{k-1}\,\sin^3\frac{x}{3^k},$$ by summing the telescoping series,
$$x-\sin x=4\,\sum^\infty_{k=1}3^{k-1}\,\sin^3\frac{x}{3^k}.$$ Since $\sin x<x$ for $x>0$, the RHS is $\le x^3/6$.
BTW, the inequality for $\sin x$ can be shown in a similar way, because
$$\frac{\sin x}{x}=\prod^\infty_{k=1}\cos\frac{x}{2^k}<1$$ for $x\neq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find value of this Triple summation Find the value of $$S=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{i+j+k}}$$ where $i \ne j \ne k$
i have tried in this way:
First fix $j$ and $k$, then we get
$$S_1=\sum_{\substack{i=0 \\ i\neq j,k}}^{\infty}\frac{1}{3^{i+j+k}}$$ where $j \ne k$
So we get $$S_1=\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}$$
So now
$$S_2=\sum_{\substack{j=0 \\ j\neq k}}^{\infty}\left(\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}\right)$$
So $$S_2=\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)-\left(\frac{3}{2}\times \frac{1}{3^{2k}}\right)+\left(\frac{2}{3^{3k}}\right)$$
Finally
$$S=\sum_{k=0}^{\infty}\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)+\sum_{k=0}^{\infty}\frac{2}{3^{3k}}-\sum_{k=0}^{\infty}\frac{\frac{3}{2}}{3^{2k}}$$
we get
$$S=\frac{27}{16}-\frac{27}{16}+\frac{54}{26}-\frac{27}{16}$$ hence
$$S=\frac{81}{208}$$
is there any other approach like using integration etc
| For all $i,j,k$,
$$A=\sum_{i=0}^\infty\frac 1{3^i}=\frac 1{1-\frac 13}=\frac 32\\
\therefore A^3=\frac {27}8$$
For $i=j(=r)\neq k$,
$$B=\sum_{k=0}^\infty\frac 1{3^k}\sum_{r=0}^\infty \frac 1{3^{2r}}=\frac 32\cdot \frac 1{1-\frac 19}=\frac 32\cdot \frac 98=\frac {27}{16}$$
For $i=j=k(=r)$,
$$C=\sum_{r=0}^\infty \frac 1{3^{3r}}=\frac 1{1-\frac 1{27}}=\frac {27}{26}$$
The required sum,
$$\sum_{i,j,k=0\\i\neq j\neq k}^\infty\frac 1{3^{i+j+k}}=A^3-3B+2C=27\left(\frac 18-\frac 3{16}+\frac 1{13}\right)=\color{red}{\frac {81}{208}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Ramanujan and his problem Following was proposed by Ramanujan:
$ \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=1+4\sin(10^o)$
Working on this I got the radical on the left equal to $(1+2\sqrt{2})$ implying that $\sin(10^o)=1/\sqrt{2}$
How is this possible? What is wrong here?
| Inspired by Ramanujan's above formula, I added some similar results here:
$ $
$ 4 \cos \left(\frac{4 \pi }{9}\right)+1=\sqrt{11-2 \sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{4 \pi }{9}\right)+1\right)}}} $
$ 4 \cos \left(\frac{2 \pi }{9}\right)+1=\sqrt{11+2 \sqrt{11-2 \sqrt{11-2 \left(4 \cos \left(\frac{2 \pi }{9}\right)+1\right)}}} $
$ 4 \cos \left(\frac{\pi }{9}\right)-1=\sqrt{11-2 \sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{9}\right)-1\right)}}} $
$ $
$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{11+2 \sqrt{11-2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)}} $
$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{11-2 \sqrt{11+2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)}} $
$ $
$ 4 \cos \left(\frac{\pi }{6}\right)+1=\sqrt{11+2 \left(4 \cos \left(\frac{\pi }{6}\right)+1\right)} $
$ 4 \cos \left(\frac{\pi }{6}\right)-1=\sqrt{11-2 \left(4 \cos \left(\frac{\pi }{6}\right)-1\right)} $
$ $
$ 4 \cos \left(\frac{\pi }{4}\right)+1=\sqrt{7+2 \left(4 \cos \left(\frac{\pi }{4}\right)+1\right)} $
$ 4 \cos \left(\frac{\pi }{4}\right)-1=\sqrt{7-2 \left(4 \cos \left(\frac{\pi }{4}\right)-1\right)} $
$ $
Other similar results can be found here:
https://math.stackexchange.com/a/4233146/954936
https://math.stackexchange.com/a/4232525/954936
http://eslpower.org
http://eslpower.org/Notebook.htm
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum\limits_{k=0}^\infty n^{2^k}\prod\limits_{m=0}^k\frac1{n^{2^m}+1}$ I found this particular series in a book which asks to evaluate:
$$S=\lim_{k \to \infty}\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
which I believe is equal to
$${\lim_{k \to \infty}\sum_{j=0}^k\frac{n^{2^j}}{\prod_{m=0}^j(n^{2^m}+1)}}$$
So, I did it like this:
$$\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
$$=\frac{(n+1)-1}{n+1}+\frac{(n^2+1)-1}{(n+1)(n^2+1)}+\frac{(n^4+1)-1}{(n+1)(n^2+1)(n^4+1)}+...+\frac{(n^{2^k}+1)-1}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
Now breaking this gives
$$1-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{(n+1)(n^2+1)}+\frac{1}{(n+1)(n^2+1)}-\frac{1}{(n+1)(n^2+1)(n^4+1)}...+\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)}-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which telescopes to
$$S=1-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which gives $$\lim_{k \to \infty}S=1$$
Is this correct? If not, Could somebody help me in the right direction?
