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How to solve this kind of equation? I have an equation (in my homework) of the form
$a=\sqrt{x^2 + b^2} + \sqrt{x^2 + c^2}$
which I would like to solve for $x$. I am not sure how best to proceed. My thought is to square both sides of the equation, which gives me
$a^2=b^2+c^2+2x^2+\sqrt{(b^2+x^2)\times(c^2+x^2)}$
and then
$a^2=b^2+c^2+2x^2+\sqrt{b^2c^2 + b^2x^2 + c^2x^2 + x^4}$
but I am not sure what to do with the square root in order to eventually solve for $x$. How do I finish solving this, or am going the wrong way?
| Suppose that our equation
$$\sqrt{x^2+b^2}+\sqrt{x^2+c^2}=a.\tag{1}$$
holds. Assume $a\ne 0$ and $b\ne c$. Flip both sides over. We get
$$\frac{1}{\sqrt{x^2+b^2}+\sqrt{x^2+c^2}}=\frac{1}{a}.$$
Multiply top and bottom on the left by $\sqrt{x^2+b^2}-\sqrt{x^2+c^2}$. We get
$$\frac{\sqrt{x^2+b^2}-\sqrt{x^2+c^2}}{b^2-c^2}=\frac{1}{a}.$$
Conclude that
$$\sqrt{x^2+b^2}-\sqrt{x^2+c^2}=\frac{b^2-c^2}{a}.\tag{2}$$
Add (1) and (2). Now we know $2\sqrt{x^2+b^2}$, and the rest is downhill.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum likelihood to throw exactly two 6s One throws a dice $n$ times. For which value of $n$ is maximum the probability to obtain exactly two 6s?
I get $$n=11 \text{ or } n=12.$$
My solution:
the probability to obtain exactly two 6s in $n$ throws is (Bernoulli distribution)
$$ \binom{n}{2} \left( \dfrac{1}{6} \right)^2 \left(\dfrac{5}{6} \right) ^{n-2}.$$
To find its maximum consider
$$ f(x) = \dfrac{x(x-1)}{2} \dfrac{1}{36} \left(\dfrac{5}{6} \right) ^{x-2}.$$
Searching for its stationary points, differentiating one obtains
$$ x^2 \log \dfrac 5 6 + x (2- \log \dfrac 5 6) -1 = 0, $$
whose solutions $x_1,x_2$ are s.t. $\lfloor x_1 \rfloor = 0, \lfloor x_2 \rfloor = 11.$
Substituing back in $f$ one finds that $$f(11) = f(12) > f(0) = f(1),$$ and therefore $n=11$ or $12.$
Is there another (faster/more interesting) way to obtain this result?
| You do not need calculus, since you can consider
$$\dfrac{f(n+1)}{f(n)} = \dfrac{P(\text{sixes}=2|\text{dice}=n+1)}{P(\text{sixes}=2|\text{dice}=n)} = \dfrac{n+1}{n-1}\times \dfrac56$$
which is greater than $1$ when $2 \le n \lt 11$, is less than $1$ when $n \gt 11$, and is equal to $1$ when $n=11$.
So $11$ and $12$ give the maximum value of $f(n)=P(\text{sixes}=2|\text{dice}=n)$.
| {
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Proving that a function is bijective I have trouble figuring out this problem:
Prove that the function $f: [0,\infty)\rightarrow[0,\infty)$ defined by $f(x)=\frac{x^2}{2x+1}$ is a bijection.
Work: First, I tried to show that $f$ is injective. $\frac{a^2}{2a+1}=\frac{b^2}{2b+1}$
I got $a^2(2b+1)=b^2(2a+1)$. However, I get stuck here and cannot simplify the equation to get $a=b$, which would prove that the function is injective.
| Notice $f(x) = \frac{x^2}{2x+1} = \frac{x}{2 + \frac{1}{x}}$. Suppose $f(a) = f(b) $, then
$$ \frac{a}{2 + \frac{1}{a} } = \frac{b}{2 + \frac{1}{b}} \iff \frac{a}{b} = \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}}$$
We want to show that $a = b$. Suppose not. then either $a > b $ or $b >a $.
Say $a > b $. Then $\frac{1}{b} > \frac{1}{a}$. and so
$$ \frac{a}{b} = \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}} < \frac{ 2 + \frac{1}{b}}{2 + \frac{1}{b}} = 1 \implies a < b \; \; \text{contradiction} $$
Now, say $b > a $. then $\frac{1}{a} > \frac{1}{b} \implies \frac{1}{a} + 2 > \frac{1}{b} + 2 \implies \frac{ 2 + \frac{1}{a}}{2 + \frac{1}{b}} > 1 \implies \frac{a}{b} > 1 \implies a > b $ contradiction. Hence
$a= b$ and $f$ must be injective.
| {
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Convergence of $ \sum_{n=1}^{\infty} (\frac{n^2+1}{n^2+n+1})^{n^2}$
Find if the following series converge: $$\displaystyle \sum_{n=1}^{\infty} \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
What I did:
$$a_n=\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
$$b_n=\frac {1}{(n+1)^{n^2}}$$
$$\forall n\ge 1 : a_n < b_n \ \Rightarrow \ \displaystyle \sum_{n=1}^{\infty}\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}<\displaystyle \sum_{n=1}^{\infty}\frac {1}{(n+1)^{n^2}} \approx \displaystyle \sum_{n=1}^{\infty} \frac 1 {n^2}$$
$\displaystyle \sum_{n=1}^{\infty}b_n$ converges like $\displaystyle \sum_{n=1}^{\infty}\frac 1 {n^2}$ so $\displaystyle \sum_{n=1}^{\infty}a_n$ converges as well.
| I would take the following approach:
$$\begin{align}\left (\frac{n^2+1}{n^2+n+1}\right )^{n^2} &= \left (1-\frac{n}{n^2+n+1}\right )^{n^2}\\ &=e^{n^2 \log{\left (1-\frac{n}{n^2+n+1}\right )}} \\ &\sim e^{-n^2 (1/n - 1/(2 n^2))}\\ &\sim e^{-n+1/2} \end{align}$$
Thus the series converges by comparison with a geometric series.
| {
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"timestamp": "2023-03-29T00:00:00",
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An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?
| How $a^2+b^2+c^2=3$, then is equivalent to proof that
$$\frac{a^3}{3-a^2}+\frac{b^3}{3-b^2}+\frac{c^3}{3-c^2}\geq\frac{3}{2}$$
Now the function $f(x)=\frac{x^3}{3-x^2}$ have $f''(x)>0$ in $(0,\sqrt{3})$ then $f(x)$ is convex in this interval, then:
$$f(a)+f(b)+f(c)\geq3f\left(\frac{a+b+c}{3}\right)$$
how $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, for the Rearrangement Inequality, $2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$, then $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\leq 3(a^2+b^2+c^2)=9\Longrightarrow a+b+c\leq3$
A hint for conclude the proof is use the "smoothing principle", ($a=b=c$), so in this case $f(a)+f(b)+f(c)$ is minimized, (but I do not understand this topic), and then:
$$f(a)+f(b)+f(c)\geq 3f(1)=\frac{3}{2}\geq3f\left(\frac{a+b+c}{3}\right)$$
God bless
| {
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"timestamp": "2023-03-29T00:00:00",
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solving the derivative of a function with cos my question is
y=cos^4(2x^2-1)
here is my work
`Dy/dx=4cos^3(2x^2-1) d/dx cos(2x^2-1)
Dy/dx=4cos^3(2x^2-1) (-d/dx(2x^2-1)sin(2x^2-1))
Dy/dx=-4cos^3(2x^2-1)sin(2x^2-1)(4x)
$$dy/dx=-16x\cos^3(2x^2-1)\sin(2x^2-1)$$
any idea on where i went wrong
| Using Chain rule,
$$\frac{d\{\cos^4(2x^2-1)\}}{dx}$$
$$=\frac{d\{\cos^4(2x^2-1)\}}{d\{\cos(2x^2-1)\}}\cdot\frac{d\{\cos(2x^2-1)\}}{dx}$$
$$=4\cos^3(2x^2-1)\cdot\frac{d\{\cos(2x^2-1)\}}{d(2x^2-1)}\cdot\frac{d(2x^2-1)}{dx}$$
$$=4\cos^3(2x^2-1)\cdot\{-\sin(2x^2-1)\}\cdot(2\cdot2x)$$
| {
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How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$,
and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.
| Note that $\tan 2x$ is discontinuous at $\frac{\pi}4$ and some other values which can be easily found.
You have to break the integral from $0$ to $\frac{\pi}4$ and so on.
| {
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Integer solutions of the equation $a^{n+1}-(a+1)^n = 2001 $ I am doing number theory and I came across that question $a^{n+1}-(a+1)^n = 2001 $. Find the integer solutions and show that they are the only solution.
I really tried hard but i am nowhere near solution.
| I'm assuming $a,n$ are positive integers.
First of all, $a^{n+1}-(a+1)^n \equiv -1 \equiv 2001 \;\; \text{mod} \; a ,$ so $a$ divides $2002=2 \times 7 \times 11 \times 13. $ Because $3|2001$ then $a^{n+1} \equiv (a+1)^n \;\; \text{mod} \; 3.$ You can easily eliminate $a \equiv 0,-1 \;\; \text{mod} \; 3,$ therefore, $a \equiv 1 \;\; \text{mod} \; 3 \;(\star)$ and $n$ has to be even.
Case 1. If $a$ is even, then $ a^{n+1}-(a+1)^n \equiv -1 \;\; \text{mod} \; 4$ but $2001 \equiv 1 \;\; \text{mod} \; 4$ and $2 \not \equiv 0 \;\; \text{mod} \; 4$ so there's no solution.
Case 2. If $a$ is odd then $a^{n+1}-(a+1)^n \equiv a \;\; \text{mod}\; 4$ as $a^n \equiv 1 \;\; \text{mod} \; 4.$ Thus, $a \equiv 2001 \equiv 1 \;\; \text{mod} \; 4 \; (\star \star).$ From $(\star), (\star \star)$ the only possibility is $a=13.$
Now, it boils down to solving $13^{n+1}-14^n=2001.$ If $n>2$ then $14^n \equiv 0 \;\; \text{mod} \; 8$ while $13^{n+1} \equiv 5^{2k+1} \equiv 25^k. 5 \equiv 5 \;\; \text{mod} \; 8$ and $5 \not \equiv 1 \;\; \text{mod} \; 8.$ Hence, $n=2$ and $(a,n)=(13,2)$ is the only solution.
| {
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Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder.
| If $n=0$, then it is trivial.
For $n>0$, we have $$(x+1)^{2n+1}+x^{n+2}\\=(x^2+2x+1)(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x((x+1)^{2n-1}+x^{n+1})$$.
By using induction, suppose $(x+1)^{2n-1}+x^{n+1}$ can be divided by $x^2+x+1$, then, $(x+1)^{2n+1}+x^{n+2}$ also can be divided by $x^2+x+1$.
| {
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Stars and bars (combinatorics) with multiple bounds Count the number of solutions to the following:
$$x_1+x_2+\cdots+x_5=45$$
when:
$1$. $x_1+x_2>0$, $x_2+x_3>0$, $x_3+x_4>0$
$2$. $x_1+x_2>0$, $x_2+x_3>0$, $x_4+x_5>1$
($x_1,\ldots,x_5$ are non-negative integers)
Thanks!
| Using the principle of inclusion-exclusion, we can derive the required generating function and extract the coefficient for the formula.
1)
Let $\mathbb{N}$ indicate the number of solutions with the only condition being that any $x_i\ge 0$.
Let the given constraints be:
$A: x_1+x_2>0\\
B: x_2+x_3>0\\
C: x_3+x_4>0
$
We require the number of solutions with all the above constraints, indicated by
$N\left({A\cap B\cap C}\right)$
which can also be written as
\begin{align}
N\left({A\cap B\cap C}\right)&=\mathbb{N}-N\left(\overline{A\cap B\cap C}\right) \tag 1\\
&=\mathbb{N}-N\left(\overline A\cup \overline B\cup \overline C\right)
\end{align}
The overlines are just compliments of the constraints, e.g. $\overline A : x_1+x_2=0$ etc.
Also, by PIE
$$
N\left(\overline A\cup \overline B\cup \overline C\right)=N(\overline A)+N(\overline B)+N(\overline C)-N(\overline A \cap \overline B) -N(\overline A \cap \overline C)-N(\overline B \cap \overline C)+N(\overline A\cap \overline B\cap \overline C) \tag 2
$$
In the above formulas, the condition for other $x_i\ge 0$ holds good simultaneously.
We can then easily write down the g.f. for $(2)$
\begin{align}
g(x)&=\frac{1}{(1-x)^3}+\frac{1}{(1-x)^3}+\frac{1}{(1-x)^3}-\frac{1}{(1-x)^2}-\frac{1}{(1-x)}-\frac{1}{(1-x)^2}+\frac{1}{(1-x)}\\
&=\frac{3}{(1-x)^3}-\frac{2}{(1-x)^2}
\end{align}
Hence, the g.f. for $(1)$ is:
\begin{align}
G_1(x)&=\frac{1}{(1-x)^5}-\frac{3}{(1-x)^3}+\frac{2}{(1-x)^2}
\end{align}
Hence, the general formula for the number of solutions is:
\begin{align}
a_n&=\binom{n+4}{4}-3\,\binom{n+2}{2}+2\, \binom{n+1}{1}\\
\therefore a_{45}&=\boxed{208725}
\end{align}
2)
This also can be done similarly, but need to be more careful with the constraints.
We will end up with the following generating function:
\begin{align}
G_2(x)&=\frac{1}{(1-x)^5}-\frac{3+2\, x}{(1-x)^3}+\frac{1}{(1-x)^2}+\frac{2\, (1+2\, x)}{(1-x)}-(1+2\, x)
\end{align}
and
\begin{align}
b_n&=\binom{n+4}{4}-3\, \binom{n+2}{2}-2\, \binom{n+1}{2}+\binom{n+1}{1}+6\\
\therefore b_{45}&=\boxed{206615}
\end{align}
| {
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integer solutions to $x^2+y^2+z^2+t^2 = w^2$ Is there a way to find all integer primitive solutions to the equation $x^2+y^2+z^2+t^2 = w^2$? i.e., is there a parametrization which covers all the possible solutions?
| For the equation:
$X^2+Y^2+Z^2+Q^2=R^2$
We can write the solution:
$X=2p^2+2(a+b+t)ps+(ab+at+bt)s^2$
$Y=2p^2+2(b+t-a)ps+(bt-a^2)s^2$
$Z=2p^2+2(a+t-b)ps+(at-b^2)s^2$
$Q=2p^2+2(a+b-t)ps+(ab-t^2)s^2$
$R=4p^2+2(a+b+t)ps+(a^2+b^2+t^2)s^2$
And more:
$X=-2p^2+2(a+b+t)ps+(4a^2+4b^2+4t^2-ab-at-bt)s^2$
$Y=-2p^2+2(b+t-5a)ps+(a^2-4a(b+t)+4b^2+4t^2-bt)s^2$
$Z=-2p^2+2(t+a-5b)ps+(b^2-4b(a+t)+4a^2+4t^2-at)s^2$
$Q=-2p^2+2(a+b-5t)ps+(t^2-4t(a+b)+4a^2+4b^2-ab)s^2$
$R=4p^2+2(a+b+t)ps+(7a^2+7b^2+7t^2-4ab-4at-4bt)s^2$
$a,b,t,p,s$ - integers asked us.
| {
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Finding the limit of $\lim_{x\to 1} (x^2-\sqrt x)/(1-\sqrt x)$ How do I evaluate
$$\lim_{x\to 1} \frac{(x^2-\sqrt x)}{(1-\sqrt x)}$$
Can someone explain the steps by steps solution to this problem?
| Notice
$$ \frac{ x^2 - \sqrt{x}}{1 - \sqrt{x} } = \frac{ x^2 - \sqrt{x}}{1 - \sqrt{x} } \cdot \frac{1 + \sqrt{x}}{1+ \sqrt{x}} = \frac{x^2 + x^2 \sqrt{x} - \sqrt{x} - x}{1-x} = \frac{x(x-1) + \sqrt{x}((x-1)(x+1)}{-(x-1)}= \frac{x + \sqrt{x}(x+1)}{-1} \to_{x \to 1} -3$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving a trigonometric relation $$\frac{1+\cos{2\alpha}}{\sin^2{2\alpha}}=\frac{1}{2}\csc^2{\alpha}$$
Here the question is that I can prove that the $LHS=RHS$ if I use the variable $x$ but if we take
($2x=\pi$ on the left side) and ($x=\dfrac{\pi}{2}$ on the other side)
then the equation is not equal,why?
Then prove:
\begin{align}
\frac{1+\cos{2\alpha}}{\sin^2{2\alpha}} & = \frac{1+\cos{2\alpha}}{1-\cos^2{2\alpha}} \\
\require{cancel}& =\frac{\cancel{(1+\cos{2\alpha})}}{\cancel{(1+\cos{2\alpha})}(1-\cos{2\alpha})} \\
& =\frac{1}{1-\cos{2\alpha}}\\
& =\frac{1}{2\sin^2{\alpha}} & \text{because}, 1-\cos{2\alpha}=2\sin^2{\alpha}\\
& =\frac{1}{2}\csc^2{\alpha}\\
\end{align}
Actually it was an indefinite integration but I just noticed this.
| $$
\frac {1+\cos 2x}{\sin^2 2x} = \frac {2\cos^2 x}{4 \sin^2 x \cos^2 x} = \frac 1{2\sin^2 x} = \frac 12 \csc^2 x
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$ $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?
| Let $x=\cfrac{3}{2}\sin\theta$, then $dx=\cfrac{3}{2}\cos\theta\,d\theta$.
\begin{align}
\require{cancel}
\int\frac{\sqrt{9-4x^2}}{x}\, dx&=\int\frac{\sqrt{9-4\left(\frac{3}{2}\sin\theta\right)^2}}{\cancel{\frac{3}{2}}\sin\theta}\, \cancel{\frac{3}{2}}\cos\theta\,d\theta\\
&=\int\frac{3\sqrt{1-\sin^2\theta}}{\sin\theta}\, \cos\theta\,d\theta\\
&=3\int\frac{\cos\theta}{\sin\theta}\, \cos\theta\,d\theta\\
&=3\int\frac{\cos^2\theta}{\sin\theta}\,d\theta\\
&=3\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta\\
&=3\int\frac{1}{\sin\theta}\,d\theta-3\int\sin\theta\,d\theta\\
&=3\int\frac{1}{\sin\theta}\,d\theta+3\cos\theta+C
\end{align}
The last integral can be seen here, and can be done using the substitution $u = \cos \theta$ and partial fractions.
\begin{align}
\int \frac{d\theta}{\sin \theta}
&= \int \frac{\sin \theta}{\sin^2 \theta} d\theta\\
&= \int \frac{\sin \theta}{1 - \cos^2 \theta} d\theta\\
&= \int \frac{-du}{1 - u^2}\\
&= \int \frac{du}{u^2 - 1}\\
&= \frac{1}{2}\left(\ln\ \left|1 - u\right| - \ln\ \left|1 + u\right|\right) + C_2\\
&= \frac{1}{2}\left(\ln\ \left|1 - \cos \theta\right| - \ln\ \left|1 + \cos \theta\right|\right) + C_2
\end{align}
Hope this helps Dan.
| {
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Focus of parabola with two tangents
A parabola touches x-axis at $(1,0)$ and $y=x$ at $(1,1)$. Find its focus.
