Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cdot 2-1)^2=-2\cdot 4$
lets take $n=4$ so $(2\cdot 4-3)^2-(2\cdot 4-1)^2\neq-2\cdot 16$
I did something wrong?
Thanks!
| If $f(m)=1^2-3^2+5^2-7^2+\dots+(2m-3)^2-(2m-1)^2$
$m=2\implies f(2)=1^2-3^2=-8=-2.2^2$
$m=4\implies f(4)=1^2-3^2+5^2-7^2=1-9+25-49=-8-24=-2\cdot4^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$ let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that
$$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0, g(0)=0$$
find $g(x)$
my idea
let $x\longrightarrow x+1$, then we have
$$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$
$$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(x+2)-6}{4}=0$$
$$\cdots\cdots $$
$$g^2(x+n)-g(x+n+1)-\dfrac{(x+n)^2+2(x+n)-6}{4}=0$$
following I can't work,Thank you
| HINT:
Let $g(x)=\sum_{n\ge i\ge 0} a_ix^i$
So, $\displaystyle \frac{x^2+2x-6}4= g^2(x)-g(x+1)=(\sum_{n\ge i\ge 0} a_ix^i)^2-\sum_{n\ge i\ge 0} a_i(x+1)^i=a^2_nx^{2n}+\cdots$
So comparing the coefficients of the highest power of $x$, $n=1$ and $a_1=\frac12$
$\implies g(x)=\frac x2+a_0$
As $g(0)=0$ and $g(0)=a_0\implies \displaystyle a_0=0\implies g(x)=\frac x2$
So, $\displaystyle g^2(x)-g(x+1)=\left(\frac x2\right)^2-\frac{x+1}2=\frac{x^2-2x-2}4$
can this be equal to $\displaystyle\frac{x^2+2x-6}4$
| {
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Question involving approximation, taylor series and proving Question:
Consider the approximation $$\ln(2)\approx 2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}} \right )$$
Prove that the error in this approximation is less than $$\frac{1}{7\times 2^{2} \times 3^{5}}$$
Attempt:
It looks like the expression comes from the taylor series expansion so:
$\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+... \text{ for }\ -1< x< 1$
$\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+...$
$\therefore \ln\left ( \frac{1+x}{1-x} \right )=2\left ( x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7} \right )$
$\text{Now let}\ x=\frac{1}{3}$
$\therefore \ln(2)=2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}}+\frac{1}{7\times 3^{7}}\right )$
So we have to prove that:
$2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right) < \frac{1}{7\times 2^{2} \times 3^{5}}$
| We can say that $2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right)<2 \left (\frac{1}{7\times 3^{7}}+\frac{1}{7\times 3^{9}}+\frac{1}{7\times 3^{11}}\cdots\right)= 2 \left(\frac{1}{7\times 3^7}\right)
\div \left(1-\frac{1}{3^2}\right)=\frac{1}{7\times 2^2\times3^5}$
| {
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Reasoning that $ \sin2x=2 \sin x \cos x$ In mathcounts teacher told us to use the formula $ \sin2x=2 \sin x \cos x$.
What's the math behind this formula that made it true? Can someone explain?
| Let's start with the power series definitions:
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots, $$
$$\cos x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots. $$
If you expand the product $$\sin x \cos x= \left( x-\frac{x^3}{3!}+\frac{x^5}{5!} -+ \dots \right) \left( 1-\frac{x^2}{2!}+\frac{x^4}{4!}-+ \dots \right), $$
and collect terms, you will find that only odd powers of $x$ appear, and the coefficient of $x^{2n+1}$ is $$(-1)^n \left(\frac{1}{(2n+1)!0!}+\frac{1}{(2n-1)! 2!}+ \dots +\frac{1}{1!(2n)!} \right).$$
You can prove (using induction for example) that the last expression equals $$(-1)^n \frac{2^{2n}}{(2n+1)!}=(-1)^n \frac{1}{2} \frac{2^{2n+1}}{(2n+1)!}.$$
Thus we have $$\sin x \cos x=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!}, $$ which is precisely $\frac{1}{2} \sin 2x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the integer part I have to calculate the integer part of this:
$$[(\sqrt{2}+\sqrt{5})^2]$$
I tried to write it like this:
$$[2+5+2\sqrt{10}]=[7+2\sqrt{10}]=7+[2\sqrt{10}]$$
Any ideas?
| A direct way of solving this problem is to realize that
$$ 0 < \sqrt{ 5} - \sqrt{2} < 1 $$
This can be seen by squaring both sides, and using that $ 6 < 2 \sqrt{10} < 7 $ which again is true by squaring.
Since $(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5}- \sqrt{2})^2 = 5 + 2\sqrt{10} + 2 + 5 - 2 \sqrt{10} + 2 = 14 $, hence this allows us to conclude that
$$ \lfloor (\sqrt{5} + \sqrt{2})^2 \rfloor = 13 $$
| {
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Using partial fractions to find an antiderivative of $(x^2+2x)/(x+1)^2$
Evaluate $$
\int\frac{x^2+2x}{(x+1)^2}dx
$$
My solution
Let $u =x+1$,
$
du=dx
$.
Then
$
du(x^2+2x)=(x^2+2x)dx
$ and
$
x=u-1
$.
We get
$$
\int\frac{(u-1)^2+2(u-1)}{u^2}du
=
\int\frac{u^2-2u+1+2u-2}{u^2}du
=
\int\frac{u^2-1}{u^2}du
$$
which simplifies to
$$
\int(1-u^{-2})du
=u-\frac{u^{-1}}{-1} +C
=u+\frac{1}{u}+C
=x+1+\frac{1}{x+1}+C
$$
... But the answer is
$$
x+\frac{1}{x+1}+C
$$
What is wrong?
| The question has been well-answered, this is just a comment. Note that $x^2+2x=(x+1)^2-1$.
So we are integrating $\dfrac{(x+1)^2-1}{(x+1)^2}$, that is, $1-\dfrac{1}{(x+1)^2}$. Easier!
| {
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Simple question about antiderivative? So this has been confusing me a lot. Let $f(x)=x^2$ and let $F(x)=\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$. Then $F(1)=0$, obviously, but the antiderivative of $f$ (which is the same as $\displaystyle \int_{1}^{x} f(x) \, \mathrm{d}x$) is $F(x)=x^3/3$, so $F(1)=1/3$, not $0$?
What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as $F(x)=x^3/3 + C$, and in this example $F(x)=x^3/3 -1/3$, $x^3/3$ is just a special antiderivative where $C=0$.
| It does not matter what C is equal to.