| What you've done is correct assuming $n\geq 1$ or $n<-1$ when $-1< n <1$ we can write $S$ as $$S=\lim_{k\to\infty}\left(1-\frac{n-1}{(n-1)(n+1)(n^2+1)\cdots(n^{2^k}+1)}\right)\\=\lim_{k\to\infty}\left(1-\frac{n-1}{n^{2^{k+1}}-1}\right)=1-\frac{n-1}{-1}=n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Null space of a hermitian matrix I have the following matrix:
$\begin{pmatrix}
4 &2-i &-3i \\
2+i &1 &1-i \\
3i &1+i &1
\end{pmatrix}$
The following vector generates its null space:
$\begin{pmatrix}
-1\\
1+2i\\
1
\end{pmatrix}$
When I tried to find the null space, I wrote the original system as a symmetric real matrix of size $6\times 6$ by separating my matrix into its real and imaginary components:
$\begin{pmatrix}
4 &2 &0 &0 &1 &3 \\
2 &1 &1 &-1 &0 &1 \\
0 &1 &1 &-3 &-1 &0 \\
0 &-1 &-3 &4 &2 &0 \\
1 &0 &1 &2 &1 &1 \\
3 &1 &0 &0 &1 &1
\end{pmatrix}$
The vector $\begin{pmatrix}
-1\\
1\\
1\\
0\\
2\\
0
\end{pmatrix}$ is supposed to be in its nullspace but wolframalpha tells me it's not:
https://www.wolframalpha.com/input/?i=%7B%7B4,2,0,0,1,3%7D,%7B2,1,1,-1,0,1%7D,%7B0,1,1,-3,-1,0%7D,%7B0,-1,-3,4,2,0%7D,%7B1,0,1,2,1,1%7D,%7B3,1,0,0,1,1%7D%7D*%7B-1,1,1,0,2,0%7D
I can't find where I went wrong. Could someone please help. Thanks in advance.
| Wolfram tells me that $\left( \begin{array}{c}
0 \\ -2 \\ 0 \\ -1 \\ 1 \\ 1 \end{array} \right)$
is in the null space of the $6 \times 6$ matrix you've done there, which is similar to yours, but with the halves switched, and the wrong sign on the two.
Are you sure you have the right $6 \times 6$ matrix? Since the null space uses left-multiplication:
$$
\left( \begin{array}{ccc}
4 & 2-i & -3i \\
2+i & 1 & 1-i \\
3i & 1+i & 1 \end{array} \right)
\left( \begin{array}{c}
x_1 \\ x_2 \\ x_3 \end{array} \right)
= \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)
$$
shouldn't the rows match your matrix and not the columns?
This would give you:
$$
\left( \begin{array}{cccccc}
4 & 2 & 0 & 0 & -1 & -3 \\
2 & 1 & 1 & 1 & 0 & -1 \\
0 & 1 & 1 & 3 & 1 & 0 \\
0 & 1 & 3 & 4 & 2 & 0 \\
-1 & 0 & 1 & 2 & 1 & 1 \\
-3 & -1 & 0 & 0 & 1 & 1 \end{array} \right)
$$
...and, in fact, a quick bash into Wolfram reveals
$\left( \begin{array}{c}
-1 \\ 1 \\ 1 \\ 0 \\ 2 \\ 0 \end{array} \right)$
to be in the null space.
I hope this is right, I have no idea what a hermitian matrix is, and I've never seen this approach used for anything before!
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c = 3$ show $9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2$
If $a, b, c$ are positive, and $a+b+c = 3$ show
$$
9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2
$$
This is yet another one of those cyclic symmetric inequalities in three positive variables, with cyclic symmetric constraints, but it has a twist: The inequality can be saturated (the equality holds) at $a=b=c=1$ but it can also be saturated at other values of $a,b,c$.
I did not get this from a contest problem, but it might make a good one at the moderate-skill high-school level.
| We know that $\cos(2x)=2\cos^2(x)-1$
$$\Rightarrow\sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=2\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]-(a+b+c)=2\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]-3$$
$$\Rightarrow9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=6\left [\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\right ]=2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)$$
And by using Cauchy-Schwarz
$$2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2$$
So,
$$9 + 3 \sum_{\mbox{cyc}}a\cos\left( \frac{2b}{c}\right)=2(a+b+c)\sum_{\mbox{cyc}}a\cos^2\left( \frac{b}{c}\right)\geq 2\left( \sum_{\mbox{cyc}}a\cos\left( \frac{b}{c}\right) \right)^2$$
| {
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"url": "https://math.stackexchange.com/questions/2320214",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove the following trigonometric identity: $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$ Prove:
$$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$$
My attempt:
LHS=
$$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$$
$$=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}$$
$$=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$
$$\text{Using componendo and dividendo}$$
$$\frac{\sin A+1-\cos A+\sin A-1+\cos A}{\sin A+1-\cos A-\sin A+1-\cos A}$$
$$\frac{2\sin A}{2-2\cos A}=\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}$$
$$\frac{\sin A(1+\cos A)}{\sin^2 A}=\frac{1+\cos A}{\sin A}$$
$$\text{Which is not equal to right hand side!}$$
Have I done something wrong in the componendo dividendo step?
I dont know how to use componendo dividendo rule . I saw it being used like this in some question and hence applied it here the same way. Maybe I am wrong in application of that rule. Please tell me the right way to use it.