My attempt : All I can say is that as angle subtended by this chord at focus is $90^\circ$ as angle between tangents is $45^\circ$. I can find equation of directrix by taking mirror image of focus in tangents and then use the fact that distance from focus is same as from directrix. But, this will get dirty.
Any help will be appreciated.
| I have a different approach to this problem. Using the general conic equation
$$ A x^2 + B x y + C y^2 + D x + E y + F =0 $$
and its derivative (for tangents)
$$ {\rm d} x ( 2 A x + B y + D ) + {\rm d} y ( B x + 2 C y + E) =0 $$
With the parabola constraint $B^2 = 4 A C$, and the following 4 constraints, all coefficients can be found
$$ \begin{align}
A + D + F & =0 & & \mbox{curve} & (x=1,y=0) \\
A + B + C + D +E + F & = 0 & & \mbox{curve} & (x=1,y=1) \\
2 A + D & =0 & & \mbox{tangent} & (x=1,y=0,{\rm d}x=1,{\rm d}y=0) \\
2 A + 2 B+ 2 C + D + E & =0 & & \mbox{tangent} & (x=1,y=1,{\rm d}x=1,{\rm d}y=1) \\
\end{align} $$
The above is solved for $$C=-B\\D=-2 A\\E=0\\F=A$$
$$ A x^2 -2 A x - B y^2 + B x y +A =0 $$ and $B^2 = 4 A C = -4 A B \} B = -4 A$
$$ A \left( x^2 - 4 x y - 4 y^2 - 2 x + 1 \right) = 0 \\ A = 1 \mbox{ arbitrary}$$
To find the center, the quadratic is at an extrema ( slope is zero ) at
$$ 2 A x_c + B y_c +D =0 \\ B x_c + 2 C y_c +E =0 $$
$$
x_c = \frac{2 C D - B E}{B^2 - 4 A C} = \frac{4 A}{4 A +B} \\
y_c = \frac{2 A E - B D}{B^2 - 4 A C} = \frac{2 A}{4 A +B}
$$
with solution for the parabolic constraint $B=0$ yielding
$$ x_c = 1 \\ y_c = \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Use mathematical induction to prove that $(3^n+7^n)-2$ is divisible by 8 for all non-negative integers. Base step: $3^0 + 7^0 - 2 = 0$ and $8|0$
Suppose that $8|f(n)$, let's say $f(n)= (3^n+7^n)-2= 8k$
Then $f(n+1) = (3^{n+1}+7^{n+1})-2$
$(3*3^{n}+7*7^{n})-2$
This is the part I get stuck. Any help would be really appreciated.
Thanks.
| If for $n=k, f(k)=3^k+7^k-2$
for $n=k+1,$
$\displaystyle f(k+1)-3f(k)= 3^{k+1}+7^{k+1}-2-3(3^k+7^k-2)$
$\displaystyle=7^k(7-3)+6-2=4(7^k+1)$ which is divisible by $8$ as $7^k$ is odd
So, $f(k+1)$ will be divisible by $8\iff f(k)$ is
We can start with $f(k+1)-7f(k)$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Real and Imaginary $$Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2$$
$$Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$$
I got that $Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2$
And, that $\Big(({\frac{1+i}{1-i})^5\Big)} = i $ , which means that $Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1$
Can you guys confirm that it's true? Thanks in advance!
| You can write
$$
1+i\sqrt{3}=2\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)=
2(\cos(\pi/3)+i\sin(\pi/3))
$$
so
$$
(1+i\sqrt{3})^4=16\cos(4\pi/3)+i\sin(4\pi/3)=
16\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right).
$$
On the other hand
$$
1-i=\sqrt{2}\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)=
\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4))
$$
so
$$
(1-i)^4=4(\cos(-\pi)+i\sin(-\pi))=-4.
$$
Therefore
$$
\left(\frac{1+i\sqrt{3}}{1-i}\right)^4=
\frac{8(-1-i\sqrt{3})}{-4}=
2+2i\sqrt{3}.
$$
Similarly, $1+i=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))$, so
$$
\frac{1+i}{1-i}=
\frac{\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))}{\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4))}=
\cos(\pi/2)+i\sin(\pi/2)=i
$$
and so its fifth power is again $i$. Perhaps more easy, in this case, is
$$
\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+2i+i^2}{1+1}=i.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Which is the minimum number of days that are required,so that all the exams are taken? In a university, the secretariat plans the examination period. There are $6$ subjects, $A,B,C,D,E,Z$ and $9$ students($1, \dots , 9$). At the subject $A$ the students $1,2,3$ are subscribed, at the subject $B$ the students $1,2,9$, at the subject $C$ the students $1,7,8$, at the subject $D$ the students $3,5,7,9$, at the subject $E$ the students $4,5,8$ and at the subject $Z$ the students $4,6,8$. Each examination lasts $2$ hours, and it can only be during the morning hours $10-12$. The only restriction at the planning is that it is not allowed that $2$ subjects, ,at which the same student is subscribed , get examinated simultaneously.
Which is the minimum number of days that are required,so that all the exams are taken?
I tried to solve the exerise,with the chromatic number,using the following graph:
So,the minimum number of days is $4$..or am I wrong??
| I added labels to the graph's edges, for anyone having trouble seeing what's going on.
In addition, here is a table of the classes and students.
$$
\newcommand{\x}{\blacksquare}
\begin{array}{|c|cccccc|}
\hline
& A & B & C & D & E & Z \\
\hline
1 & \x & \x & \x & & & \\
2 & \x & \x & & & & \\
3 & \x & & & \x & & \\
4 & & & & & \x & \x \\
5 & & & & \x & \x & \\
6 & & & & & & \x \\
7 & & & \x & \x & & \\
8 & & & \x & & \x & \x \\
9 & & \x & & \x & & \\
\hline
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that the following series do not converge absolutely Prove that the following series do not converge absolutely:
$$\sum_{x=1}^\infty \frac{\sin x}{x}$$
| The idea behind the proof below is that for any two consecutive integers $x$ and $x+1$, at least one of $|\sin(x)|$ and $|\sin(x+1)|$ is "big."
We use the fact that $\sin(x+1)=\sin x\cos 1 +\cos x\sin 1$. Suppose that $|\sin x|\le \frac{1}{3}$. Then $|\cos x|\ge \sqrt{1-\frac{1}{9}}$.
Thus if $|\sin x\le \frac{1}{3}$ then
$$|\sin(x+1)|\ge \sqrt{1-\frac{1}{9}}\sin 1-\frac{1}{3}\cos 1\approx 0.6.$$
It follows that in $n$ and $n+1$ are a pair of consecutive integers, then at least one of $|\sin(n)|$ and $|\sin(n+1)$ is greater than $\frac{1}{3}$. Thus
$$\frac{|\sin(2k+1)|}{2k+1}+\frac{|\sin(2k)|}{2k} \ge \frac{1}{6k}.$$
But $\sum_{k=1}^\infty \frac{1}{6k}$ diverges, so $\sum_{x=1}^\infty \frac{|\sin x|}{x}$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
radicals and rational exponents. simplifying $4^{5/3}$ This question is not for homework.
The book says:
$$4^{5/3} = (\sqrt4)^5 = (\sqrt{2^2})^5 = 2^5 = 32$$
when a bunch of us tried it:
$$4^{5/3} = \sqrt[3]{4^5} = \sqrt[3]{2^3 \cdot 2^2} = 2 \cdot \sqrt[3]{2^2} = 2 \cdot \sqrt[3]{4}$$
I get the sense we missed something. Where did we mess it up? We don't know how to get to 32.
| There is probably a typo in your book, since
$$(\sqrt 4)^5 = 4^{5/2}\neq 4^{5/3}.$$
However, you still made a mistake in your calculation. You replaced $4^5$ with $2^3\cdot 2^2$, when in reality, $4^5 = 4^3\cdot 4^2 = 2^6 \cdot 2^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ Let $P=\left\{\theta:\sin \theta-\cos \theta = \sqrt{2}\cos \theta\right\}$ and $Q=\left\{\theta:\sin \theta+\cos \theta=\sqrt{2}\sin \theta\right\}$ be two sets , Then
which one is Right.
$(a)\;\;\; P\subset Q$ and $Q-P\neq \phi\;\;\;\;\;\; (b)\;\;\; Q\not\subset P\;\;\;\;\;\; (c)\;\;\;P\not\subset Q\;\;\;\;\;\; (d)\;\;\; P=Q$
$\bf{My\; Try::}$ Given $\left(\sin \theta - \cos \theta\right) = \sqrt{2}\cos \theta\Rightarrow \left(\sin \theta - \cos \theta\right)^2 = 2\cos^2 \theta$
$\displaystyle = \sin^2 \theta +\cos^2 \theta -2\sin \theta \cos \theta = 2\cos^2 \theta \Rightarrow \sin^2 \theta -\cos^2 \theta -2\sin \theta \cos \theta = 0$
$\Rightarrow \cos^2 \theta+2\sin \theta \cdot \cos \theta = \sin^2 \theta$
Now Add $\sin^2 \theta$ on both side, we get
$\Rightarrow \cos^2 \theta+2\sin \theta \cdot \cos \theta +\sin^2 \theta = \sin^2 \theta+\sin^2 \theta$
So $\left(\sin \theta +\cos \theta\right) = \left(\sqrt{2}\cdot \sin \theta\right)^2$
So $\left(\sin \theta +\cos \theta\right) = \sqrt{2}\sin \theta$ and $\left(\sin \theta +\cos \theta\right) = -\sqrt{2}\sin \theta$
So $Q = \left\{\theta:\sin \theta +\cos \theta = \pm \sqrt{2}\sin \theta\right\}$
My doubt is How can we say $P = Q$ because here set $Q$ contain $\pm $ sign.
and answer is also given $Q$, please explain me,
Thanks
| Unfortunately, you cannot get rid of the $\pm$ sign. When you squared the set expression for $P$, the resulting set was not equivalent to $P$; in general, you cannot assume $\{x \mid f(x) = g(x)\} = \{x \mid f(x)^2 = g(x)^2\}$. Usually the second set will be larger.
Here is an approach that avoids squaring by dividing instead:
$$\begin{align}
P &= \{\theta \mid \sin \theta - \cos \theta = \sqrt 2 \cos \theta\}\\
&= \{\theta \mid \tan \theta - 1 = \sqrt 2 \} \quad\quad\text{ see note 1}\\
&= \{\theta \mid \tan \theta = 1 + \sqrt 2 \} \\
\end{align}$$
$$\begin{align}
Q &= \{\theta \mid \sin \theta + \cos \theta = \sqrt 2 \sin \theta\}\\
&= \{\theta \mid 1 + \cot \theta = \sqrt 2\}\\
&= \{\theta \mid \tan \theta = \frac{1}{\sqrt 2 - 1}\}\\
&= \{\theta \mid \tan \theta = \frac{1}{\sqrt 2 - 1}\frac{\sqrt 2 + 1}{\sqrt 2 + 1}\}\\
&= \{\theta \mid \tan \theta = \sqrt 2 + 1\}\\
\end{align}$$
Nicely, they are the same set. Note 1 : dividing by $\cos \theta$ can potentially lose any solutions of the form $\{\theta \mid \cos \theta = 0\}$ from $P$, so we have to double check those. Applying $\cos \theta = 0$ to $P$, you get $P|_{\cos \theta = 0} = \{\theta \mid \sin \theta = 0\}$, and since $\sin \theta = 0$ and $\cos \theta = 0$ don't occur at the same time, the division doesn't lose any elements of the set $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for the value of $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$
Unlike the other infinite series, this infinite series only converges conditionally.
| In the same spirit as this answer, note that
$$
\log\left(\frac{n+i}n\right)=\frac12\log\left(1+\frac1{n^2}\right)+i\arctan\left(\frac1n\right)
$$
Furthermore, using Gautschi's Inequality
$$
\begin{align}
\prod_{k=1}^{n-1}\frac{k+x}{k}
&=\frac{\Gamma(n+x)}{\Gamma(1+x)\Gamma(n)}\\
&\sim\frac{n^x}{\Gamma(1+x)}
\end{align}
$$
Therefore, we get
$$
\begin{align}
\sum_{k=1}^{2n}(-1)^{k-1}\arctan\left(\frac1k\right)
&=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2}{2+i}\frac{3+i}{3}\frac{4}{4+i}\cdots\frac{2n}{2n+i}\right)\right)\\
&=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2+i}{2}\frac{3+i}{3}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\
&-2\,\mathrm{Im}\left(\log\left(\frac{2+i}{2}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\
&=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2+i}{2}\frac{3+i}{3}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\
&-2\,\mathrm{Im}\left(\log\left(\frac{1+\frac i2}{1}\frac{2+\frac i2}{2}\cdots\frac{n+\frac i2}{n}\right)\right)\\
&\sim\mathrm{Im}\left(\log\left(\frac{(2n)^i}{\Gamma(1+i)}\right)-2\log\left(\frac{n^{i/2}}{\Gamma(1+\frac i2)}\right)\right)\\
&=\log(2)+\mathrm{Im}\left(\log\left(\frac{\Gamma(1+\frac i2)^2}{\Gamma(1+i)}\right)\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^{2n}(-1)^{k-1}\arctan\left(\frac1k\right)
&=\log(2)+\mathrm{Im}\left(\log\left(\frac{\Gamma(1+\frac i2)^2}{\Gamma(1+i)}\right)\right)\\[6pt]
&=\log(2)-\mathrm{Im}\left(\log\binom{i}{i/2}\right)\\[9pt]
&=\log(2)-\arg\binom{i}{i/2}\\[12pt]
&\doteq0.506670903216622981985255804784
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 2
} |
Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper half-plane and integrating around the standard semicircle. However, the sum of the two residues becomes a complicated expression involving nested square roots, which magically simplifies to the concise expression above.
Sometimes such 'magical' cancellations indicate that there is a faster, more elegant method to reach the same result.
Is there a faster or more insightful way to compute the above integral?
| Note
\begin{align} \int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2}
&= 2\int_{0}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} \overset{x\to\frac b x}=
2\int_{0}^{\infty} \frac{\frac{x^2}b\ dx}{x^4 + a x^2 + b ^2}\\
&= \int_{0}^{\infty} \frac{1+ \frac{x^2}b}{x^4 + a x^2 + b ^2}dx=\frac1b \int_{0}^{\infty} \frac{d(x-\frac b x)}{(x-\frac b x)^2 +a+2b}\\
&= \frac1{b\sqrt{a+2b}}\tan^{-1}\frac{x-\frac b x}{\sqrt{a+2b}}\bigg|_0^\infty= \frac{\pi}{b\sqrt{a+2b}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
Roots of Polynomial Equation?
$y=1/x$ so I plugged in $x=1/y$ into the equation above and got $y^{4}+y^{3}+y^{2}/c+y/4-1/2$, but apparently it's wrong, when I looked up the answer below.
What am I missing?
| $$\begin{align}
x^4 + x^3 + cx^2 + 4x - 2 & = 0 \\
\left(\frac{1}{y}\right)^4 + \left(\frac{1}{y}\right)^3 + c\left(\frac{1}{y}\right)^2 + 4\left(\frac{1}{y}\right) - 2 & = 0 \\
\frac{1}{y^4} + \frac{1}{y^3} + c\frac{1}{y^2} + 4\frac{1}{y} - 2 & = 0 \\
\text{multiply by }y^4\text{ since y=0 isn't a solution} \\
1 + y + cy^2 + 4y^3 - 2y^4 & = 0 \\
\text{multiply by -1 and rearrange to get the same answer as the book}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equation of tangent perpendicular to line I've got the homework question which I cannot solve.
Find the equation of the tangents to $4x^2+y^2=72$ that are
perpendicular to the line $2y+x+3=0$.
What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation perpendicular to the tangent of $4x^2+y^2=72$ is $m_2 = 2$, since $m_1m_2 = -1$.
I've found the derivative of $4x^2+y^2=72$ which is $\dfrac{dy}{dx} = -\dfrac{4x}{y}$.
The next thing I have done is set $m_2=\dfrac{dy}{dx}$.
Now what?
| HINT :
You found $m_1=-\dfrac{1}{2}$ and $m_2=2$. This is correct. Then
$$
\begin{align}
4x^2+y^2&=72\\
y^2&=72-4x^2\\
y&=\sqrt{72-4x^2}\\
\frac{dy}{dx}&=\frac{d}{dx}(72-4x^2)^{\Large\frac12}\\
m_2&=\frac12(72-4x^2)^{-\Large\frac12}(-8x)\\
2&=-\frac{4x}{\sqrt{72-4x^2}}\\
\sqrt{72-4x^2}&=-2x\\
72-4x^2&=(-2x)^2\\
72-4x^2&=4x^2\\
x^2&=9\\
x&=\pm3
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the power-series of $\frac{1}{(2-x)^2}$ I am going through some old Calculus-tasks in preparation for an upcoming exam, and a seemingly simple task is being stubborn with me. We are simply to find the power-series of the function $$\frac{1}{(2-x)^2}$$
Now, we have our basic power-series:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
So we have
$$\frac{1}{2-x} = \frac{1}{2(1-\frac{1}{2}x)} = \frac{\frac{1}{2}}{1-\frac{1}{2}x} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots$$
Let $f(x) = \frac{1}{2-x}$. Using the chain-rule, we set $u = 2-x$ and $g(u) = \frac{1}{u}$. Then we have $f'(x) = g'(u) * u' = -\frac{1}{u^2}*(-1) = \frac{1}{(2-x)^2}$
So we have $$\frac{1}{(2-x)^2} = \frac{1}{4} + \frac{1}{4}x + \frac{3}{16}x^2 \cdots$$
The text does not agree on this answer. In particular, it claims that the first term should be $\frac{1}{2}$
| Using Newton's generalised binomial theorem or the Binomial Series and assuming the convergence ,
$$\frac1{(2-x)^2} =\frac14\left(1-\frac x2\right)^{-2}$$
$$=\frac14\left(1+\frac x2\cdot2+\frac{-2(-2-1)}{2!}\left(\frac x2\right)^2+\cdots\right)=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\sum_{k=1}^n \left(k \sum_{i=0}^{k-1} {n \choose i}\right)$ How do I calculate the following summation?
$$\sum_{k=1}^n \left[k \sum_{i=0}^{k-1} {n \choose i}\right]$$
| Here is a slightly different take on this that emphasises the use of
formal power series.
