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Find integer a,b > 1 such that $2^a + 3^b = 2^{a+b} +1$ I would like to know if it is possible to find an integer solution to
$2^a + 3^b = 2^{a+b} +1$
with $a,b > 1$
|
We have
$$2^a(2^b - 1) = (3-1)(1 + 3 + 3^2 + \ldots + 3^{b-1}),$$ so
$$2^{a-1}(2^b - 1) = (1 + 3 + 3^2 + \ldots + 3^{b-1}).$$
Considering the second equation mod $3$ we thus see that $2^{a-1}$ and $2^b - 1$ aren't zero. As $2^b$ isn't zero mod $3$ either, it must be $2$ mod $3$, i.e. $b$ must be odd.
But this implies that $1 + 3 + 3^2 + \ldots + 3^{b-1}$ is $2$ mod $4$ and hence (from the first equality in this answer) that $a = 2$.
But there are no solutions with $a = 2$, since then we have $4 - 1 + 3^b = 2^{a+b}$ and the two sides are different modulo $3$.
|
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|
Improper fraction to proper fraction in algebra This question is in regards to the very first part of a question titled 'Express the following in partial fractions'
From my lecture notes I have got the following written down:
$$\frac{x(x+3)}{x^2+x-12}=1+\frac{2x+12}{x^2+x-12}$$
I cant understand how the numerator goes from $x(x+3)$ , to, $1 + (2x+12)$.
I understand that the $x * x$ makes the $2x$, but how is $12$ derived from the $3$ part ?
I can do the rest of the question (factoring, cross multiply, compare numerators), just cant understand to $3$ to $12$ transition.
Any help is greatly appreciated.
-Connor
|
Working backwards:
$$1+\frac{2x+12}{x^2+x-12}= \frac{x^2+x-12}{x^2+x-12}+\frac{2x+12}{x^2+x-12}=\frac{x^2+x-12+2x+12}{x^2+x-12}=\frac{x^2+3x}{x^2+x-12}=\frac{x(x+3)}{x^2+x-12}$$
|
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How to solve inequalities with more than one absolute value expression There are two parts to this question.
1. I'm seeing the correct method to solve these types of inequalities as something to do with "transition points". I don't quite understand this method. How do we find the specific inequalities, and how do we work with the boundaries once we set them?
*I personally thought of a more systematic approach: Simplify one absolute value expression (convert to positive/negative cases), then, for each of those cases, split the second expression to get a total of 4 cases. Then, I string together the inequalities that I get after solving them all, via union or intersection as the inequality sign marks.
Edit: I've noticed the mistake in my process. In the Case 2 line, I forgot to switch the inequality sign. Still, I'm wondering why this isn't a viable strategy for solving these types of inequalities. Why is this method inferior to the method described in (1)?
Example problem:
|x-3|+|2x+5| > 6
|x-3| > 6-|2x+5|
Case 1: x-3 > 6-|2x+5|
|2x+5| > 9-x
Case 1a: 2x+5 > 9-x, 3x > 4, x > 4/3
Case 1b: 2x+5 < x-9, x < -14
Case 2: x-3 > |2x+5|-6
|2x+5| < x+3
Case 2a: 2x+5 < x+3, x < -2,
Case 2b: 2x+5 < -x-3, 3x < -8, x > -8/3
String together the 4 resulting inequalities:
((x < 4/3)U(x < -14))U((x < -2)U(x > -8/3))
So my answer would be x < -2 U x > -8/3
This is wrong. The answer would actually be x < -8/3 U x > -2.
What did I do wrong?
|
We need to rewrite the inequation without using the absolute value. Let's recall the definition of the absolute value:
$$
\lvert x \rvert =
\begin{cases}
-x & \text{if $x < 0$,} \\
x & \text{if $x \geq 0$.}
\end{cases}
$$
The transition points are the solutions of the equations $x-3 = 0$ and $2x+5 = 0$. It can be covenient to use a table like this one.
$$
\begin{array}{r|cccccccc}
x & -\infty & & -5/2 & & 3 & & +\infty\\
\text{sign of $x-3$} & & - & & - & & + & \\
\lvert x-3 \rvert & & -x+3 & & -x+3 & & x-3 & \\
\text{sign of $2x+5$} & & - & & + & & + & \\
\lvert 2x+5 \rvert & & -2x-5 & & 2x+5 & & 2x+5 & \\
\lvert x-3 \rvert + \lvert 2x+5 \rvert > 6 & & -3x-2 > 6 & & x+8 > 6 & & 3x+2 > 6 & \\
\text{Solutions} & & x<-8/3& & x>-2 & & x > 4/3 & & \\
\end{array}
$$
The set of solutions of the inequation is $(-\infty, -8/3] \cup (-2,+\infty)$.
|
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Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$
But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to
$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$
I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right ) $$
But I cannot see how the numerator becomes
$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$
What have I done wrong here?
|
Another approach:
Put: $\tan (\alpha/2)=a$ and $ \tan (\beta/2)=b$.
Than:
$$
\sin \alpha= \dfrac{2a}{1+a^2} \qquad \cos \alpha=\dfrac{1-a^2}{1+a^2}
$$
$$
\sin \beta= \dfrac{2b}{1+b^2} \qquad \cos \beta=\dfrac{1-b^2}{1+b^2}
$$
and:
$$
\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \dfrac{2a(1+b^2)+2b(1+a^2)}{(1-a^2)(1+b^2)+(1-b^2)(1+a^2)}
$$
that, after a bit of algebra, becomes:
$$
=\dfrac{a+b}{1-ab}=\dfrac{\tan \alpha/2+\tan \beta/2}{1-\tan \alpha/2 \tan \beta/2}= \tan \left(\dfrac{\alpha +\beta}{2} \right)
$$
|
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Evaluating the indefinite integral $\int\frac{\sqrt{1-x}}{x\sqrt{1+x}}\,dx$ $$\int\frac{\sqrt{1-x}}{x\sqrt{1+x}}\,dx$$
Looking at the term under the square root, I tried out the subsitution $x = \cos\theta$. This reduced the expression to $\int(1 - \sec\theta)d\theta$. This doesn't tally with the answer provided though. What is happening?
|
Let $$\displaystyle I = \int\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot \frac{1}{x}dx = \int\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot \frac{\sqrt{1-x}}{\sqrt{1-x}}\cdot \frac{1}{x}dx$$
So $$\displaystyle I = \int\frac{1-x}{x\sqrt{1-x^2}}dx = \int\frac{1}{x\sqrt{1-x^2}}dx-\int\frac{1}{\sqrt{1-x^2}}dx$$
Now Put $1-x^2=t^2$ Then $xdx = -tdt$ in $1^{st}$ and $x=\sin \theta$ and $dx = \cos \theta d\theta $ in $2^{nd}$ Integral
So $$\displaystyle I = \int\frac{1}{(t^2-1)}dt-\sin^{-1}(x)+\mathcal{C}$$
So $$I = \ln \left|\frac{t-1}{t+1}\right|-\sin^{-1}(x)+\mathcal{C}$$
So $$I = \ln \left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|-\sin^{-1}(x)+\mathcal{C}$$
|
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Find the the number of functions $f:A\to B$ whose domain is $A$ such that if $x_1\geq x_2$ then $f(x_1)\geq f(x_2)\forall x_1,x_2\in A$ Let set $A=\left\{1,2,3,4,5\right\}$ and $B=\left\{-2,-1,0,1,2,3,4,5\right\}$
Find the the number of functions $f:A\to B$ whose domain is $A$ such that if $x_1\geq x_2$ then $f(x_1)\geq f(x_2)\forall x_1,x_2\in A$.
How can i solve it,i learnt a question in which number of strictly increasing functions were to count but now this is different and i could not solve it.Please help me.
|
Here we have to find number of non-decreasing function from $A = \left\{1,2,3,4,5\right\}$ to
$B=\left\{-2,-1,0,1,2,3,4,5\right\}.$
Means $f(1)\leq f(2)\leq f(3)\leq f(4)\leq f(5).$
Where $f(1),f(2),f(3),f(4),f(5)\in \left\{-2,-1,0,1,2,3,4,5\right\}$
Now we will break into Different cases.
$\bullet\; $ If $f(1)<f(2)<f(3)<f(4)<f(5)\;,$ So we will get $\displaystyle \binom{8}{5}$ ways
$\bullet\; $ If $f(1)=f(2)<f(3)<f(4)<f(5)\;,$ So will get $\displaystyle \binom{8}{4}\times 4$ ways
$\bullet\; $ If $f(1)=f(2)=f(3)<f(4)<f(5)\;,$ So will get $\displaystyle \binom{8}{3}\times 6$ ways
$\bullet\; $ If $f(1)=f(2)=f(3)=f(4)<f(5)\;,$ So will get $\displaystyle \binom{8}{2}\times 4$ ways
$\bullet\; $ If $f(1)=f(2)=f(3)=f(4)=f(5)\;,$ So will get $\displaystyle \binom{8}{1}$ ways
So Total number of ways $\displaystyle = \binom{8}{5}+\binom{8}{4}\times 4 +\binom{8}{3}\times 6+\binom{8}{2}\times 4+\binom{8}{1}$
$\displaystyle = \left[\binom{8}{5}+\binom{8}{4}\right]+3\cdot [\binom{8}{4}+\binom{8}{3}]+ 3\cdot \left[\binom{8}{3}+\binom{8}{2}\right]+\left[\binom{8}{2}+\binom{8}{1}\right]$
Now Using $\displaystyle \bullet\; \binom{n}{r}+\binom{n}{r-1} = \binom{n+1}{r}$
So we get $\displaystyle = \binom{9}{5}+3\cdot \left[\binom{9}{4}+\binom{9}{3}\right]+\binom{9}{2} = \binom{12}{5}.$
|
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Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$ Prove that the sum of the moduli of the roots of $x^4-5x^3+6x^2-5x+1=0$ is $4$.
I tried various methods to find its roots,but no luck.Is there any method to solve it without actually finding the roots of the equation.Please help me.
Thanks.
|
\begin{align}
x^4-5x^3+6x^2-5x+1&=(\color{red}1\times x^4-\color{red}4\times x^3+\color{red}1\times x^2)\\
&-(\color{red}1\times x^3-\color{red}4\times x^2+\color{red}1\times x)\\
&+(\color{red}1\times x^2-\color{red}4\times x+\color{red}1)\\
&=x^2(x^2-4x+1)-x(x^2-4x+1)+(x^2-4x+1)\\
&=(x^2-4x+1)(x^2-x+1)
\end{align}
In general we have
$$1 \pm n x + (1 + n) x^2 \pm n x^3 + x^4=\pm\left(x^2\pm x+1\right) \left(n x\pm x^2-x\pm 1\right)$$
for example we have $$x^4+\sqrt{2} x^3+\left(1+\sqrt{2}\right) x^2+\sqrt{2} x+1=\left(x^2+x+1\right) \left(x^2+\sqrt{2} x-x+1\right)$$
Something similar you may find for $1 \pm n x + (1 - n) x^2 \pm n x^3 + x^4$. You could find similar rules for higher order polynomials, but the rules quickly get complicated :-)
|
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|
Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2}x-1)}+\frac{x+\sqrt{2}}{2\sqrt{2}(x^2+\sqrt{2}x+1)}$$
BTW, are there any easier method to solve this integral? As seen on the answer, it seems not very possible. ;)
|
Notice, $$\int\frac{dx}{1+x^4}=\int\frac{dx}{x^2\left(\frac{1}{x^2}+x^2\right)}$$
$$=\int\frac{\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx=\frac{1}{2}\int\frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}}dx$$
$$=\frac{1}{2}\int\frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}dx$$
$$=\frac{1}{2}\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx-\frac{1}{2}\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$
$$=\frac{1}{2}\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+(\sqrt2)^2}dx-\frac{1}{2}\int\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-(\sqrt2)^2}dx$$
$$=\frac{1}{2}\int\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt2)^2}dx-\frac{1}{2}\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-(\sqrt2)^2}dx$$
$$=\frac{1}{2\sqrt 2}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt 2}\right)-\frac{1}{4\sqrt 2}\ln\left|\frac{x+\frac{1}{x}-\sqrt 2}{x+\frac{1}{x}+\sqrt 2}\right|+C$$
|
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Number Theory: Mod Simultaneous Congruences I have this problem assigned for homework. I completed it but I wanted to see if there was a simpler way of solving it. Here it is:
Problem: Solve the set of simultaneous congruences:
$2x\equiv 1\pmod{5}$,
$3x\equiv 9\pmod{6}$,
$4x\equiv 1\pmod{7}$,
$5x\equiv 9\pmod{11}$.
Solution:
$2x\equiv 1\equiv 6\pmod{5}\implies x\equiv 3\pmod{5}$
$\implies x=3+5k$, some $k\in \mathbb{Z}$.
$3x\equiv 9\pmod{6}\implies x\equiv 3\pmod{2}\implies3+5k\equiv 3\pmod2\implies 5k\equiv 0\pmod2$
$\implies k\equiv 0\pmod2\implies k=2i$
$\implies x=3+5(2i)=3+10i$, some $i\in \mathbb{Z}$.
$4x\equiv 1\pmod7 \implies 4(3+10i)\equiv 1\equiv 8\pmod7\implies 3+10i\equiv 2\equiv 23\pmod7$
$\implies 10i\equiv 20\pmod7\implies i\equiv 2\pmod7\implies i=2+7j$
$\implies x=3+10(2+7j)=23+70j$, some $j\in \mathbb{Z}$.
$5x\equiv 9\equiv 20\pmod{11}\implies x\equiv 4\pmod{11}\implies 23+70j\equiv 4\equiv 653\pmod{11}$
$\implies 70j\equiv 630\pmod{11}\implies j\equiv 9\pmod{11}\implies j=9+11t$
$\implies x=23+70(9+11t)=653+770t$, some $t\in \mathbb{Z}$.
$\implies x\equiv 653\pmod{770}$.
|
There is a classic formula to "write down" the solution after solving some auxiliary congruences. In your case, solve for the following inverses using your favorite algorithm for modular inversion:
$b_1 \equiv (2 \cdot 2 \cdot 7 \cdot 11)^{-1} \mod 5$
$b_2 \equiv (5 \cdot 7 \cdot 11)^{-1} \mod 2$
$b_3 \equiv (4 \cdot 5 \cdot 2 \cdot 11)^{-1} \mod 7$
$b_4 \equiv (5 \cdot 5 \cdot 2 \cdot 7)^{-1} \mod 11$
We find, by inspection, solutions $b_1 = 2$, $b_2 = 1$, $b_3 = 6$, and $b_4 = 5$.
A solution to your system is then
$b_1 \cdot 2 \cdot 7 \cdot 11 + b_2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 + b_3 \cdot 5 \cdot 2 \cdot 11 + b_4 \cdot 9 \cdot 5 \cdot 2 \cdot 7$
which, using the above values, is $5273$. It is easy to see this is congruent to 653 modulo 770.
|
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Show that $\ln(a+b) =\ ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$ Show that $\ln(a+b) = \ln(a) + \ln(b)$ when $a = \frac{b}{b-1}$
My attempt at a solution was
$$
\ln(a+b) = \ln \left(a(1+\frac{b}{a})\right)
$$
and by setting $a = \frac{b}{b-1}$ we get
$$
\ln \left(\frac{b}{b-1}(1+(b-1))\right) = \ln \left(\frac{b}{b-1}+b \right)
$$
But how can I continue from here?
The original question in my textbook asked if the equality $\ln(a+b) = \ln(a) + \ln(b)$ was true.
My initial answer was that it was not true, but I missed this special case, which I think was very hard to "see". Any help would be much appreciated.
|
$$\begin{align}\ln (a + b) &= \ln \left(\frac{b}{b - 1} + b\right)\\&=\ln\left(\frac{b + b^2 - b}{b - 1}\right)\\&=\ln\left(\frac{b^2}{b - 1}\right)\\&= \ln \left(\frac{b}{b - 1} \cdot b\right)\\&= \ln\left(\frac{b}{b - 1}\right) + \ln b\\&= \ln a + \ln b\end{align}$$
The trick is that the $b$ terms in the numerator gets eliminated after merging both $a$ and $b$ into a single fraction.
|
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Rationalise the denominator and simplify $\frac {3\sqrt 2-4}{3\sqrt2+4}$ Does someone have an idea how to work $\dfrac {3 \sqrt 2 - 4} {3 \sqrt 2 + 4}$ by rationalising the denominator method and simplifying?
|
$\dfrac{3\sqrt 2 - 4}{3\sqrt 2 + 4} \cdot \dfrac{3\sqrt 2 - 4}{3\sqrt 2 - 4}=$
$=\dfrac{18-24\sqrt 2 + 16}{(3\sqrt 2)^2-4^2}=\dfrac{34-24\sqrt 2}{18-16}=\dfrac{2(17-12\sqrt 2)}{2}=$
$=17-12\sqrt 2.$
|
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|
Infinite inequality with logarithm Solve inequality
$3 - \log_{0.5}x - \log^2_{0.5}x - \log^3_{0.5}x - \cdots \ge 4\log_{0.5}x$
Any suggestions how to start?
|
We have
$$\begin{align}
3-\log_{0.5}x-\log^2_{0.5}x-\log^3_{0.5}x-\cdots&=4-\sum_{n=0}^{\infty}\log^n_{0.5}x\\\\
&=4-\frac{1}{1-\log_{0.5}x}\\\\
&=\frac{3-4\log_{0.5}x}{1-\log_{0.5}x}
\end{align}$$
for $|\log_{0.5}x|<1\implies 0.5<x<2$. Now note that the inequality
$$\frac{3-4\log_{0.5}x}{1-\log_{0.5}x}\ge 4\log_{0.5}x$$
is equivalent to
$$\left(2\log_{0.5}x-3\right)\left(2\log_{0.5}x-1\right)\ge 0$$
which implies $\log_{0.5}x\le 0.5$ or $x\ge \sqrt{2}/2$. Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{\sqrt{2}/2\le x<2}$$
|
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Suppose $a,b$ are integers. If $4\mid(a^2 +b^2)$, then $a$ and $b$ are not both odd. I've tried using direct proof and contrapositive proof to prove this.
I'm stuck. Is anyone able to help solve or give me some hints?
|
If $4|a^2+b^2$ then $a^2+b^2=4m$ for some integer $m$.
Now suppose that $a$ and $b$ are both odd. Then $a=2i+1$ and $b=2j+1$ for some integers $i,j$.
Then $a^2+b^2=4i^2+4i+1+4j^2+4j+1=4(i^2+i+j^2+j)+2$, which is a contradiction.
If you are comfortable with modular arithmetic, we would say that we can't have both $a^2+b^2\equiv 0$ mod $4$ AND $a^2+b^2\equiv 2$ mod $4$.
|
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How to solve recurrence $T(n) = nT(n - 1) + 1$ Assume $T(n) = \Theta(1)$ for $n \leq 1$. Using iterative substitution.
So far I have:
\begin{align*}
&T(n) = nT(n - 1) + 1\\
&= n((n - 1)T(n - 2) + 1) + 1\\
&= n(n - 1)T(n - 2) + n + 1
\end{align*}
I'm stuck on how I'm supposed to get the asymptotic value from this. Or how would I keep expanding? Thanks!
|
Method 1: Iteration
What you have so far is good. Just keep going!
Here are the next two iterations, to help you:
\begin{align*}
T(n) &= n(n-1)(n-2)T(n-3) + n(n-1) + n + 1 \\
T(n) &= n(n-1)(n-2)(n-3)T(n-4) + n(n-1)(n-2) + n(n-1) + n + 1
\end{align*}
Do you see the pattern? (It's not so obvious!)
Method 2: Generating Functions
Using the exponential generating function $F(x) = \sum_n T(n) x^n / n!$, from
$$
T(n)\frac{x^n}{n!} = n T(n-1) \frac{x^n}{n!} + \frac{x^n}{n!}
$$
we get
$$
F(x) = xF(x) + e^x + T(0) - 1.
$$
Therefore, with $c = T(0) - 1$,
$$
F(x) = \frac{e^x + c}{1-x}.
$$
It follows that the coefficient of $x^n$ is
$$
c + \sum_{i=0}^n \frac{1}{i!}
$$
so that
$$
T(n) = cn! + \sum_{i=0}^n \frac{n!}{i!}.
$$
In fact, for $n \ge 1$ the latter sum is less than $e \cdot n!$ but bigger than $e \cdot n! - 1$, and is also an integer. So we get
$$
T(n) = c \cdot n! + \lfloor{e \cdot n!\rfloor},
$$
for $n \ge 1$. For $n = 0$, the floor formula does not work, and we just have $T(0) = c + 1$.
|
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|
How to show that $\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$? How to show that $$\sum_{n=0}^\infty\frac{(2n)!}{(n+1)!n!}(\frac{a}{2})^{2n}\sum_{m=0}^n b^{2m}=\frac{1}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}$$
I substitute $\sum_{m=0}^n b^{2m}=\frac{1-b^{2n+2}}{1-b^2}$ into the equation. I tried really hard and I have no clue about what to do to derive the square root.