The antiderivative of f is $F(x) = \int _0 ^x f(t) dt = \int _0 ^x t^2 dt = (\frac{t^3}{3} + C) |_0^x = (\frac {x^3}{3}+ C) - (\frac {0^3}{3} +C) = \frac{x^3}{3}$
$\therefore F(1) = \int _0 ^1 f(t) dt = 1/3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given three integers in $\{0,\ldots,100\}$ which sum up to $100$. What is the probabilty that two of them are the same? We pick $3$ numbers (one by one) from set $\{0,1,...,100\}$.
What is probabilty that two numbers are the same if sum of those $3$ numbers is $100$?
My solution:
Which two are the same we can pick in $\binom {3}{2}$ ways. Suggest $x_2=x_3$- we need to find compositon $x_1+x_2+x_2=100 \implies x_1+2x_2=100$ which implies that $x_1$ is even so we can divide this by $2$. Now we get $y_1+y_2=50$ , and using formula there are $$\binom{50+2-1}{2-1}=51$$ compositions. So, probability is $$\frac{51*3}{\binom{100+3-1}{3-1}}$$
Is this right answer?
P.S.$\binom{100+3-1}{3-1}$ is number of compositions of 100 into 3 parts (allowing $0$)
| First, we count the triples $(a,b,c)$ such that $a + b + c = 100$.
For the choice of $a,b$ we need $a + b \leq 100$. Then $c = 100 - a - b$ is uniquely determined.
For $a + b \leq 100$, the number of possibilities is
$$101 + 100 + 99 + \ldots + 2 + 1 = \frac{102\cdot 101}{2} = 5151.$$
Now we count the triples with the additional condition that two numbers are the same. Note that $a = b = c$ is not possible (otherwise, $100 = a + b + c = 3a$ would be a multiple of $3$).
So there are $3$ possibilities to choose the identical pair of numbers.
For a single such choice, say $a = b$, we have the $51$ possibilities $a\in\{0,\ldots,50\}$ s.t. $a + b = 2a \leq 100$. Again, $c$ is uniquely determined.
So in total, there are $3 \cdot 51 = 153$ triples $(a,b,c)$ such that two numbers are the same and $a + b + c = 100$.
This gives the resulting probability of
$$\frac{153}{5151} = \frac{3}{101} \approx 3.0\%.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}$=
$\lim_{x\to 0} \frac {x+49-49}{9-x+9}.\frac {\sqrt{x+49}+7}{3+\sqrt{x+9}}= \frac {x(\sqrt{x+49}-7)}{x(3-\sqrt{x+9})}=\frac{14}{3}=\frac{7}{3}$ i don't know what is the error!
| At $x=0$, both numerator and denominator are $0$, so one way to do this is to apply L'Hospital's rule :
$$
\lim_{x\to 0} \frac{\sqrt{x+49}-7}{3-\sqrt{x+9}} = \lim_{x\to 0} \frac{1/2}{-1/2}\frac{(x+49)^{-1/2}}{(x+9)^{-1/2}}
$$
Now just plug in $x=0$.
| {
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Easy trigonometry question How is
$$\frac{2 \sin x}{(1+ \cos x)^2}= \tan\left(\frac{x}{2}\right)\sec^2\left(\frac{x}{2}\right)\;?$$
It should be easy.
But somehow I don't get it.
Can you help me with this?
| $$\frac{2 \sin x}{(1+ \cos x)^2}=\frac{4 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\left[1+ \cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\right]^2}=\frac{4 \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{\left[2 \cos^2\left(\frac{x}{2}\right)\right]^2}=\tan\left(\frac{x}{2}\right)\sec^2\left(\frac{x}{2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u = \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v = x^2$ and $v= \frac{x^3}{3}$.
\begin{align}
&\int x^2 \cos \frac{x}{2} \\
&\cos \frac{x}{2} \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x
\end{align}
So now I integrate $\int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:
\begin{align}
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot \frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2}
\end{align}
Now, this is where I get stuck. I know if I continue, I will end up with $\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos \frac{x}{2}$. So, where do I go from here?
Thanks!
| Hint: Switch the functions you originally set equal to $u$ and $dv$.
General Strategy: Assume you're trying to compute $\int p(x) f(x) dx$ where $p(x) = a_0 + a_1 x + \ldots + a_nx^n$ is a polynomial and $f(x)$ is a function that you know how to integrate $n$ times. (For example, in your problem $p(x) = x^2$, $f(x) = \cos x$, and we note that it's easy to integrate $\cos x$ twice.) Then you can just keep integrating by parts ($n$ times), setting $u = p(x)$ (or $p^{(k)}(x)$) and $dv = \lbrack\text{whatever is leftover}\rbrack dx$ to eventually arrive at
$$
\int p(x) f(x) dx = \lbrack\text{a bunch of terms not involving integrals}\rbrack + \int F(x) dx
$$
where $F(x)$ is a function such that $F^{(n)}(x) = f(x)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finesse vs. brute force in solving quadratic equations In Higher Algebra by Hall and Knight, the following "artifice" for solving a certain type of equations is given:
Solve: $\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x-11} = 9$
They make use of the fact that $(3x^2-4x+34) - (3x^2-4x-11) = 45$, and utilizing the formula $a^2 - b^2 = (a-b)(a+b)$, obtain a neat answer that $x = 3, -5/3$.
My attempt, however, results in something weird:
$\sqrt{3x^2-4x+34} - \sqrt{3x^2-4x+11} = 9$
Let $3x^2 - 4x = t$
$\Rightarrow \sqrt{t+34} + \sqrt{t+11}$ = 9
Squaring both sides,
$t + 34 + t + 11 + 2 \sqrt{(t+34)(t+11)} = 81$
$\Rightarrow 18-t = \sqrt{(t+34)(t+11)}$
Squaring again,
$t^2 - 36t + 324 = t^2 + 45t + 374$
$\Rightarrow t = -50/81$
$\Rightarrow 3x^2 - 4x = -50/81$
$\Rightarrow 243x^2 - 324x + 50 = 0$
The discriminant here is $\sqrt{324^2 - 200.243} = 56376$, which is not a perfect square and hence the roots are irrational.