Suppose we seek to evaluate
$$Q_n = \sum_{k=1}^n k \sum_{q=0}^{k-1} {n\choose q}.$$
We have $$\sum_{q=0}^n {n\choose q} z^q = (1+z)^n$$
and therefore
$$\sum_{q=0}^{k-1} {n\choose q}
= [z^{k-1}] \frac{1}{1-z} (1+z)^n$$
because multiplication by $1/(1-z)$ sums coefficients.
The sum now becomes
$$\sum_{k=1}^n k [z^{k-1}] \frac{1}{1-z} (1+z)^n
= \sum_{k=0}^{n-1} (k+1) [z^k] \frac{1}{1-z} (1+z)^n$$
Once more deploying multiplication by $1/(1-z)$ this is equivalent to
$$[z^{n-1}] \frac{1}{1-z}
\sum_{k=0}^{n-1} z^k (k+1) [z^k] \frac{1}{1-z} (1+z)^n.$$
The operator sequence that extracts the coefficient on $z^k$,
multiplies by $k+1$ and thereafter by $z^k$ is a generalized
annihilated coefficient extractor and represents multiplication by
$z$ followed by differentiation, so we get
$$Q_n = [z^{n-1}] \frac{1}{1-z}
\left(\frac{z}{1-z} (1+z)^n\right)'.$$
Actually computing the derivative we obtain
$$[z^{n-1}] \frac{1}{1-z}
\left(\frac{1}{(1-z)^2} (1+z)^n
+ \frac{z}{1-z} \times n \times (1+z)^{n-1}\right)$$
which is
$$[z^{n-1}]
\frac{1}{(1-z)^3} (1+z)^n
+ n [z^{n-2}] \frac{1}{(1-z)^2} (1+z)^{n-1}.$$
This is
$$\frac{1}{2}
\sum_{q=0}^{n-1} {n\choose q} (n-1-q+1)(n-1-q+2)
+ n \sum_{q=0}^{n-2} {n-1\choose q} (n-2-q+1)$$
which is
$$\frac{1}{2}
\sum_{q=0}^{n-1} {n\choose q} (n-q)(n-q+1)
+ n \sum_{q=0}^{n-2} {n-1\choose q} (n-q-1)
\\ = \frac{1}{2} \times n \times 2 +
\frac{1}{2}
\sum_{q=0}^{n-2} {n\choose q} (n-q)(n-q+1)
+ n \sum_{q=0}^{n-2} {n-1\choose q} (n-q-1)$$
which finally simplifies to
$$n +
\frac{1}{2}
\sum_{q=0}^{n-2} {n\choose q} (n-q)(n-q+1)
+ \sum_{q=0}^{n-2} {n\choose q} (n-q-1) (n-q)
\\ = n + \frac{3}{2}
\sum_{q=0}^{n-2} {n\choose q} (n-q)^2
- \frac{1}{2} \sum_{q=0}^{n-2} {n\choose q} (n-q)
\\ = n - \frac{3}{2} n + \frac{1}{2} n
+ \frac{3}{2} \sum_{q=0}^n {n\choose q} (n-q)^2
- \frac{1}{2} \sum_{q=0}^n {n\choose q} (n-q)
\\ = \frac{3}{2} \sum_{q=0}^n {n\choose q} (n-q)^2
- \frac{1}{2} \sum_{q=0}^n {n\choose q} (n-q).$$
Recall the well-known identities (not difficult to prove)
$$\sum_{q=0}^n q {n\choose q} = n 2^{n-1}
\quad\text{and}\quad
\sum_{q=0}^n q^2 {n\choose q} = n(n+1) 2^{n-2}$$
to finally obtain
$$\frac{3}{2} n(n+1) 2^{n-2}
-\frac{1}{2} n 2^{n-1}
= n (3n+3) 2^{n-3}
- 2n 2^{n-3} = n(3n+1) 2^{n-3}.$$
There is more on annihilated coefficient extractors at this MSE link.
Addendum. These last two identities can be proved with the
operator given by $z\frac{d}{dz}.$ For the first identity we have
$$\left.\left(z\frac{d}{dz}\right)
(1+z)^n\right|_{z=1} =
\left.nz(1+z)^{n-1}\right|_{z=1} = n2^{n-1}
=\sum_{q=0}^n q {n\choose q}$$
and for the second one
$$\left.\left(z\frac{d}{dz}\right)^2
(1+z)^n\right|_{z=1} =
\left.\left(z\frac{d}{dz}\right)
nz(1+z)^{n-1}\right|_{z=1}
\\= \left.zn(1+z)^{n-1} + n(n-1)z^2(1+z)^{n-2}\right|_{z=1}
= n2^{n-1} + n(n-1) 2^{n-2}
\\= n(n+1) 2^{n-2}
= \sum_{q=0}^n q^2 {n\choose q}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried:
$f(0)=c$
$f(1)=a+b+c$
$f(-1)=a-b+c$
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Solving for $a$ and $b$ respectively we get $$a=\frac{f(1)+f(-1)-2f(0)}{2}$$ and $$b=\frac{f(1)-f(-1)}{2}$$ Then
$$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$
$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$. Now individually the maximum of $|x^2+x|$ happens at $x=1$ , $|x^2-x|$ at $x=-1$ and $|1-x^2|$ at $x=0$. So All I get $|f(x)| \le 3$. But I need $\frac{3}{2}$.
Thanks for the help!!
| Though this has been answered before, to answer the question, I am posting it here. The answer credit goes to Tom Collinge who answered my question.
The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (0, -1), (1, -1) with a minimum at (1/2, -5/4), or some reflection of this. Back to 3/2 ...
If $f$ is monotonic in [-1, 1] then trivially, $|f| \le 1$, so assume not.
Then $f$ is differentiable and (being quadratic) can only have one extreme point at $f'(x_0) = 0$, i.e $2ax_0 + b$ = 0, so that $ax_0 = -b/2$. The extreme value of the function at this point is therefore $f(x_0) = x_0(-b/2) + bx_0 + c = bx_0/2 + c$.
Taking the possible values at -1, 0, and 1 gives
$|f(-1)| = |a - b + c| \le 1$; $|f(0)| = | c| \le 1$; $|f(1)| = |a + b + c| \le 1$
So $|c| \le 1 $ and it follows from these inequalites that $|b| \le 1$
Substituting in the value of $f(x_0)$,
$|f(x_0)| = |bx_0/2 + c| \le |b|.|x_0|/2 + |c| \le 1.1/2 + 1 = 3/2$
Continuation of proof to show that $|f(x)| \le 5/4$
(1) If $|a| \ge 1$
At the extremum, $x_0 = -b/2a$ so by substitution $f(x_0) = ab^2/4a^2 -b^2/2a +c$
$|f(x_0)| = |c - b^2/4a| \le |c| + |b|^2/4|a|$. We already proved $|c|, |b| \le 1$ and by assumption $|a| \ge 1$ so $|f(x_0)| \le 1 + 1/4 = 5/4$
(2) If $|a| \le 1$
Take the Taylor expansion of $f(x)$ about $x_0$: $f(x) = f(x_0) + (x-x_0) f'(x_0) + (x-x_0)^2 f''(z)/2$ for $z$ between $x$ and $x_0$. Note that $f'(x_0) = 0$ at the extremum, and $f''(z) = 2a $ for all $z$, so
$f(x) = f(x_0) + a(x-x_0)^2$, and $|f(x_0)| = |f(x) - a(x - x_0)^2| \le |f(x)| + |a|(x-x_0)^2$. By assumption, $|a| \le 1$, so
$|f(x_0)| \le |f(x) | + (x-x_0)^2$
For $x = -1, 0, or +1$ and $x_0$ in [-1, 1], one of these values makes $|x - x_0| \le 1/2$ (take 0 if $-1/2 \le x_0 \le +1/2 $ ; +1 if $x_0 > 1/2$; -1 if $x_0 \le -1/2$), and in all cases the corresponding function value satisfies $|f(x)| \le 1$, so at this x,
$|f(x_0)| \le 1 + (1/2)^2 = 5/4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given perimeter of triangle and one side, find other two sides In triangle ABC, all three sides have integer lengths. If AB = 21, the perimeter is 54, and the area is a positive integer, what are the lengths of BC and AC?
I tried using Heron's Formula, but I couldn't really get anywhere. Any suggestions? This question was in my Math Challenge II Number Theory packet.
| Just to show that Heron's formula isn't so bad ...
In
$$\text{area}^2 = \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
you know $c=21$ and $a+b+c=54$, so that $b=33-a$. So, the above becomes
$$\begin{align}
\text{area}^2 &= \frac{1}{16}(54)(54-2a)(2a-12)(12) \\[4pt]
&= 2\cdot 3^4 \cdot (27-a)(a-6)
\end{align}$$
Thus, $a$ is somewhere between $6$ and $27$. Since the right-hand side needs to be a perfect square, the factors $(27-a)$ and $(a-6)$ need to contribute an odd number of $2$s, and an even number of any other prime, to the factorization. You can tick through cases pretty quickly, using the latter fact to instantly weed-out cases with obvious singleton prime factors before expending any real mental energy counting other primes:
$$\begin{array}{rccll}
a & (27-a) & ( a - 6 ) & & \\
7 & 20 & 1 & \text{single } 5\\
8 & 19 & 2 & \text{single } 19 \\
9 & 18 & 3 & - & 3^\text{odd} \to \text{nope!}\\
10 & 17 & 4 & \text{single } 17\\
11 & 16 & 5 & \text{single } 5\\
12 & 15 & 6 & \text{single } 5 \\
\color{red}{13} & \color{red}{14} & \color{red}{7} & \color{red}{-} & \color{red}{2^\text{odd} \cdot 7^\text{even} \to \text{yes!}} \\
14 & 13 & 8 & \text{single } 13\\
15 & 12 & 9 & - & 3^\text{odd}, 2^\text{even} \to \text{nope!}\\
16 & 11 & 10 & \text{single } 11 \\
17 \dots 26 & - & - & \text{don't need to check} & \text{(why?)} \\
\end{array}$$
Therefore, the other sides have length $13$ and $33-13=20$. $\square$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Infinite Sum of products What is the infinite sum
$$S = {1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+ ....}?$$
I attempted messing around with the $n$ th term in the series but didnt see any solution.
How should I proceed?
| The $n+1$ th term is $$\frac{1\cdot 3\cdot 5\cdots(2n-1) }{3^{n}(n)!}=\frac{(2n)!}{6^{n}(n!)^2}$$ Hence the series is $$\lim_{x\to 1/6}\sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4/6}}=\sqrt{{3}}$$ This type of sum appears while calculating probability of return to origin in $1D$ random walks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Maximium value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ For $b>a$ what is the maximum possible value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ ?
| By partial differentiation:
Assume $b$ fixed, differentiate with respect to $a$, we get the condition
$$ a^2 + a - \frac {3}{4} = 0 $$
Assume $a$ fixed, differentiate with respect to $b$, we get the condition
$$ b^2 -b - \frac {3}{4} = 0 $$
This gives us $(a, b) = ( - \frac{3}{2}, \frac{1}{2} ) $. It remains to verify if this is indeed a local maximum, and if it is the global maximum. This gives us the value $ \frac{4}{3} $.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Let $a=\dfrac{3+\sqrt{5}}{2}.$ Show that $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
Let $a=\dfrac{3+\sqrt{5}}{2}.$
Show that for all $n\in\mathbb N$, $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
My teacher solve this problem with induction, I am just curious if we can do this exercise without it ?
| It is in fact the case that,
for $a = \frac{3 + \sqrt{5}}{2}$,
$$
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\boldsymbol{\floor{a \floor{ a n}} + n = 3 \floor{a n}.}
$$
Proof. Let $r = a n - \floor{a n}$, $0 < r < 1$.
Then $\floor{a n} = (a n - r)$, so
we need to show that
\begin{align*}
3(a n - r) - n &\le a (a n - r) < 3(a n - r) - n + 1 \\
3 a n - 3 r - n &\le a^2 n - r < 3 a n - 3 r - n + 1 \\
\end{align*}
Notice that $a$ satisfies $a^2 = 3 a - 1$.
This reduces the above to
$$
3 a n - 3 r - n \le 3 a n - a r - n < 3 a n - 3 r - n + 1
$$
Adding $3r + n - 3a n$,
$$
0 \le (3 - a) r < 1
$$
which is true since $0 < r < 1$ and
$$
3 - a = \frac{3 - \sqrt{5}}{2} < \frac{3 - 2}{2} < 1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Find the value of this combinatorial sum. $\sum_{k=4}^{100}\binom{k-1}{3}$ How to compute this sum without laboring?
$$\sum_{k=4}^{100}\dbinom{k-1}{3}.$$
Is it possible to reduce this to a single combinatorial term?
| $$\sum_{k=4}^{100}\dbinom{k-1}{3} = \sum_{k=4}^{100} \frac{(k-1)(k-2)(k-3)}{3!} = \sum_{k=4}^{100}\frac{k^3 - 6k^2 + 11k - 6}{6} \\ = \frac{1}{6} \sum_{k=4}^{100} \left(k^3 - 6k^2 + 11 k - 6\right) = \frac{1}{6} \sum_{k=1}^{100} \left(k^3 - 6k^2 + 11 k - 6\right)- \frac 1 6 \sum_{k=1}^{3} \left(k^3 - 6k^2 + 11 k - 6\right)$$
then perhaps look into sum of cubes and squares formula??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
simplify $\sqrt[3]{11+\sqrt{57}}$ I read in a book (A Synopsis of Elementary Results in Pure and Applied Mathematics) that the condition to simplify the expression $\sqrt[3]{a+\sqrt{b}}$ is that $a^2-b$ must be a perfect cube.
For example $\sqrt[3]{10+6\sqrt{3}}$ where $a^2-b
=(10)^2-(6 \sqrt{3})^2=100-108=-8$ and $\sqrt[3]{-8} = -2$
So the condition is satisfied and $\sqrt[3]{\sqrt{3}+1}^3=\sqrt{3}+1$.
But the example $\sqrt[3]{11+\sqrt{57}}$ where $a^2-b =
(11)^2-57=121-57=64$ and $\sqrt[3]{64}=4$ so the condition is satisfied.
But I can’t simplify this expression. Please help us to solve this problem. Note: this situation we face it in many examples
| I imagine you want to write, with integer $u$ and $v$,
$$11+\sqrt{57}=(u+\sqrt{v})^3$$
So that the cube root would simplify.
Hence
$$u^3+3u^2\sqrt{v}+3uv+v\sqrt{v}=11+\sqrt{57}$$
So you would need
$$\begin{eqnarray}
u(u^2+3v)&=&11\\
v(3u^2+v)^2&=&57
\end{eqnarray}$$
But $57=3 \cdot 19$ has no square factor, so $3u^2+v=\pm1$, and $v=57$. But then the first equation can't hold, since $u^2+3v\geq 3\cdot57=171$.
So it's in fact not possible to simplify this way.
| {
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"url": "https://math.stackexchange.com/questions/790738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
If $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$, how can it imply $x\ge1$? I encountered this problem in a book. Solve the equation $\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\frac{3}{2}\sqrt{\frac{x}{x+\sqrt{x}}}$.
According to the book, $x\ge1$ from observation. And after some algebra, they get $\sqrt{x}(2\sqrt{x}-1-2\sqrt{x-1})=0$. But since $\sqrt{x}\ge1>0$, so $2\sqrt{x}-1-2\sqrt{x-1}=0$ and therefore $x=\frac{25}{16}$.
What I didn't get is why is $x\ge1$? From what we can tell is $x+\sqrt{x}>0$ and $x-\sqrt{x}\ge0$ but from these two how can we conclude $x\ge1$?
Any help is appreciated, thanks!
| By itself, the expression on the left is also defined at $x=0$, but then the expression on the right is not.
If $0\lt x\lt 1$, then $\sqrt{x}\gt x$, so $x-\sqrt{x}$ is negative. Presumably we are working in the reals, so there is no such thing as $\sqrt{x-\sqrt{x}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of a sequence defined by a sum: $\lim_{n\to \infty} \frac1{2^n}\sum_{k=1}^n \frac1{\sqrt k}\binom nk$ $\lim_{n\to \infty} \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k}$
Could it have a connection to Euler summability?
| $$
\frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} = \frac{1}{2^n}\sum_{k=1}^{\log n-1} \frac{1}{\sqrt k}\binom{n}{k} + \frac{1}{2^n}\sum_{k=\log n}^{n} \frac{1}{\sqrt k}\binom{n}{k}
$$
The second term is upper bounded by
$$
\frac{1}{\sqrt{\log n}}\cdot\frac{1}{2^n}\sum_{k=\log n}^{n} \binom{n}{k} \leq \frac{1}{\sqrt{\log n}}\cdot\frac{1}{2^n}\sum_{k=0}^{n} \binom{n}{k} = \frac{1}{\sqrt{\log n}}\cdot 1 \xrightarrow[n\to\infty]{} 0
$$
and the first one by
$$
\frac{1}{2^n}\sum_{k=1}^{\log n-1} \binom{n}{k} \leq \log n\cdot\frac{1}{2^n} \binom{n}{\log n}
$$
What can you say about this last thing?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dirichlet series Suppose that the series $\sum \limits_{n=1}^{\infty}\dfrac{a_n}{n^{\sigma}} \quad(\sigma>0)$ converges. Prove that $\lim \limits_{n\to \infty}\dfrac{a_1+a_2+\dots+a_n}{n^{\sigma}}=0$.