Thank you.
| Note
\begin{eqnarray}
LHS&=&\frac{\tan A+\sec A-1}{\tan A-\sec A+1}\\
&=&\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}\\
&=&\frac{\sin A+1-\cos A}{\sin A-1+\cos A}\\
&=&\frac{\sin A+1-\cos A}{\sin A-1+\cos A}\cdot\frac{\sin A+1+\cos A}{\sin A+1+\cos A}\\
&=&\frac{(\sin A+1)^2-\cos^2 A}{(\sin A+\cos A)^2-1}\\
&=&\frac{\sin^2 A-\cos^2 A+2\sin A+1}{2\sin A\cos A}\\
&=&\frac{2\sin^2 A+2\sin A}{2\sin A\cos A}\\
&=&\frac{1+\sin A}{\cos A}\\
&=&RHS
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Rational Equation
I got this question and I'm stuck on one of the steps.
And this is where I'm stuck.I'm not sure what to do next. Do I multiply everything by 2(a-2)(a+2) over 1 or something entirely different?
| $$\frac{2}{(a-2)(a+2)} = \frac{1}{2(a-2)(a+2)} - \frac{1}{a+2}$$
You can arrange the equation so that all terms have a common denominator. In this case, the simplest such denominator will be $2(a-2)(a+2)$.
$$\frac{4}{2(a-2)(a+2)} = \frac{1}{2(a-2)(a+2)} - \frac{2(a-2)}{2(a-2)(a+2)}$$
Now, if $a \neq \pm 2$, we can multiply both sides by the denominator to get
$$4 = 1 - 2(a-2)$$
$$4 = 1 - 2a + 4$$
$$2a = 1$$
$$a = \boxed{\frac{1}{2}\,}$$
Since the $a = \pm 2$ possibilities would not work with your original equation, $1/2$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
why does $3^{10^n} \text{ mod } 10^{n+2} = 2 \cdot 10^{n+1}+1$? For integers n > 1 it seems that: $$3^{10^n} \text{ mod } 10^{n+2} = 2 \cdot 10^{n+1}+1$$
For example, the last 12 digits of $3^{10^{10}}$ are 200000000001.
I have verified that is true at least up to n = 250 and want a way to prove that it is true for all integers n > 1.
It also appears that many more of the last digits of $3^{10^n}$ are the same.
$3^{10^{10}} = \quad ...228442786552200000000001$
$3^{10^{20}} = \quad ...8084427865522000000000000000000001$
| You can try induction on $n$.
$$3^{10^n} = k_n10^{n+2}+2\cdot 10^n+1$$
for some integer $k_n$.
It's true for $n=2$, by your calculation. Now assume true for some $n\geq 2$. Then applying the binomial theorem:
Then $$3^{10^{n+1}}=\left(3^{10^n}\right)^{10}=(k_n10^{n+2}+2\cdot 10^n +1)^{10}= C10^{n+3}+(2\cdot 10^n+1)^{10}$$
But $$(2\cdot 10^n+1)^{10}=1+\binom{10}{1}2\cdot 10^n+ \binom{10}{2}2^2\cdot (10^{2n})+\cdots=1+2\cdot 10^{n+1}+C_210^{n+3}$$
Basically, when $n\geq 2$, then $10^{n+3}\mid \binom{10}{i}2^{i}10^{ni}$ for $i\geq 2$. This is because when $i>2$, then $ni\geq n+3$ and when $i=2$, $ni\geq n+2$ and $10\mid\binom{10}{2}\cdot 2^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $n$ be any positive integer and let $x \in (0, \pi)$. Prove that $\sin x+ \frac{\sin 3x}{3} + \cdots +\frac{\sin(2n-1)x}{2n-1}$ is positive. I came across the following problem in a high school calculus exam paper. I do have the solution, but it took me quite a while to work it out and I think it's very clumsy. I'm curious to know if there is a simpler solution.
Problem
Let $n$ be any positive integer and let $x \in (0, \pi)$. Prove that
$$\sin{x} + \frac{\sin{3x}}{3} + \frac{\sin{5x}}{5} + \cdots +\frac{\sin{(2n-1)x}}{2n-1}$$ is positive.
| Let the required sum be $S(x)$,
$$\begin{align*}
S(x) &= \sum_{k=1}^{n} \frac{\sin(2k-1)x}{2k-1}\\
S'(x) &= \sum_{k=1}^{n} \cos(2k-1)x\\
&= \Re \left[\sum_{k=1}^ne^{i(2k-1)x}\right]\\
&= \Re \left[e^{ix}\cdot\frac{1-e^{i2nx}}{1-e^{i2x}}\right]\\
&= \Re \left[e^{inx} \cdot \frac{e^{-inx} -e^{inx}}{e^{-ix} - e^{ix}}\right]\\
&= \Re\left[(\cos nx +i\sin nx)\cdot\frac{\sin nx}{\sin x}\right]\\
&= \frac{\sin 2nx}{2\sin x}
\end{align*}$$
Since $S(x) = S(\pi-x)$, it is sufficient to consider only $x \in (0,2\pi/4]$.