Have I made a mistake somewhere or it is really the case that finesse leads to correct answers?
| Assuming the real, actual question has a plus sign between the square roots and $\,-11\,$ in the second square root, and doing your substitution:
$$\sqrt{3x^2-4x+34}+\sqrt{3x^2-4x-11}=9\;,\;\;\color{red}{t:=3x^2-4x}\implies$$
$$\sqrt{t+34}+\sqrt{t-11}=9\implies 2t+23+2\sqrt{(t+34)(t-11)}=81\implies$$
$$29-t=\sqrt{(t+34)(t-11)}\implies t^2-58t+841=t^2+23t-374\implies$$
$$81t=1,215\implies t=15\implies$$
$$15=\color{red}{t=3x^2-4x}\implies 0=3x^2-4x-15=3(x-3)\left(x+\frac53\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopic.php?p=362838&sid=00aa42b316d41e251e24e658594fcc51#p362838)
for 2005 china west problem we have two methods (at least)
solution 1：
note
$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$$
$$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)$$
then
$$\Longleftrightarrow 10[1-3(a+b)(b+c)(a+c)]-9[1-5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)]\ge 1$$
$$\Longleftrightarrow 3(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)-2(a+b)(b+c)(a+c)\ge 0$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ac)\ge 2=2(a+b+c)^2$$
$$ a^2+b^2+c^2-ab-bc-ac\ge 0$$
It's Obviously.
solution 2:
$$10(a+b+c)^2(a^3+b^3+c^3)-9(a^5+b^5+c^5)-(a+b+c)^5\ge0$$
it is equivalent to
$$15(a+b)(b+c)(c+a)(a^2+b^2+c^2-ab-bc-ca)\ge0$$
But for
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
and for this equality I think
$$10a^3-9a^5\le p (a-1/3)+q$$
and let
$$f(x)=10x^3-9x^5\Longrightarrow f'(x)=30x^2-45x^4\Longrightarrow p=f'(1/3)=\dfrac{25}{9}$$
$$q=f(1/3)=\dfrac{1}{9}$$
so if we can prove this
$$10a^3-9a^5\le\dfrac{25}{9}(a-1/3)+\dfrac{1}{9}$$
These methods I can't work, can someone help deal it. Thank you
| EDIT: The original proof contained an error. It is (hopefully) fixed now.
Consider the function $f(x) = 10x^3 - 9x^5$. Then our goal is to show that $f(a) + f(b) + f(c) \leq 9/4$.
We will need the following two claims
Claim1: $f(x) + f(1-x) \leq 9/4$ for all $x \in [0,1]$.
Proof: A straightforward calculation says that the local maximum is obtained in $x = 0.5 \pm \frac{1}{2\sqrt{3}}$ and is equal to $9/4$.
Claim2: For all $0 \leq a \leq b$ such that $a+b \leq 2/3$ we have $f(a)+f(b) \leq f(a+b)$
Proof: The claim is trivial if $a=0$. Therefore, we shall assume that $a>0$.We need to prove that
$$
10a^3 - 9a^5 + 10b^3 - 9b^5 \leq 10(a+b)^3 - 9(a+b)^5.
$$
Opening the parenthesis on the RHS, and reducing we get that the above is equivalent to
$$
0 \leq 10(3a^2b + 3ab^2) - 9(5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4)
$$
Since $a,b > 0$, we can divide by $15ab$, and so, by moving sides it is enough to show that
$$
3(a^3 + 2a^2b + 2ab^2 + b^3) \leq 2(a + b).
$$
Adding $3(a^2b+ab^2)$ to both sides we get
$$
3(a+b)^3 \leq (2+3ab)(a + b).
$$
Since $a,b \geq 0$, we can divide by $a+b$ to get
$$
3(a+b)^2 \leq 2+3ab,
$$
or equivalently
$$
3a^2+3b^2 + 3ab \leq 2
$$
It is easy to check that the inequality holds if $a+b \leq 2/3$.
We now turn to the proof. Let's s assume that $a \leq b \leq c$.
Since $a+b \leq 2/3$, by Claim2 we have $f(a)+f(b) \leq f(a+b)$, and therefore, by Claim1 we get $f(a)+f(b)+f(c) \leq f(a+b) + f(c) = f(1-c) + f(c) \leq 9/4$, as required.
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$4^\text{th}$ power of a $2\times 2$ matrix $$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I tried to use some trigonometric expressions but it gets very complicated and couldn't solve it.
| Here is a hint:
We have
$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$
Use matrix multiplication to compute (I'll do it for you) ...
$B=A^2 = \left(\begin{array}{cc}\cos^2 x -\sin^2 x& -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x -\sin^2 x\end{array}\right)$
Now simplify using the trigonometry you know, and compute $B^2=A^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$.
Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$.
Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$.
Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.
Is my proof correct? Is there a shorter way to prove this?
| I would do it this way:
We have $x^n=y^n$ and $n=2k$ so we have $(y^2)^k=(x^2)^k$, taking $k$-th root of the both sides we obtain $y^2=x^2$ (beacuse $k$-th root function is an injection) so it follows $(x-y)(x+y)=0$ from which it follows that either $x=y$ or $x=-y$, short enough?
| {
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What is the remainder? What is the remainder when $7^{2002}+3^{2002}+2002$ is divided by $29$ ?
I tried out the problem using congruent modulo but cannot help my cause.
| These kind of problems can always be solved in a similar matter. The goal is to find powers that are congruent $1$ or $-1$ modulo $29$. So you start with $7$ and calculate it's powers modulo $29$:
$$\begin{align*}
7^1&\equiv 7\\
7^2&\equiv 7\cdot 7\equiv 49 \equiv 20\\
7^3&\equiv 20\cdot 7\equiv 140\equiv 24\\
7^3&\equiv 24\cdot 7\equiv 168\equiv 23\\
7^3&\equiv 23\cdot 7\equiv 161\equiv 16\\
&\vdots\\
7^7&\equiv 25\cdot 7\equiv 175\equiv 1
\end{align*}$$
The same for powers of $3$ gives you $3^{14}\equiv 28\equiv -1$. Since $1^k\equiv 1$ and $(-1)^{2k}\equiv 1$ for all $k\in\mathbb Z$, you can reduce the exponent of $7$ modulo $7$ (i.e. $2002\equiv 0\mod 7$) and the exponent of $3$ modulo $28$ (i.e. $2002\equiv 14\mod 28$ and $3^{14}\equiv -1\mod 29$). Now you are left with $1-1+2002$. All you need to do is calculate the remainder of $2002$.
If you know a little number theory you can even use Fermat's little theorem $a^{p-1}\equiv 1\mod p$ to begin with and reduce all exponents modulo $28$, or only try divisors of $28$ to find smaller exponents giving $1$ or $-1$.
| {
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Determining the general form of $10^x \bmod 210$ While solving a problem I came across solving $10^x\bmod 210$ for various values of $x$. It seems that the values repeat after an interval of 6 for $x\geq4$. Can any one explain how can solve this equation for some value of $x$ and show that it repeats at an interval of 6.