I applied the Abel transformation, but unsuccessfully.
| For each $n$, let $R_n = \sum\limits_{k=n}^\infty \frac{a_k}{k^\sigma}$. Then $a_k = k^\sigma(R_k-R_{k+1})$. Since by assumption the series converges, we have $R_n \to 0$, thus $S_n = \sup\limits_{k\geqslant n} \lvert R_k\rvert$ is finite for all $n$ and $S_n \to 0$. Then we have
$$\begin{align}
\left\lvert \frac{a_1 + \dotsc + a_n}{n^\sigma}\right\rvert
&= n^{-\sigma}\left\lvert \sum_{k=1}^n k^\sigma(R_k - R_{k+1})\right\rvert\\
&= n^{-\sigma}\left\lvert \sum_{k=1}^n k^\sigma R_k - \sum_{k=2}^{n+1} (k-1)^\sigma R_k\right\rvert\\
&= n^{-\sigma}\left\lvert R_1 - n^\sigma R_{n+1} + \sum_{k=2}^n \left(k^\sigma - (k-1)^\sigma\right)R_k\right\rvert\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + \sum_{k=2}^n \frac{k^\sigma - (k-1)^\sigma}{n^\sigma}S_k\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + S_2\sum_{k=2}^{\lfloor\sqrt{n}\rfloor} \frac{k^\sigma - (k-1)^\sigma}{n^\sigma} + S_{\lfloor\sqrt{n}\rfloor+1}\sum_{k=\lfloor\sqrt{n}\rfloor+1}^n\frac{k^\sigma-(k-1)^\sigma}{n^\sigma}\\
&= \frac{S_1}{n^\sigma} + S_{n+1} + S_2\frac{\lfloor\sqrt{n}\rfloor^\sigma-1}{n^\sigma} + S_{\lfloor\sqrt{n}\rfloor+1} \frac{n^\sigma - \lfloor\sqrt{n}\rfloor^\sigma}{n^\sigma}\\
&\leqslant \frac{S_1}{n^\sigma} + S_{n+1} + \frac{S_2}{n^{\sigma/2}} + S_{\lfloor\sqrt{n}\rfloor+1}.
\end{align}$$
Now, given $\varepsilon > 0$, one can choose $n_\varepsilon$ so large that $S_{\lfloor\sqrt{n_\varepsilon}\rfloor+1} < \frac{\varepsilon}{4}$ and $\frac{S_1}{n_\varepsilon^{\sigma/2}} < \frac{\varepsilon}{4}$. Then, one has
$$\left\lvert\frac{a_1+\dotsc + a_n}{n^\sigma}\right\rvert < \varepsilon$$
for all $n \geqslant n_\varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Arc Length polar curve $$r=a\sin^3\left(\frac{\theta}{3}\right) $$
I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.
| This solution borders on trivial when solved in the complex plane. First, the range of $\theta$ must be determined. It was pointed out in an earlier that $\theta\in[0,3\pi]$. I found the same thing by plotting the equation. Now, the equation for the arc length in the complex plane is given by
$$s=\int|\dot z|du$$
Thus, for this problem we have (here we take $a=1$ without any loss in generality)
$$
z=\sin^3(\theta/3)e^{i\theta}\\
\dot z=\left[3\sin^2\left(\frac{\theta}{3}\right)\cos\left(\frac{\theta}{3}\right)\frac{1}{3}+i\sin^3\left(\frac{\theta}{3}\right)\right]e^{i\theta}\\
\begin{align}
|\dot z|
&=\sqrt{\sin^4\left(\frac{\theta}{3}\right)\cos^2\left(\frac{\theta}{3}\right)+\sin^6\left(\frac{\theta}{3}\right)}\\
&=\sin^2\left(\frac{\theta}{3}\right)\sqrt{\cos^2\left(\frac{\theta}{3}\right)+\sin^2\left(\frac{\theta}{3}\right)}\\
&=\sin^2\left(\frac{\theta}{3}\right)\\
\end{align}\\
$$
And finally,
$$
s=\int_0^{3\pi}\sin^2\left(\frac{\theta}{3}\right)d\theta=\frac{3\pi}{2}
$$
I have verified this result by numerical integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show how to compute $2^{343}$ using the least multiplication. Show how to compute $2^{343}$ using the least multiplication.
| One way to do this (and probably the way the prof wants) is exponentiation by squaring. Note that $x^{2n} = (x^n)^2$ and $x^{2n + 1} = (x^n)^2 \cdot x$. We can either break down $343$ into smaller pieces this way, but it's easier to think about starting at the bottom:
$$
\begin{array}{ccc}
Value & Computation & Multiplications & Total Multiplications \\
2^2 & (2^1)^2 & 1 & 1 \\
2^5 & (2^2)^2 \cdot 2 & 2 & 3 \\
2^{10} & (2^5)^2 & 1 & 4 \\
2^{21} & (2^{10})^2 \cdot 2 & 2 & 6 \\
2^{42} & (2^{21})^2 & 1 & 7 \\
2^{85} & (2^{42})^2 \cdot 2 & 2 & 9 \\
2^{171} & (2^{85})^2 \cdot 2 & 2 & 11 \\
2^{343} & (2^{171})^2 \cdot 2 & 2 & 13 \\
\end{array}
$$
How did I pick these numbers? Look at the binary expansion of $343 = 101010111_2$. Starting at the second-left digit, for each $1$, square and multiply, and for each zero, just square. Note that each computation only uses $2$ and the previous computation, so you don't have to store any auxilary results. If you really wanted to, you could write $2^{343}$ as (this is what starting from the top looks like):
$$ \left( 2^{171} \right)^2 \cdot 2 $$
$$ \left( \left( 2^{85} \right)^2 \cdot 2 \right)^2 \cdot 2 $$
$$ \left( \left( \left(2^{42} \right)^2 \cdot 2 \right)^2 \cdot 2 \right)^2 \cdot 2 $$
$$\vdots$$
$$ \left( \left( \left( \left( \left( \left( \left( 2^2 \right)^2 \cdot 2 \right)^2 \right)^2 \cdot 2 \right)^2 \right)^2 \cdot 2 \right)^2 \cdot 2 \right)^2 \cdot 2 $$
Because we don't store any auxilary results, this method is one multiplication slower than mjqxxxx's method, but this one generalizes much more cleanly. Which you should use depends on if you want a general method or just speed (in which case, perhaps a lookup table is appropriate?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the integral of primitive $\frac{1}{x(x+2)}$ I am doing this integral by example:
$$\int_1^\infty\frac{1}{x(x+2)}\ dx.$$
The example in the book, starts with $\dfrac{1}{x(x+2)} \leq \dfrac{1}{x^2}.$
Why is this important? Why does the example in the book start like that?
Thanks.
| If $\dfrac{1}{x(x+2)}\leq\dfrac{1}{x^2}$, then
$$\int^\infty_1\dfrac{1}{x(x+2)}dx\leq\int^\infty_1\dfrac{1}{x^2}dx=\left.-\dfrac{1}{x}\right|^\infty_1=1$$
i.e. $\displaystyle\int^\infty_1\dfrac{1}{x(x+2)}dx\leq1$. This shows convergence of the integral.
Now, split $\dfrac{1}{x(x+2)}$ as a partial fraction:
$$\dfrac{1}{x(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+2}=\dfrac{(A+B)x+2A}{x(x+2)}\\
\implies (A+B)x+2A=1.$$
Solve for $A$ and $B$, then substitute them into $\dfrac{1}{x(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+2}$, and integrate it term by term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
| Real Part
Substituting $x\mapsto1/x$ says
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\int_1^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{1}
$$
Therefore,
$$
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{2}
$$
Contour Integration Part
Putting the branch cut for $\log(z)$ along the positive real axis, and using the contour
$$
\gamma=[0,R]e^{i\epsilon}\cup Re^{i[\epsilon,2\pi-\epsilon]}\cup[R,0]e^{-i\epsilon}\tag{3}
$$
as $R\to\infty$ and $\epsilon\to0$, $\log(z)=\log(x)$ on the outbound segment and $\log(z)=\log(x)+2\pi i$ on the inbound segment and he integral around huge circular arc vanishes. Therefore,
$$
\begin{align}
\int_\gamma\frac{\log(z)^3}{z^2-z+1}\mathrm{d}z
&=\int_0^\infty\frac{-6\pi i\log(x)^2+12\pi^2\log(x)+8\pi^3i}{x^2-x+1}\mathrm{d}x\\
&=\frac{124\pi^3}{27\sqrt3}\cdot2\pi i\tag{4}
\end{align}
$$
where $\dfrac{124\pi^3}{27\sqrt3}$ is the sum of the residues of $\dfrac{\log(z)^3}{z^2-z+1}$ at $e^{i\pi/3}$ and $e^{i5\pi/3}$.
Combining Real and Complex Analysis
Therefore, using $(2)$ and the imaginary part of $(4)$, we get
$$
\begin{align}
\int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x
&=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\\
&=\frac12\left[-\frac{124\pi^3}{81\sqrt3}
+\frac{4\pi^2}{3}\int_0^\infty\frac1{x^2-x+1}\mathrm{d}x\right]\\
&=\frac12\left[-\frac{124\pi^3}{81\sqrt3}
+\frac{4\pi^2}{3}\frac{4\pi}{3\sqrt3}\right]\\
&=\frac{10\pi^3}{81\sqrt3}\tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 0
} |
Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$
Prove the following
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation}
I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌
Original question:
And of course, for the sadist with a background in differential
equations, I invite you to try your luck with the last integral of the
group.
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx\end{equation}
Source: Integration: The Feynman Way
| Here is my attempt:
\begin{align}
\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\
&=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\
&=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\
&=\frac{\pi}{4}+\frac{\pi}{4e^2}
\end{align}
where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.
Unfortunately, this is not the Feynman way but I still love this method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 3,
"answer_id": 0
} |
Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
| Let $y=\arctan x\;\Rightarrow\;\tan y=x\;\Rightarrow\;\sec^2y\ dy=dx$, then the integral turns out to be
\begin{align}
\int\frac{\arctan x}{x^2}\ dx&=\int\frac{y}{\tan^2y}\cdot\sec^2y\ dy\\
&=\int\frac{y}{\sin^2y}\ dy.
\end{align}
The last integral can be solved by using IBP. Taking $u=y$ and $dv=\dfrac1{\sin^2x}\ dx\;\Rightarrow\;v=-\cot y$.
\begin{align}
\int\frac{\arctan x}{x^2}\ dx&=\int\frac{y}{\sin^2y}\ dy\\
&=-y\cot y+\int\cot y\ dy\\
&=-y\cot y+\int\frac{\cos y}{\sin y}\ dy\\
&=-y\cot y+\int\frac{d(\sin y)}{\sin y}\\
&=-y\cot y+\ln|\sin y|+C\\
&=-\frac{\arctan x}{x}+\ln\left|\frac{x}{\sqrt{x^2+1}}\right|+C.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find sum of $\frac{1}{\sin\theta\cdot \sin2\theta} + \frac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$ $$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$$ up to $n$ terms.
I tried but in vain
| Note
$$\dfrac{\sin{\theta}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\dfrac{\sin{((k+1)\theta-k\theta)}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\cot{(k\theta)}-\cot{(k+1)\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the sum of the infinite geometric series? $$2/3-2/9+2/27-2/81+\cdots$$
The formula is $$\mathrm{sum}= \frac{A_g}{1-r}\,.$$
To find the ratio, I did the following:
$$r=\frac29\Big/\frac23$$
Then got:
$$\frac29 \cdot \frac32= \frac13=r$$
and $$A_g= \frac23$$
Then I plug it all in and get:
$$\begin{align*}
\mathrm{sum} &= \frac23 \Big/ \left(1-\frac13\right)\\
&= \frac23 \Big/ \left(\frac33-\frac13\right)\\
&= \frac23 \Big/ \frac23\\
&= \frac23 \cdot \frac32\\
&= 1\,.
\end{align*}$$
But the real answer is $\frac12$.
What did I do wrong?
| Hint: the terms are alternating in sign. $\displaystyle r = \frac{\frac{2}{9}}{-\frac{2}{3}} = -\frac{1}{3}$.
Note the minus sign.
Hence the sum is $\displaystyle \frac{\frac{2}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| All that was done was factoring the numerator. Notice that
$$
x^2+x-6=(x+3)(x-2)
$$
because $(x+3)(x-2)=x^2-2x+3x-6=x^2+x-6$. Then it follows that
$$
\frac{x^2+x+6}{x-2}=\frac{(x+3)(x-2)}{x-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$
I ask
$$
f\left( z \right) = - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}}
$$
is to :
$$\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}} = \left( {\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$
I found:
\begin{array}{l}
\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\
\mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\
\end{array}
\begin{array}{l}
\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)4n + 7}}} = - \frac{\pi }{4} = - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\
\Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\
\end{array}
I have a question for the result
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$
thank you in advance
| (Posted as an answer in case my earlier comment is removed)
It is relatively easy to prove (either through elementary means or via complex analysis) the well-known identity $$1 + 2 \sum_{n=1}^\infty \frac{z^2}{z^2 - (n\pi)^2} = z \cot z.$$ Then the given series (not the one written in summation notation, but the one that was actually written out) $$1 - \frac{1}{7} + \frac{1}{9} + \frac{1}{15} - \frac{1}{17} + \cdots$$ is simply the special case $z = \frac{\pi}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\
\ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\
\Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\
\mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\
\end{array}
$
help me what you please
| Let $y= \dfrac{x-1-\ln x}{x\ln x - 1+ x} \to yx\ln x - y + xy = x - 1 - \ln x \to (xy + 1)\ln x = (y-1)(1-x) \to (xy+1)\cdot \dfrac{\ln x}{x-1} = 1 - y$.
Now using a well-known fact that: $\dfrac{\ln x}{x-1} \to 1$ when $x \to 1$. Taking limit as $x \to 1$ both sides of the above equation we have:
$y + 1 = 1 - y \to y = 0$.
Note: The answer obtained by L'hopital rule is $y = 0$ , not $1$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}
$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
| Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.
Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
A difficult integral evaluation problem How do I compute the integration for $a>0$,
$$
\int_0^\pi \frac{x\sin x}{1-2a\cos x+a^2}dx?
$$
I want to find a complex function and integrate by the residue theorem.
| I'll first consider the case $0<a<1$.
Let $\displaystyle f(z) = \frac{z}{a-e^{-iz}}$ and integrate around a rectangle with vertices at $\pm \pi$ and $\pm \pi + iR$.
The function $f(z)$ has poles where $a-e^{-iz}=e^{\ln a + 2 \pi i n}-e^{-iz}= 0$.
That is, when $z= i \ln a - 2 \pi n$.
If $ 0<a< 1$, all of those points are in the lower half-plane.
Letting $ R \to \infty$ and going counterclockwise around the contour,
$$ \int_{- \pi}^{0} \frac{z}{a-e^{-ix}} \ dx + \int_{0}^{\pi} \frac{x}{a-e^{-ix}} \ dx + i \int_{0}^{\infty}f(\pi + i t) \ dt + \lim_{R \to \infty} \int_{-\pi}^{\pi} f(t + iR) \ dt $$
$$+ \ i \int_{\infty}^{0} f(-\pi+it) \ dy =0$$
Combing the first two integrals,
$$ \begin{align} \int_{- \pi}^{0} \frac{x}{a-e^{-ix}} \ dx + \int_{0}^{\pi} \frac{x}{a-e^{-ix}} \ dx &= -\int_{0}^{\pi} \frac{u}{a-e^{iu}} \ du + \int_{0}^{\pi} \frac{x}{a-e^{-ix}} \ dx \\ &= \int_{0}^{\pi} \frac{-x(a-e^{-ix})+ x(a-e^{ix}) }{(a-e^{ix})(a-e^{-ix})} \ dx \\ &= - 2 i \int_{0}^{\pi} \frac{x \sin x}{1-2a \cos x +a^{2}} \ dx \end{align}$$
Combining the third and fifth integrals,
$$ \begin{align} i \int_{0}^{\infty} f(\pi + it) \ dt - i \int_{0}^{\infty} f(-\pi + it) \ dt &= i \int_{0}^{\infty} \frac{\pi + it}{a-e^{-i(\pi + it)}} \ dt - i \int_{0}^{\infty} \frac{- \pi + it}{a-e^{-i( -\pi + it)}} \ dy \\ &= i \int_{0}^{\infty} \frac{\pi + it}{a +e^{t}} \ dt - i \int_{0}^{\infty} \frac{- \pi + it}{a+ e^{t}} \ dt \\ &= 2 \pi i \int_{0}^{\infty} \frac{1}{a+e^{t}} \ dt \\ &= 2 \pi i \int_{0}^{\infty} \frac{e^{-t}}{1+ae^{-t}} \ dt \\ &= - \frac{2 \pi i}{a} \ln (1+ae^{-t}) \Big|^{\infty}_{0} \\ &= \frac{2 \pi i}{a} \ln(1+a) \end{align}$$
And the fourth integral vanishes.
So we have
$$ - 2i \int_{0}^{\pi} \frac{x \sin x}{1-2a \cos x +a^{2}} \ dx + \frac{2 \pi i}{a} \ln(1+a) = 0$$
or
$$ \int_{0}^{\pi } \frac{x \sin x}{1-2a \cos x +a^{2}} \ dx = \frac{\pi \ln (1+a)}{a}$$
The case for when $a >1$ is similar.
The only difference is that there is now a pole inside of the contour at $z=i \ln a$ with residue $ \displaystyle \frac{\ln a}{a}$.
And therefore,
$$- 2i \int_{0}^{\pi} \frac{x \sin x}{1-2a \cos x +a^{2}} \ dx + \frac{2 \pi i}{a} \ln(1+a) = 2 \pi i \frac{\ln a}{a} $$
or
$$ \int_{0}^{\pi} \frac{x \sin x}{1-2a \cos x+a^{2}} \ dx = \frac{\pi \ln (1+a)}{a} - \frac{\pi \ln a}{a} = \frac{\pi \ln \left(\frac{1+a}{a} \right)}{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?
| Let $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\cdot \begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd\\ ac + cd & bc + d^2\end{bmatrix}$. Thus:
$A - A^2 = \begin{bmatrix} a-a^2-bc & b-ab-bd\\ c-ac-cd & d-bc-d^2\end{bmatrix} = \begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$. Thus:
$a-a^2-bc = 3$
$d-bc-d^2 = 4$
$b-ab-bd = 1$
$c-ac-cd = 1$.
Subtract the last $2$ equations to each other:
$b-c - a(b-c) - d(b-c) = 0 \to (b-c)(1-a-d) = 0$.
And subtract the first $2$ equations to obtain:
$a-d -(a^2-d^2) = -1 \to (a-d)(1-a-d) = -1$. Thus: $b-c = 0$, and $b = c$. Let's rewrite the system:
$a-a^2-b^2 = 3$
$d-b^2-d^2 = 4$
$b-ab-bd = 1$. Using Wolframalpha:
$a \approx 0.5 - 0.479i$
$b \approx 1.588i$
$d \approx 0.5 + 1.109i$.