In the range $x\in(0,2\pi/4]$, $S'(x)$ has the same sign as $\sin 2nx$. Local minima of $S(x)$ (including the boundary case $x=0$) are at
$$\begin{align*}
2nx &= 2\pi k\\
x &= \frac{m}{2n}\cdot 2\pi, &&0\le m\le \frac n2,\quad m\in \mathbb N
\end{align*}$$
The difference between two adjacent local minima is
$$\begin{align*}
D &= S\left(\frac{m}{2n}\cdot 2\pi\right) - S\left(\frac{m-1}{2n}\cdot 2\pi\right)\\
&= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}S'(x)\ dx\\
&= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx\\
&= \int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx + \int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\frac{\sin 2nx}{2\sin x}dx\\
\end{align*}$$
The numerator $\sin 2nx$ is positive in the first integral, and negative in the second integral. Using the fact that $\sin x$ is increasing in $x\in(0,2\pi/4]$,
$$\begin{align*}
D &> \frac{\int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin \frac{m-1/2}{2n}\cdot 2\pi} + \frac{\int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin\frac{m-1/2}{2n}\cdot 2\pi}\\
&= \frac{\int_{\frac{m-1}{2n}\cdot 2\pi}^{\frac{m-1/2}{2n}\cdot 2\pi}\sin 2nx\ dx + \int_{\frac{m-1/2}{2n}\cdot 2\pi}^{\frac{m}{2n}\cdot 2\pi}\sin 2nx\ dx}{2\sin\frac{m-1/2}{2n}\cdot 2\pi}\\
&= 0
\end{align*}$$
In other words, the values of $S$ at $x = 0, \frac{2\pi}{2n}, \frac{2\cdot 2\pi}{2n}, \cdots, \frac{\lfloor n/2\rfloor\cdot 2\pi}2$ are strictly increasing compared with the previous one, so they are all greater than $S(0) = 0$. Also, since these $x$'s are local minima, every other $x$ between them will also be greater than $S(0) = 0$.
| {
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"question_score": "5",
"answer_count": 1,
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Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| if the square side is x, then the width of area 2 is 4/x and the reminder of that side is (x^2-4)/x. Similiarly the width of the 4 area is 8/x, and the other side of the 3 area is (x^2-8)/x.
Then from the 3 area (x^2-4)/x * (x^2-8)/x = 3*2. Let A = x^2, then (A-4)(A-8) = 6A. A^2 -18A +32 = 0 ... A = 16, and blue = 7
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 5
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What is my mistake in this integral: $\int \tan^5(x) dx$? I'm doing an exercise from Stewart's Calculus textbook in which I have to evaluate the following integral:
$\int \tan^5(x) dx$
I start by rewriting the integral this way:
$\int \tan^5(x) dx = \int \frac{\sin^5(x)}{\cos^5(x)} dx = \int \frac{\sin^4(x)}{\cos^5(x)} \sin(x) dx$
And here I make a substitution:
$u = \cos(x)$
$-du = \sin(x)dx$
So the integral with the substitution becomes:
$\int \frac{(1-u^2)^2}{u^5} (-1) du$
$\int \frac{(1-2u^2+u^4)}{u^5} (-1) du$
$\int \frac{(2u^2-u^4-1)}{u^5} du$
$\int \frac{(2u^2)}{u^5} - \frac{(u^4)}{u^5} - \frac{1}{u^5} du = 2 \frac{u^{-2}}{(-2)} - \ln\vert u \vert - \frac{u^{-4}}{(-4)} + constant$
$= \frac{-1}{u^2} - \ln\vert u \vert + \frac{1}{4u^4} + constant $
$= \frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant $
This antiderivative seems to be correct, because if I differentiate it I get:
$\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant] $
$-(-2)\cos^{-3}(x)(-\sin(x))-\frac{1}{\cos(x)}(-\sin(x))+\frac{1}{4}(-4)\cos^{-5}(x)(-\sin(x))$
$ \frac{-2\sin(x)}{\cos^3(x)} +\tan(x)+\frac{\sin(x)}{\cos^5(x)}$
$\tan(x)[1-\frac{2}{\cos^2(x)} + \frac{1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-2\cos^2(x)+1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+1-\cos^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(\cos^2(x)-1)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(-\sin^2(x))+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(1-\cos^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(\sin^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^4(x)}{\cos^4(x)}] = \tan^5(x)$
However, when I graph $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $ and $\tan^5(x)$ at a first glance they look like they are the same, but when I start to zoom in I notice that these curves are actually not equal:
The magenta curve is $\tan^5(x)$ and the black curve is $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $
The answer that Stewart provides at the end of the book is the following:
$ \int \tan^5(x) dx = \frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C$
Which is confusing me even more, because if I try to differentiate his answer I get:
$\frac{d}{dx}[\frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln\vert1/\cos(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln(1) - \ln\vert\cos(x)\vert+ C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)- \ln\vert\cos(x)\vert+ C]$
$= \frac{1}{4}(-4)\cos^{-5}(x)-2\tan(x)\sec^2(x)-\frac{1}{\cos(x)}(-\sin(x))$
$= \frac{-1}{\cos^5(x)}-2\tan(x)\sec^2(x)+\tan(x)$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-2\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-\sec^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(-\tan^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}-\tan^3(x)-\tan(x)\sec^2(x)$
$= \frac{-1}{\cos^5(x)}-\frac{\sin^3(x)}{\cos^3(x)}-\frac{\sin(x)}{\cos^3(x)}$
$=\frac{-1-\cos^2(x)\sin^3(x)-\cos^2(x)\sin(x)}{\cos^5(x)}$
$=\frac{-1-(1-\sin^2(x))\sin^3(x)-(1-\sin^2(x))\sin(x)}{\cos^5(x)}$
$=\frac{-1-(\sin^3(x)-\sin^5(x))-(\sin(x)-\sin^3(x))}{\cos^5(x)}$
$=\frac{-1-\sin^3(x)+\sin^5(x)-\sin(x)+\sin^3(x)}{\cos^5(x)}$
$=\frac{-1-\sin(x)+\sin^5(x)}{\cos^5(x)}$
$=\tan^5(x)-\frac{(1+\sin(x))}{\cos^5(x)}$
If I plot the derivative of Stewart's answer along the derivative of my answer and the original function $\tan^5(x)$ this is how the the graph looks like:
*
*Magenta curve: $\tan^5(x)$
*Black curve: $\frac{d}{dx}[\text{my answer}]$
*Blue curve: $\frac{d}{dx}[\text{Stewart's answer}]$
What's going on? Is my answer correct? If it isn't, what am I doing wrong? And what about Stewart's answer? Am I making a mistake when I differentiate it or is it correct?