Note that $210 = 2\times3\times5\times7$ is a product of the first four primes.
Thanks
| Using Carmitcheal function, $\lambda(21)=$lcm$(\lambda(3),\lambda(7))=$lcm$(2,6)=6$
and $(10,21)=1,10^6\equiv1\pmod{21}$
$10^0\equiv1\pmod{210}$
As $a\equiv b\pmod m\implies a\cdot k\equiv b\cdot k\pmod{m\cdot k} $
$10^0\equiv1\pmod{21}\implies 10\equiv10\pmod{210}$
$10^1\equiv10\pmod{21}\implies 10^2\equiv100\pmod{210}$
$\displaystyle 10^2\equiv-5\implies 10^3\equiv-50\pmod{210}\equiv160$
$\displaystyle 10^3=10\cdot10^2\equiv10(-5)\equiv-8\implies 10^4\equiv-80\pmod{210}\equiv130$
$\displaystyle 10^4=(10^2)^2\equiv(-5)^2\equiv4\implies 10^5\equiv40\pmod{210}$
$\displaystyle 10^5=10\cdot10^4\equiv10\cdot4\equiv-2\implies 10^6\equiv-20\pmod{210}\equiv190$
$\displaystyle 10^6\equiv1\pmod{21}\implies 10^7\equiv10\pmod{210}$
So, for $10'$s power$(>0)$ the cycle is $\cdots,10,100,160,130,40,190,\cdots$
| {
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Find the domain of $\sqrt{x^2-9}$ $$\sqrt{x^2-9}$$
I know that the domain of square root is greater than or equal to zero. I solve for when $x^2-9<0$ and get $x^2<9$. Now I get $x<-3$ and $x<3$. I know that the domain is $(-\infty,-3] \cup [3,\infty)$ The problem is I do not understand how the $x<3$ gets flipped to $x>3$, am I doing this step properly or is there another way to do it?
| You need to have $x^2-9\ge 0$, not $x^2-9<0$. This is $(x-3)(x+3)\ge 0$, which is the case when
*
*one of $x-3$ and $x+3$ is $0$,
*both $x-3$ and $x+3$ are positive, or
*both $x-3$ and $x+3$ are negative.
In all other cases $(x-3)(x+3)$ is negative, which is precisely what you don’t want.
*
*The first of these happens when $x=3$ or $x=-3$.
*The second happens when $x>3$ and $x>-3$; of course if $x>3$, then automatically $x>-3$, so this happens when $x>3$.
*And the third happens when $x<3$ and $x<-3$; this time we notice that if $x<-3$, then automatically $x<3$, so this happens when $x<-3$.
Putting the pieces together, we see that $x^2-9\ge 0$ when $x\le -3$ or $x\ge 3$.
With practice, however, you should come to realize that if $a\ge 0$, then $x^2\ge a^2$ if and only if $|x|>\sqrt{a}$. Here that means that $|x|\ge 3$, which means that $x$ is at least as far away from $0$ as $3$ is, i.e., that $x\le -3$ or $x\ge 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the given answer of $-n(2n+1)$
| $${ 1 }^{ 2 }-{ 2 }^{ 2 }=-\left( 1+2 \right) \\ { 3 }^{ 2 }-{ 4 }^{ 2 }=-\left( 3+4 \right) \\ { 5 }^{ 2 }-{ 6 }^{ 2 }=-\left( 5+6 \right) \\ \vdots \\ { \left( 2n-1 \right) }^{ 2 }-{ \left( 2n \right) }^{ 2 }=-\left( 2n-1+2n \right) $$
Sum up, you get the answer that O.L. has written.
| {
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How do you evaluate $\int \frac{y^2}{y^2+d^2}dy$? $\color{green}{question}$:
How do you evaluate this integral?
$$\int \frac{y^2}{y^2+d^2}dy=y-d\,{\tan}^{-1}\left ( \frac{y}{d} \right )+\mathrm{constant}$$
$\color{green}{I~know}$ I should use the change of variables, But I do not know how to do.
Thank you for any hint.
| Hint
*
*$$\frac{y^2}{y^2+d^2}=\frac{y^2+d^2-d^2}{y^2+d^2}=1-\frac{d^2}{y^2+d^2}$$
*$$\frac{d^2}{y^2+d^2}=\frac{1}{(y/d)^2+1}=\frac{1}{t^2+1}$$
*$$\int\frac{dt}{t^2+1}=\arctan t+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| Here's another solution for the identity of Antonio Cagnoli :
We want to show that :
$\tan A + \tan B + \tan C = \tan A\times \tan B \times \tan C\ $ with $A+B+C=180^\circ=\pi$.
By definition we have : $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ so here, we want to prove that :
$\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}=\frac{\sin A}{\cos A}\times\frac{\sin B}{\cos B}\times\frac{\sin C}{\cos C}$
$\Leftrightarrow$ $\frac{\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos C}{\cos A \cos B \cos B}=\frac{\sin A \sin B \sin C}{\cos A \cos B \cos C}$
$\Leftrightarrow$ $\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos B=\sin A \sin B \sin C$
However,
$\sin A \cos B \cos C+ \sin B \cos A \cos C + \sin C \cos A \cos B=\cos C [\sin A \cos B + \sin B \cos A]+\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C [\sin A \cos B + \sin B \cos A]+\sin C \cos A \cos B=\cos C \sin(A+B)+\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin(A+B)+\sin C \cos A \cos B=\cos C \sin(\pi - C) +\sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin(\pi - C) +\sin C \cos A \cos B=\cos C \sin C + \sin C \cos A \cos B$
$\Leftrightarrow$
$\cos C \sin C + \sin C \cos A \cos B=\sin C[\cos C + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[\cos C + \cos A \cos B]=\sin C[\cos(\pi - (A+B)) + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[\cos(\pi - (A+B)) + \cos A \cos B]=\sin C[-\cos(A+B) + \cos A \cos B]$
$\Leftrightarrow$
$\sin C[-\cos(A+B) + \cos A \cos B]=\sin C \sin A \sin B$.
We finally proved the equality !
Here https://play.google.com/books/reader?id=Lj51QYK1fIcC&printsec=frontcover&output=reader&hl=fr&pg=GBS.PA27 you can see the fastest answer !!!
| {
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Finding the minimum value of a rational function. Prove that if $x$ is real and $a>c$ & $b>c$ the minimum value of $$\frac{(a+x)(b+x)}{(c+x)} ;Given\space( x>-c)$$ is $$({\sqrt{a-c}+\sqrt{b-c \space}})^2$$
I tried using minima condition but the expression was too complicated to arrange and solve .How can this be solved algebraically? Please give answers/suggestions.