Thus all roots are imaginary numbers, which means there is no matrix $A$ with real entries for the equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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On combining $n$ and $n^2$ into one number Consider the sequence $T_n$ formed by combining $n$ and $n^2$ into one number.
ie. (A053061)
$$T_n=\{11,24,39,416,525,636,749 \cdots\}$$
It is easy to see
$$T_n= 10^{\lceil 2 \log_{10}(n) \rceil } n+ n^2$$
I looked at the sequence closely trying to find if there are any perfect squares in the sequence but wasn't able to upto $n=100$.
I also was able to prove that if $n^{th}$ term is a perfect square then :
1) $n \equiv 8 \text{ or } 0 (\text{ mod } 9)$
2)In the case where $n=9m$ , m is not a square free number.
But I am unable to attack the question
Do there exist any perfect square in the sequence?
| There are no squares in this sequence.
Elements of this sequence are of the form $b^2+10^n b$ for $10^{n-1} \leq b^2 < 10^n$. So we are looking for a special solution to the equation $a^2 =b^2+ 10^n b$. This is an easy Diophantine equation to solve: We have $a^2 - (b+10^n/2)^2 = - 10^{2n}/4$, or
$(a+b+10^n/2)(a-b-10^n/2) = -10^{2n}/4$
The two factors on the left side have the same parity. Clearly we may assume $n>1$, so these factors both must be even. If those factors are both multiples of $20$, then $a$ and $b$ are both multiples of $10$, and we may divide $a$ and $b$ by $10$ and subtract $1$ from $n$ without changing the equation. However, it does modify the inequality. So each factor is either a power of $2$ or $2$ times a power of $5$, hence the two factors must be $2 \cdot 5^{2n}$ and $2^{2n-3}$. This gives:
$a+b+10^n/2 = 2\cdot 5^{2n}$
$a-b-10^n/2 = 2^{2n-3}$
$b= 5^{2n} - 2^{2n-4} -10^n/2$
For $n=2$, we get $575$, but $575^2>100$. For higher $n$, it's even worse: $b^2$ is much greater than $10^n$. Similarly, multiplying $a$ and $b$ by $10$ and adding $1$ to $n$ also makes $b^2$ larger, relative to $10^n$. So there are no solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that a given ideal is not maximal in $\mathbb C[x,y,z]$ I'm trying to prove this ideal:
$$(x^2+y^2+z^2+x+y+z,\ x^5+y^5+z^5+2(x+y+z),\ x^7+y^7+z^7+3(x+y+z))\subset
\mathbb C[x,y,z]$$
can't be maximal.
In order to do so, I'm using the Nullstellensatz theorem and showing this ideal is not of this form: $(x-a_1,y-a_2,z-a_3)$, where $a_i\in \mathbb C$.
The problem I don't how to do this.
| To show your ideal is not maximal, I'll show that the set of common zeros of your ideal contains more than one point. Taking advantage of the symmetry of your ideal, we will look for solutions that even satisfy the extra condition $x+y+z=0.$ Then (using Newton's Identities and remember $e_1=x+y+z=0$) the first equation is $$x^2+y^2+z^2+x+y+z = p_2 + e_1 = p_2 = e_1^2 - 2e_2.$$
The next one is $$ x^5+y^5+z^5 + 2(x+y+z) = p_5 + 2e_1 = -5e_3e_2 $$ and the third is
$$ x^7+y^7+z^7 + 3(x+y+z) = p_7 + 3e_1 = c e_3 e_2^2$$ where $c$ is a non-zero constant.
This system is satisfied if we further impose $e_2=0.$ Hence, any common solutions of $e_1=x+y+z=0$ and $e_2 = xy+yz+xz=0$ are in the set of common zeros of your ideal. It is easy to find solutions to this that aren't the trivial $(0,0,0)$ solution. For example, pick $x=1,$ so $y+z=-1$ and $y+yz+z =0,$ which easily solves for $y= \dfrac{-1}{2} - i \dfrac{\sqrt{3}}{2} \ , \ z = \dfrac{-1}{2} + i \dfrac{\sqrt{3}}{2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Massive expression mod $2^{2013}$ Find $$1^1+3^3+5^5 + \cdots +(2^{2012}-1)^{(2^{2012}-1)}$$ modulo $2^{2013}$. By checking small cases I am pretty sure that it is $2^{2012}$. I tried applying induction but you "lose" powers of two at each step. So instead I used binomial theorem to reduce it to $$\sum_{i=1,i \text{ odd}}^{2^{2011}-1} i^i-i^{2^{2012}-i}$$ again it looks very much like induction, but I can't make it work. Any help please.
EDIT: Also
$\phi({2^{2013}})=2^{2012}$ so really, by euler's theorem the above becomes $\sum_{i=1,i \text{ odd}}^{2^{2011}-1} i^i-i^{-i}$
| I think induction does work to show that for $k\ge2$,
$$
\sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} \equiv 2^k \pmod{2^{k+1}}.
$$
One fact we'll use is that for $n\ge3$, we actually have $a^{2^{n-2}}\equiv1\pmod {2^n}$ for any odd $a$; this is one power of $2$ stronger than Euler's theorem.
Checking the $k=2$ case is easy. Suppose that it holds for some particular $k$. Then
\begin{align*}
\sum_{i=1}^{2^k} (2i-1)^{2i-1} &= \sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} + \sum_{i=2^{k-1}+1}^{2^k} (2i-1)^{2i-1} \\
&= \sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} + \sum_{i=1}^{2^{k-1}} (2^k+2i-1)^{2^k+2i-1} \\
&\equiv \sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} + \sum_{i=1}^{2^{k-1}} (2^k+2i-1)^{2i-1} \pmod{2^{k+2}}
\end{align*}
by the above fact. Now note that $(2^k+j)^m \equiv j^m + m2^kj^{m-1} \pmod{2^{k+2}}$ by the binomial theorem (since $k\ge2$); in particular, only $j\pmod4$ matters, so if $j$ and $m$ are odd then $(2^k+j)^m \equiv j^m + m2^k \pmod{2^{k+2}}$. Therefore
\begin{align*}
\sum_{i=1}^{2^k} (2i-1)^{2i-1} &\equiv \sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} + \sum_{i=1}^{2^{k-1}} \big( (2i-1)^{2i-1} + (2i-1)2^k \big) \\
&= 2\sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} + 2^{2k-2} 2^k \\
&= 2\sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1} \pmod{2^{k+2}}
\end{align*}
(again since $k\ge2$). The induction hypothesis that $\sum_{i=1}^{2^{k-1}} (2i-1)^{2i-1}\equiv 2^k\pmod {2^{k+1}}$ now implies that $\sum_{i=1}^{2^k} (2i-1)^{2i-1} \equiv 2^{k+1}\pmod {2^{k+2}}$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Stronger version of AM-GM with condition If $a,b,c>0$ and $a^2+b^2+c^2=3$, prove that
$$a+b+c\ge 3(abc)^{13/76}.$$
I was thinking of using AM-GM :P but clearly some clever trick is needed. I was thinking about expanding but that is not feasible too. Note I would like a non-computational proof which can be done by hand. Also this was given to me by my teacher who has solved it and so I suppose it is correct. So any help?
| First we check that $abc \leq 1$, it's easy by AM-GM:
$1=\frac{a^2+b^2+c^2}{3} \leq (abc)^{\frac{2}{3}}$
Next we prove that $(a+b+c)^2 \geq 9(abc)^{\frac{13*2}{76}}$. If we prove this the inequality $a+b+c \geq 3(abc)^{\frac{13}{76}}$ will be proven, because $x \leq y$ implies $\sqrt{x} \leq \sqrt{y}$ for $x,y \geq 0$.Next:
$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=1+1+1+ab+ab+ac+ac+bc+bc$
Apply AM-GM to this sum. We have:
$ 1+1+1+ab+ab+ac+ac+bc+bc \geq 9(abc)^{\frac{2}{9}}$,
but:
$9(abc)^{\frac{2}{9}} \geq 9(abc)^{\frac{2*13}{76}}$, because it's equal:
$1 \geq (abc)^{\frac{2*13}{76}-\frac{2}{9}}=(abc)^{\frac{41}{342}}$ and it's true, because $abc \leq 1$, so
$(a+b+c)^2 \geq 9(abc)^{\frac{2*13}{76}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $n^2(n^2+16)$ is divisible by 720 Given that $n+1$ and $n-1$ are prime, we need to show that $n^2(n^2+16)$ is divisible by 720 for $n>6$.
My attempt:
We know that neither $n-1$ nor $n+1$ is divisible by $2$ or by $3$, therefore $n$ must be divisible by both $2$ and $3$ which means it must be divisible by $6$.
So, $n = 6k$ and since $n>6$ we must have $k>3$. So the expression becomes $36k^2(36k^2+16) = 144k^2(9k^2+4)$
...
But then I get stuck.
Could someone please guide me towards a solution.Thanks.
| $n=6k$ implies $n\equiv k \bmod 5$. If $k\equiv 0 \bmod 5$, then you're done.
Othwerwise, $n-1$ and $n+1$ prime implies $k\equiv n\equiv \pm 2 \bmod 5$ and so $9k^2+4\equiv -k^2+4 \equiv 0 \bmod 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$ The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$.
We can consider a similar nested radical where the degree of the radicals increases:
$$\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$$
which converges to a constant $C=1.8695973...$. Is there a closed-form expression for this limit?
| One of the well known formula by Indian mathematician Ramanujan is this one:
$$f(x)=x+1=\sqrt { \left( x+1 \right) ^{2}}=\sqrt {1+{x}^{2}+2\,x}=\sqrt {1+x\sqrt {(x+2)^2}}$$
$$=\sqrt {1+x\sqrt {1+{x}^{2}+4\,x+3}}=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {(x+3)^2}}}$$
$$
=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {1+ \left( x+2 \right)
\sqrt {1+ \left( x+3 \right) \sqrt {1+ \left( x+4 \right) \sqrt {\cdots}}}}}}
$$
So it is interesting to show that why your mentioned limit tends to $f(0.86959730667536)$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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Find a closed expression for a formula including summation Let:
$$\sum\limits_{k = 0}^n {k\left( {\matrix{
n \cr
k \cr
} } \right)} \cdot {4^{k - 1}} \cdot {3^{n - k}}$$
Find a closed formula (without summation). I think I should define this as a "series" which generated by $F(x)$. I don't really have a lead here.
Any ideas? Thanks.
| The binomial expansion is given by
\begin{align}
(1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}.
\end{align}
Differentiating both sides with respect to $x$ yields
\begin{align}
n (1+x)^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} x^{k-1}.
\end{align}
Multiplying this last expression by $3^{n} 4^{-1}$ leads to
\begin{align}
\frac{3^{n} n}{4} \ x(1+x)^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} 4^{-1} 3^{n} x^{k}.
\end{align}
Now let $x = 4/3$ to obtain the desired expression
\begin{align}
n \cdot 7^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} 4^{k-1} 3^{n-k}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A formula for $\pi$ and an inequality For any $n\in \mathbb{N}$ prove the identity :
$$\pi =\sum_{k=1}^{n}\frac{2^{k+1}}{k\dbinom{2k}{k}}+\frac{4^{n+1}}{\dbinom{2n}{n}}\int_{1}^{\infty}\frac{\mathrm{d}x}{(1+x^2)^{n+1}}\tag{1}$$
and thus derive the inequality :
$$\frac{2}{2^n\sqrt{n}}<\pi -\sum_{k=1}^{n}\frac{2^{k+1}}{k\dbinom{2k}{k}}<\frac{4}{2^n\sqrt{n}}\tag{2}$$
How to prove these two? I was given them in my calculus exam. I wasted 2 hours of the 4 hour allotted on them but nothing came to my mind. Can someone help? Thanks a lot.
| Let
$$I_n = \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}.$$
Then $I_0 = \frac{\pi}{4}$ is well-known, and $(1)$ holds for $n = 0$. Integration by parts yields
$$\begin{align}
I_n &= \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}\\
&= \left[\frac{x}{(1+x^2)^{n+1}}\right]_1^\infty + 2(n+1) \int_1^\infty \frac{x^2\,dx}{(1+x^2)^{n+2}}\\
&= -\frac{1}{2^{n+1}} + 2(n+1) \left[I_n - I_{n+1}\right].
\end{align}$$
Rearranging yields
$$I_n = \frac{2(n+1)}{2n+1}I_{n+1} + \frac{1}{(2n+1)2^{n+1}},$$
and so
$$\begin{align}
\frac{4^{n+1}}{\binom{2n}{n}} I_n &=\frac{4^{n+1}}{2^{n+1}(2n+1)\binom{2n}{n}}
+ \frac{4^{n+1}}{\binom{2n}{n}\frac{2n+1}{2(n+1)}}I_{n+1}\\
&= \frac{2^{n+2}}{\frac{(n+1)2(n+1)(2n+1)}{(n+1)(n+1)}\binom{2n}{n}}
+ \frac{4^{n+2}}{\binom{2n}{n}\frac{(2n+1)2(n+1)}{(n+1)(n+1)}}I_{n+1}\\
&= \frac{2^{n+2}}{(n+1)\binom{2(n+1)}{n+1}} + \frac{4^{n+2}}{\binom{2(n+1)}{n+1}}I_{n+1},
\end{align}$$
which completes the induction on $(1)$.
Then $(2)$ is equivalent to
$$\frac{2}{2^n\sqrt{n}} < \frac{4^{n+1}}{\binom{2n}{n}}I_n < \frac{4}{2^n\sqrt{n}}.$$
Here, we must assume $n > 0$ of course.
We can estimate
$$I_n < \int_1^\infty \frac{x\,dx}{(1+x^2)^{n+1}} = \left[-\frac{1}{2n(1+x^2)^n}\right]_1^\infty = \frac{1}{n2^{n+1}}$$
and
$$I_n > \int_1^\infty \frac{2x\,dx}{(1+x^2)^{n+2}} = \left[-\frac{1}{(n+1)(1+x^2)^{n+1}}\right] = \frac{1}{(n+1)2^{n+1}}.$$
The former estimate shows that it is enough to prove
$$\frac{2^{2n-1}}{\sqrt{n}} \leqslant \binom{2n}{n}\tag{$\ast$}$$
to obtain the upper bound. For $n = 1$, we have equality in $(\ast)$, and then
$$\binom{2n+2}{n+1} = \binom{2n}{n}\frac{4(n+\frac{1}{2})}{n+1} \geqslant \frac{2^{2n-1}}{\sqrt{n}} \frac{4(n+\frac{1}{2})}{n+1} = \frac{2^{2n+1}}{\sqrt{n+1}}\frac{n+\frac{1}{2}}{\sqrt{n(n+1)}} > \frac{2^{2n+1}}{\sqrt{n+1}}$$
completes the induction for $(\ast)$.
For the lower bound, it suffices to show
$$\binom{2n}{n} \leqslant \frac{\sqrt{n}2^{2n}}{n+1}\tag{$\ast\ast$}.$$
For $n = 1$, we have equality in $(\ast\ast)$, and then
$$\begin{align}
\binom{2(n+1)}{n+1} & \leqslant \frac{\sqrt{n}2^{2n}}{n+1}\frac{(2n+1)2(n+1)}{(n+1)^2}\\
&= \frac{2^{2n+2}\sqrt{n}(n+\frac{1}{2})}{(n+1)^2}\\
&= \frac{2^{2n+2}\sqrt{n+1}}{n+2}\frac{\sqrt{n}(n+2)(n+\frac{1}{2})}{(n+1)^{5/2}}\\
&< \frac{2^{2n+2}\sqrt{n+1}}{n+2}
\end{align}$$
finishes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability Puzzle: Mutating Loaded Die Take an (initially) fair six-sided die (i.e. $P(x)=\frac{1}{6}$ for $x=1,…,6$) and roll it repeatedly.
After each roll, the die becomes loaded for the next roll depending on the number $y$ that was just rolled according to the following system:
$$P(y)=\frac{1}{y}$$
$$P(x)=\frac{1 - P(y)}{5} \text{, for } x \ne y$$
i.e. the probability that you roll that number again in the next roll is $\frac{1}{y}$ and the remaining numbers are of equal probability.
What is the probability that you roll a $6$ on your $n$th roll?
NB: This is not a homework or contest question, just an idea I had on a boring bus ride. Bonus points for calculating the probability of rolling the number $x$ on the $n$th roll.
| The transition matrix is given by $$\mathcal P = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ \tfrac{1}{10} & \tfrac{1}{2} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} & \tfrac{1}{10} \\ \tfrac{2}{15} & \tfrac{2}{15} & \tfrac{1}{3} & \tfrac{2}{15} & \tfrac{2}{15} & \tfrac{2}{15} \\ \tfrac{3}{20} & \tfrac{3}{20} & \tfrac{3}{20} & \tfrac{1}{4} & \tfrac{3}{20} & \tfrac{3}{20} \\ \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{4}{25} & \tfrac{1}{5} & \tfrac{4}{25} \\ \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}.$$ It is fairly easy to get numerical values for the probability distribution of being in state $6$ after $n$ steps, but a closed form solution appears difficult.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid y^2$ then
$ 7\mid x^2+y^2 $
*
*Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$
Now let's show
$7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid y$
$7\mid x^2+y^2 \Longleftrightarrow x^2+y^2=0 \pmod 7 $
for
\begin{array}{|c|c|c|c|c|} \hline
x& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline
y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline
\end{array}
which means we have one possibility that $x=y= 0 \pmod 7 $
*
*Am I right and are there other ways?
| Let $x,y \in \mathbb{F}_p$ be with $x^2+y^2=0$. If $x=0$, then $y=0$. Now assume $x \neq 0$. Let $z:=y/x$, then $z^2=-1$. If $p=2$, this means $z = 1$. If $p > 2$, this means that $z$ has order $4$ in $\mathbb{F}_p^*$, which happens iff $4|p-1$ i.e. $p \equiv 1 \bmod 4$. Hence, for every odd prime $p$ with $p \not\equiv 1 \bmod 4$ the quadratic form $x^2+y^2=0$ has only the trivial solution.
(While your method for $p=7$ is fine, try it with $p=67$!)
| {
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"timestamp": "2023-03-29T00:00:00",
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Find an unknown coefficient in a line equation... So, I have to find the unknown coefficient in this line: $$x+y+C=0$$ so that it is a tangent to this circle: $$x^2+y^2-5x-7y+6=0$$
I've transformed the circle equation to this form:
$$(x-\frac{5}{2})^2+(y-\frac{7}{2})^2=\frac{25}{2}$$
But I can't see how it can help me lol :D
| Substitute $y = C-x$ into the equation of the circle to get:
$x^2+(C-x)^2-5x-7(C-x)+6 = 0$
$2x^2+(2-2C)x+(C^2-7C+6) = 0$
If the line is tangent to the circle, this equation has only one solution for $x$. So the discriminant of the quadratic should be $0$, i.e.