Thanks for your help!
| To double check:
Notice that $$\tan'x=\tan^2x+1.$$
Then
$$\int\tan^5x\,dx=\int((\tan^2x+1)\tan^3x-(\tan^2x+1)\tan x+\tan x)dx\\
=\frac{\tan^4x}4-\frac{\tan^2x}2-\log|\cos x|.$$
In terms of the secant (and dropping the constant),
$$\frac{(\sec^2x-1)^2}4-\frac{\sec^2x-1}2+\log|\sec x|=\frac{\sec^4x}4-\sec^2x+\log|\sec x|.$$
You can also replace the second term $\sec^2x$ by $\tan^2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Three different result for the same indefinite integral I was killing time solving some indefinite integrals, when I found this one:
\begin{equation}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x
\tag{1}\label{integral}
\end{equation}
Not a particularly difficult one, I'll post my solution here:
\begin{equation}
\begin{split}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x = \int\frac{1}{(x+1)\sqrt{(x+1)^2-1}}\ \mathrm{d}x
\end{split}
\tag{2}\label{calculus}
\end{equation}
by substituting $t = x+1$, d$t = \mathrm{d}x$, and then $s = \sqrt{t^2-1} \rightarrow \mathrm{d}s = \frac{t}{\sqrt{t^2-1}}\mathrm{d}t$
\begin{equation}
\begin{split}
= \int\frac{1}{t\sqrt{t^2-1}}\ \mathrm{d}t &= \int\frac{1}{(x+1)\sqrt{(x+1)^2-1}}\ \mathrm{d}x =\\
&= \int\frac{\mathrm{d}s}{1+s^2} =\\
&= \arctan{s} + \mathrm{cost}\\
&= \arctan{\sqrt{t^2-1}} + \mathrm{cost}\\
&= \arctan{\sqrt{\left(x+1\right)^2-1}} + \mathrm{cost}
\end{split}
\tag{3}\label{calculus2}
\end{equation}
I then derived (I recommend, if you wanna check my results to use this online derivative calculator which actually shows steps...) my solution finding the starting function:
$$\frac{\mathrm{d}}{\mathrm{d}x}(\arctan{\sqrt{(x+1)^2-1}} + \mathrm{cost}) = \frac{1}{(x+1)\sqrt{x^2+2x}}$$
The graph (plotted with Grapher from Mac Os X), with $\color{red}{\text{function}}$ and $\color{blue}{\text{integral}}$ :
Now this integral comes from the exercise book Problems in Mathematical Analysis by Boris Demidovich and it's the number 1271, and even if I'm pretty sure this mine is the correct solution I lost quite time to understand the proposed solution of the book, which, if you don't have it and you can't check for yourself, is
1271. $\qquad-\arcsin(\frac{1}{1+x})$
deriving this function you'll find:
\begin{equation}
\begin{split}
\frac{\mathrm{d}}{\mathrm{d}x}&\left[-\arcsin(\frac{1}{1+x})\right] =\\
&= \dfrac{1}{\left(x+1\right)^2\sqrt{1-\frac{1}{\left(x+1\right)^2}}}\\
&= \dfrac{1}{\frac{\left(x+1\right)^2}{\sqrt{(x+1)^2}}\sqrt{\left(x+1\right)^2-1}}\\
&= \dfrac{1}{\frac{\left(x+1\right)^2}{|x+1|}\sqrt{\left(x+1\right)^2-1}}\\
&= \dfrac{1}{{\left(x+1\right)}\mathrm{sgn}(x+1)\sqrt{\left(x+1\right)^2-1}}\\
\end{split}
\tag{4}\label{calculus3}
\end{equation}
Plotting the result will give you an idea about the mistakte he could have done, in cyan the $\color{cyan}{Demidovich's~~primitive}$:
So I was pretty sure I was right and he was not, so I tried integrate the \eqref{integral} with Mathematica, with another unsatisfying outcome:
\begin{equation}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x = \frac{\sqrt{2}\sqrt{x}\sqrt{x+2}\arctan(\sqrt{\frac{x}{x+2}})}{\sqrt{x(x+2)}}
\end{equation}
This solution, in $\color{orange}{orange}$, is almost like mine, even if it does not comprehend the negative values of the function...
Also, if I try to derive (I've done it with calculator, as it's quite long to do for yourself...) you get
$$\left(\frac{\sqrt{2}\sqrt{x}\sqrt{x+2}\arctan(\sqrt{\frac{x}{x+2}})}{\sqrt{x(x+2)}}\right)' = $$
$$-\dfrac{\left(\sqrt{x+2}\left(-x^\frac{7}{2}-3x^\frac{5}{2}-2x^\frac{3}{2}\right)+\left(x+2\right)^\frac{3}{2}\left(x^\frac{5}{2}+x^\frac{3}{2}\right)\right)\arctan\left(\frac{\sqrt{x}}{\sqrt{x+2}}\right)-x^3-4x^2-4x}{\left(x\left(x+2\right)\right)^\frac{3}{2}\left(x^2+3x+2\right)}$$
Now, since I've found that in my range of definition $x^2-2x>0$ the primitive function that i found, derived, gives me the starting function, this should tell me that I am right and others (computer, this case) have bugs or they encounter problems deriving such a function. So which of the three solution $\color{blue}{mine}$, $\color{orange}{Mathematica's}$ or $\color{cyan}{Demidovich's}$ is the correct one? Why are them wrong, if they are? Does it depend on calculator's bug or it's my problem?