Could anyone kindly help? Thanks!
|
The right-hand side of the equation should actually be
$$\frac{\color{red}{2}}{\sqrt{1 - a^2} + \sqrt{1 - a^2b^2}}.\tag{*}$$
I'll assume $a$ and $b$ are real numbers such that $|a| < 1$ and $|ab| < 1$. Since
$$(2n)! = 2^{2n}n!\left(n - \frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)$$
and
$$\left(n - \frac{1}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = (-1)^n\left(\frac{1}{2} - n\right)\cdots \left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right) = n!\binom{-1/2}{n},$$
then
$$\sum_{n = 0}^\infty \frac{(2n)!}{(n+1)!n!}\left(\frac{a}{2}\right)^{2n}\sum_{m = 0}^n b^{2m} = \sum_{n = 0}^\infty \frac{(-1)^n}{n+1}\binom{-1/2}{n}a^{2n}\frac{1 - b^{2n+2}}{1-b^2}.\tag{**}$$
Let $F(a,b)$ represent the series in (**). If $a = 0$, then $F(a,b) = 1$, which agrees with formula (*) when $a = 0$. If $a\neq 0$, then
$$\begin{align}F(a,b) &= \frac{1}{a^2(1 - b^2)}\sum_{n = 0}^\infty \frac{(-1)^n}{n+1}\binom{-1/2}{n}[(a^2)^{n+1} - (a^2b^2)^{n+1}]\\
&= \frac{1}{a^2(1-b^2)}\sum_{n = 0}^\infty (-1)^n\binom{-1/2}{n}\int_{a^2b^2}^{a^2} x^n\, dx\\
&= \frac{1}{a^2(1-b^2)}\int_{a^2b^2}^{a^2}\sum_{n = 0}^\infty \binom{-1/2}{n}(-x)^n\, dx\\
&= \frac{1}{a^2(1-b^2)}\int_{a^2b^2}^{a^2} (1 - x)^{-1/2}\, dx\\
&= \frac{1}{a^2(1-b^2)}[2\sqrt{1-a^2b^2} - 2\sqrt{1 - a^2}]\\
&= \frac{2}{a^2(1-b^2)}\frac{a^2 - a^2b^2}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}\\
&= \frac{2}{\sqrt{1-a^2}+\sqrt{1-a^2b^2}}.
\end{align}$$
|
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|
Problem related to integration I tried using partial fractions but I am not sure whether it is the right approach or not. I need help with this problem
Evaluation of Integral $$\int\frac{3x^4-1}{(x^4+x+1)^2}dx $$
|
Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4+4x^3-4x^3-1}{(x^4+x+1)^2}dx $$
So $$I= \underbrace{\int\frac{3x^4+4x^3}{(x^4+x+1)^2}dx}_{J}-\int\frac{4x^3+1}{(x^4+x+1)^2}dx$$
for second Integral Put $x^4+x+1 = t\;,$ Then $(4x^3+1)dx = dt$
Now for calculation of $\displaystyle J = \int\frac{3x^4+4x^3}{(x^4+x+1)^2}dx$
For these type of integral, we will take highest power of $x$ from denominator,
and then adjust it into Numerator
So we get $$\displaystyle J = \int\frac{3x^4+4x^3}{x^8(1+x^{-3}+x^{-4})}dx = \int\frac{3x^{-4}+4x^{-5}}{(1+x^{-3}+x^{-4})^2}dx$$
Now put $(1+x^{-3}+x^{-4}) = u\;,$ Then $(3x^{-4}+4x^{-3})dx = -du$
So we get $$\displaystyle I = -\int\frac{1}{u^2}du-\int\frac{1}{t^2}dt = \frac{1}{u}+\frac{1}{t}+\mathcal{C} = \frac{1}{1+x^{-3}+x^{-4}}+\frac{1}{x^4+x+1}+\mathcal{C}$$
So we get $$I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \frac{x^4+1}{x^4+x+1}+\mathcal{C} = -\left[\frac{x}{x^4+x+1}\right]+\mathcal{C'}$$
Bcz $$\frac{x^4+1}{x^4+x+1} = \frac{(x^4+x+1)-1}{x^4+x+1} = 1-\frac{x}{x^4+x+1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Division of polynomials to find unknown coefficients The polynomial $-x^4 -x^3 + ax^2 -17x +b$, where $a$ and $b$ are constants, is denoted by $p(x)$. It is given that $x^2 + 4x + 1$ is a factor of $p(x)$.
(i) Find the values of $a$ and $b$.
(ii) With theses values of $a$ and $b$, show that the equation $p(x)=0$ has exactly two roots.
|
by using the long division of $\frac{-x^4-x^3+ax^2-17x+b}{x^2+4x+1}$
$$\frac{-x^4-x^3+ax^2-17x+b}{x^2+4x+1}=(-x^2+3x+a-11)+\frac{(24-4a)x+b-a+11}{x^2+4x+1}$$
the reminder should be zero
$$(24-4a)x+b-a+11=0$$
$$24-4a=0$$
$$a=6$$
$$b-a+11=0$$
$$b-6+11=0$$
$$b=-5$$
$$-x^2+3x+a-11=-x^2+3x+6-11=-x^2+3x-5$$
or
$$x^2-3x+5$$
the roots of this equation are not real,but the $x^2+4x+1$ has two real roots
|
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|
Finding right inverse matrix Given a $3\times 4$ matrix $A$ such as
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
\end{pmatrix}
,$$
find a matrix $B_{4\times 3}$ such that
$$AB =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$
Apart from simply multiplying $A$ with $B$ and generating a $12$ variable system of equations, is there any simpler way of finding $B$ ?
|
Consider the following case:
$$\begin{bmatrix}1&1&1&1\\0&1&1&0\\0&0&1&1\end{bmatrix}\times \begin{bmatrix}0&0&0\\?&?&?\\?&?&?\\?&?&?\end{bmatrix}=I$$
Removing the first row, the remaining matrix is square:
$$\begin{bmatrix}?&?&?\\?&?&?\\?&?&?\end{bmatrix}=\begin{bmatrix}1&1&1\\1&1&0\\0&1&1\end{bmatrix}^{-1}=\begin{bmatrix}1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$
$$B=\begin{bmatrix}0&0&0\\1&0&-1\\-1&1&1\\1&-1&0\end{bmatrix}$$
Of course there are infinite number of solutions
|
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|
Divisibility by 41 and 5-digit number How to prove that if a $5$-digit number is divisible by $41$,then all the numbers generated from it by cyclic shift are also divisible by $41$
|
The reason is that $10^5 \equiv 1 \bmod 41$:
$$
0 \equiv 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0 \bmod 41
\implies\\
0 \equiv 10(10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0) \equiv 10^4 a_3 + 10^3 a_2 + 10^2 a_1 + 10 a_0 + a_4 \bmod 41
$$
|
{
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|
Help with a Differential Equation / Variation of Parameters - Wrong Answer Problem:
Solve the following differential equation.
\begin{eqnarray*}
y'' + y &=& \cot x \\
\end{eqnarray*}
Answer:
The solution I seek is $y = y_c + y_p$ where $y_c$ is the solution to
the corresponding homogeneous differential equation. To find $y_c$ I setup
the following characteristic equation.
\begin{eqnarray*}
m^2 + 1 &=& 0 \\
m^2 &=& -1 \\
m &=& \pm i \\
y_c &=& c_1 \sin x + c_2 \cos x \\
\end{eqnarray*}
Now to find $y_p$, I use the technique of variation of parameters.
\begin{eqnarray*}
y_p &=& v_1(x) \sin x + v_2(x) \cos x \\
y'_p &=& v_1(x) \cos x - v_2(x) \sin x + v'_1(x) \sin x + v'_2(x) \cos x \\
\text{ Now, I impose the condition } &&
v'_1(x) \sin x + v'_2(x) \cos x = 0 \\
y'_p &=& v_1(x) \cos x - v_2(x) \sin x \\
y''_p &=& -v_1(x) \sin x - v_2(x) \cos x
+ v_1'(x) \cos x - v_2'(x) \sin x \\
\end{eqnarray*}
\begin{eqnarray*}
y'' + y &=& -v_1(x) \sin x - v_2(x) \cos x + v_1'(x) \cos x - v_2'(x) \sin x
+ v_1(x) \sin x + v_2(x) \cos x \\
y'' + y &=& v_1'(x) \cos x - v_2'(x) \sin x = \cot x \\
\end{eqnarray*}
Now, I have a system of two linear equations. I apply Cramer's rule to solve
it.
\begin{eqnarray*}
v'_1(x) \sin x + v'_2(x) \cos x &=& 0 \\
v_1'(x) \cos x - v_2'(x) \sin x &=& \cot x \\
\begin{vmatrix}
\sin x & \cos x \\
\cos x & - \sin x \\
\end{vmatrix} &=& - { \sin ^ 2 x } - { \cos ^ 2 x } = -1 \\
\begin{vmatrix}
0 & \cos x \\
\cot x & - \sin x \\
\end{vmatrix} &=& - \cot x ( \cos x ) = -\frac{\cos ^ 2 x}{\sin x} \\
v'_1(x) &=& \frac{ -\frac{\cos ^ 2 x}{\sin x} } { -1 } =
\frac{\cos ^ 2 x}{\sin x} \\
%
v_1(x) &=&
\cos(x) + \ln ( \sin( \frac{x}{2} ) ) - \ln ( \cos ( \frac{x}{2} ) ) \\
%
\frac{\cos ^ 2 x}{\sin x}(\sin x) + v'_2(x) \cos x &=& 0 \\
\cos ^2 x + v'_2(x) \cos x &=& 0 \\
\cos x + v'_2(x) &=& 0 \\
v'_2(x) &=& - \cos x \\
v_2(x) &=& - \sin x \\
\end{eqnarray*}
Hence, the solution is
\begin{eqnarray*}
y &=& c_1 \sin x + c_2 \cos x +
( \cos(x) + \ln ( \sin( \frac{x}{2} ) ) - \ln ( \cos ( \frac{x}{2} ) ) )
(\sin x) + - \sin x (\cos x) \\
y &=& c_1 \sin x + c_2 \cos x +
( \ln ( \sin( \frac{x}{2} ) ) - \ln ( \cos ( \frac{x}{2} ) ) ) (\sin x) \\
\end{eqnarray*}
However, the book gets the following answer and I have reason to believe the
book is right.
\begin{eqnarray*}
y &=& c_1 \sin x + c_2 \cos x + (\sin x)( \ln | \csc x - \cot x | ) \\
\end{eqnarray*}
I am hoping that somebody can tell me what I did wrong.
|
You have the correct answer. (You merely forgot to account for the absolute values when integrating to get the logarithm function.) Combine the two logarithms in your answer, and use the trigonometric identity:
$$\tan(x/2)=\csc(x)-\cot(x)$$
|
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|
Verify $\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C$ Question says it all, how can I verify the following?
$$\int\sec x\ dx=\frac12 \ln \left|\frac{1+\sin x}{1-\sin x}\right| + C$$
|
You can verify this by integrating $\sec (x)$ and applying some trig identities:
\begin{align}
\displaystyle
LHS &=
\int \sec(x) \; \mathrm{d}x \\
&= \int \sec(x) \left ( \frac{\sec (x) + \tan (x)}{\sec (x) + \tan (x)} \right ) \; \mathrm{d}x \\
&= \int \frac{\sec^2 (x) + \sec (x) \tan (x)}{\sec (x) + \tan (x)} \; \mathrm{d}x \\
&= \int \frac{1}{u} \; \mathrm{d}u \tag{using $u=\sec (x) + \tan (x)$ as substitution} \\
&= \ln \left \lvert u \right \rvert + C \\
&= \ln \left \lvert \sec (x) + \tan (x) \right \rvert + C \\
&= \ln \left \lvert \frac{1+\sin (x)}{\cos (x)} \right \rvert + C \\
&= \ln \left \lvert \left ( \frac{\left (1+\sin (x) \right )^2}{\cos^2 (x)} \right )^{\frac{1}{2}}\right \rvert + C \\
&= \frac{1}{2} \ln \left \lvert \frac{\left (1+\sin (x) \right )^2}{\cos^2 (x)} \right \rvert + C \\
&= \frac{1}{2} \ln \left \lvert \frac{\left (1+\sin (x) \right )^2}{ 1- \sin^2 (x)} \right \rvert + C \\
&= \frac{1}{2} \ln \left \lvert \frac{\left (1+\sin (x) \right )^2}{\left ( 1- \sin (x) \right ) \left ( 1 + \sin (x) \right )} \right \rvert + C \\
&= \frac{1}{2} \ln \left \lvert \frac{1+\sin (x)}{ 1- \sin (x) } \right \rvert + C = RHS \tag*{Q.E.D.}
\end{align}
|
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|
What is the maximum value of $ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ f(x)=\frac{2 \sin (3 x)}{3 \sin (x)+3 \sqrt{3} \cos (x)} (\frac{\pi}{3}<x<\frac{2\pi}{3})$$
I have done my work here
$$f'(x)=\frac{2 \left(2 \sqrt{3} \cos (2 x)+\sqrt{3} \cos (4 x)-8 \sin ^3(x) \cos (x)\right)}{3 \left(\sin (x)+\sqrt{3} \cos (x)\right)^2}=0$$
I tried to solve problems but I couldn't make further progress
|
HINT: Notice, $$\frac{2\sin(3x)}{3\sin(x)+3\sqrt{3}\cos (x)}=\frac{1}{3}\left(\frac{\sin(3x)}{\frac{1}{2}\sin(x)+\frac{\sqrt{3}}{2}\cos (x)}\right)=\frac{1}{3}\left(\frac{\sin(3x)}{\sin(x)\cos\frac{\pi}{3}+\cos (x)\sin \frac{\pi}{3}}\right)=\frac{1}{3}\frac{\sin(3x)}{\sin \left(x+\frac{\pi}{3}\right)}$$
Now, $$f'(x)=\frac{1}{3}\frac{\sin \left(x+\frac{\pi}{3}\right)3\cos(3x)-\sin (3x)\cos \left(x+\frac{\pi}{3}\right)}{\sin^2 \left(x+\frac{\pi}{3}\right)}=0$$
$$2\sin \left(x+\frac{\pi}{3}\right)\cos(3x)+\left(\sin \left(x+\frac{\pi}{3}\right)\cos(3x)-\sin (3x)\cos \left(x+\frac{\pi}{3}\right)\right)=0$$
$$\sin \left(4x+\frac{\pi}{3}\right)-2\sin\left(2x-\frac{\pi}{3}\right)=0$$
|
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|
how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt{x^2+1}+x}\right)$
$= \left(\frac{x}{\left(\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}+\frac{x}{x}\right)}\right)\:$
$= \:\left(\frac{x}{\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\right)\:$
|
For the edited and corrected question, i.e. the limit at $\infty$ of
$$
x\left(\sqrt{x^2+1} - x\right)
$$
indeed you can deal with this multiplying by the conjugate (cf. jeantheron's answer). However, I would advocate for a more systematic approach, simple enough and highly generalizable. Assuming you have heard or will soon hear about Taylor series, you can do the following (detailing a lot each step):
$$\begin{align}
x\left(\sqrt{x^2+1} - x\right)
&=
x\left(x\sqrt{1+\frac{1}{x^2}} - x\right)
=
x\left(x\left(1+\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)\right) - x\right)
\\
&=
x\left(x+\frac{1}{2x}+o\left(\frac{1}{x}\right) - x\right)
=
x\left(\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)
\\
&=
\frac{1}{2}+o\left(1\right)
\end{align}$$
using that $(1+t)^\alpha = 1+\alpha t+o(t)$ when $t\to0$.
|
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|
What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
|
$$\small\begin{align}&\frac{x+1}{(x-1)(x+2)}\gt\frac{x}{(x-1)(x-3)}\\\\&\iff \frac{x+1}{(x-1)(x+2)}\cdot (x-1)^2(x+2)^2(x-3)^2\gt\frac{x}{(x-1)(x-3)}\cdot (x-1)^2(x+2)^2(x-3)^2\\\\&\iff (x+1)(x-1)(x+2)(x-3)^2\gt x(x-1)(x+2)^2(x-3)\\\\&\iff (x-1)(x+2)(x-3)((x+1)(x-3)-x(x+2))\gt 0\\\\&\iff (x-1)(x+2)(x-3)(-4x-3)\gt 0\\\\&\iff (x-1)(x+2)(x-3)(4x+3)\lt 0\\\\&\iff -2\lt x\lt -\frac 34\quad\text{or}\quad 1\lt x\lt 3\end{align}$$
|
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|
Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods would be welcome. I was solving a couple of AM GM inequality questions and I'm assuming this should be solved in a similar way. Correct me.
|
Let $\displaystyle \frac{x}{y} =a \;\;,\frac{y}{z} = b\;\;,\frac{z}{x} = c\;,$ Then $a+b+c=7$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$
and $abc=1$. Then we have to calculate $a^3+b^3+c^3 = $
Now Using $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9\Rightarrow \frac{ab+bc+ca}{abc} = 9\Rightarrow ab+bc+ca = 9$$
Now Let $t=a\;,t=b\;,t=c$ be the roots of cubic equation in terms of $t\;$ Then
Using factor Theorem, We get $(t-a)\;,(t-b)\;,(t-c)$ are the factor of that equation.
So we get $$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$
So we get $$t^3-7t^2+9t-1=0\Rightarrow t^3=7t^2-9t+1$$
Now above we have take $t=a\;,t=b\;,t=c$ be the root of that cubic equation
So we get $$a^3+b^3+c^3=7(a^2+b^2+c^2)-9(a+b+c)+3 = 7(31)-9(7)+3=157$$
|
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|
What is the smallest possible value of $\lfloor (a+b+c)/d\rfloor+\lfloor (a+b+d)/c\rfloor+\lfloor (a+d+c)/b\rfloor+\lfloor (d+b+c)/a\rfloor$? What is the smallest possible value of
$$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor$$
where $a,b,c,d>0$?
|
Since $\lfloor x\rfloor \gt x-1$, we have
$$\begin{align}&\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\frac{d+b+c}{a}\right\rfloor\\\\&\gt \frac{a+b+c}{d}-1+\frac{a+b+d}{c}-1+\frac{a+d+c}{b}-1+\frac{d+b+c}{a}-1\\\\&=\left(\frac ab+\frac ba\right)+\left(\frac ac+\frac ca\right)+\left(\frac ad+\frac da\right)+\left(\frac bc+\frac cb\right)+\left(\frac bd+\frac db\right)+\left(\frac cd+\frac dc\right)-4\\\\&\ge 2\sqrt{\frac ab\cdot\frac ba}+2\sqrt{\frac ac\cdot\frac ca}+2\sqrt{\frac ad\cdot\frac da}+2\sqrt{\frac bc\cdot\frac cb}+2\sqrt{\frac bd\cdot\frac db}+2\sqrt{\frac cd\cdot\frac dc}-4\\\\&=8\end{align}$$
By the way,
$$\left\lfloor\frac{1+1+1}{0.9}\right\rfloor+\left\lfloor\frac{1+1+0.9}{1}\right\rfloor+\left\lfloor\frac{1+0.9+1}{1}\right\rfloor+\left\lfloor\frac{0.9+1+1}{1}\right\rfloor=9.$$
So, the smallest possible value is $\color{red}{9}$.
|
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|
Powers of a root in terms of basis of field extension I'm wondering if my solution correct for these types of problems
Problem: given irreducible $$f(x)=x^3-2x-2 \in \mathbb{Q}[x]$$
Let $\theta$ be a root of $f$, and $K=\mathbb{Q}(\theta)$ extension over $\mathbb{Q}$. $K$ has $\{1,\theta,\theta^2\}$ as a $\mathbb{Q}$ basis. Express $\theta^5$ in the given basis.
Solution: take powers of $\theta$ in terms of $\theta,\theta^2$, and $\theta^3 = 2\theta + 2$, as $0=\theta^3-2\theta-2$. Therefore
$$\theta^0 =1$$
$$\theta^1 =\theta$$
$$\theta^2 =\theta^2$$
$$\theta^3 =2\theta+2$$
$$\theta^4 =(2\theta+2)\theta = 2\theta^2+2\theta$$
$$\theta^5 = (2\theta^2+2\theta)\theta = 2(2\theta+2)+2\theta^2 = 2\theta^2+4\theta+4$$
Forgive redundant computations, I've merely added it for clarity. edited s.t. $2\cdot 2 = 4$ and not 5
|
Divide the polynomial $P=X^5$ by $T=X^3-2X-2$.
First step : $P=X^5=\color{red}{X^2}\,T+2X^3+2X^2$.
Second step : $P=\color{red}{X^2}T+\color{blue}{2}T+2X^2+4X+4$.
Conclusion
$$P=\underbrace{(X^2+2)}_QT+\underbrace{2X^2+4X+4}_{R}.$$
This is the Euclidian division of polynomials, the remainder $R$ is what you want because
$$P(\theta)=Q(\theta)\underbrace{T(\theta)}_{0}+R(\theta).$$
Therefore, in $\mathbb Q(\theta)$ we have $\theta^5=2\theta^2+4\theta+4$.
|
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|
Find the angle of complex number Let
$$
z = \frac{a-jw}{a+jw}.
$$
Then the angle of $z$ is
$$
-\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right).
$$
How is that so?
|
I suppose that it could be easier to see writing $$z = \frac{a-jw}{a+jw}= \frac{a-jw}{a+jw} \times \frac{a-jw}{a-jw}=\frac{(a-jw)^2}{a^2+w^2}=\frac{a^2-w^2}{a^2+w^2}-\frac{2 a w}{a^2+w^2}j$$ which makes the angle to be such that $$\tan(\theta)=-\frac{2aw}{a^2-w^2}=-\frac{2\frac aw}{1-(\frac aw)^2}$$ and to remember the development of $\tan(2x)$.
|
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|
Linear equations with parameters
For which $a$ the following had one solution, no solution, infinite solutions:
$$\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
4a-8 & 16 & 4a-2 & a+14 \\
10a-20 & -2a+36 & 8a-4 & 2a+30 \\
\end{array}
\right]$$
I manage to come to this:
$$\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
0 & 2a+4 & 2a-2 & a+6 \\
0 & 3a+6 & 3a-4 & 2a+10 \\
\end{array}
\right]$$
Now I should try to create zero elements?
|
What about that: divide the second row by $2$ and the third row by $3$
$$
\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
0 & 2a+4 & 2a-2 & a+6 \\
0 & 3a+6 & 3a-4 & 2a+10 \\
\end{array}
\right]=\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a & 4 \\
0 & a+2 & a-1 & \frac12 a+3 \\
0 & a+2 & a-\frac43 & \frac23a+\frac{10}{3} \\
\end{array}
\right]
$$
Is it clearer now what to do?
|
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|
Evaluate $P(1)/P(-1)$
The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.