$(2-2C)^2 - 4 \cdot 2 \cdot (C^2-7C+6) = 0$
This will give you two solutions for $C$.
($C=1$ and $C=11$)
| {
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"timestamp": "2023-03-29T00:00:00",
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What does this log notation mean? Can someone please explain what $^2\log x$ means? Is it the same as saying $\log x^2$ or is it something completely different? Here is an image of it as an example:
| If $\log_2 \dfrac{x - y}{3} = 0$, then $\dfrac{x - y}{3} = 1 \Rightarrow x - y = 3$.
Now, $\log_4 y = \dfrac{\log_2 y}{\log_2 4} = \dfrac{1}{2}\log_2{y}$.
Then $\log_2 x + 2 \log_4 y = \log_2 x + \log_2 y = \log_2 xy$.
Thus, $\log_2 x + 2 \log_4 y = 2 \Rightarrow \log_2 xy = 2 \Rightarrow xy = 2^2 = 4$.
Now we have, $xy = 4$ and $x - y = 3$, and $x = 4, y = 1$ is an obvious solution.
Therefore, $\boxed{x + y = 5}$.
Note: To solve it in a more general manner-
$x - y = 3 \Rightarrow\\
y = x - 3 \Rightarrow\\
x(x - 3) = 4 \Rightarrow\\
x^2 - 3x - 4 = 0 \Rightarrow\\
(x - 4)(x + 1) = 0 \Rightarrow\\
x = 4, -1
$
We reject $x = -1$ as $\log (-1)$ is not a real number.
| {
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"timestamp": "2023-03-29T00:00:00",
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Double Integral I need help calculating these double integrals (in order to show they are not equal):
$$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy\,dx \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx\,dy$$
| Consider the double integral over the rectangle $[0,1]\times[0,1]$
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x.$$
Partial fraction decomposition yields
$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}y \, \mathrm{d}x,$$
$$=\int\limits_{[0,1]} \! \frac{y}{x^2+y^2} \Bigg|_{y=0}^{y=1} \, \mathrm{d}x = \int\limits_{[0,1]} \! \frac{\mathrm{d}x}{x^2+1} = \frac{\pi}{4}.$$
We then consider the double integral over the rectangle $[0,1]\times[0,1]$
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y.$$
$$=\iint\limits_{[0,1]\times[0,1]} \! \frac{2x^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \, \mathrm{d}x \, \mathrm{d}y,$$
$$=-\int\limits_{[0,1]} \! \frac{x}{x^2+y^2} \Bigg|_{x=0}^{x=1} \, \mathrm{d}y = -\int\limits_{[0,1]} \! \frac{\mathrm{d}y}{y^2+1} = -\frac{\pi}{4}.$$
Therefore,
$$\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}y \, \mathrm{d}x \neq\iint\limits_{[0,1]\times[0,1]} \! \frac{x^2-y^2}{(x^2+y^2)^2} \, \mathrm{d}x \, \mathrm{d}y$$
| {
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Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive.
\begin{equation*}
\big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a
\end{equation*}
How can I obtain some positive lower-bound?
| Factor out an $a^2$ from the radical to get $a\sqrt{1+\frac{b}{a^2}}-a=a\left(\sqrt{1+\frac{b}{a^2}}-1\right)$
Which can then be expanded for $\left|\frac{b}{a^2}\right|<1$, which is true for $a \gg b > 0$.
This expansion, to first order, is $a\left(1+\frac{b}{2a^2}-1\right)=\frac{b}{2a}$.
EDIT: Forgot that you were looking for meaningful bounds.
For a lower bound, you need to take the expansion to second order, so $a\left(1+\frac{b}{2a^2}-\frac{b^2}{8a^4}-1\right)=\frac{b}{2a}-\frac{b^2}{8a^3}$.
For an upper bound, you need the third order expansion, so $a\left(1+\frac{b}{2a^2}-\frac{b^2}{8a^4}+\frac{b^3}{16a^6}-1\right)=\frac{b}{2a}-\frac{b^2}{8a^3}+\frac{b^3}{16a^5}$
So overall, we have $\frac{b}{2a}-\frac{b^2}{8a^3}<\sqrt{a^2+b}-a<\frac{b}{2a}-\frac{b^2}{8a^3}+\frac{b^3}{16a^5}$
These inequalities are true for $0<\frac{b}{a^2}<1$, which is true in this case.
| {
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Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answer. :/
| $$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}=\frac{3(x^2 + x -2)}{2(x^2 + 3x + 2)}=$$
$$=\frac{3(x^2 -1+ x -1)}{2(x^2 + 2x+x + 2)}=\frac{3((x-1)(x+1)+(x-1))}{2(x(x+2)+(x+ 2)}=$$
$$=\frac{3(x-1)(x+1+1)}{2(x+2)(x+ 1)}=\frac{3(x-1)(x+2)}{2(x+2)(x+1)}=\frac{3(x-1)}{2(x+1)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Powers and differences of positive integers Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5=b^4$, $c^3=d^2$, and $c-a=19$. Determine $d-b$.
I know this question isn't particularly hard, but I've been having trouble getting any substantial ground. I've tried substituting that $c=a+19$ and $a=c-19$ into the first two equations, but it hasn't looked very useful.
A path to a solution or a few hints would be great. Thanks!
| From $a^5 = b^4 \implies a = \left(\frac{b}{a}\right)^4$, we can deduce that $\frac{b}{a}$ is an integer. I will let $m = \frac{b}{a}$.
Similarly, from $c^3 = d^2 \implies c = \left(\frac{d}{c}\right)^2$, we can deduce that $\frac{d}{c}$ is an integer. I will let $n = \frac{d}{c}$.
Since $a = m^4$ and $c = n^2$, we have
$$n^2 - m^4 = 19$$
$$(n + m^2)(n - m^2) = 19$$
Since $19$ is a prime, we only have to check the cases for which $19 = 1\cdot19$ and for which $19 = -1\cdot-19$. If we let $n + m^2 = 19$ and $n - m^2 = 1$, then we can get $n = 10, m = 3$. I suppose it is trivial to show that all other cases do not give valid results.
Hence, $\frac{d}{c} = n = 10 \implies d = 10c$, and $\frac{b}{a} = m = 3 \implies b = 3a$. It should be easy to continue on from here!
(Just in case you wanna check: $a = 81, b = 243, c = 100, d = 1000$)
| {
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If $f(x) = \frac{x}{x+1}$ and $g(x) = 2x-1$, find $(g\circ f) (x)$. If $f(x) = \frac{x}{x+1}$ and $g(x) = 2x-1$, find $(g\circ f)(x)$.
My answer is $\frac{x-1}{x+1}$. However, the answer key in the book states $\frac{2x}{x+1}$. How is that? Is the book wrong?
| Wolframalpha and me agree with you, if
$$f(x)=\frac{x}{x+1}, g(x)=2x-1 $$ then
$$g \circ f(x)=g(f(x))= 2f(x)-1 = \frac{2x}{x+1}-1= \frac{2x-x-1}{x+1}= \frac{x-1}{x+1}.$$
| {
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Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
| It is easier to break into two limits, $x-\sqrt[3]{x^3+x^2+1}$ and $x-\sqrt[3]{x^3-x^2+1}$ then we shall see the first limits to $-\frac{1}{3}$ and the second $\frac{1}{3}$ so the sum limits to zero.
Note the more general result,
$$x-\sqrt[n]{x^n +ax^{n-1}+\cdots}\rightarrow -\frac{a}{n}$$
If we write $A=\sqrt[n]{x^n +ax^{n-1}+\cdots}$ then
$$x-A=\frac{x^n-A^n}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}=\frac{-ax^{n-1}+\cdots}{x^{n-1}+x^{n-2}A +\cdots +A^{n-1}}$$
and since $\frac{A}{x}\rightarrow 1$ the rest is easy.
| {
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Proving an expression is composite I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.
I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.
| I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.
*
*We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
*
*$n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
*Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
*$(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
*$(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
*$(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
*$(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
*We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
*
*Treat $n^2+2^n$ as $a$
*Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
*So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
*$2^{2m}n^2=(n2^m)^2$ associative law
*Treat $n2^m$ as $b$
*It implies that both $a$ and $b$ are both positive integer
*From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
*Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of the series $\sum \limits_{n=3}^{\infty} \dfrac{1}{n^5-5n^3+4n}$ Feel free to skip obvious steps, or use a calculator when required.
I just want to understand the theme of the solution.
Any help is appreciated
EDIT :
We can write$$ \dfrac{1}{n^5-5n^3+4n} = -\dfrac{1}{6 (n-1)}+\dfrac{1}{4 n}-\dfrac{1}{ 6(n+1)}+\dfrac{1}{ 24(n+2)}+\dfrac{1}{ 24(n-2)}$$
How do I do telescopic sum !
| $$\sum_{n=3}^{\infty}\dfrac{1}{n^5-5n^3+4n} = \sum_{n=3}^{\infty}\dfrac{1}{(n-2)(n-1)n(n+1)(n+2)}$$
$ $
Rewrite this as $$ \dfrac{1}{4}\sum_{n=3}^{\infty}\dfrac{(n+2)-(n-2)}{(n-2)(n-1)n(n+1)(n+2)}$$
$ $
$$ \dfrac{1}{4}\sum_{n=3}^{\infty}\left[\dfrac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\dfrac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)} \right]$$
$ $
$$ \dfrac{1}{4}\sum_{n=3}^{\infty}\left[\dfrac{1}{(n-2)(n-1)n(n+1)}-\dfrac{1}{(n-1)n(n+1)(n+2)} \right]$$
$ $
$$\textbf{This is a telescoping sum}$$
$ $
$ $
$$\lim\limits_{n \to \infty}\dfrac{1}{4} \sum_{i=3}^{n}\left[\dfrac{1}{(i-2)(i-1)i(i+1)}-\dfrac{1}{(i-1)i(i+1)(i+2)} \right]$$
$ $
$$\require{cancel} =\lim\limits_{n \to \infty} \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} \cancel{-\dfrac{1}{2\cdot 3\cdot 4\cdot 5} } \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{2\cdot 3\cdot 4\cdot 5}}\cancel{ -\dfrac{1}{3\cdot 4\cdot 5\cdot 6} } \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{3\cdot 4\cdot 5\cdot 6}}\cancel{ -\dfrac{1}{4\cdot 5\cdot 6\cdot 7} }\right]$$
.
$ $
.
$ $
........
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{(n-4)\cdot (n-3)\cdot (n-2)(n-1)} }\cancel{-\dfrac{1}{(n-3)\cdot (n-2)\cdot (n-1)\cdot(n)} } \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{(n-3)\cdot (n-2)\cdot (n-1)\cdot (n)}} -\dfrac{1}{(n-2)\cdot (n-1)\cdot n\cdot (n+1)} \right]$$
$ $
$ $
$$ =\lim\limits_{n \to \infty} \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} -\dfrac{1}{(n-2)\cdot (n-1)\cdot n\cdot (n+1)} \right]$$
$ $
$$ = \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} -0 \right] = \dfrac{1}{1\cdot 2\cdot 3\cdot 4^2} = \dfrac{1}{96}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what is a smart way to find $\int \frac{\arctan\left(x\right)}{x^{2}}\,{\rm d}x$ I tried integration by parts, which gets very lengthy due to partial fractions.
Is there an alternative
| It is not so lenghty. If $$I =\int \frac{\tan^{-1}x}{x^2}\,dx$$ let start with $u=\tan^{-1}x$ and $v'=\frac{\tan^{-1}x}{x^2}\,dx$. So $u'=\frac{\text{dx}}{x^2+1}$ and $v=-\frac{1}{x}$ and then $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=-\frac{\tan ^{-1}(x)}{x}+\int \frac{dx}{x \left(x^2+1\right)}$$ Now, using partial fraction decomposition $$\frac{1}{x \left(x^2+1\right)}=\frac{1}{x}-\frac{x}{x^2+1}$$ and, finally $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=\log (x)-\frac{1}{2} \log \left(x^2+1\right)-\frac{\tan ^{-1}(x)}{x}+C$$
Added later after André Nicolas comment
If you really want to avoid partial fraction decomposition, use the substitution $x=\frac{1}{y}$; this gives $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=-\int \tan ^{-1}\left(\frac{1}{y}\right)\,dy $$ and a single integration by parts will give $$I=-\frac{1}{2} \log \left(y^2+1\right)-y \tan ^{-1}\left(\frac{1}{y}\right)+C$$
| {
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Is this procedure for $5^{300} \bmod 11$ correct? I'm new to modular exponentiation. Is this procecdure correct?
$$5^{300} \bmod 11$$
$$5^{1} \bmod 11 = 5\\
5^{2} \bmod 11 = 3\\
5^{4} \bmod 11 = 3^2 \bmod 11 = 9\\
5^{8} \bmod 11 = 9^2\bmod 11 = 4\\
5^{16} \bmod 11 = 4^2 \bmod 11 = 5\\
5^{32} \bmod 11 = 5^2 \bmod 11 = 3$$
$$5^{300} = 3 + 3 + 3 + 3 +3 +3 + 3 + 3 +3 +4 + 9$$
| I see what you were trying to do with adding the numbers at the end.
Multiply them instead, and take that product $\mod 11$.
| {
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?
| Note that this answer is not completely rigorous, but it was too fun to pass up.
$$K^2 = 101 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots \cdot 100 \cdot 100}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots \cdot 99 \cdot 101}$$
Now, $$\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots} = \pi / 2$$
(This is known as the Wallis product)
So $K$ is approximately $\sqrt{101 \pi / 2}$ and $\lfloor K \rfloor = 12$
| {
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Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. The problem asks to show that $$f(x,y) = \left\{ \begin{align} \frac{x^3y^2}{x^4+y^4}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{align} \right.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and therefore $$\left\vert \frac{x^3y^2}{x^4+y^4} \right\vert \leq \left\vert \frac{x^3}{y^2} \right\vert$$ and similar strategies but they have failed. Trying to disprove it and see if it's discontinuous has only strengthened the belief that it's continuous.
| To apply the squeeze theorem, notice that if $x\neq 0$ and $y\neq 0$ then
$$\begin{align*}
\left|\frac{x^3y^2}{x^4+y^4}\right|&=\left|x\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
&=\left|x\frac{x^{-2}}{x^{-2}}\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^{-2}}{y^{-2}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
& =\left|x\frac{1}{\sqrt{1+y^4/x^4}}\frac{1}{\sqrt{x^4/y^4+1}}\right|\\
&=|x|\left|\frac{1}{\sqrt{1+y^4/x^4}}\right|\left|\frac{1}{\sqrt{x^4/y^4+1}}\right|\\
&\leq |x|\cdot 1\cdot 1.
\end{align*}$$
So, for all $(x,y)\in\mathbb{R}^2$ we have $0\leq |f(x,y)|\leq|x|$. It follows that
$$\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0)$$
and thus $f$ is continuous at origin.
| {
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Solve $a$ and $b$ for centre of mass in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Given ellipse:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
What length do $a$ and $b$ have to be so the centre of mass is $S(4;2)$?
I've tried steps to solve the equation to $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ and integrate
$$A=b\int_0^a{\sqrt{1-\frac{x^2}{a^2}}}$$
But I'm not achieving a satisfying result. There must be an easier way . Enlighten me please
| Combining the results in the previous discussion and answers, we set $x=a t$, then $y=b\sqrt{1-t^2}$ and $dx=adt$. So:
$$\bar{x}=\frac{\int_0^a xy(x)\,\mathrm{d}x}{\int_0^a y(x)\,\mathrm{d}x}=\frac{a^2b}{ab}\frac{\int_0^1 t \sqrt{1-t^2}\,\mathrm{d}t}{\int_0^1 \sqrt{1-t^2}\,\mathrm{d}t}=a\frac{1/3}{\pi/4}=\frac{4a}{3\pi}$$
Similarly we obtain:
$$\bar{y}=\frac{4b}{3\pi}$$
Thus
$$\bar{x}=\frac{4a}{3\pi}=4 \implies a=3\pi$$
$$\bar{y}=\frac{4b}{3\pi}=2 \implies b=\frac{3\pi}{2}$$
| {
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Chinese Remainder Theorem $\ x\equiv m_k-2\pmod{m_k}$ I am trying to find all integers that give remainders 1,2,3 when divided by 3,4,5 respectively. So I start defining
$$a_1=1, a_2=2, a_3=3,$$$$ m_1=3, m_2=4, m_3=5,$$$$ m_1m_2=12, m_1m_3=15, m_2m_3=20,$$$$ m=60$$
All the moduli are pairwise coprime, so by the Euclidean Algorithm
$$12x+5y=1\Rightarrow x=-2, y=5.$$
$$15x+4y=1\Rightarrow x=-1, y=4.$$
$$20x+3y=1\Rightarrow x=-1, y=7.$$
Therefor a solution should be
$$20\cdot(-2)\cdot1 +15\cdot (-1)\cdot 2+12\cdot (-1)\cdot 3=-40-30-36=-106$$
But under mod 60, this implies a solutions should be of the form, $60t+14$. But it is not. THis is not right. Why did this technique not work?
| Another, more direct way of solving this runs as follows. You are looking at
$$x \equiv 1 \text{ mod } 3$$
$$x \equiv 2 \text{ mod } 4$$
$$x \equiv 3 \text{ mod } 5.$$
Now put $x=1+3k$ and mod this by $4$. Then $2\equiv 1 +3k \text{ mod } 4$, so $k\equiv -1 \text{ mod } 4$, say $k=4l-1$, hence $x=-2+12l$. Now take the third equation into account. Then $3 = -2+12l\text{ mod } 5$, which is equivalent to $l \equiv 0 \text{ mod } 5$, say $l=5m$. So, finally $x=-2+60m$, whence $x\equiv -2\equiv58 \text{ mod } 60$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that
$$ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013
let $$P(x)=\sum_{i=0}^{n}a_{i}x^i,a_{i}\in R$$
then
$$\sum_{i=0}^{n}[a_{i}x^3(x+1)^i-mx^2(x+1)^i+(x+1)^i]+\sum_{i=0}^{n}[x^3(x-1)^i+mx^2(x-1)^i+(x-1)^i]=2\sum_{i=0}^{n}[x^{3+i}-mx^{i+1}+x^i]$$
then I can't
| EDIT (by abiessu): the original is wrong, but as it has been accepted, it seems that it would be good to make the accepted answer correct. The original approach is generally used but corrected where in error.