Thanks.
| Expand and simplify (or rather, do partial fraction decomposition), we get that
$$ \frac{ (a+x)(b+x)}{c+x} = (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c $$
Apply AM-GM to the first two terms to conclude that
$$ (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c \geq 2 \sqrt{ (a-c)(b-c) } + a+b-2c = \left( \sqrt{a-c} + \sqrt{b-c} \right)^2$$
| {
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Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$
reduces to $2p^2+p-10=0$.
I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and finding areas to substitute but have come with no progress. Could someone offer a slight hint on how to proceed?
Thanks
| $$
\begin{align}
2x^4+x^3-6x^2+x+2 & = x^2\Big(2x^2 + x - 6 + \frac1x + \frac{2}{x^2}\Big) \\[12pt]
& = x^2 \Big(2\left(x+\frac1x\right)^2 + \left(x+\frac1x\right) - 10 \Big) \\[12pt]
& = x^2(2p^2 +p-10).
\end{align}
$$
This equals $0$ only if $x=0$ or the second factor equals $0$. But clearly $x=0$ is not one of the solutions.
| {
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Creating a degree $n$ Taylor polynomial for $\sqrt{1+x}$ I have been asked to produce a general formula for the degree $n$ Taylor polynomial for $\sqrt{1+x}$ using a=0 as the point of approximation.
Given that $p_n(x)=f(a)+(x-a)f^\prime(a)+\frac{(x-a)^2}{2!}f^{(2)}(a)+...+\frac{(x-a)^n}{n!}f^{(n)}(a)=\sum_{j=0}^{n}\frac{(x-a)^j}{j!}f^{(j)}(a)$
$f(x)=\sqrt{1+x}=(1+x)^{1/2} \\
f^\prime (x)=\left ( \frac{1}{2} \right )(1+x)^{-1/2} \\
f^{(2)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )(1+x)^{-3/2} \\
f^{(3)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )(1+x)^{-5/2} \\
f^{(4)}(x)=\left ( \frac{1}{2}\right )\left ( \frac{-1}{2} \right )\left ( \frac{-3}{2} \right )\left ( \frac{-5}{2} \right )(1+x)^{-7/2} \\
f^{(j)}(x)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)(1+x)^{(2j-1)/2} \\
f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$
I'm thinking I have calculated the j-th derivative correctly, however I can't seem to figure out how to simplify it and the rest of the steps into a compact summation.
Added: It's been pointed out to me that the $(1\cdot 1\cdot 3\cdot 5\cdot ... \cdot 2j-3)$ above should be $\prod_{i=0}^{j-1}2i-1$
Thus $f^{(j)}(0)=(-1)^{j+1}\left ( \frac{1}{2^{j}} \right )\prod_{i=0}^{j-1}2i-1$.
If anyone can give me a hint on how to simplify this further it would be appreciated.
| Hint: Using the Binomial Theorem, prove by induction that
$$
(1+x)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}(-x)^k
$$
Truncate as needed. Note that $n!!$ is the Double Factorial:
$$
\begin{array}{rll}
(2k-1)!!&=1\cdot3\cdot5\cdots(2k-1)&=\frac{(2k)!}{2^kk!}\\[6pt]
(2k)!!&=2\cdot4\cdot6\cdot8\cdots(2k)&={2^kk!}
\end{array}
$$
| {
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Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
| The values of $2(\frac 12 +\frac 12 \sqrt{-7})^{n-2}$ give $x+y\sqrt{-7}$, which are a solution to the equation $x^2+7y^2 = 2^n$, where $x$ and $y$ are always odd.
These solutions are thence unique, because all representations of $x^2+7y^2$ reduce to a complex number $x+y\sqrt{-7}$, and that $\frac 12 \pm \frac 12 \sqrt{-7}$ is the unique primes of this form. $2$ is the product of these numbers, and since it requires both primes to give an even value of $x$ and $y$, powers of a single prime must lead always to odd numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $\int_{a}^{b} \frac{(x-x_1)(x-x_2)}{(x_0 - x_1)(x_0-x_2)} dx = \int_{-1}^{1} \frac{t(t-1)}{2} \frac{b-a}{2} dt$ I'm a second year math student, and I have some trouble understanding this equality in my book. I would appreciate a little help :)
Let $x_0=a,x_1=(a+b)/2,x_2=b$. \begin{align*} \int_{a}^{b}
\frac{(x-x_1)(x-x_2)}{(x_0 - x_1)(x_0-x_2)} dx &= \int_{-1}^{1}
\frac{t(t-1)}{2} \frac{b-a}{2} dt\\ \end{align*}
| The substitution $t = \frac{x-x_1}{x_1-x_0}$ converts the left integral into the right:
We have $\frac{dx}{dt} = x_1 - x_0 = \frac{b-a}{2}$, and $\frac{x - x_2}{x_2 - x_0} = \frac{(x-x_1) - (x_2-x_1)}{2(x_1-x_0)} = \frac{1}{2}\left(t - \frac{x_2-x_1}{x_1-x_0}\right) = \frac{t-1}{2}$ since $x_2 - x_1 = x_1 - x_0$. Further, $\frac{a-x_1}{x_1-x_0} = \frac{x_0-x_1}{x_1-x_0} = -1$ and $\frac{b-x_1}{x_1-x_0} = \frac{x_2-x_1}{x_1-x_0} = 1$, so
\begin{align}
\int_a^b \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}\,dx
&= \int_a^b \underbrace{\frac{x-x_1}{x_1-x_0}}_t \underbrace{\frac{x-x_2}{x_2-x_0}}_{\frac{t-1}{2}}\,dx\\
&= \int_{-1}^1 \frac{t(t-1)}{2} \underbrace{\frac{b-a}{2}\,dt}_{dx}.
\end{align}
| {
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Factoring an element in $\mathbb{Z}[\sqrt6]$ I want to know if $1+\sqrt6 $ factors in $\mathbb{Z}[\sqrt6]$.
$1+\sqrt6=(a+b\sqrt6)(c+d\sqrt6)$ implies $a^2-6b^2\not=0,c=\frac{a-6b}{a^2-6b^2},a\not=0, d=\frac{a-b}{a^2-6b^2}$
How do I determine if there are integer solutions?
| Use the multiplicativity of the norm : if you define
$N(a+b\sqrt{6})=a^2-6b^2$, then $N(w_1w_2)=N(w_1)N(w_2)$ for any $w_1,w_2$.
So if $1+\sqrt{6}=(a+b\sqrt{6})(c+d\sqrt{6})$, then
$-5=(a^2-6b^2)(c^2-6d^2)$, so that one of $a^2-6b^2$ or $c^2-6d^2$ is $\pm 1$.