Let $P(x) = a_{2007}x^{2007} + ... + a_0$.
$r_j + \frac{1}{r_j} = \frac{r_j^2 + 1}{r_j}$
So we need the map: $x + \frac{1}{x} \to x$.
We need a $u \to x$. Where $u(x + 1/x) = x$.
The inverse function is: $u = \frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4})$ [USED WOLFRAM ALPHA]
So,
$P(x) = \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2007} + 17 \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2006} + 1$
But letting $x= 1$ or $-1$ makes it so that the answer is imaginary!
|
Continuing after step $(2)$ in @Jack D'Aurizio's answer.
$$\begin{align}\frac{P(1)}{P(-1)}&=\prod_{i=1}^{2007}\frac{r_i^2-r_i+1}{r_i^2+r_i+1}=\prod_{i=1}^{2007}\frac{(r_i+\omega)(r_i+\omega^2)}{(r_i-\omega)(r_i-\omega^2)}=\prod_{i=1}^{2007}\frac{(-\omega-r_i)(-\omega^2-r_i)}{(\omega-r_i)(\omega^2-r_i)}\\ &=\prod_{i=1}^{2007}\frac{f(-\omega)\ f(-\omega^2)}{f(\omega)\ f(\omega^2)}\\ &=\frac{(17\omega^2)(17\omega)}{(2+17\omega^2)(2+17\omega)}=\frac{289}{259}\end{align}$$
|
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|
Finding more critical points If we have the function $f$ defined:
$$f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$$ for $-\pi \le x\le \pi$ and $-\pi \le y \le \pi$
Find the critical points and determine the nature of each.
I'm a bit stuck on this.
I've found:
$\frac {\partial f}{\partial x}=2\cos(x)+\cos(x+y)$
$\frac {\partial f}{\partial y}=2\cos(y)+\cos(x+y)$
$\frac {\partial^2 f}{\partial x^2}=-2\sin(x)-\sin(x+y)$
$\frac {\partial^2 f}{\partial y^2}=-2\sin(y)-\sin(x+y)$
$\frac {\partial^2 f}{\partial x\partial y}=-\sin(x+y)$
Solving $f_x=0$ and $f_y=0$ I get $x=\arccos\left(\frac{-1\pm\sqrt 3}{2}\right)$ which has only one real solution, but looking at the graph there should be more than one critical point in the region. How should I find the others?
|
Based on your comment the function is $f(x,y)=2\sin(x)+2\sin(y)+\sin(x+y)$, to solve critical points,
$f_x=2cos(x)+cos(x+y)=0$ and
$f_y=2cos(y)+cos(x+y)=0$
Subtract equations we have $\cos(x)=\cos(y)$, when $x$ and $y$ is constrained on given range, that means $x=y$ or $x=-y$.
When $x=y$, $2\cos(x)+\cos(2x)=0\Rightarrow 2\cos(x)+2\cos^2(x)-1=0\Rightarrow \cos(x)=\frac{-1+\sqrt{3}}{2}$ ($\frac{-1-\sqrt{3}}{2}<-1$ therefore outside the range of $\cos(x)$)
That means $x=y=\pm\arccos(\frac{-1+\sqrt{3}}{2})$.
When $x=-y$, $\cos(x+y)=1\Rightarrow 2cos(x)=-1\Rightarrow cos(x)=-\frac{1}{2}$
$x=\pm 2\pi/3$, $y=-x=\mp2\pi/3$.
So critical points are $(\arccos(\frac{-1+\sqrt{3}}{2}),\arccos(\frac{-1+\sqrt{3}}{2}))$, $(-\arccos(\frac{-1+\sqrt{3}}{2}), -\arccos(\frac{-1+\sqrt{3}}{2}))$, $(2\pi/3,-2\pi/3)$, $(-2\pi/3,2\pi/3)$
|
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|
$ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $
For positive variables $ x+y+z=3 $, show that $ \displaystyle \sum_{cyc} \dfrac{x}{2x^2+x+1} \leq \dfrac{3}{4} $.
Apart from $ (n-1)$ EV, I could not prove this inequality. I've tried transforming it into a more generic problem - it looks fairly more interesting to me.
Consider a continuous function $ f(x) \geq 0, \; 0 < x < \alpha $, with a unique value of $ \gamma $ such that $ f'(\gamma) = 0 $ and a unique value of $ \theta $ such that $ f''(\theta) = 0 $, being $ \gamma < \theta $.
If $ \displaystyle \sum_{i=1}^{n} x_i = \alpha, \; f(0)=0$ and $ f \left ( \frac{\alpha}{n} \right ) = \frac{\beta}{n} $, is it always correct to conclude that $ \displaystyle \sum_{i=1}^{n} f (x_i) \leq \beta $, being $ f(\gamma) > \frac{\beta}{n} $? Provide an example other than $ f(x) = \frac{x}{2x^2+x+1} $ and a counter-example.
|
$\left(\frac{x}{2x^2+x+1}\right)''=\frac{2(4x^3-6x-1)}{(2x^2+x+1)^3}<0$ for all $0<x<1$.
Hence, by Vasc's LCF Theorem it remains to prove our inequality for $y=x$ and $z=3-2x$,
which gives $(x-1)^2(8x^2-14x+9)\geq0$. Done!
|
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|
What is the remainder of $31^{2008}$ divided by $36$? What is the remainder of $31^{2008}$ divided by $36$?
Using Euler's theorem, we have:
$$
\begin{align*}
\gcd(31,36) = 1 &\implies 31^{35} \equiv 1 \pmod{36} \\
&\implies 31^{2008} \equiv 31^{35(57) + 13} \equiv (31^{35})^{57+13} \equiv 1^{57}*31^{13} \\
&\implies 31^{13} \equiv 31\pmod{36}
\end{align*}
$$
So the remainder is $31$, is this correct?
|
Euler's Theorem states $a^{\phi (n)} \equiv 1 \left(\mod \, \, n\right)$ if $a \perp n$, or $\gcd(a,n)=1$. Since $\gcd(31,36)=1$,
\begin{equation}
31^{\phi (36)} = 31^{12} \equiv 1 (\mod\,\,36)
\end{equation}
Then,
\begin{align}
(31^{12})^2 &= 31^{144} \equiv 1 (\mod \, \, 36) \\
31^{144} \cdot (31^{12})^{155} &= 31^{2004} \equiv 1 (\mod \, \, 36)
\end{align}
So,
\begin{equation}
31^{2008} = 31^{2004} \cdot 31^4 \equiv 1 \cdot 31^4 (\mod \, \, 36)
\end{equation}
But $31 \equiv 31 (\mod \, \, 36) \equiv -5 (\mod \, \, 36)$ so
\begin{equation}
31^2 \equiv (-5)^2 (\mod \, \, 36) \equiv 25 (\mod \, \, 36) \equiv -11 (\mod \, \, 36)
\end{equation}
and then,
\begin{equation}
31^4 \equiv (-11)^2 (\mod \, \, 36) \equiv 121 (\mod \, \, 36) \equiv 13 (\mod \, \, 36)
\end{equation}
Thus,
\begin{equation}
31^{2008} \equiv 1 \cdot 31^4 (\mod \, \, 36) \equiv 13 (\mod \, \, 36)
\end{equation}
|
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|
Proving a binomial identity Prove that ($k\le m$)
$$\sum_{j=k}^m(_{2m+1}^{2j+1})(_k^j)=\frac{2^{2(m-k)}(2m-k)!}{(2m-2k)!k!} , (k\le m)$$
Please help me with this identity, I've spent a lot of time on it but didn't solve the problem.
|
According to OPs expression we show:
The following is valid
\begin{align*}
\sum_{j=k}^{m}\binom{2m+1}{2j+1}\binom{j}{k}=4^{m-k}\binom{2m-k}{k}\qquad\qquad 0\leq k \leq m
\end{align*}
In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^{k}$ in a series $\sum_{j=0}^{\infty}a_jx^j$.
We start with the left hand side and transform the sum into a coefficient of $x^k$ of a polynomial in $x$.
\begin{align*}
\sum_{j=k}^{m}\binom{2m+1}{2j+1}\binom{j}{k}&=\sum_{j=0}^{m}\binom{2m+1}{2j+1}\binom{j}{k}\tag{1}\\
&=[x^k]\sum_{j=0}^{m}\binom{2m+1}{2j+1}(1+x)^j\tag{2}
\end{align*}
Comment:
*
*In (1) we start the index $j$ from $0$. Note that $\binom{j}{k}=0$ if $j<k$. We also write $\binom{j}{k}=[x^k](1+x)^j$.
*In (2) we use the linearity of the coefficient of operator.
Note, that (2) contains binomial coefficients with odd $2j+1$. We consider $(1\pm y)^{2m+1}$ and split it into odd and even parts.
\begin{align*}
(1\pm y)^{2m+1}&=\sum_{j=0}^{m}\binom{2m+1}{2j}y^{2j}\pm \sum_{j=0}^{m}\binom{2m+1}{2j+1}y^{2j+1}\\
\end{align*}
We get
\begin{align*}
\sum_{j=0}^{m}\binom{2m+1}{2j+1}y^{2j}=\frac{(1+y)^{2m+1}-(1-y)^{2m+1}}{2y}
\end{align*}
Substituting: $y\rightarrow\sqrt{1+x}$ we obtain from (2)
\begin{align*}
\sum_{j=k}^{m}\binom{2m+1}{2j+1}\binom{j}{k}
=[x^k]\frac{(1+\sqrt{1+x})^{2m+1}-(1-\sqrt{1+x})^{2m+1}}{2\sqrt{1+x}}\tag{3}
\end{align*}
Note, the expression on the RHS is a polynomial, since the factor $\sqrt{1+x}$ cancels out.
Next we transform the RHS of OPs expression. We consider it as coefficient of the polynomial
\begin{align*}
4^{m-k}\binom{2m-k}{k}&=4^{m-k}[x^k]\sum_{j=0}^{m}\binom{2m-k}{k}x^{j}\\
&=4^m[x^k]\sum_{j=0}^{m}\binom{2m-k}{k}\left(\frac{x}{4}\right)^{j}\tag{4}
\end{align*}
In order to prove (4) we refer to a useful identity stated as formula (2.1) in H.W.Goulds Combinatorial Identities, Vol. 4. With $n=2m+1$ we get the following expression
\begin{align*}
\sum_{j=0}^{m}(-1)^j\binom{2m-j}{j}(t_1t_2)^j(t_1+t_2)^{2m-2j}=\frac{t_1^{2m+1}-t_2^{2m+1}}{t_1-t_2}
\end{align*}
Setting $t_1=1+\sqrt{1+x}$ and $t_2=1-\sqrt{1+x}$ we get
\begin{align*}
t_1t_2&=-x\\
t_1+t_2&=2\\
t_1-t_2&=2\sqrt{1+x}
\end{align*}
It follows
\begin{align*}
\sum_{j=0}^{m}\binom{2m-j}{j}x^j2^{2m-2j}
=\frac{(1+\sqrt{1+x})^{2m+1}-(1-\sqrt{1+x})^{2m+1}}{2\sqrt{1+x}}
\end{align*}
and we obtain
\begin{align*}
4^{m-k}\binom{2m-k}{k}=[x^k]\frac{(1+\sqrt{1+x})^{2m+1}-(1-\sqrt{1+x})^{2m+1}}{2\sqrt{1+x}}\tag{5}
\end{align*}
The RHS of (3) and (5) coincide and OPs identity follows.
|
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|
Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?
|
Given any $(a,b)$ such that $a^3 + 4a^2b = 4a^2 + b^4$.
(1) If $a = 0$, then $b$ is clearly 0, $\sqrt{a} - 1 = -1$ is not a rational square.
(2) If $b = 0$, then $a^3 = 4a^2 \implies a = 0, 4$.
*
*In the first sub case $a = 0 \implies \sqrt{a}-1 = -1$ is not a square.
*In the second sub case, $a = 4 \implies \sqrt{a} - 1 = 1$ is a square.
Assuming $a, b \ne 0$, we have
$$a(a+2b)^2 = ( a^3 + 4a^2b ) + 4ab^2 = (4a^2 + b^4) + 4a^2b = (2a+b^2)^2\tag{*1}$$
(3) Assuming $a + 2b \ne 0$, this leads to
$$a = \left(\frac{2a + b^2}{a+2b}\right)^2
\implies \sqrt{a} = \left|\frac{2a + b^2}{a+2b}\right| \quad\text{ is rational }$$
Divide both sides of $(*1)$ by $a^2$ and taking square root, we get
$$\sqrt{a} + \frac{2b}{\sqrt{a}} = \pm \left( \frac{b^2}{a} + 2 \right)$$
The sign on RHS cannot be -ve. otherwise, we will have
$$\sqrt{a} + \frac{2b}{\sqrt{a}} = - \left( \frac{b^2}{a} + 2 \right)
\implies \sqrt{a} + \left(\frac{b}{\sqrt{a}}+1\right)^2 = -1
\quad\text{ which is absurb! }$$
So the sign on RHS is +ve and we have
$$\sqrt{a} - 1 = \frac{b^2}{a} - \frac{2b}{\sqrt{a}} + 1
= \left(\frac{b}{\sqrt{a}} - 1\right)^2
= \left(b\left|\frac{a+2b}{2a + b^2}\right| - 1\right)^2$$
which is the square of a rational number.
(4) This leaves us with the case $a + 2b = 0$. In this case,
$$( 2a + b^2 )^2 = 0 \implies 2a + b^2 = 0 \implies b(b-4) = 0$$
Since we have assume $a, b \ne 0$, we get $b = 4$ and hence $(a,b) = (-8,4)$.
where $\sqrt{a} - 1$ is not defined.
Summary
Aside from the special case $a = 0$ and $(a,b) = (-8,4)$, $\sqrt{a}-1$ is a the square of a rational number.
|
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|
The Diagonal Elements Of A Special Symmetric Matrix
A $n \times n$ matrix $M$ is a symmetric matrix,where $n$ is odd($i.e.n=2k+1,k\in \mathbb{Z}^{+}\cup{\{0\}}$). Every row of $M$ is a permutation of $\{1,2,\cdots,n\}$.
Show that the diagonal elements of $M$ is also a permutation of $\{1,2,\cdots,n\}.$
$e.g.$ when $n=3, $ all possible matrices satisfying $M$'s requirements as following:
$$\begin{pmatrix}
{\color{red} 3} & 2 & 1\\
2& {\color{red}1}& 3\\
1 & 3& {\color{red}2}
\end{pmatrix},\begin{pmatrix}
{\color{red}3} & 1 & 2\\
1& {\color{red}2}& 3\\
2 & 3& {\color{red}1}
\end{pmatrix},\begin{pmatrix}
\color{red}2 & 1 & 3\\
1& \color{red}3& 2\\
3 & 2& \color{red}1
\end{pmatrix},\begin{pmatrix}
\color{red}2 & 3 & 1\\
3& \color{red}1& 2\\
1 & 2& \color{red}3
\end{pmatrix},\begin{pmatrix}
{\color{red}1} & 2 & 3\\
2& {\color{red}3}& 1\\
3 & 1& {\color{red}2}
\end{pmatrix},\begin{pmatrix}
{\color{red}1} & 3 & 2\\
3& {\color{red}2}& 1\\
2 & 1& {\color{red}3}
\end{pmatrix}.$$
Obviously,the diagonal elements of each one is a permutation of $\{1,2,3\}.$
For $n=2k+1,k\geq 2 ,k\in \mathbb{N}.$
I consider the characteristic polynomial of $M$: $$f_{\mathbf{M}}(\lambda)=(\lambda-\frac{n(n+1)}{2})(\lambda^{2}+a_{1}\lambda+b_{1})\cdots(\lambda^{2}+a_{k}\lambda+b_{k}).$$
If we can prove $a_{1}=a_{2}=\cdots=a_{k}=0,$then $\mathbb{trace}(M)=\frac{n(n+1)}{2}.$
Additionally,if we can prove the product of diagonal elements $\prod_{i=1}^{n}a_{11} a_{22}\cdots a_{nn}=n!,$then the question will be sloved.
But both of them are not easy to be proved.If you have some good ideas, please give me some hints !
|
A simple parity argument will do. By assumption, every element of $S=\{1,2,\ldots,n\}$ must occur an odd number of times (or more precisely, $n$ times). Yet, if some element of $S$ does not appear on the diagonal, it must appear in $M$ an even number of times, because the matrix is symmetric and off-diagonal elements occur in pairs.
|
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|
Matrices such that $(A+B)^n = A^n + B^n$ How can I prove that $(A+B)^n = A^n + B^n$ for all integers $n \geq 1$ ?
I have been thinking about induction? Start for example with basecase 2 and then assume it's true for n = k, which would imply it's true for n = k + 1. But exactly how would this look, could someone help me please? I guess something like $(A+B)^{k+1} = (A+B) \cdot (A+B)^k$
Oh yes...and A is a matrix which looks like $ \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}$
B looks like $ \begin{bmatrix}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2 \\
\end{bmatrix}$
|
Once you establish $AB = BA = 0$ the induction is easy:
We will take as our base case $n = 2$ (so we may assume later $n$ is at least $2$ -the case $n = 1$ is trivial).
$(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2 = A^2 + 0 + 0 + B^2 = A^2 + B^2$.
Suppose $(A + B)^{n-1} = A^{n-1} + B^{n-1}$.
Then $(A + B)^n = (A + B)(A + B)^{n-1} = A(A + B)^{n-1} + B(A + B)^{n-1}$
$= A(A^{n-1} + B^{n-1}) + B(A^{n-1} + B^{n-1})$ (by our induction hypothesis)
$= A^n + AB^{n-1} + BA^{n-1} + B^n$ (the distributive law for matrices)
$= A^n + AB(B^{n-2}) + BA(A^{n-2}) + B^n$ (perhaps now you see why we want $n \geq 2$)
$= A^n + 0(B^{n-2}) + 0(A^{n-2}) + B^n = A^n + B^n$.
There is no need to use the binomial theorem, although this is indeed a "special case" of when it applies (namely, when $A$ and $B$ commute).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely contradictory to what is asked to prove
|
For $E=(a+1/a)^2+(b+1/b)^2=a^2+b^2+1/a^2+1/b^2+4$ you have $1=(a+b)^2=a^2+b^2+2ab\leq 2(a^2+b^2)$, so $a^2+b^2\geq 1/2$. Moreover, $\frac{a+b}{2}\geq 2\sqrt{ab}$ so $\frac{1}{(ab)^2}\geq 16$. This implies $$E=a^2+b^2+\frac{a^2+b^2}{a^2b^2}+4\geq 9/2+8=\frac{25}{2},$$
because $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot 16=8$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
solve complex equation for $x$ and $y$ How to solve this?
I have tried to put all $z$s on one side,
but I don't have an idea to continue.
$$z^3-i(z-2i)^3=0$$
|
$$
z^3 = i\cdot(z-2i)^3.
$$
Therefore $z$ is one of the three cube roots of $i(z-2i)^3$.
Observe that $(-i)^3 = i$ so $z$ is one of the three cube roots of $(-i)^3 (z - 2i)^3 = (-2-iz)^3$.
One of those three cube roots is $-2-iz$.
So either $z=-2-iz$ or $z$ is one of the other two cube roots.
If $z=-2-iz$ then $z(1+i) = -2$ so $z=-2/(1+i) = i-1$.
Since $z$ is one of the three cube roots of its cube and it's one of the three cube roots of $-2-iz$, the three cube roots of $i-1$ are what we seek.
What are the three cube roots of $i-1 = \sqrt{2}\left(\cos(135^\circ) + i\sin(135^\circ)\right)$. Here we invoke de Moivre's theorem and seek the three numbers that are $135^\circ/3$ and this division is done modulo $360^\circ$. That gives us
$$
45^\circ \text{ or } {45}^\circ \pm \frac{360^\circ} 3 = 45^\circ \text{ or } {-75}^\circ \text{ or } {+165}^\circ.
$$
Modulo $360^\circ$, that is the same as
$$
45^\circ \text{ or } 165^\circ \text{ or } 285^\circ,
$$
all differing from each other by a third of a circle. So we have:
\begin{align}
\sqrt 2\left(\cos45^\circ + i \sin45^\circ\right) & = 1+i \\[10pt]
\sqrt2 \left(\cos165^\circ + i\sin165^\circ\right) & = \cdots \\[10pt]
\sqrt 2\left(\cos 185^\circ + i \sin185^\circ\right) & = \cdots
\end{align}
|
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|
How to find solutions to $x^2 \equiv 4 \pmod{91}$? As the title says, I'm looking to find all solutions to $$x^2 \equiv 4 \pmod{91}$$ and I am not exactly sure how to proceed.
The hint was that since 91 is not prime, the Chinese Remainder Theorem might be useful.