Thanks for attention.
| Noting that $\sqrt{x^2+2x}=\sqrt{(x+1)^2-1}$, it seem sensible to try $x+1=\sec(u)$:
$$
\begin{align}
\int\frac1{(x+1)\sqrt{x^2+2x}}\,\mathrm{d}x
&=\int\frac1{\sec(u)\tan(u)}\sec(u)\tan(u)\,\mathrm{d}u\\
&=u+C\\[8pt]
&=\sec^{-1}(x+1)+C
\end{align}
$$
Of course, the branch of $\sec^{-1}$ needs to be chosen to be in the third quadrant when its argument is negative (since $\sec(u)=x+1\le0$ and $\tan(u)=\sqrt{x^2+2x}\ge0$). If we want to use the standard choice for $\sec^{-1}$ for negative arguments, we can alter the answer to be
$$
\int\frac1{(x+1)\sqrt{x^2+2x}}\,\mathrm{d}x=\operatorname{sgn}(x+1)\sec^{-1}(x+1)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
all possible Taylor's and Laurent series of $\frac{2z-3}{z^2-3z+2} $ about $z=0$ Find all possible Taylor's and Laurent series of $\frac{2z-3}{z^2-3z+2} $ about $z=0$
I solved it as
$$\frac{2z-3}{z^2-3z+2} = \frac{1}{z-2} + \frac{1}{z-1}$$
$$= \frac{-1}{2}{(1+\frac{z}{2}+\frac{z^2}{4}+.....)} - (1+z+z^2+z^3)$$
Is this answer correct ? Is the above series both Taylor as well as Laurent ?
| Your result is only for $|z|<1$ and you are missing two other cases.
To expand $$\frac{2z-3}{z^2-3z+2}=-\left(\frac{1}{1-z} +\frac{1}{2-z}\right)$$
there are two singularities at $z=1$ and $z=2$. So we can have three different cases.
It is all about the region of convergence (RoC) of the geometric series, since $$\displaystyle\sum_{n=a}^\infty r^n=\frac{r^a}{1-r},\color{red} {\text{when } |r|<1}$$
*
*If $|z|<1$, then
$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$
$$\frac{1}{2-z}=\frac{1/2}{1-z/2}=\frac12\sum_{n=0}^{\infty}\left(\frac z2\right)^n$$
*
*If $1<|z|<2$, then
$$\frac{1}{1-z}=-\frac{1/z}{1-1/z}=-\sum_{n=1}^{\infty}\left(\frac 1z\right)^n$$
$$\frac{1}{2-z}=\frac{1/2}{1-z/2}=\frac12\sum_{n=0}^{\infty}\left(\frac z2\right)^n$$
*
*If $|z|>2$, then
$$\frac{1}{1-z}=-\frac{1/z}{1-1/z}=-\sum_{n=1}^{\infty}\left(\frac 1z\right)^n$$
$$\frac{1}{2-z}=-\frac{1}{2}\cdot\frac{2/z}{1-2/z}=-\frac12\sum_{n=1}^{\infty}\left(\frac 2z\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of power of Markov matrix Recently, a friend gave me the following problem:
Let $$M = \begin{pmatrix} p & q & r \\ q & r & p \\ r & p & q \end{pmatrix}$$ where $p, q, r > 0$ and $p + q + r = 1$. Prove that $$\lim_{n \rightarrow \infty} M^n = \begin{pmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{pmatrix}$$
Some of my observations include that $M^{2^n}$ remains cyclic, so I tried to bound (with no luck) the maximum and minimum to see if they both converged to $1/3$.
It would be nice if someone could post an elementary solution to this problem.
| Let
$$ \sigma = p^2 + q^2 + r^2 - qr - rp - pq = \frac{1}{2} \left( (q-r)^2 + (r-p)^2 + (p-q)^2 \right) \geq 0. $$
Note $ \sigma = (p+q+r) \sigma = p^3 + q^3 + r^3 - 3 pqr.$
Which is why the characteristic polynomial of your matrix is
$$ \lambda^3 - \lambda^2 - \sigma \lambda + \sigma = (\lambda - 1)(\lambda^2 - \sigma) $$
On the other hand
$$ 1 = p^2 + q^2 + r^2 +2 qr +2 rp +2 pq, $$ so
$$ \sigma = 1 - 3 (qr + rp + pq) < 1. $$
The eigenvalues are $1, \sqrt \sigma, -\sqrt \sigma.$ The diagonalized matrix $ D = P^T MP$ (where $P$ is orthogonal) raised to an arbitrarily large power, converges to the diagonal matrix with entries $1,0,0.$ Reverse the diagonalization, we get a symmetric rank one matrix.. give me a minute
Where was I?
$$
P =
\left(
\begin{array}{rr}
\frac{1}{\sqrt 3} & a & b \\
\frac{1}{\sqrt 3} & c & d \\
\frac{1}{\sqrt 3} & e & f
\end{array}
\right)
$$
$$
D_\infty =
\left(
\begin{array}{rr}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}
\right)
$$
We then just multiply out
$$
M_\infty = P D_\infty P^T =
\left(
\begin{array}{rr}
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3} \\
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3} \\
\frac{1}{ 3} & \frac{1}{ 3} & \frac{1}{ 3}
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 0
} |
Vectors in a rectangle Let $ABCD$ be a rectangle , $K$ a point on $AC$ such that $BK$ is perpendicular to $AC$, and $M$, $N$ are the midpoints of $AK, CD$. Use vectors to show that $MB$ is orthogonal to $MN$.