(original) I suspect that there is a typo in your OP, because as it is stated now it is rather uninterestingly easy and there are only trivial (linear) solutions.
Definitely not IMO Shortlist stuff.
We can assume that $n$ is the degree of $P$, so that $a_n\neq 0$. Then,
if $n \neq 0$ the coefficient before $x^{n+2}$ in
$D=(x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) -2(x^3 - mx +1 ) P(x)$
is given by considering
$$x^3\sum_{j=0}^n a_j\left((x+1)^j+(x-1)^j-2x^j\right)\tag 1$$
$$-mx^2\sum_{j=0}^na_j(x+1)^j+mx^2\sum_{j=0}^na_j(x-1)^j\tag 2$$
$$2mx\sum_{j=0}^na_jx^j\tag 3=2m\sum_{j=0}^na_jx^{j+1}$$
We see that $j=n$ results in $0=0$ from $(1)$, and this is the only $x^{n+3}$ source. The same thing happens when we try to find $x^{n+2}$ terms, and so non-zero $x^{n+1}$ terms must be considered, and these are
$$a_{n}n(n-1)x^{n+1}-2ma_{n}nx^{n+1}+2ma_nx^{n+1}=0$$
The result of this is that $a_{n}(n(n-1) - 2m(n-1))=0$. This is not fully conclusive, but it does mean that we can claim one of three scenarios:
*
*$a_n = 0$
*$n = 2m$
*$n-1 = 0$
Of these, $a_n = 0$ contradicts our assumption, so we have either $n = 2m$ or $n = 1$. Considering the first possibility, note that $m\ne 0$ is an integer, and therefore $n=2m \to m \ge 1$ and $n=2m \to n \ge 2$.
Now considering terms having $x^n$, we find terms
$$a_{n-1}n(n-1)x^n-2mna_{n-1}x^n+2ma_{n-1}x^n=0$$
This exactly repeats our previous scenario, but now $a_{n-1}=0$ is an acceptable possibility, and $n=2m$ or $n=1$ are simply confirmed as alternate options. This pattern will almost certainly repeat itself ad infinitum, except that considering terms with $x^{n-1}$ will bring in the following extra terms:
$$\frac 1{12}a_nn(n-1)(n-2)(n-3)x^{n-1}-\frac 13ma_nn(n-1)(n-2)x^{n-1}=0$$
which brings us to $\frac 14n(n-1)(n-2)(n-3)-mn(n-1)(n-2)=0$. This narrows our possibilities to either $n=1$ or $n=2m$ with $n=2$ or $n-3=4m$.
Now we can see that $n=1,2$ are the only possibilities when $a_n$ is non-zero, since $n=4m+3$ is incompatible with $n=2m$ for positive $n$.
If we assume that $n=2$ and use $P(x) = ax^2 + bx + c$, we get
$$ax^2(x^3-mx^2+1)+(2a+b)x(x^3-mx^2+1)+(a^2+b+c)(x^3-mx^2+1)\\
+ax^2(x^3+mx^2+1)+(-2a+b)x(x^3+mx^2+1)+(a^2-b+c)(x^3+mx^2+1)\\
=2ax^2(x^3-mx+1)+2bx(x^3-mx+1)+2c(x^3-mx+1)$$
resulting in
$$-2amx^3+2a^2x^3+2a^2=-2cmx$$
The only way this can be true for all $x$ is if $a = 0$ and $c=0$, thus we have that $n=1$ is the only source of polynomial solutions for our original equation, and every such solution is of the form $P(x)=ax$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An inequality about a sequence Let $(a_n)$ be a sequence such that $a_0=1 , a_1=2 , a_{n+1}=a_n+\dfrac {a_{n-1}}{1+ a_{n-1}^2} , \forall n \ge1 $ , then is it true that $52 < a_{1371} < 65$ ?
$ EDIT:-$ I am posing another question , so I'm not able to accept Oleg567 ' s very correct answer :
Let $(a_n)$ be a sequence such that $a_0=1 , a_1=2 , a_{n+1}=a_n+\dfrac {a_{n-1}}{1+ (a_n-1)^2} , \forall n \ge1 $ , then is it true that $52 < a_{1371} < 65$ ?
| First, note that next recurrent relation is true for sequence $(a_n)$:
$$
a_0=1,\\
a_{n+1} = a_n+\frac{1}{a_n}, \qquad n\ge 0.\tag{1}
$$
Yes, $a_0=1$, $a_1 = 1+\frac{1}{1}=2$ for both definitions; if $a_{n+1}$ is defined by $(1)$, then
$$
a_{n+1} = a_n+\dfrac{1}{a_n} = a_n + \dfrac{1}{a_{n-1}+\frac{1}{a_{n+1}}} =
a_n + \dfrac{a_{n-1}}{1+a_{n-1}^2}, \qquad n\ge 1.\tag{2}
$$
Hmm, nice recurrent formula as finite continued fraction:
$$
a_{n+1} = a_n+\cfrac{1}{a_{n-1}+\cfrac{1}{a_{n-2}+\cfrac{1}{\cdots + \cfrac{1}{a_1+\cfrac{1}{a_0+1}}}}}\tag{3}
$$
Looking at $(1)$, one can show that
$$
\sqrt{2n}<a_{n-1}<\sqrt{\left(2+\frac{1}{6}\right)n}, \qquad n\ge 3.\tag{4}
$$
Proof of $(4)$ is here.
For $a_{1371}$ we get:
$$
52<52.383\approx\sqrt{2744}<a_{1371}<\sqrt{2972\frac{2}{3}}\approx 54.522<65.$$
Note:
looking at bounds $52$ and $65$, I think that there is enough to show weaker (than $(4)$) inequality: just that
$$
\sqrt{2n}<a_{n-1}<\sqrt{3n},\qquad n\ge 3.\tag{5}
$$
| {
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$ \int \frac{1}{(x-a)(x+b)} dx $ Could you please explain how to integrate this integral:
$$ \int \frac{1}{(x-a)(x+b)} dx $$
| Since there is linear factor in denominator,
$$\int \frac{1}{(x-a)(x+b)} dx=\int \frac{1}{(x-a)(a+b)}+\frac{1}{(-b-a)(x+b)} dx$$
Factor out $\frac{1}{a+b}$,
$$=\frac{1}{a+b}\int \frac{1}{x-a}-\frac{1}{x+b} dx$$
The anti-derivatives of $\ln{(x+k)}$ is $\frac{1}{x+k}$
$$=\frac{1}{a+b}\bigg[\ln{(x-a)}-\ln{(x+b)}\bigg]+C$$
Simplify them will give you,
$$=\frac{1}{a+b}\ln{\frac{x-a}{x+b}}+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$
This is what I did.
Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\frac{2}{x+1}$ by $(x-1)$ as following and I think this part might be wrong:
*
*$x(x-1) = x^2-x$. Times $x+1$ cause that's the denominator =
*$x^3+x^2-x^2-x = x^3-x$.
*After this I added the $+ 2$
*$\frac{x^3-x+2}{(x-1)(x+1)}$
Step 4: I canceled out the denominator $(x-1)(x+1)$ on both sides.
Step 5: And I'm left with: $\frac{x+2}{x^3-x+2}$
Step 6: Removed $(x+2)$ from both sides I got my UN-correct answer: $\frac{1}{x^3}$
Please help me. What am I doing wrong?
| You need to multiply the +2 by (x-1) before you add it to $x^3-x$ because
$$\frac{2}{x+1}=\frac{2(x-1)}{(x-1)(x+1)}$$
Also, in step 6, you can't remove (x+2) from top and bottom. Firstly, $x^3-x+2=x^3-(x-2)$, so there is no $x+2$ in the denominator. Secondly, you need (x+2) to be a factor of the whole denominator, not just part.
| {
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What is the count of the strict partitions of n in k parts not exceeding m? Lets say we had a $k,m,n \in \mathbb{N}$ where $k < m \le n$. How many different sets $X_1,..,X_m$ with $|X_i|=k$ for $i=1,..,m$, where the sets do not include duplicates, for which the sum of their elements is equal to n?
$$n = X_{sum}^i := \sum_{j=1}^k x_j^i, \; x_j^i \in X_i$$
It holds, that all elements $x_j^i \le m$ and for each pair $|X_i \cap X_j| < k$, where $i=1,..,m$ and $j=1,..,m$ but $i\neq j$.
As example, we take $k=3$, $n=15$ and show all possibilities for m:
m = 6 : 6+5+4 (1)
m = 7 : 7+5+3, 7+6+2 (2)
m = 8 : 8+4+3, 8+5+2, 8+6+1 (3)
m = 9 : 9+4+2, 9+5+1 (2)
m = 10: 10+3+2, 10+4+1 (2)
m = 11: 11+3+1 (1)
m = 12: 12+2+1 (1)
Is there a general formula to calculate this values for $n,k,m$ or at least for $k=3$?
Finally my goal was if $C_{n,k,m}$ is the count of the values, the calculation of
$$C_{n,k} = \sum_{i=1}^n (C_{n,k,i} - 1).$$
I found out a formula for $C_{n,2}$ which was:
$$C_{n,2} = {n - 2 \choose 2},$$
but I was not able to generalize it for k, since it does not hold, that
$$C_{n,k} = {n - 2 \choose k},$$
for all k.
| I'm assuming from your list of examples that what you are looking for is the number of partitions of $n$ into $k$ distinct parts with largest part equal to $m$ (this is different from the question in the title, but seems to fit your examples best).
The number of partitions of $n$ with at most $k$ parts, each of length at most $m$, is the coefficient of $q^n$ in the Gaussian binomial coefficient $\binom{m+k}{k}_q$. To have only those with exactly $k$ parts, one of which is exactly of length $m$, we first get rid of all partitions in $k-1$ parts or less, and then of all those whose largest part is at most $m-1$, and add back those in at most $k-1$ parts of length at most $m-1$, which we have deleted twice. This number is then the coefficient of $q^n$ in
$$\binom{m+k}{k}_q -\binom{m+k-1}{k}_q - \binom{m+k-1}{k-1}_q + \binom{m+k-2}{k-1}_q.$$
Now, we want partitions of $n$ having these conditions and having distinct parts. The trick here is to see that such partitions are in bijection with partitions of $n-\binom{k}{2}$ with exactly $k$ (not necessarily distinct) parts and largest part of length $m-k+1$. This trick is explained for instance in this answer.
Therefore, the number you are looking for is the coefficient of $q^{n-\binom{k}{2}}$ in
$$\binom{m+1}{k}_q -\binom{m}{k}_q - \binom{m}{k-1}_q + \binom{m-1}{k-1}_q.$$
In the particular example where $m=12$ and $k=3$, this expression is the polynomial
$$q^{30} + q^{29} + 2q^{28} + 2q^{27} + 3q^{26} + 3q^{25} + 4q^{24} + 4q^{23} + 5q^{22} + 5q^{21} + 5q^{20} + 4q^{19} + 4q^{18} + 3q^{17} + 3q^{16} + 2q^{15} + 2q^{14} + q^{13} + q^{12},$$
so for $n=15$, you look at the coefficient in front of $q^{15-\binom{k}{2}} = q^{12}$, which is $1$ as expected. For $n=23$, you would take the coefficient in front of $q^{20}$, which is $5$; this corresponds to the five partitions $12+10+1 = 12+9+2 = 12+8+3 = 12+7+4 = 12+6+5$.
| {
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Seeking concise proof: $\frac18(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge\frac1{27}(ab+bc+ca)^3$, where $a$, $b$, $c$ are positive numbers I was just encountered an inequality in AoPs, Here it is:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=4569&view=next,
that is:
If $a$, $b$ and $c$ are positive numbers, then we have inequality: $$\dfrac{(a^2+b^2)(b^2+c^2)(c^2+a^2)}{8}\ge\dfrac{(ab+bc+ca)^3}{27}$$
If the above ineq is succeed, it's easy to generalize the variables into real number.
Here I've got an terrible solution:
just expand(I think it is complicated to deal with this ineq using expand and easy to make mistake) and the inequality is equivalent to the following symmetric Ineq.
$$6a^2b^2c^2+27\sum_{sym} a^4b^2\ge4\sum_{sym} a^3b^3+24\sum_{sym} a^3b^2c,$$
then by the A-G Ineq,
$$6a^2b^2c^2+\sum\limits_{sym} b^4c^2=\sum\limits_{sym}(a^4b^2+a^2b^2c^2)\ge2\sum\limits_{sym}a^3b^2c, $$
and Muirhead's theorem implies
\begin{equation}
\sum_{sym}a^4b^2\ge\sum_{sym}a^3b^2c,\quad
\sum_{sym}a^4b^2\ge\sum_{sym}a^3b^3.
\end{equation}
finally the ineq can be proved by three appropriate coeffients multiplied to the previous three ineq and total them up.
Is there a concise proof without complicated calculating?(also it swept the buffalo way away). this is what I interested.
| We'll prove that your inequality is true for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+b^2)=\prod\limits_{cyc}(a^2+b^2+c^2-c^2)=-w^6+A(u,v^2)w^3+B(u,v^2)$,
we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function,
which says that it's enough to prove our inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Since our inequality is even degree, it's enough to assume $b=c=1$,
which gives $27(a^2+1)^2\geq4(2a+1)^3$ or $(a-1)^2(27a^2+22a+23)\geq0$.
Done!
| {
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Trouble evaluating the sum involving logarithm I was trying to solve this problem: Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
In the procedure I followed, I came across the following sum:
$$\sum_{k=1}^{\infty} (-1)^{k-1}k\left(\frac{\ln(2k+1)}{2k+1}-\frac{\ln(2k-1)}{2k-1}\right)$$
I cannot think of any approaches which would help me in evaluating the sum.
Any help is appreciated. Thanks!
| On the limit
Experimenting a bit I find
$$\lim_{m\to\infty} S_m = \ln\left(\frac{\Gamma(\frac14)}{2\Gamma(\frac34)}\right) = 0.391594392706836...$$
To be the exact limit
On convergence
Let $a_k := \frac{\ln(2k-1)}{2k-1}$ then
$$\begin{align*}
S_m & = \sum_{k=1}^m (-1)^{k-1} k (a_{k+1} - a_k) \\
& = \sum_{k=2}^{m+1} (-1)^k (k-1) a_k + \sum_{k=1}^m (-1)^k k a_k \\
& \stackrel{a_1 = 0}= \sum_{k=2}^m (-1)^k ((k-1) a_k + k a_k) + (-1)^{m+1} ma_{m+1} \\
& = \sum_{k=2}^m (-1)^k \ln(2k-1) + (-1)^{m+1} \frac{m}{2m+1} \ln(2m+1)
\end{align*}$$
The problem all answerers overlooked is the factor of the last summand not being $\frac1{2m+1}$ but $\frac{m}{2m+1}$.
Let's first try to prove that the sequence is cauchy:
$$\begin{align*}
|S_{m+1}-S_m| & = |(-1)^{m+1} \ln(2m+1) \\
&\left. \qquad + (-1)^m \frac{m+1}{2m+3} \ln(2m+3) - (-1)^{m+1} \frac m{2m+1} \ln(2m+1) \right| \\
& = \left| (m+1) \left(\frac1{2m+3} \ln(2m+3) - \frac1{2m+1} \ln(2m+1)\right) \right| \\
& \le (m+1) \ln(2m+3) \left(\frac1{2m+1} - \frac1{2m+3}\right) \\
& = \ln(2m+3) (m+1) \frac2{4m(m+2) + 3} \\
& \le \ln(2m+3) \frac1{2m} \to 0 & (m\to\infty)
\end{align*}$$
This proves convergence. The actual value seems a tad bit harder to prove.
| {
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How do I reduce radian fractions? For example, I need to know $\sin (19π/12)$.
I need to use the subtraction formula. How do I get $(\text{what}) - (\text{what}) = 19π/12$? I am stuck at what are the radians
Do I divde it by something? What is the process?
| $$\frac{19\pi}{12}=\pi+\frac{3\pi}4-\frac{\pi}6$$
Since $\sin(\pi+x)=-\sin x$, $\sin\left(\frac{19\pi}{12}\right)=-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)$.
And since $\sin(A-B)=\sin A\cos B-\cos A\sin B$:
$$\begin{align}
-\sin\left(\frac{3\pi}4-\frac{\pi}6\right)&=-\left(\sin\left(\frac{3\pi}4\right)\cos\left(\frac{\pi}6\right)-\cos\left(\frac{3\pi}4\right)\sin\left(\frac{\pi}6\right)\right) \\
&=-\left(\frac{\sqrt{2}}2*\frac{\sqrt{3}}2-\left(-\frac{\sqrt{2}}2 * \frac12\right)\right) \\
&=-\frac{\sqrt{2}+\sqrt{6}}4
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem on nested radicals Find the value of $x$ for all $a>b^2$ if:
$$\large x=\sqrt{a-b\sqrt{a+b\sqrt{a-b{\sqrt{a+b.......}}}}}$$
My attempt
$$\large x=\sqrt{a-b\sqrt{(a+b)x}}$$
$$\large x^4=(a-b)^2(a+b)x$$
$$\large x=((a-b)^2(a+b))^{1/3}$$ (real root)
Question: Is my solution correct??
| Here's another one:
$$\text{Let} x=\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\cdots}}}}$$
$$\text{Let}y=\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{a-b\cdots}}}}$$
Now $$x^2=a-by---1$$ and $$y^2=a+bx----2$$ $$\implies x^2-y^2=-b(x+y)$$ $$\implies(x+y)(x-y+b)=0$$(Considering only positive solutions of $x$) $$y=x+b$$, putting it in eq $2$, $$x^2+bx+b^2-a=0$$, on Solving:
$$\large x=\frac{-b+\sqrt{4a-3b^2}}{2}$$
| {
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if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good partial fractions methods by hand?
because I take an hour to solve this problem.
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
| The $n$-th term is obtained from $(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)(1+x^4+x^8+\cdots)$ which is the number of triples $(i,jk)$ that are solutions to $2i+3j+4k=n$. Hence, $$a_n=\sum_{2i+3j+4k=n}1=\sum_{j=0}^{\lfloor n/3\rfloor }\sum_{2i+4k=n-3j}1$$
Now we want to obtain $$b_m=\sum_{2i+4k=m}1$$
It is clear $b_{2m+1}=0$, so we look at $$b_{2m}=\sum_{i+2k=m}1$$ Now $$\sum_{2k=l}1= [2\mid l]$$ hence $$b_{2m}=\left\lfloor\frac m2\right \rfloor+1$$
For example, $2i+4k=10$ has three solutions, $(5,0),(3,1),(1,2)$, i.e $\lfloor 5/2\rfloor+1=3$.