So I've started by separating into two separate congruences:
$$x^2 \equiv 4 \pmod{7}$$ $$x^2 \equiv 4 \pmod{13}$$
but now I'm confused about how to apply the CRT so I'm a bit stuck, and I'd appreciate any help or hints!
|
We begin with your system of equations: $$\begin{cases} x^2 \equiv 4 \pmod{7} \\ x^2 \equiv 4 \pmod{13} \end{cases}$$
Then, solving each congruence, we obtain the system: $$\begin{cases} x \equiv \pm 2 \pmod{7} \\ x \equiv \pm 2 \pmod{13} \end{cases}$$
We therefore have four systems of linear congruences, each with a unique solution by the Chinese Remainder Theorem:
$$\begin{cases} x \equiv 2 \pmod{7} \\ x \equiv 2 \pmod{13} \end{cases} \tag{1}$$
$$\begin{cases} x \equiv 2 \pmod{7} \\ x \equiv - 2 \pmod{13} \end{cases} \tag{2}$$
$$\begin{cases} x \equiv - 2 \pmod{7} \\ x \equiv 2 \pmod{13} \end{cases} \tag{3}$$
$$\begin{cases} x \equiv - 2 \pmod{7} \\ x \equiv -2 \pmod{13} \end{cases} \tag{4}$$
These four solutions will be distinct modulo $91$, it remains to find them!
|
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|
Show that if $\gcd(a,b)=1$, then $\gcd(a+b,a^2+b^2-ab)=1\text{ or } 3$. I come across things like the following but I don't quite know how to use them.
$$\gcd(a+b,a^2+b^2-ab)|3ab$$
$$\gcd(a,b)=1\Rightarrow \gcd(a+b,ab)=1$$
|
$a^2 + b^2 - ab = (a + b)^2 - 3ab$
so
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， 3ab)$$
If $3$ does not divide $ a + b$ then
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， ab) = \gcd(a,b) = 1$$
otherwise let $a + b = 3k,\,\, k\in \mathbb{Z}$, then $b = 3k - a$ and
$$\gcd(a+b,a^2 + b^2 - ab) = 3\gcd(k， a(3k - a)) = 3\gcd(k,a^2) = 3$$
since $3$ does not divide $a$ and $\gcd(a,k) = \gcd(a,3k) = \gcd(a,3k - a) = \gcd(a,b) = 1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
A good way to solve this trigonometric equation $$\sin x+\cos x=\frac{1}{2}$$
What is the value of $\tan x$? I tried using
$$\sin2 x=\frac{2\tan x}{1+\tan^2x}$$ and $$\cos2 x=\frac{1-\tan^2x}{1+\tan^2x}$$ but we get a quadratic for $\tan\left(\frac{x}{2}\right)$ . So any better approach would be much appreciated. Thanks!
|
HINT:
$$\cos(x)+\sin(x)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{2}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{2\sqrt{2}}\Longleftrightarrow$$
$$\frac{\pi}{4}+x=\pi-\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow\space\space\vee\space\space\frac{\pi}{4}+x=\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right)+2\pi n_2\Longleftrightarrow$$
$$x=2\left(\pi n_1+\tan^{-1}\left(\frac{1}{3}\left(2-\sqrt{7}\right)\right)\right)\space\space\vee\space\space x=2\left(\pi n_2+\tan^{-1}\left(\frac{1}{3}\left(2+\sqrt{7}\right)\right)\right)$$
With $n_1,n_2\in\mathbb{Z}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating manually?
|
$$I=\int\frac{\sqrt{4x+1}}{2\sqrt{x}}dx$$ Use $$\sqrt{x}=t$$ $\implies$ $$\frac{dx}{2\sqrt{x}}=dt$$
we get $$I=\int \sqrt{4t^2+1}dt=2\int\sqrt{t^2+0.5^2}dt$$
now use standard integral $$\int\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln|x+\sqrt{x^2+a^2}|$$
|
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|
Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?
Steps I took:
*
*Let $A= \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right]$ and let $B= \left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right]$
*$\left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right] \cdot \left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right] = \left[\begin{array}{rr}
aw+by & ax+bz \\
cw+dy & cx+dz \\
\end{array}\right] $ (Which is also equal to A)
*$\left[\begin{array}{rr}
w & x \\
y & z \\
\end{array}\right] \cdot \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right] = \left[\begin{array}{rr}
wa+xc & wb+xd \\
ya+zc & yb+zd \\
\end{array}\right] $ (Which is also equal to B)
At this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.
My proof after consultation with answerers below:
Proof:
1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$
2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)
3) Then, $AB=A$ (since $AB=A$ was given)
4) Therefore, $AA$ is equal to $A$
5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$
6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)
7) Then, $BA=B$
8) Therefore, $BB$ is equal to $B$
9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$
Q.E.D.
|
$$A \cdot B = A \\
\implies (A \cdot B) \cdot A
= A \cdot A \\
\implies A \cdot (B \cdot A) = A^2 \\
\implies A \cdot B = A^2 = A$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1519239",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
|
Your proof is correct, but I feel that this could be proved by contradiction.
Assume for contradiction $\exists x>0$ such that the equation $\sqrt{x+2}-\sqrt{x+1}=\sqrt{x+1}-\sqrt{x}$ is true. Then,
\begin{align}
\sqrt{x+2}-\sqrt{x+1}&=\sqrt{x+1}-\sqrt{x}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&=\frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &=\sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &=\sqrt{x+2}\\
x&=x+2
\end{align}
This is not true and we have reached a contradiction. Thus the equation does not hold.
|
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|
Proof (cases & induction): Find the set of positive integers such that $n! \geq n^3$ I need to find the set of positive integers such that $n! \geq n^3$, and then prove my answer is true using cases and induction on $n$.
There is a lemma that I will need to prove and use for this proof.
The lemma is :
$n^2+2n+1\leq n^3$ when $n\geq$ ??
Here is my outline of the proof:
Claim: The set of all positive integers that satisfy $n! \geq n^3$ is $\{n\in \mathbb{Z} | n=1$ or $n\geq 6\}$
Let us prove the following Lemma : $n^2+2n+1\leq n^3$ when $n\geq 3$
Let $n \in \mathbb{Z}^+$
$n^2+2n+1 \leq n^2+2n+n$ when $n\geq 1$
$n^2+2n+1 \leq n^2+n^2+n$ when $n\geq 2$
$n^2+2n+1 \leq n^2+n^2+n^2$ when $n \geq 3$
$n^2+2n+1 \leq 3n^2 \leq n^3$ when $n \geq 3$
Thus, we have $n^2+2n+1\leq n^3$ when $n\geq 3$
This lemma is important because when we use induction on $n$, we want to eventually show that
$n!\geq n^3$ implies $(n+1)!\geq(n+1)^3$
In order to show this, we need to multiply both sides of $n!\geq n^3$ by $(n+1)$
Our result would be $(n+1)n!\geq (n+1)n^3$
Now we need to show that $(n+1)n^3\geq (n+1)(n+1)^2$ so that we can get to our goal.
Using our lemma, we have proved that $(n+1)^2\leq n^3$ when $n\geq 3$, so we know that we can use induction on our original claim so long as $n\geq 3$.
So using our lemma, we can assume that if $(n+1)n!\geq (n+1)n^3$ is true, then $(n+1)!\geq(n+1)^3$ is also true, so long as $n\geq 3$
Now we can use cases and induction to prove our claim.
Case 1: $n=1$, $1!\geq 1^3$
Case 2: $n=2$, $2!\ngeq 2^3$
Case 3: $n=3$, $3!\ngeq 3^3$
Case 4: $n=4$, $4!\ngeq 4^3$
Case 5: $n=5$, $5!\ngeq 5^3$
Case 6: $n=6$, $6!\geq 6^3$
Let us use induction on $n$ to prove that $n! \geq n^3$ for all positive integers greater than or equal to $6$.
Base case: $n=6$. $6!\geq 6^3$, so we have proved the base case.
Inductive step: Suppose we have $(\star)$ $n!\geq n^3$. We want to show that $(n+1)!\geq(n+1)^3$. Let us multiply both sides of $(\star)$ by $(n+1)$.
We have $(n+1)n!\geq (n+1)n^3$ This can be rewritten as $(n+1)!\geq (n+1)n^3$
Using our lemma, we can assume that if $(n+1)!\geq (n+1)n^3$ is true, then $(n+1)!\geq(n+1)^3$ is also true, so long as $n\geq 3$.
This completes the induction, and we have proved the claim.
My questions are:
1) Have I completely proved the lemma?
2) Are there any errors in the proof?
3) How can I improve this proof?
|
Let $f(x) = x^3 - (x + 1)^2$
$f'(x) = 3x^2 - 2x + 2$. $f'(x) = 0$ when $x = \frac{2 \pm \sqrt(4 - 24}{6}$ so $f'(x)$ never equals $0$ for any real x. So $f(x)$ has no extrema and thus only one real root.
$ f(2) = -1; f(3) = 18$ so $f(x)$ has only one real root is between 2 and 3. So $f(n) > 0$ iff $n^3 > (n+1)^2$ iff $n \ge 3$.
=====
D'oh! The obvious easiest way:
Let $ m = n+1;$ $\frac{n^3}{n^2 + 2n + 1} = \frac{(m - 1)^3}{m^2}$
$= \frac{m^3 - 3m^2 + 3m -1}{m^2} = m - 3 + (\frac 3 m - \frac 1 {m^2})> m -3$
so $\frac{n^3}{n^2 + 2n + 1} > n -2$
so $\frac{n^3}{n^2 + 2n + 1} > 1 \iff n^3 > n^2 + 2n + 1$ when $n \ge 3$.
|
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|
Evaluate thus limit using series: $\lim_{x\to0} (\sin x-\tan x)/x^3$ Evaluate thus limit using series:
$$\lim_{x\to0} \frac{\sin x-\tan x}{x^3}$$
I know the value of this limit is -1/2, and I also know the series expansion for $\sin x$ is $$x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$
I am having trouble being able to compute the limit. Is there another series that needs to be expanded out?
|
Expand both $\sin$ and $\tan$: around $0$,
$$\begin{align}
\sin x &= x - \frac{x^3}{6} + o(x^3) \\
\tan x &= x + \frac{x^3}{3} + o(x^3)
\end{align}$$
See below for more.
$$\begin{align}\frac{\sin x - \tan x}{x^3} &= \frac{x - \frac{x^3}{6} - \left(x + \frac{x^3}{3}\right) + o(x^3)}{x^3} = \frac{-\frac{x^3}{3} - \frac{x^3}{6} + o(x^3)}{x^3} = -\frac{1}{2} + o(1)\end{align}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1525034",
"timestamp": "2023-03-29T00:00:00",
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|
Proving the limit of a radical using epsilon delta Define $f:(0,1)->R$ by $f(x)=\frac{\sqrt{9-x}-3}{x} $
I know that $f(x)=\frac{\sqrt{9-x}-3}{x}*\frac{\sqrt{9-x}+3}{\sqrt{9-x}+3}=\frac{9-x-9}{\sqrt{9-x}+3}=-\frac{1}{\sqrt{9-x}+3} $
Since $f$ is only defined in $(0,1)$, $-\frac{1}{5}<-\frac{1}{\sqrt{9-x}+3}<-\frac{1}{6} $
So can I just pick some $\delta$ close to 0?
Therefore $\lvert -\frac{1}{\sqrt{9-x}+3}+\frac{1}{6} \rvert =\lvert \frac{-6+\sqrt{9-x}+3}{6(\sqrt{9-x}+3)} \rvert =\lvert \frac{\sqrt{9-x}-3}{6\sqrt{9-x}+18}\rvert=\lvert \frac{1}{6}*\frac{\sqrt{9-x}-3}{\sqrt{9-x}+3} \rvert < \epsilon$
Take $\delta = \frac{\epsilon}{6}$
$\therefore \lvert -\frac{1}{\sqrt{9-x}+3}+\frac{1}{6} \rvert < \epsilon $ if $0<\lvert x- 0 \rvert < \delta $ for all $x \in (0,1)$
|
From what you wrote I do not see by what you make at the choice. Instead:
If $0 < x < 1$, then
$$
\bigg| \frac{-1}{\sqrt{9-x}+3} + \frac{1}{6} \bigg|
=
\frac{3 - \sqrt{9-x}}{6\sqrt{9-x}+18} < 3 - \sqrt{9-x} =: s_{x} \leq 3;
$$
given any $0 < \varepsilon \leq 3$, we have $s_{x} < \varepsilon$ if $x < 9 - (3-\varepsilon)^{2}$; putting the above together, we may take $\delta := 9 - (3 - \varepsilon)^{2}$.
|
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|
How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? I can show for any given value of n that the equation
$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$
is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction and converting the cosine to a sum of complex exponential functions
$$\frac{1}{2}\sum_{k=1}^n [\exp(\frac{i 2 \pi k}{n})+\exp(\frac{-i 2 \pi k}{n})] = 0$$
and using the conversion for finite geometric sequences
$$S_n = \sum_{k=1}^n r^k = \frac{r(1-r^n)}{(1-r)}$$
I have even tried this this suggestion I have seen on the net by pulling out a factor of $\exp(i \pi k)$ but I have still not gotten zero.
Please assist.
|
One approach is to write
$$\cos\left(\frac{2\pi k}{n}\right)=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(k+1)}{n}\right)-\sin\left(\frac{2\pi(k-1)}{n}\right)\right)$$
Now, we have converted the sum into a telescoping sum, which we can evaluate directly as
$$\begin{align}
\sum_{k=1}^n\cos\left(\frac{2\pi k}{n}\right)&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)+\sin\left(\frac{2\pi(n)}{n}\right)\right)\\\\
&-\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(2-1)}{n}\right)+\sin\left(\frac{2\pi(1-1)}{n}\right)\right)\\\\
&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi(n+1)}{n}\right)-\sin\left(\frac{2\pi(2-1)}{n}\right)\right)\\\\
&=\frac{1}{2\sin\left(\frac{2\pi}{n}\right)}\left(\sin\left(\frac{2\pi}{n}\right)-\sin\left(\frac{2\pi}{n}\right)\right)\\\\
&=0
\end{align}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Random Variables, Die toss A fair die is tossed twice. Let $X$ = the ssm of the faces, $Y$= the maximum of the two faces, and $Z$=|face 1 - face 2|.
write down the value of $X,Y,$ and $W=XZ$ for each outcome $w\in\ S$
I already found the value and range of $X,Y$ but I'm not sure how to find $W=XZ$.
I saw someone post a similar question already answered however, it wasn't explained how to find $W=XZ$.
|
Since you said you've enumerated the outcomes for $X$, do the same for $Z$. Below I made a table for the values of both $X$ and $Z$. Can you now make the corresponding table for $W = XZ$?
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
X& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 \\
4 & 5 & 6 & 7 & 8 & 9 & 10 \\
5 & 6 & 7 & 8 & 9 & 10 & 11 \\
6 & 7 & 8 & 9 & 10 & 11 & 12\\
\hline
\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|c|c|}
\hline
Z & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 0 & 1 & 2 & 3 & 4 & 5 \\
2 & 1 & 0 & 1 & 2 & 3 & 4 \\
3 & 2 & 1 & 0 & 1 & 2 & 3 \\
4 & 3 & 2 & 1 & 0 & 1 & 2 \\
5 & 4 & 3 & 2 & 1 & 0 & 1 \\
6 & 5 & 4 & 3 & 2 & 1 & 0 \\
\hline
\end{array}
$$
|
{
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|
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
|
More unorthodoxly, we first take the midpoint of $x-\frac{\pi}{6}$ and $x + \frac{\pi}{4}$ before applying the angle sum and difference identities:
Let $A = x + \frac{\pi}{24}$ (the midpoint). Then we need
$$\sin\left(A - \frac{5\pi}{24}\right) + \cos \left(A + \frac{5\pi}{24}\right) = 0$$
$$\sin A\cos\frac{5\pi}{24} - \cos A\sin\frac{5\pi}{24} + \cos A\cos\frac{5\pi}{24} - \sin A\sin\frac{5\pi}{24} = 0$$
$$\sin A\left(\cos \frac{5\pi}{24} - \sin \frac{5\pi}{24}\right) + \cos A\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right) = 0$$
$$\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right)\left(\sin A + \cos A\right) = 0$$
so we need
$$\sin A + \cos A = 0$$
$$\tan A = -1$$
$$\tan \left( x + \frac{\pi}{24}\right) = -1$$
which can be solved directly as per normal.
|
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|
How to calculate $\lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$. Let $s = \lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$.
$$st = t\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ...}}}} = \sqrt{t^{2} + \sqrt{2t^{4} + \sqrt{4t^{8} + \sqrt{8t^{16} + ...}}}}$$
Let $2t^{4} = t^{2}$:
$$2t^{2} = 1$$
$$t^{2} = \frac{1}{2}$$
$$t = \frac{1}{\sqrt{2}}$$
$$\frac{s}{\sqrt{2}} = \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + ...}}}}$$
It can be shown that:
$$\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + ...}}}} = \frac{1 \pm \sqrt{1 + 4x}}{2}$$
Therefore:
$$\frac{s}{\sqrt{2}} = \frac{1 \pm \sqrt{3}}{2}$$
$$s = \frac{\sqrt{2} \pm \sqrt{6}}{2}$$
Is this correct?
|
If $t=\frac1{\sqrt2}$, then $t^2=\frac12$, and $2t^4=\frac12$, but $4t^8=\frac14$. Your expression for $\frac s{\sqrt2}$ isn't composed entirely of $\frac12$s.
|
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|
Simultaneous equations How can I find the solution $(3, 3, \frac{-3}{2})$ from the following system of equations:
$x+(y-1) z = 0$
$(x-1) z+y = 0$
$x (y-1)-y+2 z = 0$
I have done eq1 - eq2 to find the other solutions. How would i get to the solution $(3, 3, \frac{-3}{2})$
|
$$x+(y-1) z = 0 \implies (y-1)=-\frac xz \tag{*1} $$
$$x (y-1)-y+2 z = 0 \implies (y-1)=\frac{y-2z}{x} $$
so $$(y-1)^2=- \frac yz +2 $$
$$ (x-1) z+y = 0 \implies \frac yz = -(x-1) \tag {*2}$$
so $$(y-1)^2= x+1 \tag{*3}$$
we can get another relation between $x$ and $y$ by dividing (*1) by (*2)
$$ -\frac xy =-\frac{y-1}{x-1 } \implies y(y-1)=x(x-1) \tag{*4} $$
now solve (*3) for $x$ and plug in to (*4)
$$ y(y-1)=((y-1)^2-1)((y-1)^2-2) $$
to simplfy the algebra let $a \equiv (y-1)$
$$a(a+1)=(a^2-1)(a^2-2) \implies a=(a-1)(a^2-2)$$
so we end up with the cubic equation ...
$$ a^3-a^2-3a+2=0 $$
That's as far as I'm willing to take it, but clearly $a=2$ does solve the equation so $y=3$ then (*3) gives you $x=3$ and (*1) gives $z=-\frac 32$
|
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|
Evaluate limit $\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}$ How to evaluate $$\displaystyle\lim_{x\rightarrow\infty} (1+\sin{x})^{\frac{1}{x}}?$$
Only idea I can think of is sandwich theorem, but then I get $0^0$.
|
This limit does not exist. We can pick two sequences $a_n$ and $b_n$ such that $f(a_n)$ has different limit than $f(b_n)$.
\begin{align}
a_n &= 2\pi n + \frac{\pi}{2} \\
b_n &= 2\pi n - \frac{\pi}{2}
\end{align}
Then
$$
\left(1+\sin\left(2\pi n + \frac{\pi}{2}\right)\right)^{\frac1{2\pi n + \frac{\pi}{2}}} = 2^{\frac1{2\pi n + \frac{\pi}{2}}} \rightarrow 1
$$
and
$$
\left(1+\sin\left(2\pi n - \frac{\pi}{2}\right)\right)^{\frac1{2\pi n - \frac{\pi}{2}}} = 0^{\frac1{2\pi n - \frac{\pi}{2} }} =0 \rightarrow 0
$$
|
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|
Explain the following equalities I am a little stuck on coming up with geometrical explanation for why the following equalities are true. I tried arguing the $\cos(\theta)$ is the projection to the x-axis of a vector $r$ inside a unit circle, so as it goes around by $2 \pi$, the projections on both the positive and negative part of the x-axis cancel out. Could someone please confirm if it is completely correct to think about it this way?
$$\cos(\theta) + \cos(\theta + 2 \pi/3) + \cos(\theta + 4\pi/3) = 0$$
$$\sin(\theta) + \sin(\theta + 2 \pi/3) + \sin(\theta + 4\pi/3) = 0$$
And what about:
$$\cos^2(\theta) + \cos^2(\theta + 2 \pi/3) + \cos^2(\theta + 4\pi/3) = \frac{3}{2}$$
$$\sin^2(\theta) + \sin^2(\theta + 2 \pi/3) + \sin^2(\theta + 4\pi/3) = \frac{3}{2}$$
|
First set of equations: Here's a geometric reason. Imagine we place $1$ kg weights at the unit circle corresponding to angles of $\theta$, $\theta+\frac{2\pi}{3}$ and $\theta+\frac{4\pi}{3}$. These points form a configuration that is a rotation of
The center of mass of this configuration is at the origin, and hence the same is true for its rotations. By looking at the $x$ and $y$ coordinates, this implies that. $$\cos(\theta)+\cos(\theta+2\pi/3)+\cos(\theta+4\pi/3)=0$$ and $$\sin(\theta)+\sin(\theta+2\pi/3)+\sin(\theta+4\pi/3)=0.$$
Second set of equations:
Our goal is to show that $$\cos^2(\theta)+\cos^2\left(\theta+\frac{2\pi}{3}\right)+\cos^2\left(\theta+\frac{4\pi}{3}\right)=\sin^2(\theta)+\sin^2\left(\theta+\frac{2\pi}{3}\right)+\sin^2\left(\theta+\frac{4\pi}{3}\right),$$ since we know that the sum of all $6$ terms equals $3$ as these are three unit vectors in the plane. Proceeding by the identity $\cos(2\theta)=2\cos^2(\theta)-1$ and using the first set of equations is the quickest way to prove the result. While I do not have a geometric proof, here is another proof that I am fond of:
For $0<\theta<2\pi/3$ define $$u=\left[\begin{array}{c}
\cos\theta\\
\cos\left(\theta+2\pi/3\right)\\
\cos\left(\theta+4\pi/3\right)
\end{array}\right],\ \ \ v=\left[\begin{array}{c}
\sin\theta\\
\sin\left(\theta+2\pi/3\right)\\
\sin\left(\theta+4\pi/3\right)
\end{array}\right].$$
Then we know that $\|u\|_{2}^{2}+\|v\|_{2}^{2}=3$
, and we are trying to show that $\|v\|_{2}=\|u\|_{2}$
. By the identity $$\sin(\theta)=\frac{\sin\left(\theta+\frac{\pi}{6}\right)+\sin\left(\theta-\frac{\pi}{6}\right)}{\sqrt{3}}=\frac{-\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)}{\sqrt{3}},$$
the operator $$T=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}
0 & -1 & 1\\
1 & 0 & -1\\
-1 & 1 & 0
\end{array}\right]$$
satisfies $T(u)=v$
. Since $u$
is in the kernel of the constant matrix by the first part, we see that $$S=T+\frac{1}{3}\left[\begin{array}{ccc}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{array}\right]$$
satisfies $S(u)=v$
, and moreover $S^{-1}=S^{T}$
so that $S$
is an orthogonal matrix. This implies that $\|u\|_{2}=\|v\|_{2}$
as $S$
is then an isometry.
|
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|
How to prove $2^{2^{6n+2}} + 21$ is a composite number? How to prove that $$ 2^{2^{6n+2}} + 21 $$ is a composite number for each $$n=1,2,3,\ldots$$
I'm stuck on a what seems like a simple problem but I don't even know where to start.
|
$ 2^{2^{6n+2}} + 21=2^{4 (2^{6n} )}-2^4+2^4+21= $
$ =37+2^4\left (2^{4(2^{6n}-1 )}-1 \right )=37+2^4\left (2^{4(2^6-1 )(2^{6(n-1)}+ \dots +1 )}-1 \right )=$
$ =37+2^4\left (2^{36 \cdot 7 \cdot (2^{6(n-1)}+ \dots +1 )}-1 \right )=$
$ =37+2^4\left (2^{36}-1\right )\left (\dots \right ) \equiv 0 \bmod 37$
|
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|
Finding the equation of a tangent line to a curve at a given point. Find the equation for the tangent (in standard form) to the curve defined by $$y = \sqrt\frac{5x}{2x-1}$$
For the procedure of this question would I simply differentiate it using the quotient rule? From which you insert $x=5$ in to, and then use $y_2-y_1 = m(x_2-x_1)$ to find the equation.