I know that if $BK$ is perpendicular to $AC$ that the dot product is $0$. I also know that I can write the magnitude of vectors in terms of other vectors. So what I keep getting to is:
$\vec{MB}=\vec{MC}+\vec{CB}=\vec{MA}+\vec{AB}=\vec{MA}+2\vec{NC}$
$\vec{MN}=\vec{MC}-\vec{NC}$
$\vec{BK}=-\vec{AB}+2\vec{AM}$
$\vec{AC}=\vec{BC}-\vec{AB}$
I can not find a connection between these vectors to show that the dot product of $\vec{MB}$ and $\vec{MN}$ equals $0$.
| Let $AB=a$, $AD=b$, $\vec{AB}=\vec{u}$ and $\vec{AD}=\vec{v}$.
Thus,
$$\vec{BM}=-\vec{u}+\frac{AK}{2AC}\cdot\vec{AC}=$$
$$=-\vec{u}+\frac{a^2}{2(a^2+b^2)}(\vec{u}+\vec{v})=-\frac{a^2+2b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2}{2(a^2+b^2)}\vec{v}$$ and
$$\vec{MN}=-\frac{a^2}{2(a^2+b^2)}(\vec{u}+\vec{v})+v+\frac{1}{2}\vec{u}=\frac{b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2+2b^2}{2(a^2+b^2)}\vec{v}.$$
Id est, $$\vec{BM}\cdot\vec{MN}=\left(-\frac{a^2+2b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2}{2(a^2+b^2)}\vec{v}\right)\left(\frac{b^2}{2(a^2+b^2)}\vec{u}+\frac{a^2+2b^2}{2(a^2+b^2)}\vec{v}\right)=$$
$$=-\frac{(a^2+2b^2)a^2b^2}{4(a^2+b^2)^2}+\frac{(a^2+2b^2)a^2b^2}{4(a^2+b^2)^2}=0$$
and we are done!
There is a nice proof without vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $I=\int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$? This integral popped up when I was trying to solve this integral : $$\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx $$
I simplified it a little bit, substituted $\tan(x)=u$ and came up with,
$$I= \int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$$
any suggestions for how I can take it from here ?
| For the fun of it :-)
\begin{align}
4(\cos^4x+2\sin^4x)+2\sin^2 2x&=4(1-\sin^2x)^2+8\sin^4x+2(2\sin x \cos x)^2\\
&=4-8\sin^2x+4\sin^4x+8\sin^4x+8\sin^2 x (1-\sin^2x)\\
&=4+4\sin^4x
\end{align}
Hence
\begin{align}
I&=\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx\\
&=\frac14\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{1+\sin^4x}\,dx\\
&=\frac12\int_{0}^{\pi/2}\frac{\sin^2x}{1+\sin^4x}\,dx\\
&=\frac12\int_{0}^{\pi/2}\sin^2x\sum_{j=0}^{\infty}(-1)^j\sin^{4j}x\,dx\\
&=\frac12\sum_{j=0}^{\infty}(-1)^j\int_{0}^{\pi/2}\sin^{2\times\frac{4j+3}{2}-1}x\cos^{2\times\frac12-1}x\,dx\\
&=\frac14\sum_{j=0}^{\infty}(-1)^j \text{Beta}\left(\frac{4j+3}{2},\frac12\right)\\
&=\frac14\sum_{j=0}^{\infty}(-1)^j \frac{\Gamma(\frac12)\Gamma(2j+\frac32)}{\Gamma(2j+2)}\\
\end{align}
I am not sure how to pin the last down. I will be happy if someone would share thoughts on this :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Area under the graph - integration
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .
Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?
| Alternatively, note that $Q=2P \Rightarrow P+Q=3P$. Make up the equation:
\begin{align}\int _{ 0 }^{ 2a } \left( 3x-{ x }^{ 2 } \right) \text dx &=3\int _{ 0 }^{ a } \left( 3x-{ x }^{ 2 } \right) \text dx \\
\left.\left(\frac{3x^2}{2}-\frac{x^3}{3}\right)\right|_{0}^{2a}&=3\left.\left(\frac{3x^2}{2}-\frac{x^3}{3}\right)\right|_{0}^{a}\\
\left(\frac{12a^2}{2}-\frac{8a^3}{3}\right)&=3\left(\frac{3a^2}{2}-\frac{a^3}{3}\right)\\
\frac{36a^2-16a^3}{6}&=\frac{27a^2-6a^3}{6}\\
10a^3-9a^2&=0\\
a^2(10a-9)&=0 \stackrel{a\ne0}\Rightarrow 10a-9=0 \Rightarrow a=\fbox{$\frac{9}{10}$}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
An infinite series for $\ln 2$ and taking limits of summations Example $1$.$$\tfrac 14\text{Log }2=\sum\limits_{k=1}^\infty\frac 1{\left\{2(2k-1)\right\}^3-2(2k-1)}\tag1$$
Proof. The right side above is equal to $\tfrac 12\{\varphi(2)-\varphi(4)\}$. Hence, the result follows from the two identities$$\text{Log }2=\tfrac 12\varphi(2)$$$$\tfrac 32\text{Log }2=\varphi(4)$$
Question: How do you prove $(1)$?
I started off with $\varphi(2,n)$ and $\varphi(4,n)$; subtracted them, and tried to show that together, the limit was equal to $\tfrac 12\text{Log }2$. However, this is all I have$$\begin{align*}\varphi(2,n)-\varphi(4,n) & =\sum\limits_{k=1}^n\left\{\frac 1{2k-1}+\frac 1{2k+1}-\frac 1k\right\}-\sum\limits_{k=1}^n\left\{\frac 1{4k-1}+\frac 1{4k+1}-\frac 1{2k}\right\}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1k-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}+\sum\limits_{k=1}^n\frac 1{2k}\\ & \\ & =\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}-\sum\limits_{k=1}^n\frac 1{4k-1}-\sum\limits_{k=1}^n\frac 1{4k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k\end{align*}$$However, it's after this step that I'm not sure what to do. If you split up twice the RHS of $(1)$, you get a $1/(4k-3)$ term, a $1/(4k-1)$ term and a $1/(2k-1)$ term. I'm puzzled on how they got there.