If, say, $a^2-6b^2= \pm 1$, then $a+b\sqrt{6}$ is a unit in ${\mathbb Z}[\sqrt{6}]$.
So there are no interesting factorizations.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find integer values for which a given expression gives a perfect square? Find the integer values for which $x^2+19x+92$ is a perfect square.
Also, How to proceed if you have to find values ( not necessarily integer)?
| Complete the square:
$$\begin{align}
x^2 + 19x + 92 &= m^2\\
\iff (2x)^2 + 2\cdot 19\cdot (2x) + 368 &= (2m)^2\\
\iff (2x+19)^2 + 7 &= (2m)^2\\
\iff (2m)^2 - (2x+19)^2 &= 7.
\end{align}$$
Since $7$ is prime, that means $2m = \pm 4$ and $2x+19 = \pm 3$.
| {
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Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| An answer using the Stolz–Cesàro theorem: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$$
| {
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Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
| As $x=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$
Using de Moivre's formula for positive integer $n$
$$x^n=\left(\cos\frac{2\pi}5+i\sin\frac{2\pi}5\right)^n=\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5$$
$$\implies x^5=\cos2\pi=1\text{ and }x^{-n}=\frac1{x^n}=\frac1{\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5}=\cos\frac{2n\pi}5-i\sin \frac{2n\pi}5$$
$$\implies x^n+x^{-n}=2\cos\frac{2n\pi}5$$
As $x^5=1,x^6=x,x^4=x^{-1},x^7=x^2,x^3=x^{-2}$
$$\implies (x^2-x^3)(x^4-x)=x^6+x^4-x^7-x^3=x+\frac1x-\left(x^2+\frac1{x^2}\right)$$
$$\implies (x^2-x^3)(x^4-x)=2\cos\frac{2\pi}5-2\cos\frac{4\pi}5=2\cos\frac{2\pi}5-2\cos(\pi-\frac\pi5)=2\cos\frac{2\pi}5+2\cos\frac\pi5>0\ \ \ \ (0)$$
Now if $y^5=1$
Using $n$th root of unity, $\displaystyle y=\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=0,1,2,3,4$
$r=0\implies y=1\implies$ the roots of $\displaystyle \frac{y^5-1}{y-1}=0\iff y^4+y^3+y^2+y+1=0\ \ \ \ (1)$ are $\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=1,2,3,4$
Observe that the last equation is Reciprocal Equation of the First type like this
So, divide either sides by $y^2,$ $$y^2+y+1+\frac1y+\frac1{y^2}=0 \implies \left(y+\frac1y\right)^2+\left(y+\frac1y\right)-1=0\ \ \ \ (2)$$
We have $y+\frac1y=2\cos\frac{2r\pi}5$
and as $\cos\frac{2(5-r)\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos \frac{2r\pi}5$
the roots of $(2)$ are
$2\cos\frac{8\pi}5=\cos\frac{2\pi}5>0$ and $\cos\frac{6\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos\frac{4\pi}5=-\cos\frac\pi5<0$
Now, $2\cos\frac{2r\pi}5=y+\frac1y=\frac{-1\pm\sqrt5}2$ where $r=(1$ or $5-1=4)$ and $r=(2$ or $5-2=3)$
So, using Vieta's formulas, $2\cos\frac{2\pi}5+\left(-2\cos\frac\pi5\right)=-1$ and $2\cos\frac{2\pi}5\left(-2\cos\frac\pi5\right)=-1$
$$\implies 2\cos\frac{2\pi}5+2\cos\frac\pi5=\sqrt{\left(2\cos\frac{2\pi}5-2\cos\frac\pi5\right)^2+4\cdot2\cos\frac{2\pi}5\cdot2\cos\frac\pi5}=\sqrt{(-1)^2+4\cdot1}=\sqrt5$$
| {
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"source": "stackexchange",
"question_score": "5",
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Smallest number with specific number of divisors Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Since $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.
As a single number it is :$2$^${23}$
as product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$,
since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers
as product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two
as product of 4 numbers $2^2*3*5*7$
$k=\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$
=$2^3*3^2*5=360$
The above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?
| This approach is not fishy at all, but can be laborious. Note that you can't use $5+3 \lt 7+2$ to conclude that $2^5*3^3 \lt 2^7*3^2$, although it is true, you have to take the ratio and use $3 \lt 2^2$. For example, if you were looking for $36$ factors, one comparison would be $2^8*3^3$ versus $2^5*3^5$. Despite the fact that $5+5 \lt 3+8, 2^5*3^5=7776 \gt 6912=2^8*3^3$. You can take logs and compare $5 \log 2 + 3 \log 3$ with $7 \log 2 + 2 \log 3$ to get it right.
I don't know an easier way. Your example shows the failure of a greedy algorithm. Factor the desired number of factors, here as $3*2^3$. Starting from the largest factor, find the cheapest way to get that many. So start with $2^2$, which has $3$ factors. Now you need to double it. You can either multiply by a new prime, clearly $3$, or increase the exponent of $2$ to $5$. Since the first has a factor $3$ and the second $8$, we choose $2^2*3$ Now we want to double again, and our choices are $2^3, 3^2, \text{ or } 5$ and we take $5$, giving $2^2*3*5$ One more doubling comes from $7$, and we come up with $2^2*3*5*7=420$, not the best.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a congruence linear equation. Solve the following congruence:
$19x\equiv 1\;(\text{mod}\;36)$
My work:
I found an inverse of $19$ and $36$ which is $9$.
$9\cdot 19x\equiv 9\cdot 1\;(\text{mod}\;36)$
$171x\equiv 9\;(\text{mod}\;36)$
$9x\equiv 9\;(\text{mod}\; 36)$
$x\equiv 1\;(\text{mod}\;4)$ is my final answer...