I got this as my answer but it looks very ugly...
$y = 0.74x + 4.633$
(yes i know it isn't in standard form)
Am I doing something wrong?
|
The point on the curve will be $(5,\frac{5}{3})$. Then differentiate $y$ using the chain rule and quotient rule
\begin{align}
\frac{dy}{dx} = \frac{1}{2}\bigg(\frac{5x}{2x - 1}\bigg)^{-\frac{1}{2}}\bigg( \frac{5(2x - 1) - (2)(5x)}{(2x - 1)^2}\bigg)
\end{align}
Then by pluggin in $x = 5$
\begin{align}
\frac{dy}{dx}\bigg|_{x=5} &= \frac{1}{2}\bigg(\frac{25}{9} \bigg)^\frac{-1}{2}\left(\frac{-5}{81}\right)\\
&= \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{-5}{81} = -\frac{1}{54}
\end{align}
Use the point slope form
\begin{align}
y - \frac{5}{3} &= -\frac{1}{54}(x - 5 )\\
y + \frac{1}{54}x - \frac{95}{54} &= 0
\end{align}
Which is the equation of the tangent line at $x = 5$
|
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|
How do you prove that $\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x$? I have the task to prove that
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x ,\left|x\right|\le 1
$$
I do not have any ideas from where I should start.
Can anyone help me solve it?
|
You may just observe that, for $x \in [0,1)$, we have
$$
\left(\arcsin\left( \frac{1}{2} \sqrt{2-\sqrt {2-2x}}\right)\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ and we have
$$
\left(\frac{\pi}{8} + \frac{1}{4} \arcsin x\right)'=\frac14\frac1{\sqrt{1-x^2}}
$$ giving
$$
\arcsin( \frac{1}{2} \sqrt{2-\sqrt {2-2x}})=\frac{\pi}{8} + \frac{1}{4} \arcsin x, \quad x \in [0,1]
$$
since both functions take the same value at $x=0$ and at $x=1$.
|
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|
Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2+4}}=\\
& = \lim_{x \to 2}\sqrt{(2-x)(2+x)}=0.
\end{align}
|
$$\frac{4-x^2}{3-\sqrt{x^2+5}}=(4-x^2)\frac{3+\sqrt{x^2+5}}{9-(x^2+5)}=3+\sqrt{x^2+5}$$
$$\lim_{x\to2}\frac{4-x^2}{3-\sqrt{x^2+5}}=6$$
|
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|
sum of solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$ What is the sum of non-negative integer solutions of $\{(x,y,z)\mid x+y+z=k\}$, $k = 1,\ldots,N$?
I know that $\{(x,y,z)\mid x+y+z=k\}$ has $\binom{k+3-1}{3-1}=\binom{k+2}{2}$ non-negative integer solutions. Thus, the answer to the question above is $$\sum_{k=1}^N \binom{k+2}{2}.$$
Can we further simplify the summation?
|
Here's another elegant way to simplify the expression. See that $\binom{n}{r}$ is the coefficient of $x^r$ in $(1+x)^n$. Hence, your summation can be interpreted as the $$\sum_{k=1}^N \text{coefficient of } x^2 \text{ in } (1+x)^{k+2} = \text{coefficient of } x^2 \text{ in }\sum_{k=1}^N (1+x)^{k+2}$$ where the last series is a Geometric Progression, which can be summed easily.
$$\sum_{k=1}^N (1+x)^{k+2} = \sum_{k=3}^{N+2} (1+x)^k = \frac{(1+x)^3((1+x)^N-1)}{1+x-1}=\frac{(1+x)^{N+3}-(1+x)^3}{x}$$
The task is now reduced to finding the coefficient of $x^3$ in $(1+x)^{N+3}-(1+x)^3$ which is $$\binom{N+3}{3}-\binom{3}{3} = \binom{N+3}{3}-1$$
|
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|
Combinatorics Inclusion - Exclusion Principle Find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 25$ with $ 1 \leq x_1 \leq 6, 2 \leq x_2 \leq 8,
0 \leq x_3 \leq 8, 5 \leq x_4 \leq 9.$
Firstly, I defined $y_i = x_i - lower bound$ so for example $y_1 = x_1 - 1$
This gives $y_1 + y_2 + y_3 + y_4 = 17$ with $$0 \leq y_1 \leq 5, 0 \leq y_2 \leq 6, 0 \leq y_3 \leq 8, 0 \leq y_4 \leq 4$$
Next I applied Inclusion-Exclusion Principle;
Total number of ways = $17+3 \choose 3$
Number of ways with just $y_i$ breaking upper bound;
$y_1;$$11+3 \choose 3$
$y_2;$ $10+3 \choose 3$
$y_3;$ $8+3 \choose 3$
$y_4;$ $12+3 \choose 3$
Similarly I calculated the number of ways with $y_i$ and $y_j$ breaking upper bound at the same time. And since there is no way in which 3 of them break their upper bounds simultaneously I was left with ;
$${20 \choose 3} - {14 \choose 3} - {13 \choose 3} - {11 \choose 3} - {15 \choose 3} + {7\choose 3} + {5 \choose 3} + { 9 \choose 3} + {4 \choose 3} + {8 \choose 3} + {6 \choose 3} $$
as my final answer.
Is this solution correct and is there a nicer way of solving it/ any insights about this type of problem?
Note, this problem is from 'An Introduction to Combinatorics and Graph Theory'
|
As noted in the comments, your solution is correct. The approach is described in Balls In Bins With Limited Capacity.
To solve the problem using generating functions, write $x+x^2+\cdots+x^6=x\frac{1-x^6}{1-x}$ for $x_1$, $x^2+x^3+\cdots+x^8=x^2\frac{1-x^7}{1-x}$ for $x_2$, $1+x+\cdots+x^8=\frac{1-x^9}{1-x}$ for $x_3$ and $x^5+x^6+\cdots+x^9=x^5\frac{1-x^5}{1-x}$ for $x_4$. The product is
$$
x^8\frac{\left(1-x^6\right)\left(1-x^7\right)\left(1-x^9\right)\left(1-x^5\right)}{(1-x)^4}\;,
$$
and the number of solutions to $x_1+x_2+x_3+x_4=25$ is the coefficient of $x^{25}$ in this generating function. With the expansion
$$
\frac1{(1-x)^4}=\sum_{k=0}^\infty\binom{k+3}3x^k\;,
$$
you can see how multiplying out the numerator and matching each resulting term with the missing powers of $x$ from the denominator yields exactly the alternating inclusion–exclusion sum of binomial coefficients that you calculated.
|
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|
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce this further to $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$. We have
$N_1 = 4 \cdot 5 = 20 \implies 20 x_1 \equiv 1 \pmod{3} \implies 2x_1 \equiv 1 \pmod{3} \implies x_1 \equiv 2 \pmod {3}$
$N_2 = 3 \cdot 5 = 15 \implies 15x_2 \equiv 1 \pmod{4} \implies -x_2 \equiv 1 \pmod{4} \implies x_2 \equiv 3 \pmod {4}$
$N_3 = 3 \cdot 4 = 12 \implies 12 x_3 \equiv 1 \pmod{5} \implies 2x_3 \equiv 1 \pmod{5} \implies x_3 \equiv 3 \pmod {5}$
Now,
\begin{align*}
\bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 \\
&= 3 \cdot 20 \cdot 2 + 4 \cdot 15 \cdot 3 + 5 \cdot 12 \cdot 3 \\
&= 480 \equiv 0 \pmod {3 \cdot 4 \cdot 5}
\end{align*}
Is this correct, or is something wrong in my work? I don't like how I have $0$ remainder.
|
It follows from the hypothesis that $3,4,5,6$ divides $(x+1) $ and therefore $x+1$ is a common multiple of these 4 numbers.
|
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|
Calculating an Exponential Integral
Calculate $\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx$ where $a,c > 0$ and $b>1$.
The best I could do about this integral is to find an upperbound for it:
$\int_0^{\infty} \frac{x^{2N+1}}{a+x^{-b}} e^{-c x^2} dx \leq \int_0^{\infty} \frac{x^{2N+1}}{x^{-b}} e^{-c x^2} dx= \int_0^{\infty} x^{2N+b+1} e^{-c x^2} dx$
Then I can use $\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x =
\begin{cases}
\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\
\frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\
\frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0)
\end{cases} $
Any idea that how I can calculate the integral, not jut an upper bound?
|
$\int_0^\infty\dfrac{x^{2N+1}e^{-cx^2}}{a+x^{-b}}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{2n+b+2N+1}}{n!(ax^b+1)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^{\frac{2n+2N+1}{b}+1}}{n!(ax+1)}~d\left(x^\frac{1}{b}\right)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^\frac{2(n+N+1)}{b}}{bn!(ax+1)}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^n\left(\dfrac{x}{a}\right)^\frac{2(n+N+1)}{b}}{bn!(x+1)}~d\left(\dfrac{x}{a}\right)$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^nx^\frac{2(n+N+1)}{b}}{a^{\frac{2(n+N+1)}{b}+1}bn!(x+1)}~dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nc^n\pi}{a^{\frac{2(n+N+1)}{b}+1}bn!\sin\left(\pi+\frac{2(n+N+1)\pi}{b}\right)}$ (according to https://en.wikipedia.org/wiki/Beta_function#Properties)
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}c^n\pi}{a^{\frac{2(n+N+1)}{b}+1}bn!\sin\frac{2(n+N+1)\pi}{b}}$
|
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|
Solve $\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$ How do I solve limits such as these? The $...$ always make it seem hard to me. From what I can understand from them, they both are $0/0$ limits, and I should be looking to write the numerator in such a way that the denominator should simplify it somehow. For the first I tried writing each element from the numerator as $(a-1)$ where $a = 1, x, x^2 ... x^n$ so that I can simplify the denominator eventually, but I didn't get much out of that. I'm guessing these two are solved in similar ways, hence why I posted them both. Any clues/hints I can get would be appreciated.
$$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}$$
$$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}$$
$n,\:m\:\in \mathbb{R}$
|
Note that:
$\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}= \frac{(x-1)+(x^2-1)+\ldots +(x^n-1)}{x-1}=\frac{(x-1)[1+(x+1)+(x^2+x+1)\ldots +(x^{n-1}+x^{n-2}+\ldots+x+1)]}{x-1}$,
after simplification we find:
$n+(n-1)x+(n-2)x^2+\ldots+x^{n-1}$.
So,
$\lim \:_{x\to \:1}\frac{1+x+x^2+...\:+x^n-\left(n+1\right)}{x-1}=n+(n-1)+(n-2)+\ldots+2+1=\frac{n(n+1)}{2}$.
For the second one, we use the following method:
$\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x+x^2+...\:+x^m-m}=\lim _{x\to 1}\frac{x+x^2+...\:x^n-n}{x-1}\times\frac{x-1}{x+x^2+...\:+x^m-m}=\frac{n(n+1)}{2}\times\frac{2}{m(m+1)}=\frac{n(n+1)}{m(m+1)}$
|
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|
A weird property of $\sum_{k = 1}^{n} \sin k$ I was playing around with the sum $\sum_{k = 1}^{n} \sin k$, and using very loose rigour I arrived at the following:
Proposition. Let $n \equiv n_0 \pmod {44}$ and $n_0 \equiv n_1 \pmod {6}$. Then $$\sum_{k = 1}^{n} \sin k \sim \frac {1} {2} \left ( 3 - |3 - n_1| \right ).$$
Does heuristics support this claim?
Note: I can provide my "proof" on request.
|
I think it appropriate to present my rigour here now that it created sufficient curiosity.
Consider $$f (n) = \sum_{k = 1}^{n} \sin k.$$ Let $a_k = \left [ \frac {2k} {\pi} \right ]$ for $1 \leqslant k \leqslant n$ and $r_1, r_2, \cdots, r_n$ be real numbers in the interval $[0, 1]$. Then, denote $$\sin k = \begin {cases}
r_k, & \text {if } a_k \equiv 0 \pmod {4} \\
1 - r_k, & \text {if } a_k \equiv 1 \pmod {4} \\
-r_k, & \text {if } a_k \equiv 2 \pmod {4} \\
r_k - 1, & \text {if } a_k \equiv 3 \pmod {4} \\
\end {cases}$$ Here $r_{\theta}$ is the side of the right triangle just in front of the angle $\theta$. By Weyl's criterion, we see that $r_k$ is equidistributed modulo $1$, so the average of real numbers $r_k$ is $\frac {1} {2}$. Hence, $$\sin k \approx \begin {cases}
\frac {1} {2}, & \text {if } a_k \equiv 0 \pmod {4} \\
\frac {1} {2}, & \text {if } a_k \equiv 1 \pmod {4} \\
- \frac {1} {2}, & \text {if } a_k \equiv 2 \pmod {4} \\
- \frac {1} {2}, & \text {if } a_k \equiv 3 \pmod {4} \\
\end {cases}$$ Since $\frac {2mk} {\pi}$ is close to an integer when $m = 11 t$ and the period of $\sin k$ with respect to $a_k$ is $4$, it is safe to assume $f (44t + n_0) \approx f (n_0)$ for any integer $t$. Let $n = 44 t + n_1$. Also, since $a_{6j + i} \equiv a_i \pmod {4}$ for any integers $i, j$, we have by setting $i = n_1$ the desired result.
This is a slightly simplified and fast version of what I'd done to arrive at the ridiculous expression that my question was about.
|
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|
Finding $\binom n0+\binom n3+\binom n6+\cdots $ How to get
$$\binom n0 + \binom n3 + \binom n6 + \cdots$$
MY ATTEMPT
$$(1+\omega)^n = \binom n0 + \binom n1 \omega^1 + \binom n2 \omega^2 + \cdots$$
$$(1+\omega^2)^n = \binom n0 + \binom n1 \omega^2 + \binom n2 \omega^4 + \cdots $$
$$(1 + 1)^n = 2^n = \binom n0 + \binom n1 + \binom n2 + \cdots$$
$$(1+\omega)^n + (1+\omega^2)^n + (1 + 1)^n = 3 \left(\binom n0 + \binom n3 + \binom n6 + \cdots\right)$$
But how to solve LHS? I got the required equation in RHS
|
Let $\omega = \dfrac{-1+ i\sqrt 3} 2 = \cos120^\circ + i\sin120^\circ$.
Then $\omega^3 = 1$, and $1+\omega = \cos60^\circ + i\sin60^\circ$, so $(1+\omega)^2 = \omega$.
A bit of arithmetic shows that $n\mapsto(1+\omega)^n + (1+\omega^2)^n$ is a periodic function with period $6$:
\begin{array}{c|c}
n & (1+\omega)^n + (1+\omega^2)^n \\
\hline
0 & \phantom{+}2 \\
1 & \phantom{+}1 \\
2 & -1 \\
3 & -2 \\
4 & -1 \\
5 & \phantom{+}1
\end{array}
Therefore
$$
\binom n 0 + \binom n 3 + \binom n 6 + \cdots = \frac{2^n + \text{a periodic term never exceeding $2$ in absolute value} } 3.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convergence of two recursive sequence how could I find out whether those recursive sequences are convergent?
Let $a_{n+1} = a_1(1-a_n-b_n) + a_n \\ b_{n+1} = b_1(1-a_n-b_n) + b_n$, where
$a_1 = a \\ b_1 = b$ ($a, b \in (0, 1)$)
|
$$
a_{n+1}=a(1-a_n-b_n)+a_n\text{ ...(1)}\\
b_{n+1}=b(1-a_n-b_n)+b_n\text{ ...(2)}\\$$
$(1)+\alpha\times(2):$
$$a_{n+1}+\alpha b_{n+1}=(1-a-\alpha b)a_n+(-a+\alpha(1-b))b_n+a+\alpha b$$
$$\frac{-a+\alpha(1-b)}{1-a-\alpha b}=\alpha$$
$$b\alpha^2+(a-b)\alpha-a=0$$
$$(\alpha-1)(b\alpha+a)=0$$
i) $(1)+(2):$
$$a_{n+1}+b_{n+1}=(1-a-b)(a_n+b_n)+(a+b)$$
$$a_{n+1}+b_{n+1}-1=(1-a-b)(a_n+b_n-1)$$
$$\therefore a_n+b_n=(1-a-b)^{n-1}(a+b)\text{ ...(3)}$$
ii) $b\times(1)-a\times(2):$
$$b\cdot a_{n+1}-a\cdot b_{n+1}=b\cdot a_n-a\cdot b_n$$
$$b\cdot a_n-a\cdot b_n=...=b\cdot a_1-a\cdot b_1=ba-ab=0$$
$$\therefore b_n=\frac{b}{a}a_n\text{ ...(4)}$$
$(3)$ & $(4)$:
$$a_n=a\cdot (1-a-b)^{n-1}$$
$$b_n=b\cdot (1-a-b)^{n-1}$$
Therefore the sequence converges if
$$\mid 1-a-b\mid < 1$$
|
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|
Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx = \int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx
$$
We can also replace $1-x$ in the denominator with $\sqrt{(1-x)^2}$
$$
\int \frac{1-x}{\sqrt{(1-x)(2x+1)}}\,dx = \int \frac{\sqrt{(1-x)^2}}{\sqrt{(1-x)(2x+1)}}\,dx
$$
If we simplify this fraction we get:
$$
\int \frac{\sqrt{1-x}}{\sqrt{2x+1}}\,dx
$$
Next we apply the following substitutions
$$
u = -x
$$
so : $-du = dx$
We can rewrite the integral as following:
$$-\int \frac{\sqrt{1+u}}{\sqrt{1-2u}}\,du$$
Then we apply another substitution:
$\sqrt{1+u} = t $ so $ \frac{1}{2\sqrt{1+u}} = dt $
We rewrite: $ \sqrt{1+u} $ to $\frac{1}{2}t^2 \,dt $
We can also replace $\sqrt{1-2u} $ as following:
$$\sqrt{-2t^2+3}=\sqrt{-2(1+u)+3}=\sqrt{1-2u}$$
With al these substitutions the integral has now the following form:
$$-\frac{1}{2}\int \frac{t^2}{\sqrt{-2t^2+3}}\,dt$$
Next we try to ''clean'' up the numerator:
$$-\frac{1}{2} \int \frac{t^2}{\sqrt{\frac{1}{2}(6-t^2)}} \, dt$$
$$-\frac{\sqrt{2}}{2} \int \frac{t^2}{\sqrt{6-t^2}} \, dt$$
And that's where I got stuck. I can clearly see that an arcsin is showing up in the integral but don't know how to get rid of the $t^2$.
|
You have:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx
$$
First we'll do a routine substitution:
$$
u = 1+x-2x^2, \qquad du = (1-4x)\,dx, \qquad \frac{-du} 4 = \left( \frac 1 4 - x \right)\, dx
$$
\begin{align}
\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx & = \int \frac{\frac 1 4-x}{\sqrt{1+x-2x^2}}\,dx + \int \frac{\frac 3 4}{\sqrt{1+x-2x^2}}\,dx \\[15pt]
& = \frac{-1} 4 \int \frac{du}{\sqrt u} + \frac 3 4 \int \frac{dx}{\sqrt{1+x-2x^2}}.
\end{align}
I expect you can handle the first integral above. The second integral should make you think of completing the square:
\begin{align}
-2x^2 + x+1 = -2\left( x^2 - \frac 1 2 x \right)^2 + 1 & = -2\left( \overbrace{x^2 - \frac 1 2 x +\frac 1 {16}}^\text{a perfect square} \right)^2 + 1 + \frac 1 8 \\[10pt]
& = -2 \left( x - \frac 1 4 \right)^2 + \frac 9 8.