Also, I should probably add the definitions of the notations$$\lim\limits_{n\to\infty}\varphi(a,n)=\varphi(a)$$$$\varphi(a,n)=1+2\sum\limits_{k=1}^n\left\{\frac 1{(ak)^3-ak}\right\}$$Where $a$ is greater than one.
Sorry for the increase in summation questions, I'm analyzing Ramanujan's works on Harmonic Series and Inverse Tangent Functions.
| Proof:
identity: $$\frac{1}{x^3 -x} = \frac{1}{2(x-1)} + \frac{1}{2(x+1)} -\frac{1}{x}$$
$$\sum\limits_{k=1}^n\frac 1{\{2(2k-1)\}^3-2(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-3}+\frac{1}{2}\sum\limits_{k=1}^n\frac 1{4k-1}-\frac{1}{2}\sum\limits_{k=1}^n\frac 1{2k-1}$$
Let RHS = $\frac{y}{2}$
The first two terms of $y$ above give the first two terms below and last term of $y$ above gives the last term below.
$$y= \sum\limits_{k=1}^{4n}\frac{1}{k} -
\sum\limits_{k=1}^{2n}\frac{1}{2k} -\left(\sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n}\frac{1}{2k}\right)$$
$$y= \left(\sum\limits_{k=1}^{4n}\frac{1}{k} - \log 4n\right) +\log 4n - \frac{1}{2}\left(\sum\limits_{k=1}^{2n}\frac{1}{k} -\log2n\right) - (1/2)\log2n -\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\log 2n\right) - \log 2n + \frac{1}{2}\left(\sum\limits_{k=1}^{n}\frac{1}{k} - \log n\right) + (1/2)\log n$$
As $n \to \infty$, each of the expressions in parentheses reduces to Euler's constant and cancel out. The remaining terms are:
$$ \log 4n - \log 2n - (1/2)\left(\log 2n - \log n\right) = \log 2 - (1/2) \log 2 = (1/2) \log 2$$
Hence original sum is $(1/4) \log 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$
| Let $y=\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)$. Then
$$y=\log _{\sqrt{2}}\left((x-2)^{\frac {1}{2}}\right)$$
$$\left( \sqrt {2} \right) ^y = (x-2)^{\frac {1}{2}}$$
$$\left( \left( \sqrt {2} \right) ^y \right) ^2= \left( (x-2)^{\frac {1}{2}} \right) ^2$$
$$\left( \sqrt {2} \right) ^{2y} = x-2$$
$$2^{y} = x-2$$
$$y=\log_2 (x-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Maximize the volume of a block inscribed in hemisphere I want to find the maximum volume block with dimensions $a$, $b$ and $c$ incribed inside the hemisphere with radius $R$. I need to use lagrange multipliers. So far I found out that:
$R^2 = c^2 + \frac{a^2}{4} + \frac{b^2}{4}$
$V_{block} = abc$
$V_{hemisphere} = \frac{4}{6}\pi R^3 = \frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$
The equation for $R$ is taken from this picture
From here
$L = abc + \lambda\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$
And now I have folowing equations:
$L_a = bc + \lambda\pi \frac{a}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$L_b = ac + \lambda\pi \frac{b}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$L_c = ac + \lambda\pi 2c(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2} = 0$
Which seems unsolveable for me. Did I make mistake somewhere?
Thx for help.
| The formula for $V_{\textrm{hemisphere}}$ is unnecessary---this information is already encoded in the radius $R$ that appears in the constraint. In any case you're aiming to maximize the quantity
$$V = a b c$$
subject to the condition $$R^2 = \frac{a^2}{4} + \frac{b^2}{4} + c^2 .$$
Note that by parameterizing our blocks in a slightly different way, namely with $a = 2 A$, $b = 2 B$, $c = C$, we get the similar problem of trying to maximize the quantity
$$V = 4 A B C$$
subject to the condition $$R^2 = A^2 + B^2 + C^2 .$$
This formulation has the advantage of being symmetric in $A, B, C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$
The solution set of the inequality $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ is.
My attempts:
For $x>0\implies 3^x-4^x<0$
$\implies \dfrac{\ln (x+2)}{x^2-3x-4}\geq 0$. Now root of numerator is $x=-1$ and of denominator is $x=4,-1$
By wavy curve method:
$\implies x\epsilon(4,\infty)$
Now for $x\leq 0\implies 3^x-4^x\geq 0$
Hence:
$\dfrac{\ln (x+2)}{x^2-3x-4}\leq 0$
$\implies x\epsilon(-1,0]$
My answer $(-1,0]\cup (4,\infty)$
But book's answer said I missed one interval, please help.
I observed something that $-1$ are common roots in both numerator and denominator, so should we put $-ve$ on LHS of $-1$? If we do this then my answer will be correct, but don't know whether it's correct thinking or not, if it is then why?
| For $x\in (-2,-1)$ we get
\begin{align*}
3^x-4^x&>0\\
\ln(x+2)&<0\\
x^2-3x-4=\left(x+1\right)\left(x-4\right)&>0
\end{align*}
Then
$$x\in (-2,-1)\quad\implies \quad \frac{(3^x-4^2)\ln(x+2)}{x^2-3x-4}<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
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