Now $2\mid n-3j\iff 2\mid n-j$, so we get $$a_n= \sum_{j=0}^{\lfloor n/3\rfloor}\left(\left\lfloor \frac{n-3j}4\right\rfloor+1 \right)[2 \mid n-j]$$
For example, $2k+3j+4k=10$ has solutions $(5,0,0), (0,2,1), (2,2,0),(1,0,2),(3,0,1)$, and the formula gives $$(\lfloor 10/4\rfloor+1)+(\lfloor 4/4\rfloor +1)=3+2=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Intuitive ways to get formula of cubic sum Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$
But how to prove it? Please help me. Grazie!
| We have
$$
\sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\
\underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\
\underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots
$$
which is
$$
\left( \sum_{k=1}^{n}k \right)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Ordered partitions of an integer (with a twist) I would like to know how to prove (preferably algebraically) that $P_1(2,n)=F_{2n+1}$, where $P_1(2,n)$ is what I define to be the number of ordered partitions of an integer, where the number $1$ has 2 possible colours. For example,
\begin{align}
2
&=2\\
&={\color\red{1}}+{\color\red{1}}\\
&={\color\green{1}}+{\color\green{1}}\\
&={\color\green{1}}+{\color\red{1}}\\
&={\color\red{1}}+{\color\green{1}}\\
\end{align}
so in this case, $P_1(2,2)=5$.
This is my (failed) attempt to derive a formula for $P_1(k,n)$. Consider a case where the number $n$ is expressed as a sum of $m$ natural numbers, out of which there are $j$ number of $1$s. The number of possible ways to do so is given by
$$\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}$$
Now we look at how many partitions we can get from this particular case. Since there are $j$ number of $1$s, we would have to multiply by $k^j$, and since $0\le j \le m$ and $1 \le m \le n$ we have
\begin{align}
P_1(k,n)
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^{n-2m+j}](1-x)^{-(m-j)}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{m-j+n-2m+j-1}{n-2m+j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
\end{align}
I have tried using the "snake-oil" method to compute this sum, that is,
$$P_1(2,n)=2^{n-1}+[x^{n-1}]\sum_{n\ge 1}\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}x^{n-1}$$
but I, embarrassingly, have trouble with changing the order of summation and determining the limits of summation.
This is where I am stuck, and I would like to seek your help in finding a closed form for this sum, and in particular, proving that
\begin{align}
P_1(2,n)
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}2^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=F_{2n+1}
\end{align}
Help will be greatly appreciated. Thank you.
| The derivation of the generating function can also done from first principles, without using the recurrence. We have straightforwardly that these partitions have the generating function
$$\sum_{q\ge 0} \left(\frac{z}{1-z} + z\right)^q$$
where we have added in the quantity $z$ to account for the fact that there are two types of one value. The index starts at zero which produces a count of one empty ordered partition. The parameter $q$ counts the number of elements in the ordered partition.
This yields
$$\sum_{q\ge 0} \left(\frac{2z-z^2}{1-z}\right)^q
= \frac{1}{1-(2z-z^2)/(1-z)}
= \frac{1-z}{1-z-2z+z^2}
= \frac{1-z}{1-3z+z^2}.$$
Now the generating function of the Fibonacci numbers is
$$\sum_{k\ge 0} F_k z^k =\frac{z}{1-z-z^2}$$
so that odd Fibonacci numbers are generated by
$$\sum_{k\ge 0} F_{2k+1} z^{2k+1} =
\frac{1}{2} \frac{z}{1-z-z^2}
+ \frac{1}{2} \frac{z}{1+z-z^2}.$$
Therefore
$$\sum_{k\ge 0} F_{2k+1} z^{2k} =
\frac{1}{2} \frac{1}{1-z-z^2}
+ \frac{1}{2} \frac{1}{1+z-z^2}
= \frac{1}{2} \frac{2-2z^2}{(1-z^2)^2-z^2}.$$
Hence $$\sum_{k\ge 0} F_{2k+1} z^k =
\frac{1-z}{(1-z)^2-z}
= \frac{1-z}{1-3z+z^2}.$$
This proves the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Evaluate $\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$ I'm trying to solve
$$\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$$
My first approach was to factorise and then do a partial integration. However the factorisation $(u+1)\left(u+\frac{1}{2}-\frac{\sqrt{5}}{2} \right)\left(u+\frac{1}{2}+\frac{\sqrt{5}}{2} \right)$ leads me to heavy calculations. How would you proceed to solve this?
This is a continued calculation of Solving $\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$ with substitution
| Hint: $$u^3+2u^2-1=(u+1)(u^2+u-1).$$ Completing the square $$ u^2+u-1= u^2+u\frac{1}{4}-\frac{1}{4}-1=(u+1/2)^2-5/4.$$
Now use partial fraction
\begin{align}
\frac{1}{(u+1)((u+1/2)^2-5/4)}=\frac{A}{u+1}+\frac{Bu+C}{( u+1/2)^2-5/4}
\end{align}
We find that $ A=-1, B=1, C=0$, so that
\begin{align}
I&=\int { \frac{-1}{u+1} du}+\int{ \frac{u}{(u+1/2)^2-5/4}du}
\\
&=-\ln ( u+1 )+J
\end{align}
now to find $ J$, since $$denominator= \frac{5}{4}\left({ (\frac {2}{\sqrt 5} (u+1/2))^2-1 }\right)$$ we assume $(\frac{2}{\sqrt5}(u+1/2))= w$ to get
\begin{align}
J=constant \int\frac{\sqrt5 w-1}{w^2-1}dw
\end{align}
Finally, setting $ w=\sec\theta $ and proceed $\cdots $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving recurrence relation: Product form Please help in finding the solution of this recursion.
$$f(n)=\frac{f(n-1) \cdot f(n-2)}{n},$$
where $ f(1)=1$ and $f(2)=2$.
| Take some large n,
$$
\begin{align}
f(n) &= \frac{f(n-1) f(n-2)}{n} \\
&= {f(n-2)^2 f(n-3) \over n (n-1)} \\
&= {f(n-3)^3 f(n-4)^2 \over n (n-1) (n-2)^2} \\
&= \frac{f(n-4)^5 f(n-5)^3}{n (n-1) (n-2)^2 (n-3)^3} \\
&= \dots \\
&=\frac{f\left(n - (n-2) \right)^{F_{n-1}} f\left(n - (n-1) \right)^{F_{n-2}}}{\prod^{n-3}_{i=0} (n-i)^{F_{i+1}}} \\
&=\frac{2^{F_{n-1}}}{\prod^{n-3}_{i=0} (n-i)^{F_{i+1}}}
\end{align}
$$
in each step until the "..." I just plugged the formula into the left factor of the numerator until I saw a pattern. The symbol $F_n$ denotes the Fibonacci sequence, which follows $F_n = F_{n-1} + F_{n-2}$ and $F(0) = 0$ and $F(1) = 1$. Sorry this is not a formal proof, but you can easily prove it by induction by showing that, if it's true for some large n then it must be true for n+1, and then simply demonstrating that it's true for 3.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
| Since the inequality is homogeneous, WLOG assume that $a+b+c = 1$.
Then, the inequality becomes $\dfrac{a}{(1-a)^2}+\dfrac{b}{(1-b)^2}+\dfrac{c}{(1-c)^2} \ge \dfrac{9}{4}$.
Since the function $f(x) = \dfrac{x}{(1-x)^2}$ is concave up for $x > 0$, by Jensen's Inequality, we have:
$f(a)+f(b)+f(c) \ge 3f\left(\dfrac{a+b+c}{3}\right) = 3f\left(\dfrac{1}{3}\right) = \dfrac{9}{4}$, as desired.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof for complex numbers and square root Use the polar form of complex numbers to show that every complex number $z\neq0$ has two square roots.
I know the polar form is $z=r(\cos(\alpha)+i \sin(\alpha))$. I'm just not sure how to do this one.
| De Moivre's formula is $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for any integer $n$. Therefore, if $w = s(\cos\theta + i\sin\theta)$ is a non-zero complex number such that $w^2 = z$, then
$$s^2(\cos(2\theta) + i\sin(2\theta)) = r(\cos\alpha + i\sin\alpha).$$
Therefore $s^2 = r$, so $s = \pm\sqrt{r}$, but $s > 0$, so $s = \sqrt{r}$.
Furthermore, $\cos(2\theta) = \cos\alpha$, so $2\theta = \alpha + 2k\pi$ where $k \in \mathbb{Z}$, or $2\theta = -\alpha + 2l\pi$ where $l \in \mathbb{Z}$.
Likewise, $\sin(2\theta) = \sin\alpha$, so $2\theta = \alpha + 2m\pi$ where $m \in \mathbb{Z}$, or $2\theta = \pi - \alpha + 2n\pi$ where $n \in \mathbb{Z}$.
The only possibility for $2\theta$ that will give both $\cos(2\theta) = \cos\alpha$ and $\sin(2\theta) = \sin\alpha$ is $2\theta = \alpha + 2k\pi$ for some $k \in \mathbb{Z}$. Therefore $\theta = \frac{\alpha}{2} + k\pi$ where there are no restrictions on $k$, other than being an integer. So the complete set of solutions to $w^2 = z$ is
$$\left\{\sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) \mid k \in \mathbb{Z}\right\}.$$
However, as $\cos$ and $\sin$ are $2\pi$-periodic, there are many different values of $k$ which gives the same complex number. If $k$ is even, then $k = 2t$ for some $t \in \mathbb{Z}$, so
\begin{align*}
\sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+2t\pi\right) + i\sin\left(\frac{\alpha}{2} + 2t\pi\right)\right)\\
&= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right).
\end{align*}
If $k$ is odd, then $k = 2t+1$ for some $t \in \mathbb{Z}$, so
\begin{align*}
\sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+(2t+1)\pi\right) + i\sin\left(\frac{\alpha}{2} + (2t+1)\pi\right)\right)\\
&= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+\pi+2t\pi\right) + i\sin\left(\frac{\alpha}{2} + \pi 2t\pi\right)\right)\\
&= \sqrt{r}\left(\cos\left(\frac{\alpha}{2} + \pi\right) + i\sin\left(\frac{\alpha}{2}+\pi\right)\right).
\end{align*}
Using the fact that $\cos(\beta + \pi) = -\cos\beta$ and $\sin(\beta + \pi) = -\sin\beta$ we can simplify this further:
$$\sqrt{r}\left(\cos\left(\frac{\alpha}{2} + \pi\right) + i\sin\left(\frac{\alpha}{2}+\pi\right)\right) = -\sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right).$$
Therefore, the two solutions to $w^2 = z$ are $\pm\sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right)$; note that these are distinct, so there are actually two different solutions.
In general, if you were trying to solve $w^n = z$ with $z \neq 0$, you could use the same method as above. When it came to determine which $k$ gave different complex numbers, you'd consider $k$ modulo $n$, i.e. write $k = nt + j$ where $j = 0, 1, \dots, n - 1$. You'd then find $n$ distinct solutions corresponding to the different values of $j$. Usually, you would have to stop here, as there won't be a corresponding trigonometric formula to simplify the complex numbers (as there was above for the phase shift $\pi$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, ...
What is the missing number?
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
| $$\frac{1}{16}, \frac{1}{8}=\frac{2}{16}, \frac{3}{16}, \frac{1}{4}=\frac{4}{16}, \frac{5}{16}$$
So the $i$th term is of the form $$\frac{i}{16}$$ Therefore, the next term is $$\frac{6}{16}=\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
What are some good questions for this trick, if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha$ then $\alpha=\frac{a+c+e+...}{b+d+f+...}$? I need some good algebra questions that are applications of this trick, often in a non obvious and elegant way: $$\text{If } \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots=\alpha \text{ then } \alpha=\frac{a+c+e+...}{b+d+f+...}$$
| *
*If $\displaystyle\frac a{b+c}=\frac b{c+a}=\frac c{a+b};$ prove that each ratio $\displaystyle=\frac12$ if $\displaystyle a+b+c\ne0$
*If $\displaystyle\frac{a-b}{x^2}=\frac{b-c}{y^2}=\frac{c-a}{z^2}$ prove that $\displaystyle a=b=c$
*If $\displaystyle\frac{a-b}{a^2+ab+b^2}=\frac{b-c}{b^2+bc+c^2}=\frac{c-a}{c^2+ca+a^2}$ prove that $\displaystyle a=b=c$ (for a special condition)
*If $\displaystyle\frac{a+b}{a^2+ab+b^2}=\frac{b+c}{b^2+bc+c^2}$ and $a\ne b\ne c$ prove that each ratio $=\displaystyle\frac{c+a}{c^2+ca+a^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$
My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore
$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}} $$
Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.
| Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Also $f(g(x))=f'(g(x)).g'(x)$ by the chain rule.
for $ g(x)=x-\frac {2}{x}$
$\Rightarrow g'(x) = 1+\frac {2}{x^2}$
for $f'(x)=\frac{1}{2} \frac {1}{\sqrt{(x-1)}}$
$f(g(x))=\left(1+\frac {2}{x^2}\right) \left(\frac {1}{2\sqrt{(x-1)}}\right)$
$\Rightarrow \frac {-5}{8}$ $\text{by differentiating and putting x=2 for already evaluated g(x) for x=1}\\$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule:
$u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to
$$\int \sqrt{\frac{2-u}{u}} \, \frac{du}2$$
or $$\frac 12 \int \sqrt{\frac 2u - 1} \, du$$
I'm a bit stuck here. May I ask for help on how to proceed?
| Another way to solve the integral is as follows. Let $u=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ so that
$$u^{2}(1+x^{2})=1-x^{2}$$
$$u^{2}-1=-x^{2}(1+u^{2})$$
$$x^{2}=\frac{1-u^{2}}{1+u^{2}}$$
so
$$2xdx=\frac{(-2u)(1+u^{2})-(2u)(1-u^{2}))}{(1+u^{2})^{2}}du=\frac{-4u}{(1+u^{2})^{2}}du$$
Going back to the integral we have:
$$\int x\sqrt{\frac{1-x^{2}}{1+x^{2}}}dx=\frac{1}{2}\int\sqrt{\frac{1-x^{2}}{1+x^{2}}}2xdx=\int u\frac{-2u}{(1+u^{2})^{2}}du=\frac{u}{1+u^{2}}-\int\frac{1}{1+u^{2}}du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| Let $\log_2 x=A$, then $\log_2 x^2=2\log_2 x=2A$ and $\log_2\frac{2}{x}=\log_22- \log_2x=1-A$. So the given inequality becomes:
$$(A-1)+\frac{2A}{1-A} \leq 1.$$
Consequently we get
$$\frac{4A-A^2-1}{1-A} \leq 1.$$
Furthermore you get
$$\frac{5A-A^2-2}{1-A} \leq 0.$$
Hopefully you can solve from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in advance.
EDIT: The first digit should not be $0$
| My answer is similar to that of @shooting-squirrel. Edit: Now it gives the same answer, silly mistakes :D
Consider the number $\overline{abc}$, we have three cases:
Case 1: $a=b+c$, then $a$ can be $1,\dots,9$. Now, for each value of $a$, there are $a+1$ values of $b$: $0,1,\dots, a$. Each of which gives one number. Total numbers in this case is
$$\sum^{9}_{a=1}(a+1)= \frac{(10+2)\cdot 9}{2}=54.$$
Case 2: $b=a+c$. Since $b\ge a$, $b$ can only be $1,\dots,9$. For each $b$, there are $b-1$ values of $a$: $1,\dots, b-1$. ($a=b$ was counted in Case 1).
Each of which gives one number. Total number in this case is
$$\sum^9_{b=1}(b-1)=\frac{8\cdot 9}{2}=36.$$
Case 3: $c=a+b$. This case is similar to Case 2, with $c$ plays the role of $b$. So we have $36$ numbers in Case 3.
Total: Case 1 + Case 2 + Case 3 = $54+36+36=\color{red}{\mathbf{126}}$ numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/889687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Error in proving of the formula the sum of squares Given formula
$$
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
$$
And I tried to prove it in that way:
$$
\sum_{k=1}^n (k^2)'=2\sum_{k=1}^n k=2(\frac{n(n+1)}{2})=n^2+n
$$
$$
\int (n^2+n)\ \text d n=\frac{n^3}{3}+\frac{n^2}{2}+C
$$
But
$$
\frac{n^3}{3}+\frac{n^2}{2}+C $$ is not equal to
$$\frac{n(n+1)(2n+1)}{6}=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$$
Where I made a mistake?
| This is from An Introduction to the Calculus of Finite Differences and Difference Equations by Kenneth S. Miller, 1960.
We have the difference operator $$ \Delta f(x) = f(x+1) - f(x) $$
Then we have the factorial polynomials,
$$ x^{(0)} =1, $$
$$ x^{(1)} =x, $$
$$ x^{(2)} =x(x-1) = x^2 - x, $$
$$ x^{(3)} = x(x-1)(x-2) = x^3 - 3 x^2 + 2 x, $$
$$ x^{(n+1)} = (x-n) x^{(n)} . $$
The pretty difference result is
$$ \Delta x^{(n)} = n x^{(n-1)} $$
Without endpoints, indefinite, the fundamental theorem applies as
$$ \sum x^{(n)} = \frac{x^{(n+1)}}{n+1}, \; \; \; \; n \neq -1 $$
and with endpoints, we get a shift from $N$ to $N+1$ corresponding to the definition of $\Delta,$ with
$$ \sum_{x=1}^N x^{(n)} = \left. \frac{x^{(n+1)}}{n+1} \right|_{x=1}^{x=N+1}. \; \; \; \; n \neq -1 $$
So now we use $$ x^2 = x^2 - x + x = x(x-1) + x = x^{(2)} + x^{(1)} $$
and find (page 26)
$$ \sum_{x=1}^N x^{(2)} + x^{(1)} = \left. \frac{x^{(3)}}{3} + \frac{x^{(2)}}{2} \right|_{x=1}^{x=N+1} = \left. \frac{x(x-1)(x-2)}{3} + \frac{x(x-1)}{2} \right|_{x=1}^{x=N+1} $$
$$ = \frac{(N+1)N(N-1)}{3} + \frac{(N+1)N}{2} $$
$$ = \frac{1}{6} (N+1)N \left( 2(N-1) + 3 \right) = \frac{1}{6} (N+1)N \left( 2N-2 + 3 \right) = \frac{1}{6} (N+1)N \left( 2N + 1 \right) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
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