\end{align}
Then
\begin{align}
\frac 9 8 -2\left( x - \frac 1 4 \right)^2 = \frac 9 8 - (2x-1)^2 & = \frac 9 8 \left( 1 - \frac 8 9 (2x-1)^2 \right) \\[10pt]
& = \frac 9 8 \left( 1 - \left( \frac{2\sqrt2} 3 (2x-1) \right)^2 \right) \\[10pt]
& = \frac 9 8 (1 - \sin^2\theta) \\[15pt]
\frac{4\sqrt2} 3 \, dx & = d\theta
\end{align}
et cetera.
|
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|
How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of squared prime reciprocals) deal with the exact value of the above sum, and so require some non-elementary math.
|
We can deduce this quickly, and without knowing the numerical value of $\pi$, from the fact that
$$\sum_{n \in \Bbb N} \frac{1}{n^2} = \frac{\pi^2}{6},$$ for which there are numerous proofs available.
Let $E$ denote the set of even numbers; the sum of the squares of all such numbers is
$$\sum_{n \in E} \frac{1}{n^2} = \sum_{k \in \Bbb N} \frac{1}{(2 k)^2} = \frac{1}{4} \sum_{k \in \Bbb N} \frac{1}{k^2} = \frac{1}{4} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{24}.$$
Now, let $X$ denote the union of $\{2\}$ and all positive odd integers $> 1$. In particular, $X$ contains the set $\Bbb P$ of all prime numbers as a subset, and so
\begin{align}
\sum_{p \in \Bbb P} \frac{1}{p^2}
&\leq \sum_{n \in X} \frac{1}{n^2} \\
&= \sum_{n \in \Bbb N} \frac{1}{n^2} - \sum_{n \in E} \frac{1}{n^2} - \frac{1}{1^2} + \frac{1}{2^2} \\
&= \frac{\pi^2}{6} - \frac{\pi^2}{24} - 1 + \frac{1}{4} \\
&= \frac{\pi^2}{8} - \frac{3}{4} .
\end{align}
So, it suffices to show that
$$\frac{\pi^2}{8} - \frac{3}{4} < \frac{1}{2},$$
but rearranging shows that this is equivalent to $\pi^2 < 10$, and $\pi < \frac{22}{7}$ implies
$$\pi^2 < \left(\frac{22}{7}\right)^2 = \frac{484}{49} < \frac{490}{49} = 10.
$$
|
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|
Generalized Harmonic numbers I'd like to be able to prove the following inequality:
$\frac{{{H_{n, - r}}}}{{{n^r}\left( {n + 1} \right)}} \le \frac{{{H_{n - 1, - r}}}}{{n{{\left( {n - 1} \right)}^r}}}$.
It's clear that as $n \to \infty$ we get equality, the limit on each side is $1/(r+1)$, and it also seems clear that this is a lower limit. But I haven't been able to find an argument for the inequality for finite $n$.
Any advice or hints greatly appreciated-- Number theory was not my strong suit when I was a mathematics student.
|
We want to show
$\frac{H_{n, - r}}{n^r\left( n + 1 \right) }
\le \frac{H_{n - 1, - r}}{n\left( n - 1 \right)^r}
$
where
$H_{n, a}
=\sum_{k=1}^n \frac1{k^a}
$,
so
$H_{n, -r}
=\sum_{k=1}^n k^r
$.
From this,
$H_{n - 1, - r}
=H_{n , - r}-n^r
$,
so we want
$\frac{H_{n, - r}}{n^r\left( n + 1 \right) }
\le \frac{H_{n , - r}-n^r}{n\left( n - 1 \right)^r}
$.
Bringing the $H$ term together,
this becomes
$H_{n , - r}(\frac1{n\left( n - 1 \right)^r}-\frac1{n^r\left( n + 1 \right) })
\ge\frac{n^r}{n\left( n - 1 \right)^r}
$
or
$H_{n , - r}(\frac1{( n - 1)^r}-\frac1{n^{r-1}\left( n + 1 \right) })
\ge\frac{n^r}{\left( n - 1 \right)^r}
$
or
$H_{n , - r}(1-\frac{(n-1)^r}{n^{r-1}\left( n + 1 \right) })
\ge n^r
$
or
$H_{n , - r}\frac{n^{r-1}(n+1)-(n-1)^r}{n^{r-1}( n + 1) }
\ge n^r
$
or
$H_{n , - r}
\ge \frac{n^{2r-1}( n + 1) }{n^{r-1}(n+1)-(n-1)^r}
$
or
$H_{n , - r}
\ge \frac{n^{r-1}( n + 1) }{(1+1/n)-(1-1/n)^r}
$.
Since
$H_{n , - r}
> \frac{n^{r+1}}{r+1}+\frac12 n^r
$,
this is true if
$\frac{n^{r+1}}{r+1}+\frac12 n^r
\ge \frac{n^{r-1}( n + 1) }{(1+1/n)-(1-1/n)^r}
$
or
$\frac{n^2}{r+1}+\frac{n}{2}
\ge \frac{( n + 1) }{(1+1/n)-(1-1/n)^r}
$
or
$\frac{n}{r+1}+\frac{1}{2}
\ge \frac{( 1 + 1/n) }{(1+1/n)-(1-1/n)^r}
$
or
$\frac{2n+r+1}{2(r+1)}
\ge \frac{( 1 + 1/n) }{(1+1/n)-(1-1/n)^r}
$
or
$(1+1/n)-(1-1/n)^r
\ge \frac{2(r+1)(1+1/n)}{2n+r+1}
$.
If $n >> r$,
$(1-1/n)^r
\approx 1-r/n
$,
so
$(1+1/n)-(1-1/n)^r
\approx (1+1/n)-(1-r/n)
=(r+1)/n
$,
so this becomes
$(r+1)/n
\ge \frac{2(r+1)(1+1/n)}{2n+r+1}
$
or
$1
\ge \frac{2(n+1)}{2n+r+1}
$
or
$2n+r+1
\ge 2n+2
$
and this is true.
(Whew!)
Therefore,
for $n$ large compared to $r$,
your inequality is true.
|
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|
Solving a differential equation by separating variables How can one solve the following differential equation by the technique of separation of variables?
$$\frac{1}{x^2}\frac{dy}{dx}=y^5\ \ \ \text{ when }, \ y(0)=-1$$
|
Notice, $$\frac{1}{x^2}\frac{dy}{dx}=y^5$$
$$\frac{1}{y^2}dy=x^2dx$$
$$\int y^{-5}\ dy=\int x^2\ dx$$
$$\frac{y^{-4}}{-4}=\frac{x^3}{3}+C$$
$$-\frac{1}{4y^{4}}=\frac{x^3}{3}+C$$
now, setting $x=0$ & $y=-1$,
$$-\frac{1}{4(-1)^{4}}=\frac{0}{3}+C\implies C=-\frac{1}{4}$$
set the value of $C$ to get the solution
|
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|
2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, but no result. For 2, I know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but the proofs i found are over the elementary level the problem is aimed for. I also proved that it's smaller than $\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{1024^2} < 1$, but that obviously is missing $\frac{1}{2015}$
|
1.
$$9p=\underbrace{99\cdots9}_\text{4030}+\underbrace{99\cdots9}_\text{2015}$$
$$=10^{4030}-1+10^{2015}-1=\left(10^{2015}\right)^2+10^{2015}-2$$
$$=\left(10^{2015}-1\right)\left(10^{2015}+2\right)=\left(10^{2015}-1\right)\left(10^{2015}-1+3\right)$$
$$\therefore p=\frac{10^{2015}-1}{3}\left(\frac{10^{2015}-1}{3}+1\right)$$
$$\therefore n=\frac{10^{2015}-1}{3}=\underbrace{33\cdots3}_\text{2015}$$
2.
$$\sum_{k=2}^{n} \frac1{k^2}<\sum_{k=2}^{n} \frac1{(k-1)k}=\sum_{k=2}^{n} \frac1{k-1}-\frac1{k}=1-\frac{1}{n}=\frac{n-1}{n}$$
|
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|
Solve system of $n$ equations of the form $2x_k^3+4=x_k^2(x_{k+1}+3)$ Solve the system of $n$ equations, $n\geq2$:
$$
\begin{cases}
2x_1^3+4=x_1^2(x_2+3)\\
2x_2^3+4=x_2^2(x_3+3)\\
\qquad \vdots\\
2x_{n-1}^3+4=x_{n-1}^2(x_n+3)\\
2x_n^3+4=x_n^2(x_1+3)\\
\end{cases}
$$
I think that there are only two solutions, but I don't know how to prove it:
$$x_1=x_2=x_3=\cdots=x_n=-1$$
$$x_1=x_2=x_3=\cdots=x_n=2$$
|
$$2x_n^3+4=x_n^2(x_{n+1}+3)$$
$$2x_n+\frac{4}{x_n^2}=x_{n+1}+3$$
$$x_{n+1}=2x_n+\frac{4}{x_n^2}-3$$
Define $f(x)$:
$$f(x)=2x+\frac{4}{x^2}-3=x+\frac{x^3-3x^2+4}{x^2}=x+\frac{(x-2)^2(x+1)}{x^2}$$
\begin{align}
&f(x)=x\quad(x=-1, \space 2)\\
&f(x)>x\quad(x>-1,\space x\ne2)\\
&f(x)<x\quad(x<-1)\\
\end{align}
If $x_1\ne-1,2$ then
*
*If $x_1<x_2$, then $x_1<x_2<\cdots<x_n<x_1\rightarrow$ contradicts.
*If $x_1>x_2$, then $x_1>x_2>\cdots>x_n>x_1\rightarrow$ contradicts.
Therefore, $x_1=x_2=\cdots=x_n=-1$ or $x_1=x_2=\cdots=x_n=2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1558487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks
|
Let's use the angle bisector theorem on $C$, we get
$$\frac{b+a}{c}=\frac{a}{EB}.$$
By the angle bisector theorem on $B$, we get
$$\frac{c+a}{b}=\frac{a}{CD}.$$
Now from the law of sines
$$EB=\frac{a\sin{C/2}}{\sin y}$$
and
$$CD=\frac{a\sin{B/2}}{\sin x}.$$
Putting all together we find
$$\frac{(b+a)\sin {C/2}}{c \sin y}=\frac{(c+a)\sin {B/2}}{b \sin x}.$$
Plugging in the equation given by the problem, we find
$$\sin x=\sin y$$
thus $x=y$ and $ABC$ is isosceles.
|
{
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"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
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|
How to Evaluate $\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$? Here is my limit to be evaluated
$\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$=?
|
Here is a simple step by step approch
$$\begin{align}
&\quad \lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \frac{x}{4} \sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \lim_{x \to \infty} \frac{\sin\left(\frac{4}{x}\right)}{\frac{4}{x}} \\
&= \lim_{x \to \infty} 2 + 8 \lim_{u \to 0} \frac{\sin(u)}{u} \\
&=2+8(1)=10
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1562458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x^8\equiv3\pmod{13}$ has exactly $4$ incongruent solutions modulo $13$.
I was able to find on my calculator (using brute force) that these solutions are $4,6,7,$ and $9$ (i.e. $\pm4,\pm6$), but how would I go about finding them without using a calculator?
|
One can easily see that $2$ is a primitive root of $13$.
Hence, we can write any number less than $13$ as some power of $2$.
$x^8\equiv3\equiv2^4\pmod{13}$.
Any solution will also be of the form $x=2^m$ for some $1\leq m\leq12$.
So, we have
$(2^m)^8=2^{8m}\equiv2^4\pmod{13}$.
Now, these powers will be congruent modulo $\phi(13)$, giving
$8m\equiv4\pmod{12}$
$8m\equiv16\pmod{12}$
$m\equiv2\pmod3$
$m\equiv2,5,8,$ or $11\pmod{12}$.
So, we have $x\equiv2^2,2^5,2^8,$ or $2^{11}\pmod{13}\implies x\equiv4,6,9,$ or $7\pmod{13}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1565487",
"timestamp": "2023-03-29T00:00:00",
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|
Limit of relation sums of exponential functions Help to solve example:
$$
\lim_{n\rightarrow\infty}\frac{\sqrt[3]{7^{3n-1}+3^n}}{3^n+4 \cdot 7^{n+1}}
$$
|
Hint.
For all $n \in \mathbb N$
$$
\frac{\sqrt[3]{7^{3n-1}+3^n}}{3^n+4*7^{n+1}}=\frac{7^n}{7^n}\frac{\sqrt[3]{\frac{1}{7}+\left(\frac{3}{7}\right)^n}}{28+\left(\frac{3}{7}\right)^n}=\frac{\sqrt[3]{\frac{1}{7}+\left(\frac{3}{7}\right)^n}}{28+\left(\frac{3}{7}\right)^n}
$$
and $\lim\limits_{n \to +\infty} \left(\frac{3}{7}\right)^n = ?$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{\binom{n}{k}}{n^k} < \frac{\binom{n+1}{k}}{(n+1)^k}$ I have this math question that I'm kind of stuck on.
Prove that for all integers $1 < k \le n$, $$\frac{\binom{n}{k}}{n^k} <
\frac{\binom{n+1}{k}}{(n+1)^k}$$
I have to use mathematical induction on $k$ to prove this. So, I first check the base case $k=2$, I get $$f(2): \frac{\binom{n}{2}}{n^2} <
\frac{\binom{n+1}{2}}{(n+1)^2}$$$$\frac{n(n-1)}{2n^2} < \frac{n(n+1)}{2(n+1)^2}$$$$= \frac{n-1}{n} < \frac{n}{n+1}$$
Now I assume that $f(z): \frac{\binom{n}{z}}{n^z} <
\frac{\binom{n+1}{z}}{(n+1)^z}$ is true.
Check $f(z+1)$ is also true.
$$f(z+1): \frac{\binom{n}{z+1}}{n^{z+1}} <
\frac{\binom{n+1}{z+1}}{(n+1)^{z+1}}$$
$$= \frac{\frac{n!}{(z+1)!(n-z-1)!}}{n^{z+1}} < \frac{\frac{(n+1)!}{(z+1)!(n-z)!}}{(n+1)^{z+1}}$$$$= \frac{n!}{n^{z+1}(z+1)!(n-z-1)!} < \frac{(n+1)!}{(n+1)^{z+1}(z+1)!(n-z)!}$$
I'm not sure what to do from here. Thanks.
|
Assume your calculation is correct up to the step
$\frac{n!}{n^{z+1}(z+1)!(n-z-1)!} < \frac{(n+1)!}{(n+1)^{z+1}(z+1)!(n-z)!}$
We can cancel the factor $(z+1)!$ in both denominators. Rewriting $(n+1)! = n! (n+1)$ and $(n-z)! = (n-z-1)!(n-z)$, we can simplify further to
$\frac{1}{n^{z+1}} < \frac{(n+1)}{(n+1)^{z+1} (n-z)}$
$\frac{1}{n} \cdot \frac{1}{n^z} < \frac{1}{(n+1)^z}\cdot \frac{1}{n-z}$
This is equivalent to showing that
$\frac{n-z}{n} \cdot \frac{(n+1)^z}{n^z} < 1.$
I do this by simplifying the LHS even further.
$\Big(1 - \frac{z}{n}\Big) \Big(1 + \frac{1}{n} \Big)^z$ (*)
At this step, I use Bernoulli's inequality to estimate
$1 - \frac{z}{n} \leq (1 - \frac{1}{n})^z$.
See here: https://en.wikipedia.org/wiki/Bernoulli%27s_inequality
Finally:
(*) $\leq \Big(1- \frac{1}{n}\Big)^z\Big(1 + \frac{1}{n}\Big)^z$
$= \Big[ \Big(1-\frac{1}{n}\Big) \Big(1+\frac{1}{n}\Big)\Big]^z = \Big(1-\frac{1}{n^2}\Big)^z < 1,$
which is the desired inequality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of the following $n \times n$ determinantes Find the value of the following $n \times n$ determinantes
*
*$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
*$$\begin{vmatrix}
a_1+x & a_2 & a_3 & \ldots & a_n \\
a_1 & a_2+x & a_3 & \ldots & a_n \\
a_1 & a_2 & a_3+x & \ldots & a_n \\
\vdots & \vdots& &\ddots& \vdots\\
a_1 & a_2 & a_3 & \ldots & a_n+x \\
\end{vmatrix}$$
Both seem to be equally complicated to solve, I reckon that it's needed to subtract the $x$ from the diagonal in each term. I tried by subtracting the $(k-1)$-th row from the $k$-th row, however that doesn't really lead me to anything more comfortable whatsoever. So help is greatly appreciated, also perhaps a link to some methods on solving these kind of problems, would also be helpful.
|
For 2) Let $$A=\begin{bmatrix} a_1 & a_2 & .. &a_n \\
a_1 & a_2 & .. &a_n \\
a_1 & a_2 & .. &a_n \\
...&...&...&...\\
a_1 & a_2 & .. &a_n \\
\end{bmatrix}$$
Then $A$ has rank $1$ and hence $\lambda=0$ is an eigenvalue with geometric multiplicity $n-1$, and hence has algebraic multiplicity at least $n-1$.
As the trace is the sum of eigenvalues, the last eigenvalue is $a_1+a_2+..+a_n$.
Thus
$$
\det(\lambda I-A) =\lambda^n -(a_1+...+a_n) \lambda^{n-1}
$$
Now replace $\lambda$ by $-x$.
For 1) Subtract the last row from each of the previous $n-1$. You get
$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}=\begin{vmatrix}
a_1 & 0 & 0 & \ldots & -a_n \\
0 & a_2 & 0 & \ldots & -a_n \\
0 & 0 & a_3 & \ldots & -a_n \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
This determinant is a linear function in $x$. Therefore
$$\begin{vmatrix}
a_1+x & x & x & \ldots & x \\
x & a_2+x & x & \ldots & x \\
x & x & a_3+x & \ldots & x \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}=ax+b$$
Now when $x=0$ we get
$$b=a_1 ... a_n$$
All you have to do next is do row expansion by the last row in
$$\begin{vmatrix}
a_1 & 0 & 0 & \ldots & -a_n \\
0 & a_2 & 0 & \ldots & -a_n \\
0 & 0 & a_3 & \ldots & -a_n \\
\vdots & \vdots& &\ddots& \vdots\\
x & x & x & \ldots & a_n+x \\
\end{vmatrix}$$
in order to find the coefficient of $x$.
For 2), without eigenvalues: Transpose the matrix, and then add all rows to the first. We get:
$$\begin{vmatrix}
a_1+x & a_2 & a_3 & \ldots & a_n \\
a_1 & a_2+x & a_3 & \ldots & a_n \\
a_1 & a_2 & a_3+x & \ldots & a_n \\
\vdots & \vdots& &\ddots& \vdots\\
a_1 & a_2 & a_3 & \ldots & a_n+x \\
\end{vmatrix}=\begin{vmatrix}
a_1+x & a_1 & a_1 & \ldots & a_1 \\
a_2 & a_2+x & a_2 & \ldots & a_2 \\
a_3 & a_3 & a_3+x & \ldots & a_3 \\
\vdots & \vdots& &\ddots& \vdots\\
a_n & a_n & a_n & \ldots & a_n+x \\
\end{vmatrix}=\begin{vmatrix}
a_1+a_2+..+a_n+x & a_1+a_2+..+a_n+x & a_1+a_2+..+a_n+x & \ldots &a_1+a_2+..+a_n+x\\
a_2 & a_2+x & a_2 & \ldots & a_2 \\
a_3 & a_3 & a_3+x & \ldots & a_3 \\
\vdots & \vdots& &\ddots& \vdots\\
a_n & a_n & a_n & \ldots & a_n+x \\
\end{vmatrix}=(a_1+a_2+..+a_n+x)\begin{vmatrix}
1 & 1 & 1 & \ldots &1\\
a_2 & a_2+x & a_2 & \ldots & a_2 \\
a_3 & a_3 & a_3+x & \ldots & a_3 \\
\vdots & \vdots& &\ddots& \vdots\\
a_n & a_n & a_n & \ldots & a_n+x \\
\end{vmatrix}$$
Now if you do each row minus $a_i$ row one you get
$$=(a_1+a_2+..+a_n+x)\begin{vmatrix}
1 & 1 & 1 & \ldots &1\\
0 & x & 0 & \ldots & 0 \\
0 & 0 & x & \ldots & 0 \\
\vdots & \vdots& &\ddots& \vdots\\
0& 0 & 0 & \ldots & x \\
\end{vmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $abc$.
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.
In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?
|
Let $M=396396$. Because $599\times 499<M$, we infer that $a>5$. Because $11|M$, we infer that $a+c-b=11$. Finally, $3|M$ and so $3|(a+b+c)$. So let's consider the possibilities for $a+b+c$:
$$
a+b+c=2b+11\in\{12,15,18,21,24,27\}
$$
which yields $b\in\{2,5,8\}$ which corresponds to $a+c\in\{13,16,19\}$. $19$ is too high because $c<a$. If $b=5$, then it must be that $a=9$ and $b=7$. You can verify that this doesn't work. Thus
$$
b=2\implies a+c=13\implies (a,c)\in\{(7,6),(8,5),(9,4)\}.
$$
Only $\boxed{abc=924}$ works and that's our answer.
Edit: I left out the case $a+c=b$. Let's look at it
$$
a+b+c=2b\in\{0,6,12,18\}\implies b\in\{0,3,6,9\}\implies a+c\in\{0,3,6,9\}.
$$
$0$ and $3$ are too small, and so $(a,c)\in\{(6,0),(6,3),(7,2),(8,1)\}$. Eliminate these by inspection.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all functions $f$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$ Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$.