Is it correct?
| Method $1:$
We need $x\equiv19^{-1}\pmod{36}$
Using Carmichael function $\lambda(36)=$lcm$(6,2)=6\implies a^6\equiv1\pmod{36}$ if $(a,6)=1$
$\implies$ord$_{36}(a)$ must divide $6$
$19^1\equiv19\pmod{36},19^2=361\equiv1\pmod{36}\implies 19\equiv19^{-1}\pmod{36}$
Method $2:$
Like this, expressing as Continued Fraction,
$$\frac{36}{19}=1+\frac{17}{19}=1+\frac1{\frac{19}{17}}=1+\frac1{1+\frac2{17}}=1+\frac1{1+\frac1{\frac{17}2}}=1+\frac1{1+\frac1{8+\frac12}}$$
The previous convergent of $\displaystyle\frac{36}{19}$ is $\displaystyle1+\frac1{1+\frac18}=\frac{17}9$
$\displaystyle\implies 36\cdot9-19\cdot17=1\implies-17\cdot19\equiv1\pmod{36}\implies 19^{-1}\equiv-17\equiv19$
Alternatively observe that the next convergent of $\displaystyle\frac{36}{19}$ is $\displaystyle1+\frac1{1+\frac1{8+\frac11}}=\frac{19}{10}$
$\displaystyle\implies 36\cdot10-19\cdot19=-1\implies-19\cdot19\equiv-1\pmod{36}\implies 19^{-1}\equiv19$
Method $3:$
We have $19x\equiv1\pmod{36}\ \ \ \ (1)$
As $36=9\cdot4$ with $(9,4)=1$
$\ \ \ \ (1)\implies 19x\equiv1\pmod9\implies x\equiv1\pmod 9$
$\ \ \ \ (1)\implies 19x\equiv1\pmod4\implies -x\equiv1\pmod 4\implies x\equiv-1\pmod 4$
Using well-known CRT, we can show $x\equiv19\pmod{36}$
| {
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"question_score": "1",
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Evaluating the area between the curves $r=2\sin\theta$ and $r=\sin\theta+\cos\theta$
So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad r=\sin\theta +\cos\theta $$
I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder to me.
$$\begin{align*}
x&=r\cos\theta=\cos\theta(\sin\theta + \cos\theta)\\
y&=\sin\theta(\sin\theta+\cos\theta)
\end{align*}$$
so $x+y=1+2\cos\theta \sin\theta$....Which gives me no help.
I'm getting used to polar coordinates, but I need some help! Thank you
| Hint: Try multiplying both sides of $r=\sin\theta + \cos\theta$ by $r$, and using the identities $r^2=x^2+y^2$, $x=r\cos\theta$, and $y=r\sin\theta$, to find an equation in terms of $x$ and $y$.
| {
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Prove $ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots + \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2} $ by induction
Prove by induction that for all $n > 0$,
$$
\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{4}}{6} + \cdots
+ \frac{\sqrt{n+1}}{2n} > \frac{\sqrt{n}}{2}
$$
I have done the basis step, where $n = 1$ and showed that L.H.S is > R.H.S
For inductive step, I assume $n = k$ & L.H.S > R.H.S is true.
However, I am stuck at showing how for $k + 1$, it is also true for L.H.S > R.H.S
Any help will be much appreciated. Thanks!
| By cross-multiplying, it can easily be seen that
$$
\begin{align}
\frac{\sqrt{k+1}}{2k}
&\gt\frac1{\sqrt{k+1}+\sqrt{k\,}}\\[6pt]
&=\sqrt{k+1}-\sqrt{k\,}
\end{align}
$$
Summing gives
$$
\sum_{k=1}^n\frac{\sqrt{k+1}}{2k}
\gt\sqrt{n+1}-1
$$
This is not the inequality given, but for $n\ge2$, $\sqrt{n+1}-1\gt\frac{\sqrt{n}}2$.
| {
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"url": "https://math.stackexchange.com/questions/502026",
"timestamp": "2023-03-29T00:00:00",
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How do I determine whether $18 \notin A$ with these premises? With $A \subseteq \mathbb{N}$, given that
A1. $(x \in A \land y \in A) \implies x^2 + y ^2 \in A$
A2. $1 \in A$
A3. $3 \notin A$
Determine whether:
$18 \notin A$
I've been unable to prove this. First, I tried to demonstrate $18 \in A$, but no matter what I try, I can't land in $18$, so I guess that $18 \notin A$ should be true - but how can I prove it? I have never used $3 \notin A$, which is probably the key, but I'm not sure how.
| \begin{align}
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)}
&\phantom{\quad\Longrightarrow\quad
1^{2} + 1^{2} =} 1 \in A
\\
\left(1 \in A\quad \wedge\quad 1 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 1^{2} = 2 \in A
\\
\phantom{\left(1 \in A\quad \wedge\quad 1 \in A\right)}
&\phantom{\quad\Longrightarrow\quad
1^{2} + 1^{2} =} \color{#ff0000}{\,3 \not\in A}
\\
\left(1 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 2^{2} = 5 \in A
\\
\left(2 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 2^{2} = 8 \in A
\\
\left(1 \in A\quad \wedge\quad 3 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 3^{2} = 10 \in A
\\
\left(2 \in A\quad \wedge\quad 3 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 3^{2} = 13 \in A
\\
\left(1 \in A\quad \wedge\quad 4 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 4^{2} = 17 \in A
\\
\left(4 \in A\quad \wedge\quad 2 \in A\right)
&\quad\Longrightarrow\quad
4^{2} + 2^{2} = 20 \in A
\\
\left(1 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 5^{2} = 26 \in A
\\
\left(2 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 5^{2} = 29 \in A
\\
\left(1 \in A\quad \wedge\quad 6 \in A\right)
&\quad\Longrightarrow\quad
1^{2} + 6^{2} = 37 \in A
\\
\left(2 \in A\quad \wedge\quad 6 \in A\right)
&\quad\Longrightarrow\quad
2^{2} + 6^{2} = 40 \in A
\\
\left(4 \in A\quad \wedge\quad 5 \in A\right)
&\quad\Longrightarrow\quad
4^{2} + 5^{2} = 41 \in A
\\&\vdots
\\[5mm]&
\end{align}
$$
\left\{%
1,2,5,8,10,13,17,20,26,29,37,40,41...
\right\}
$$
| {
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How to prove $\lim_{n\to \infty} \prod_{i=0}^{n-1} $ $(2+\cos \frac{i \pi}{n})^{\frac{\pi}{n}}$=$\sqrt{3}$ by the sum form of a integration? The integral seems like
$$
\exp\left(\int_{0}^{\pi}\left[\ln(2+\cos\left(x\right)\right)]\,{\rm d}x\right)
$$
but the above will be a number bigger than $5$ by appraisement.
| Taking log,
\begin{align*}
\log \left( \prod_{i=0}^{n-1} \left( 2 + \cos \left( \frac{i \pi}{ n } \right ) \right )^{\frac \pi n}\right ) &=
\sum_{i=0}^{n-1} \frac \pi n \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) \\
&= \pi \cdot \frac {1}{n } \sum_{i=0}^{n-1} \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right )\\
\end{align*}
Taking limit
$$\lim_{n\to\infty } \frac \pi n \sum_{i=0}^{n-1} \log \left( 2 + \cos \left( \frac{ i \pi}{n}\right ) \right ) = \int_0^\pi \log(2 + \cos(x))dx = \pi \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) $$
Taking back to $e$, we get
$$\lim_{n \to \infty } \prod_{k=0}^\infty \left( 2 + \cos \left( \frac{i \pi }{n } \right ) \right )^{\frac \pi n } = e^{\pi \log \left(\frac{1}{2} \left(2+\sqrt{3}\right)\right) } = \left( \frac 1 2 ( 2 + \sqrt 3) \right )^\pi $$
Putting
N[Product[(2 + Cos[k Pi/10^3])^(Pi/10^3), {k, 0, 10^3}]]
on Mathematica produces the value of $7.10988$ which is close to our calculated value so I think what you claim on title must be false.