So far, I've managed to prove that if $f$ is linear, then either $f(x) = x + 1$ or $f(x) = -1$ must be true. I did this by plugging in $x=0$ to the above equation, which yields
$$ f(-f(y)) = f(f(0)) - f(y) - 1$$
and plugging in $x$ instead of $y$ and subtracting, this becomes
$$ f(-f(y)) - f(-f(x)) = f(x) - f(y)$$
Assuming $f(x) = ax + b$ then gives
\begin{align*}
f(-ay-b) - f(-ax-b) = ax - ay &\Rightarrow -a^2y - ab + a^2x + ab = a(x-y) \\&\Rightarrow a^2(x-y) = a(x-y) .
\end{align*}
Thus $a=1$ or $a=0$. If $a=0$, then the original equation becomes $b = b - b - 1$, thus $b=-1$. If $a=1$, the original equation becomes
$$ x-y-b+b = x+2b-y-b-1 \Longrightarrow b=1.$$
I briefly tried finding a quadratic function that works but didn't find anything. So my question is: how can I either show that $f$ must be linear or find all other representations?
|
If the functions attains $0$ at some point, it is linear and $f(n)=n+1$.
Suppose $f(y)=0$, for some $y \in \mathbb{Z}$ then we find for $x \in \mathbb{Z}$
$$f(x)=f(x-f(y))=f(f(x))-f(y)-1 = f(f(x))-1$$
So in particular $x=y$ gives $f(0)= 1$.
Now, if $f(n)=n+1$, then $x=n$ gives $$n+1 = f(n) =f(f(n))-1= f(n+1)-1$$
So $f(n+1) = n+2$, thus for $n \in \mathbb{N}$ we have $f(n) = n+1$.
For the other way, let $x=0$ and $y=n-1$ then
$$f(-n) = f(x - f(n-1)) = f(f(x)) -f(n-1) -1 =f(1) - n-1 =-n+1$$
So we see that if $f(y) =0$ for some $y \in \mathbb{Z}$, we have $f(x) = x+1$.
So we can now assume $f(x) \neq 0$ for all $x \in \mathbb{Z}$.
If there exists an $y \in \mathbb{Z}$ such that $y=f(y)$, then $f(x) =-1$.
Suppose there exists an $y \in \mathbb{Z}$ such that $f(y)=y$, then with $x=y$
$$f(0) = f(y-f(y)) = f(f(y)) - f(y) - 1 = y-y-1 =-1$$
So we have $f(0)=-1$. Then we have if $f(y+n)=y$ for $n \ge 0$ that
$$f(y+n+1) = f(y+n-f(0)) = f(f(y+n)) -f(0)-1 = f(y) =y$$
So we see that $f(y+n) = y$ for every $n \in \mathbb{N}$.
Now, if $y>0$, we would find with $x=2y$
$$y = f(y)=f(2y -f(y)) = f(f(2y))-f(y)-1 = f(y)-f(y)-1 =-1$$
So $y <0$, and then we have $-1 = f(0) = f(y +(-y)) =y$ and so we find $y=-1$.
Now, we have for all $x \in \mathbb{Z}$
$$-1 = f(0) = f(f(x) -f(x)) = f(f(f(x))) - f(x)-1$$
And so $f(x) = f(f(f(x)))$, and so if $f(-1-k)=-1$ for $k\ge 0$ we have
$$ f(-1-k) = f(-1-k-1 -f(0)) = f(f(-1-k-1)) -f(0)-1 = f(f(-1-k-1))$$
Then we let $f$ act on both sides and we get
$$-1 = f(-1) = f(f(-1-k)) = f(f(f(-1-(k+1)))) = f(-1-(k+1))$$
So we see $f(-1-n)=-1$ for all $n \in \mathbb{N}$.
So we have $f(x) = -1$ for all $x \in \mathbb{Z}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1Find $\int _{0}^{2\pi}{1\over a\sin t+b \cos t +c}$ where $\sqrt{a^2+b^2}=1<c$. I am lead to believe I should be using curves but I really don't understand what curves to choose and how to properly use them. I would appreaciate some help on the issue.
|
We have
$$a\sin(t)+b\cos(t) = \sqrt{a^2+b^2}\sin(t+\phi) = \sin(t+\phi)$$
Hence, we have
\begin{align}
I & = \int_0^{2\pi} \dfrac{dt}{\sin(t+\phi)+c} = \int_0^{2\pi} \dfrac{dt}{\sin(t)+c} = \dfrac1c \int_0^{2\pi} \dfrac{dt}{1+\dfrac{\sin(t)}c}\\
& = \dfrac1c \sum_{k=0}^{\infty}\left(-\dfrac1c \right)^k \int_0^{2\pi} \sin^{k}(t)dt = \dfrac1c \sum_{k=0}^{\infty} \left(\dfrac1c\right)^{2k} \int_0^{2\pi} \sin^{2k}(t)dt\\
& = \dfrac4c \sum_{k=0}^{\infty} \left(\dfrac1c\right)^{2k} \int_0^{\pi/2} \sin^{2k}(t)dt
\end{align}
From here, we have
$$\int_0^{\pi/2} \sin^{2k}(t)dt = \dfrac{\pi}{2^{2k+1}}\dbinom{2k}k$$
Hence,
\begin{align}
I & = \dfrac{2\pi}c \sum_{k=0}^{\infty} \left(\dfrac1{2c}\right)^{2k} \dbinom{2k}k = \dfrac{2\pi}c \cdot \dfrac1{\sqrt{1-1/c^2}} = \dfrac{2\pi}{\sqrt{c^2-1}}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the set of real values of $p$ for which the equation $ | 2x + 3 | + | 2x - 3 | = px + 6 $ has more than two solutions .
Options : A ) ( 4 , 0 )
B) R-{ 4 , 0 , -4 }
C) {0}
D) None
How to find p = 0 , without having to analyse the graph ?
|
let $p \in \mathbb R$, $2x-3 < 2x+3$, so if $0\leq2x-3$, $x \in [\frac{3}{2},\infty)$, the equation turns to be:
$$4x=px+6$$
this is linear equation so there is at most one solution.
if $2x+3 \leq 0 $,$x \in (-\infty,-\frac{3}{2}]$ the equation turns to be:
$$-4x=px+6$$
this is linear equation so there is at most one solution.
if $2x-3<0$ and $0 <2x+3$, $x \in [-\frac{3}{2},\frac{3}{2}]$ we have:
$$ px=0$$
so we need $p$ such that we will have at least three solution, $0$ is always a solution,and if $p=0$, we have that the last equation has many solution(more than 3 , $x \in (-1.5,1.5)$ is a solution), so $p=0$ has more than two solutions,if $ p\neq 0$ and we have that $x=\frac{6}{(4-p)}$ and $x \in [\frac{3}{2},\infty)$ is a solution (there is one solution if $0 \leq p < 4$) and $x=-\frac{6}{(4+p)}$ and $x \in (-\infty,-\frac{3}{2}]$ (there is one solution if $-4<p\leq0$) when $p \neq 4,-4$, so we have that the set of real values $p$ that the equation has more than two solutions is $\{0\}$, the set of real valued $p$ that have two or more solutions is $(4,-4)$.
|
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|
I want to show that $\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty \sin{\left(n\frac{\pi}{2}\right)}\cdot\frac{n^2+2}{n^3+n} $$ converges (absolutly?) or diverges.
My idea was: $n=2k+1$ and then it becomes: $$\sum_{n=1}^\infty (-1)^n\cdot\frac{(2k+1)^2+2}{(2k+1)^3+2k+1} $$
Then I somehow try to show whether the limit is $0$ and whether it is decreasing. But I am not sure how I would go about that?
|
You can show the terms are decreasing by finding $a_{n}-a_{n+1}$ and showing it's positive:
$\dfrac{n^2+2}{n^3+n}-\dfrac{(n+1)^2+2}{(n+1)^3+(n+1)} = \dfrac{n^4+2n^3+6n^2+5n+4}{n(n+1)(n^2+1)(n^2+2n+2)}>0$
Regarding the limit:
$\dfrac{n^2+2}{n^3+n} = \dfrac{1+\frac{2}{n^2}}{n+\frac{1}{n}}$. So as n goes to infinity the numerator goes to 1, and the denominator goes to infinity. So the limit is 0. So the series converges by the alternating series test.
As far as absolute vs conditional convergence:
$\dfrac{n^2+2}{n^3+n} = \dfrac{n^2}{n^3+n} + \dfrac{2}{n^3+n}$
So we can think of this as the sum of 2 series. The second converges.
Regarding the first:
$\dfrac{n^2}{n^3+n} = \dfrac{1}{n+\frac{1}{n}} >= \dfrac{1}{n+1}$ for n>=1
$\dfrac{1}{n+1}$ diverges by the integral test. Our series is greater so it also diverges.
Sum of a divergent and convergent series is divergent. So the absolute series diverges. So our original series converges conditionally.
|
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|
Find the solution for this differential equation Solve the differential equation;
$(xdx+ydy)=x(xdy-ydx)$
L.H.S. can be written as $\frac{d(x^2+y^2)}{2}$ but what should be done for R.H.S.?
|
I think that there is actually a general solution to a general kind of ODE. So I give here its general form and its solution. After that, the solution to the ODE in the thread above is also given.
Consider the following ODE
\begin{gather*}
A(x,y)(xd x+yd y)=B(x,y)(xd y-yd x)\tag{1}
\end{gather*}
where $A$ and $B$ are homogeneous functions of degree $\alpha$ and $\beta$ respectively, that is to say, for all $t>0,$ we have
\begin{gather*}
A(tx,ty)=t^{\alpha}A(x,y),\\
B(tx,ty)=t^{\beta}B(x,y).
\end{gather*}
We claim that for $x>0,$ ODE (1) can be transformed into a separable ODE, by a suitable change of variables. For the case of $x<0,$ similar argument works.
Since (a) can be manipulated as
\begin{align*}
&A(x,y)(xd x+yd y)=B(x,y)(xd y-yd x) \\
\iff & 2A(x,y)(xd x+yd y)=2B(x,y)(xd y-yd x) \\
\iff & A(x,y)d (x^2+y^2)=2B(x,y)x^2d \left(\frac{y}{x}\right),\tag{2}
\end{align*}
introduce the change of variables
\begin{align*}
x^2+y^2=&u, \\
\frac{y}{x}=&v,
\end{align*}
and then
\begin{gather*}
1+v^2=1+\frac{y^2}{x^2}=\frac{x^2+y^2}{x^2}=\frac{u}{x^2},
\end{gather*}
which leads to, in the case of $x>0,$
\begin{gather*}
x=\sqrt{\frac{u}{1+v^2}},\tag{3}\\
y=xv=v\sqrt{\frac{u}{1+v^2}}. \tag{4}
\end{gather*}
By homogeneity condition and (4) we have
\begin{gather*}
A(x,y)=A(x,xv)=x^{\alpha}A(1,v),\\
B(x,y)=B(x,xv)=x^{\beta}B(1,v).
\end{gather*}
Thus (2) turns out to be
\begin{align*}
&\quad x^{\alpha} A(1,v)d u=2x^{\beta+2}B(1,v)d v \\
&\iff x^{\alpha-\beta-2}A(1,v)d u=2B(1,v)d v\\
&\iff\frac{u^{\frac{\alpha-\beta}{2}-1}}{(1+v^2)^{\frac{\alpha-\beta}{2}-1}} A(1,v)d u=2B(1,v)d v\qquad \text{(by (3))}\\
&\iff u^{\frac{\alpha-\beta}{2}-1}d u=\frac{2B(1,v)(1+v^2)^{\frac{\alpha-\beta}{2}-1}}{A(1,v)}d v.\tag{5}
\end{align*}
It is apparent that (5) is separable.
Now consider the particular case
\begin{gather*}
x d x+yd y=x(xd y-yd x).\tag{6}
\end{gather*}
This corresponds to $A(x,y)=1, B(x,y)=x,$ and so the degrees of $\alpha=0, \beta=1.$ Thus, under the change of variables (3) and (4) we arrive at
\begin{gather*}
u^{-3/2}d u=2(1+v^2)^{-3/2}d v,
\end{gather*}
whose solution is
\begin{gather*}
\frac{1}{\sqrt{u}}+\frac{v}{\sqrt{1+v^2}}=C.
\end{gather*}
Revering to original variables, we have the solution to (6) is, in the case of $x>0,$
\begin{gather*}
\frac{1+y}{\sqrt{x^2+y^2}}=C.
\end{gather*}
|
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|
Simplify $2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$ . I am trying to simplify
$$2\cos(6\pi/7) + 2\cos(2\pi/7) + 2\cos(4\pi/7) + 1$$
However if I plug this in the calculator the answer is zero. Is there a way to keep on simplifying without the calculator?
I know the identity $2\cos(\theta) = (e^{i\theta} +e^{-i\theta}) $ but I think that might make it worst.
Thanks!
|
$$2\cos\left(\frac{6\pi}{7}\right)+2\cos\left(\frac{2\pi}{7}\right)+2\cos\left(\frac{4\pi}{7}\right)+1=$$
$$2\left(\cos\left(\frac{6\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{2\pi}{7}\right)\right)+1=$$
$$2\left(\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)\right)+1=$$
$$2\left(\sum_{n=1}^{3}\cos\left(\frac{2n\pi}{7}\right)\right)+1=$$
$$2\left(\frac{\csc\left(\frac{\pi}{7}\right)\sin\left(\frac{\pi+6\pi}{7}\right)-1}{2}\right)+1=$$
$$2\left(\frac{0-1}{2}\right)+1=$$
$$2\left(-\frac{1}{2}\right)+1=$$
$$-1+1=0$$
|
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|
Implicit Solution to a differential equation I'm looking at the ODE:
$\frac{dY}{dX} - \frac{ X^2 + 2 Y^2 - 1 }{ ( Y - 2 X )X } = 0$
I'm looking for an $implicit$ solution to the above. Meaning, I want to find a relation $F(X,Y)=0$, where $\frac{\partial}{\partial x} (F(x,y)) = ( Y - 2 X )X \frac{dY}{dX} - ( X^2 + 2 Y^2 - 1) $.
$\ $
For example, if we would consider the simpler DE $dY/dX = -X/Y$, the implicit solution is $X^2 + Y^2 = \mathrm{constant}$.
$\ $
I think that the implicit solution will be a cubic polynomial of $X$ and $Y$, but I am troubling identifying the function.
$\ $
I thought about setting $F(x,y)= a Y^3 + b Y^2 X + c Y X^2 + d X^3 + e Y^2 + f X Y + g X^2 + h Y + j X + k$, and differentiating, but this gets complicated too fast.
$\ $
How can I solve this problem?
|
$\dfrac{dY}{dX}-\dfrac{X^2+2Y^2-1}{(Y-2X)X}=0$
$\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{(Y-2X)X}$
$(Y-2X)\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{X}$
Let $U=Y-2X$ ,
Then $Y=U+2X$
$\dfrac{dY}{dX}=\dfrac{dU}{dX}+2$
$\therefore U\left(\dfrac{dU}{dX}+2\right)=\dfrac{X^2+2(U+2X)^2-1}{X}$
$U\dfrac{dU}{dX}+2U=\dfrac{2U^2+8XU+9X^2-1}{X}$
$U\dfrac{dU}{dX}=\dfrac{2U^2}{X}+6U+9X-\dfrac{1}{X}$
This belongs to an Abel equation of the second kind.
Follow the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0126.pdf:
Let $U=X^2W$ ,
Then $\dfrac{dU}{dX}=X^2\dfrac{dW}{dX}+2XW$
$\therefore X^2W\left(X^2\dfrac{dW}{dX}+2XW\right)=2X^3W^2+6X^2W+9X-\dfrac{1}{X}$
$X^4W\dfrac{dW}{dX}+2X^3W^2=2X^3W^2+6X^2W+9X-\dfrac{1}{X}$
$X^4W\dfrac{dW}{dX}=6X^2W+9X-\dfrac{1}{X}$
$W\dfrac{dW}{dX}=\dfrac{6W}{X^2}+\dfrac{9}{X^3}-\dfrac{1}{X^5}$
Let $T=-\dfrac{6}{X}$ ,
Then $X=-\dfrac{6}{T}$
$\dfrac{dW}{dX}=\dfrac{dW}{dT}\dfrac{dT}{dX}=\dfrac{6}{X^2}\dfrac{dW}{dT}$
$\therefore\dfrac{6W}{X^2}\dfrac{dW}{dT}=\dfrac{6W}{X^2}+\dfrac{9}{X^3}-\dfrac{1}{X^5}$
$W\dfrac{dW}{dT}=W+\dfrac{3}{2X}-\dfrac{1}{6X^3}$
$W\dfrac{dW}{dT}-W=\dfrac{T^3}{1296}-\dfrac{T}{4}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
|
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|
Prove $\frac{x}{\sqrt{1+x^2}} \lt \arctan x$ for every $x \gt 0$
Prove $$\frac{x}{\sqrt{1+x^2}} \lt \arctan x$$ for every $x \gt 0$.
I proved half of it with lagrange rule but that I got stuck. Any ideas?
I can upload my work if you want.
|
METHODOLOGY $1$:
The integral definition of the arctangent function is given by
$$\arctan(x)=\int_0^x \frac{1}{1+t^2}\,dt$$
Noting that for $t\in [0,x]$, $x\ge 0$, $1+t^2\le (1+t^2)^{3/2}$, we can assert
$$\arctan(x)\ge \int_0^x \frac{1}{(1+t^2)^{3/2}}\,dt=\frac{x}{\sqrt{1+x^2}}$$
and we are done!
METHODOLOGY $2$:
Recall from basic geometry the inequality for the sine function
$$\sin y\le y \tag 1$$
for $y\ge 0$. From $(1)$ it is easy to see that for $0\le y<1$, we have
$$\cos y\ge \sqrt{1-y^2} \tag 2$$
Taking $(1)$ and $(2)$ together reveals that for $0\le y<1$
$$\tan y\le \frac{y}{\sqrt{1-y^2}} \tag 3$$
Now, let $x=\frac{y}{\sqrt{1-y^2}}$. Note that for $0\le y<1$, $x>0$ and
$$y=\frac{x}{\sqrt{1+x^2}} \tag 4$$
Using $(4)$ in $(3)$, we obtain
$$\tan\left(\frac{x}{\sqrt{1+x^2}}\right)\le x$$
Taking the inverse function yields the desired inequality
$$\arctan(x)\ge \frac{x}{\sqrt{1+x^2}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
|
As this is in the form of $\frac{0}{0}$,so apply L Hospital rule
$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$
$=\lim_{x \to \frac{\pi}{6}} \frac{4 \sin x\cos x+\cos x}{4 \sin x\cos x-3 \cos x}=\frac{\sqrt3+\frac{\sqrt3}{2}}{\sqrt3-\frac{3\sqrt3}{2}}=-3$
|
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|
What is the probability that $x^4-y^4$ is divisible by 5? Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3...,5n\}$.
What is the probability that $x^4-y^4$ is divisible by $5$?
I divided the numbers into groups of 5 $(1,2,3,4,5),(6,7,8,9,10),...$.The probability in the first group would itself be the answer.But how to find that?
|
$x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)$.
If $x = 5k + j; 0 \le j < 5$ then $5|x^4 - y^4$ if $y = 5l + m; 0 \le m < 5$ where $m = j$; $m = 5 - j$; $m^2 + $j^2 = 5V$.
If $j = 0$ $5|x^4 - y^4 \iff m = 0$.
As $1 = 1 = 5 -4; 1^2 + 2^2 = 5; 1^2 + 3^2 = 10; 2 = 2 = 5 -3; 2^2 + 4^2 = 20; 3^2 + 4^2 = 25$. If $j \ne 0$ $5|x^2 - x^4 \iff m \ne 0$.
So... the ways it can be divisible by 5:
The first element is of form $a = 5k$ and the second is of form $b = 5m$. The probability of this is: 1/5*(n-1)/(5n - 1).
The first element is not of the form $a = 5k$ and the second is not of the form $b = 5m$. The probability of this is: 4/5*(4n -1)/(5n - 1).
These two possibilities are mutually exclusive so Probability is: [(n-1) + 4(4n - 1)]/5(5n - 1) = [17n -5]/[25n - 5]
|
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|
Integral $\int_0^\infty\frac{\tanh^2(x)}{x^2}dx$ It appears that
$$\int_0^\infty\frac{\tanh^2(x)}{x^2}dx\stackrel{\color{gray}?}=\frac{14\,\zeta(3)}{\pi^2}.\tag1$$
(so far I have about $1000$ decimal digits to confirm that).
After changing variable $x=-\tfrac12\ln z$, it takes an equivalent form
$$\int_0^1\frac{(1-z)^2}{z\,(1+z)^2 \ln^2z}dz\stackrel{\color{gray}?}=\frac{7\,\zeta(3)}{\pi^2}.\tag2$$
Quick lookup in Gradshteyn—Ryzhik and Prudnikov et al. did not find this integral, and it also is returned unevaluated by Mathematica and Maple. How can we prove this result? Am I overlooking anything trivial?
Further questions: Is it possible to generalize it and find a closed form of
$$\mathcal A(a)=\int_0^\infty\frac{\tanh(x)\tanh(ax)}{x^2}dx,\tag3$$
or at least of a particular case with $a=2$?
Can we generalize it to higher powers
$$\mathcal B(n)=\int_0^\infty\left(\frac{\tanh(x)}x\right)^ndx?\tag4$$
Thanks to nospoon's comment below, we know that
$$\mathcal B(3)=\frac{186\,\zeta(5)}{\pi^4}-\frac{7\,\zeta(3)}{\pi^2}\tag5$$
I checked higher powers for this pattern, and, indeed, it appears that
$$\begin{align}&\mathcal B(4)\stackrel{\color{gray}?}=-\frac{496\,\zeta(5)}{3\,\pi^4}+\frac{2540\,\zeta(7)}{\pi^6}\\
&\mathcal B(5)\stackrel{\color{gray}?}=\frac{31\,\zeta(5)}{\pi^4}-\frac{3175\,\zeta(7)}{\pi^6}+\frac{35770\,\zeta(9)}{\pi^8}\\
&\mathcal B(6)\stackrel{\color{gray}?}=\frac{5842\,\zeta(7)}{5\,\pi^6}-\frac{57232\,\zeta(9)}{\pi^8}+\frac{515844\,\zeta(11)}{\pi^{10}}\end{align}\tag6$$
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Let $\psi_{1}(z)$ be the trigamma function.
Similar to the answer here, we can integrate the function $$g(z) = \psi_{1}\left(\frac{z}{\pi i } \right) \tanh^{2}(z) $$ around an rectangular contour with vertices at $\pm R, \pm R + \pi i $.
The function $\psi_{1}\left(\frac{z}{\pi i} \right) $ has a double pole at the origin, but it's canceled by the double zero of $\tanh^{2}(z)$.
Also, as $\Re(z) \to \pm \infty$, the magnitude of $\tanh^{2}(z)$ tends to $1$, while the magnitude of $\psi_{1}\left(\frac{z}{\pi i} \right) $ decays like $\frac{\pi}{z}$. This, coupled with the fact the height of the contour is fixed, causes the integral to vanish on the vertical sides of the contour as $R \to \infty$.
We therefore get
$$\begin{align}\lim_{R \to \infty}\oint g(z) \, \mathrm dz &=\int_{-\infty}^{\infty} \psi_{1}\left(\frac{x}{\pi i } \right) \tanh^{2}(x) \, \mathrm dx - \int_{-\infty}^{\infty}\psi_{1}\left(\frac{x}{\pi i} + 1 \right) \tanh^{2}(x) \, \mathrm dx \\ &= - \pi^{2} \int_{-\infty}^{\infty} \frac{\tanh^{2}(x)}{x^{2}} \, \mathrm dx \tag{1} \\ &= 2 \pi i \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right] \\ &= 2 \pi i \left(\frac{\psi_{2} \left(\frac{1}{2} \right)}{\pi i}\right)\\& = -28 \zeta(3) \tag{2}.\end{align}$$
The result then follows.
The fastest way to determine the residue at $z = \frac{\pi i}{2}$ is to multiply the Laurent series expansion $$\tanh^{2}(z) = \frac{1}{\left(z-\frac{\pi i}{2}\right)^{2}} + \mathcal{O}(1)$$ with the Taylor expansion of $\psi_{1} \left(\frac{z}{\pi i} \right)$ at $z= \frac{\pi i}{2}$.
$(1)$ https://en.wikipedia.org/wiki/Polygamma_function#Recurrence_relation
$(2)$ https://mathworld.wolfram.com/PolygammaFunction.html (16)
To evaluate the integral $$\mathcal A(3)= \int_{0}^{\infty} \frac{\tanh (x) \tanh(3x)}{x^{2}} \, \mathrm dx,$$ we can integrate the function $$g(z) = \psi_{1} \left(\frac{z}{\pi i} \right) \tanh (z) \tanh(3z) $$ around the same contour.
We get $$\begin{align} \mathcal{A}(3) &= \frac{1}{\pi i } \left(\operatorname{Res}\left[g(z), \frac{\pi i }{6}\right] + \operatorname{Res}\left[g(z), \frac{5 \pi i}{6}\right] + \operatorname{Res}\left[g(z), \frac{\pi i}{2} \right]\right) \\ &= \frac{1}{\pi i} \left(\frac{i}{3} \, \psi_{1} \left(\frac{1}{6} \right) \tan \left(\frac{\pi}{6} \right) + \frac{i}{3} \, \psi_{1} \left(\frac{5 }{6} \right) \tan \left(\frac{5 \pi }{6} \right)- \frac{i}{3 \pi} \, \psi_{2} \left(\frac{1}{2} \right)\right) \\ &= \frac{1}{3 \sqrt{3} \, \pi} \left(\psi_{1} \left(\frac{1}{6} \right) - \psi_{1} \left(\frac{5}{6} \right) + \frac{14\sqrt{3}}{\pi} \, \zeta(3) \right). \end{align} $$
To show that $$\operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] = \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right),$$ we can use the formula for the residue of a double pole derived here.
$$\begin{align} \operatorname{Res} \left[g(z), \frac{\pi i}{2} \right] &= \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \sinh(z) \sinh(3z)}{\cosh(z) \cosh(3z)}, \frac{\pi i}{2} \right] \\ &= \operatorname{Res} \left[\frac{\psi_{1} \left(\frac{z}{\pi i }\right) \left(\cosh(4z) - \cosh(2z) \right)}{\cosh(2z) +\cosh(4z)}, \frac{\pi i}{2} \right] \\ &= \frac{6 \left( \frac{2}{\pi i} \, \psi_{2} \left(\frac{1}{2} \right)(12) \right) - 2 \left( 2 \, \psi_{1} \left(\frac{1}{2} \right)(0) \right)}{3(12)^{2}} \\ &= \frac{1}{3 \pi i} \, \psi_{2} \left(\frac{1}{2} \right) \end{align}$$
|
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|
Tangent identity given $a + b + c = \pi$ Given that $a + b + c = \pi$, that is, three angles in a triangle - then prove that $$\tan a + \tan b + \tan c = \tan a \tan b \tan c$$
Is my solution below completely rigorous? Can I justify taking the tangent of both sides of my equation (I think not, since tangent isn't an injective function).
|
We can write $a + b = \pi -c$ then taking the tangent of both sides, this yields $$\tan (a +b) = \tan(\pi -c) \iff \frac{\tan a + \tan b}{1 - \tan a \tan b} = -\tan c$$
So $$\tan a + \tan b = \tan a \tan b \tan c - \tan c$$
Hence we arrive at $$\bbox[10px, border: blue 1px solid]{\tan a + \tan b + \tan c = \tan a \tan b \tan c} \quad \square$$
as required.
|
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|
$\operatorname{spectrum}(AB) = \operatorname{spectrum}(BA)$? Suppose we have two $n \times n$ matrices $A, B$. It seems like $\delta(AB)=\delta(BA)$, but I can't generally poove it.
If $\det(A) \neq 0$ then $\det(AB - \lambda I) = 0 \Leftrightarrow $ $ \det(AB - \lambda I) \cdot \det(A) = 0 \Leftrightarrow \det(ABA - \lambda A) = 0 \Leftrightarrow $ $ \det(A) \cdot \det(BA - \lambda I) = 0 \Leftrightarrow \det(BA - \lambda I) = 0$, so $$\delta(AB)=\delta(BA)$$
Same if $\det(B) \neq 0$.
But how to prove it for $\det(A) = \det(B) = 0$? Is it still true?
|
I've seen the following nice proof credited to Paul Halmos. Assume that $A$ is not invertible. By performing row and column operations on $A$ and encoding them with invertible matrices, we can write
$$ A = P \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix} Q $$
where $P, Q$ are invertible and $r = \mathrm{rank}(A)$. Write also
$$ QBP = \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} $$
where $B_{11} \in M_{r}(\mathbb{F}), B_{22} \in M_{(n-r)\times(n-r)}(\mathbb{F})$, etc.
Then we have
$$ AB = P \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix} Q Q^{-1} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} P^{-1} = P \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix}P^{-1} = P \begin{pmatrix} B_{11} & B_{12} \\ 0 & 0 \end{pmatrix} P^{-1}. $$
This shows that $AB$ is similar to a block upper triangular matrix and so $\chi_{AB}(\lambda) = \lambda^{n-r} \chi_{B_{11}}(\lambda)$ (where $\chi$ is the characteristic polynomial).
Similarly,
$$ BA = Q^{-1} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} P^{-1} P \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix} Q = Q^{-1} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{pmatrix} \begin{pmatrix} I_{r \times r} & 0 \\ 0 & 0 \end{pmatrix} Q = Q^{-1} \begin{pmatrix} B_{11} & 0 \\ B_{21} & 0 \end{pmatrix} Q $$
which shows that $BA$ is similar to a block lower triangular matrix and so $\chi_{BA}(\lambda) = \lambda^{n-r} \chi_{B_{11}}(\lambda)$.
We have shown that $\chi_{AB}(\lambda) = \chi_{BA}(\lambda)$ which shows that the eigenvalues of $AB$ and $BA$ (and even their algebraic multiplicity) coincide.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1584486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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|
Is a symmetric matrix characterized by the diagonal of its resolvent? The resolvent of a square matrix $A$ is defined by $R(s) = (A-sI)^{-1}$ for $s \notin \operatorname{spect}(A)$.
Is knowing the diagonal of $R(s)$ for all $s$ sufficient to recover $A$ when $A$ is symmetric?
edit: a counter-example of two matrices $A,B$ whose resolvent have the same diagonal has been found by Robert Israel. In the counter example, $A = P B P^T$ for some permutation matrix $P$. Now the question is, it is possible to recover $A$ up to permutations of rows and columns?
|
No. For example, consider $$\left[ \begin {array}{cccc} 1&0&1&1\\ 0&0&1&1
\\ 1&1&0&0\\ 1&1&0&1\end {array}
\right] \ \text{and}\
\left[ \begin {array}{cccc} 1&1&0&1\\ 1&0&1&0
\\ 0&1&0&1\\ 1&0&1&1\end {array}
\right] $$
which both have diagonal of $R(s)$
$$ (s^4 - 2 s^3 - 3 s^2 + 4 s - 1) \left[ \begin {array}{c} -{s}^{3}+{s}^{2}+2\,s-1\\
-{s}^{3}+2\,{s}^{2}+s-1\\ -{s}^{3}+2\,{s}^{2}+s-1
\\ -{s}^{3}+{s}^{2}+2\,s-1\end {array} \right]
$$
EDIT: For the second question, try
$$ \left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&1&0
\\ 0&1&1&0\\ 1&0&0&1\end {array}
\right]
\ \text{and}\ \left[ \begin {array}{cccc} 1&0&-\sin \left( t \right) &\cos \left( t
\right) \\ 0&1&\cos \left( t \right) &\sin \left( t
\right) \\ -\sin \left( t \right) &\cos \left( t
\right) &1&0\\ \cos \left( t \right) &\sin \left( t
\right) &0&1\end {array} \right]
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1585007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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|
Comparing $\int \frac{\sqrt{1+x}}{1-x} dx$ with answer from book and sage [Edit: It turns out the answer from my book was incorrect and a later revision of the book has the correct answer:
$$
-2\sqrt{1+x} + \sqrt{2}\ln{\left| \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1+x} - \sqrt{2}} \right|} + C \tag{0}
$$
]
I'm trying to integrate this function but I don't understand how my answer is different from my book and different from sage too:
$$
\int \frac{\sqrt{1+x}}{1-x} dx \tag{1}
$$
Substitute $u = 1+x$:
$$
\int \frac{\sqrt{u}}{2-u} du \tag{2}
$$
Substitute $v = \sqrt{u}$:
$$
\int \frac{2v^2}{2-v^2} dv \tag{3}
$$
Use partial fractions:
$$
\int -2 dv + \int \frac{\sqrt{2}}{\sqrt{2}+v} dv + \int \frac{\sqrt{2}}{\sqrt{2}-v} dv \tag{4}
$$
Integrate with v:
$$
-2v + \sqrt{2} \ln{\left| \frac{\sqrt{2} + v}{\sqrt{2} - v} \right|} + C \tag{5}
$$
Replace $v$ with $u$:
$$
-2\sqrt{u} + \sqrt{2} \ln{\left| \frac{\sqrt{2} + \sqrt{u}}{\sqrt{2} - \sqrt{u}} \right|} + C \tag{6}
$$
Replace $u$ with $x$:
$$
-2\sqrt{1+x} + \sqrt{2} \ln{\left| \frac{\sqrt{2} + \sqrt{1+x}}{\sqrt{2} - \sqrt{1+x}} \right|} + C \tag{7}
$$
The answer in my book is:
$$
-2\sqrt{1+x} + 2\sqrt{2}\ln{\left| \frac{\sqrt{1+x} + 2}{1-x} \right|} + C \tag{8}
$$
But I don't understand how to get that numerator when I multiply by the conjugate of the denominator, I get:
$$
-2\sqrt{1+x} + \sqrt{2} \ln{\left| \frac{x + 2\sqrt{2}\sqrt{1+x} + 3}{1-x} \right|} + C \tag{9}
$$
The answer in sage is (latex(simplify(integral(sqrt(1+x)/(1-x))))):
$$
- 2 \, \sqrt{x + 1} -\sqrt{2} \log\left(-\frac{\sqrt{2} - \sqrt{x + 1}}{\sqrt{2} + \sqrt{x + 1}}\right) + C \tag{10}
$$
But I don't understand how to get that negative in the log, however I do understand how the negative of the log should simply inverse the numerator and denominator.
So, I would very much appreciate if someone could shed some light on what I don't understand both from my book, which might be incorrect, and from sage, which is less likely to be incorrect :)
Thank you!
|
setting $u= \sqrt {1+x}$ so we get
$x=u^2-1$ and $dx=2udu$ and our integral will be $$\int\frac {2u^2}{2-u^2}du$$ and is easy to solve the right result is $$-2\,\sqrt {1+x}+2\,\sqrt {2}{\rm arctanh} \left(1/2\,\sqrt {1+x}\sqrt
{2}\right)
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1586221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If two integer triples have the same sum of 6th powers, then their sums of squares agree $\bmod 9$ Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$
prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$
I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really confusing..
|
$(9n+a)^6=a^6\bmod27$,
so only worry about numbers between $-4$ and $4$.
Their sixth powers are
$19,0,10,1,0,1,10,0,19\pmod{27}$
and their squares are $7,0,4,1,0,1,4,7\pmod{9}$
The sixth powers are either $0$ or $9A+1$;
the squares are either $0$ or $3A+1$, the same $A$.
|
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"url": "https://math.stackexchange.com/questions/1587425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
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|
How to handle indices with fractional degree? An algebra problem ate my head!!!
$x$ and $y$ are positive real numbers such that
$$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
Find $x^{2/3} + y^{2/3}$.
It would be a great help if anybody helps me in solving this problem. I tried taking conjugates and all but I didn't get any answer.
thank you
|
Let,
\begin{align}
& x^{2/3} + y^{2/3} = z \newline
\implies & \left(x^{2/3} + y^{2/3}\right)^{3/2} = z^{3/2}\newline
\implies & z^{3/2} = 512 = 8^3 \tag{1}
\end{align}
To remove the fractional index $3/2$ on the LHS of (1) to obtain the base $z$, we take root $2/3$ (i.e. two-third root) of both sides, i.e.
\begin{align}
& \left(z^{3/2}\right)^{2/3} = \left(8^3\right)^{2/3} \newline
\implies & z^{(3/2) \times (2/3)} = 8^{3 \times 2/3} \newline
\implies & z = 8^2 = 64
\end{align}
Hence, $x^{2/3} + y^{2/3} = z = 64$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1587536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Xmas Maths 2015 Simplify the expression below into a seasonal greeting using commonly-used symbols in commonly-used formulas in maths and physics. Colours are purely ornamental!
$$
\begin{align}
\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{orange}{2^{\frac 23}}
\color{green}{c^2}
\color{red}{e^{i\theta}}
\color{orange}{v^2}
\color{green}{(x^2+xy+y^2)}}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\int_{-\infty}^\infty e^{-x^2} dx\right)^{\frac 23}}
\end{align}$$
NB: Knowledge of the following would be helpful:
Basic Maths:
*
*Taylor series expansion
*Normalizing factor for the integral of a normal distribution
*Rectangular and polar forms for complex variables
*Volume of a sphere
Basic Physics:
*
*Kinematics formulae for motion under constant acceleration
*Einstein's equation
*One of the energy equations
|
$$
\begin{align}
&\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{orange}{2^{\frac 23}}
\color{green}{c^2}
\color{red}{e^{i\theta}}
\color{orange}{v^2}
\color{green}{(x^2+xy+y^2)}}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\int_{-\infty}^\infty e^{-x^2} dx\right)^{\frac 23}}\\
&=
\frac{
\color{green}{(x+iy)}
\color{red}{(y^3-x^3)}
\color{orange}{(v^2-u^2)}
\color{red}{(3V_{\text{sphere}})^{\frac 13}}
\color{orange}{E\cdot}
\color{green}{\text{KE}}
}
{
\color{red}{e^{i\theta}}
\color{green}{(x^2+xy+y^2)}
\color{orange}{\cdot2^{\frac 23}}
\color{green}{c^2}
\color{orange}{v^2}
}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\color{orange}{\bigg/}
\color{orange}{\left(\sqrt{\pi}\right)^{\frac 23}}\\
&=
\color{green}{\left(\frac{x+iy}{e^{i\theta}}\right)}
\color{red}{\left(\frac{y^3-x^3}{x^2+xy+y^2}\right)}
\color{orange}{(v^2-u^2)}
\color{red}{\left(\frac {(3V_\text{sphere})^\frac 13}{\left(2\sqrt{\pi}\right)^{\frac 23}}\right)}
\color{orange}{\left(\frac{E}{c^2}\right)}
\color{green}{\left(\frac{\text{KE}}{v^2}\right)}
\color{red}{\sum_{n=0}^{\infty}\frac 1{n!}}
\\
&=
\color{green}{\left(\frac{re^{i\theta}}{e^{i\theta}}\right)}
\color{red}{\left(\frac{(y-x)(y^2+xy+x^2)}{x^2+xy+y^2}\right)}
\color{orange}{(v^2-u^2)}
\color{red}{\left(\frac {3\cdot \frac 43 \pi r^3}{4\pi}\right)^\frac 13}
\color{orange}{\left(\frac{mc^2}{c^2}\right)}
\color{green}{\left(\frac{\frac 12 mv^2}{v^2}\right)}
\color{red}{(e)}
\\
&=
\color{green}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{orange}{(2as)}
\color{red}{\left(r^3\right)^\frac 13}
\color{orange}{\left(m\right)}
\color{green}{\left(\frac 12m\right)}
\color{red}{(e)}
\\
&=
\color{green}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{orange}{(as)}
\color{red}{\left(r\right)}
\color{orange}{\left(m\right)}
\color{green}{\left( m\right)}
\color{red}{(e)}
\\
&=
\color{orange}{\left(m\right)}
\color{red}{(e)}
\color{green}{\left(r\right)}
\color{red}{\left(r\right)}
\color{red}{\left(y-x\right)}
\color{green}{\left(m\right)}
\color{orange}{(as)}
\end{align}$$
Merry Christmas, everyone!!
The following links might be helpful.
- Complex numbers and polar coordinates
- Difference of two cubes
- Kinematics formulae for constant acceleration in a straight line
- Volume of a sphere
- Einstein's mass-energy equivalence
- Kinetic energy
- Taylor/Maclaurin series expansion of $e$
- Gaussian integral (normalizing factor for the normal distribution)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1587620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 2,
"answer_id": 0
}
|
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?
|
The following proof might be longer than necessary, but it illustrates a general method that is useful for many similar problems.
Let $$A(k) = 1^2 - 2^2 + 3^2 - 4^2 + \ldots + (-1)^{k-1} k^2$$ and
$$B(k)=(-1)^{k-1} \cdot \frac{k(k+1)}2.$$ It suffices to prove that $A(0)=B(0)$ and $$A(k)-A(k-1)=B(k)-B(k-1)$$ for all $k\ge1$.
It is clear that $A(0)=B(0)=0$.
$$\begin{align*}
A(k) - A(k-1) &= (-1)^{k-1} k^2 \\
B(k) - B(k-1) &= (-1)^{k-1} \cdot \frac{k(k+1)}2
- (-1)^{k-2} \cdot \frac{(k-1)k}2 \\
&= (-1)^{k-1} \left(\frac{k(k+1)}2 + \frac{(k-1)k}2 \right) \\
&= (-1)^{k-1} \left(\frac{k^2+k}2 + \frac{k^2-k}2 \right) \\
&= (-1)^{k-1} k^2
\end{align*}
$$
To complete the proof, note that for all $k \ge 0$,
$$
\begin{align*}
A(k) &= A(0) + \sum_{i=1}^k (A(i) - A(i-1)) \\
&= B(0) + \sum_{i=1}^k (B(i) - B(i-1)) \\
&= B(k).
\end{align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
}
|
Solving this indefinite integral: $\int{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $ I don't know how to solve this integral
I tried change of variable but had no results.
$$\int_{0}^{1}{\frac{x\:dx} {(\sqrt{1+x} +\sqrt[3]{1+x} )}} $$
|
Let $1+x=t^6\implies dx=6t^5\ dt$ $$\int \frac{x}{\sqrt {1+x}+\sqrt[3]{1+x}}\ dx$$
$$=\int \frac{(t^6-1)}{t^3+t^2}(6t^5\ dt)$$
$$=6\int \frac{t^3(t^6-1)}{t+1}\ dt$$
$$=6\int \frac{t^3(t^3-1)(t^3+1)}{t+1}\ dt$$
$$=6\int \frac{(t^6-t^3)(t+1)(t^2-t+1)}{t+1}\ dt$$
$$=6\int (t^6-t^3)(t^2-t+1)\ dt$$
$$=6\int (t^8-t^7+t^6-t^5+t^4-t^3)\ dt$$
$$=6\left(\frac{t^9}{9}-\frac{t^8}{8}+\frac{t^7}{7}-\frac{t^6}{6}+\frac{t^5}{5}-\frac{t^4}{4}\right)+C$$
$$=6\left(\frac{(1+x)^{3/2}}{9}-\frac{(1+x)^{4/3}}{8}+\frac{(1+x)^{7/6}}{7}-\frac{(1+x)}{6}+\frac{(1+x)^{5/6}}{5}-\frac{(1+x)^{2/3}}{4}\right)+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1589448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.