To evaluate the integral, we proceed in the following way.
$$\int_0^{\pi} \log \left( 2 + \cos(\theta) \right )d\theta = \frac 1 2 \int_{-\pi}^{\pi} \log \left( 2 + \cos(\theta) \right )d\theta = \frac{1}{2} \int_0^{2\pi} \log(2 + \cos(\theta))d\theta $$
From Gauss MVT we get,
$$\int_0^{2\pi } \left(\log ( \sqrt 3 + 2 + e^{i\theta})+\log ( \sqrt 3 + 2 + e^{-i\theta}) \right ) d\theta = 4 \pi \log (\sqrt 3 + 2 )$$
\begin{align*}
4 \pi \log (\sqrt 3 + 2 ) &=\int_0^{2\pi } \left(\log (\sqrt 3 + 2 + e^{i\theta})+\log (\sqrt 3 + 2 + e^{-i\theta}) \right ) d\theta \\
&= \int_0^{2\pi} \log \left( ( \sqrt 3+2 )^2 + 1 + ( \sqrt 3 +2)2 \cos \theta \right )d\theta\\
&= \int_0^{2\pi } \log \left( 4(2 + \sqrt 3) + 2 (2 + \sqrt 3) \cos (\theta) \right ) d\theta \\
&= \int_0^{2\pi }\log(2(2 + \sqrt 3))d\theta + \int_0^{2\pi } \log(2 + \cos\theta)d\theta\\
&= 2 \pi \log (2 (2 + \sqrt 3 )) + \int_0^{2\pi } \log(2 + \cos\theta)d\theta \\
\end{align*}
Which gives
$$2\pi \log \left( \frac{(2 + \sqrt 3 )^2}{2 (2 + \sqrt 3 )}\right )= \int_0^{2\pi } \log(2 + \cos\theta)d\theta $$
Taking half of it give the value of our integral as evaluated by Mathematica earlier.
$$\pi \log \left( \frac{2 + \sqrt 3 }{2 }\right )= \int_0^{\pi } \log(2 + \cos\theta)d\theta $$
| {
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Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square. Integer values of $x$ for which $\bf{x^4+x^3+x^2+x+1}$ is a Perfect Square.
$\underline{\bf{My\; Try}}$:: Let $\bf{x^4+x^3+x^2+x+1 = k^2}$, where $k\in \mathbb{Z}$
$4x^4+4x^3+4x^2+4x+1 = 4k^2 = (2k)^2$
Now How can I proceed after that
Help Required,
Thanks
| We can just change this into a quadratic equation (or cubic equation, but that becomes very highly complicated).
$$\because x^4 + x^3 + x^2 + x + 1 = k^2$$
then this means that $x^2 + x + (1 - k^2 - x^3 - x^4) = 0$. This means that:
$$x = \frac{-1 \pm \sqrt {1 - 4(1 - k^2 - x^3 - x^4)}}{2}$$
And this means that:
$2x + 1 = \pm \sqrt {1 - 4(1 - k^2 + x^3 + x^4)}$ meaning that this root is odd if $x$ is an integer. But if that is so, then, the expression:
$$-1 \pm \sqrt{1 - 4(1 - k^2 - x^3 - x^4)}$$
is even $(2x)$. But is $x$ itself an even integer? We can test to see if $x$ is odd or even, but I will only be testing if it is even. So let's call this even number $2r$ and assume that:
$$(2r)^4 + (2r)^3 + (2r)^2 + 2r + 1 = k^2$$
$$16r^4 + 8r^3 + 4r^2 + 2r + 1 = k^2$$
Since all other values on the left hand side of the equation are even (except $1$), then this means that $k^2$ is odd because the sum of all the even numbers is another even number $2n$. And we know that $2n + 1$ is odd. And since $k^2$ is odd, then $k$ is odd. Where:
$$2n = r(16r^3 + 8r^2 + 4r + 2)$$
$$\implies 2n = r(r(16r^2 + 8r + 4) + 2)$$
$$\implies 2n = r(r(r(16r + 8) + 4) + 2)$$
$$\therefore \frac{2n}{r^3} = 16r + 8 + 4 + 2 = 16r + 14$$
Now we can form a real quadratic equation. If $2n/r^3 = 16r + 14$ then:
$$4r^2 + 14^{1/3}r - 2n = 0$$
$$r = \frac{-14^{1/3} \pm \sqrt{14^{2/3} - 32n}}{8}$$
$$r = - \frac{14^{1/3}}{2} \pm \sqrt {2n + \frac{14^{2/3}}{4}}$$
The equation just above this sentence is given by "completing the square". By simplifying this, we will be able to simplify the quadratic expression much easier. If so, we get that:
$$r = \frac{-14^{1/3} \pm \sqrt {8n + 14^{2/3}}}{2}$$
$\therefore$ $x$ cannot be an even integer since:
$$x = 2r = -14^{1/3} \pm \sqrt {8n + 14^{1/3}} = \frac{-1 \pm \sqrt {1 - 4(1 - k^2 - x^3 - x^4)}}{2}$$
$$\implies 2x = 4r = 2(-14^{1/3} \pm \sqrt {8n + 14^{1/3}}) = -1 \pm \sqrt {1 - 4(1 - (x^2 + x + 1))}$$
$$\therefore 2x = 4r = -112^{1/3} \pm \sqrt {32n + 896^{1/3}} = -1 \pm \sqrt {4(x^2 + x + 1) - 3}$$
To conclude, the equation $x^4 + x^3 + x^2 + x + 1 = k^2$ has no solutions in integers if $x$ is even and $k$ is odd, so we know that there is no value for $n$ to satisfy the equation. And just from the looks of it, I don't think there would be solutions if $x$ was odd and $k$ is even, but that is just out of pure assumption. For equations like yours, you can use quadratic formulas and "complete the squares" to solve for $x$ and use the example provided. Hope I helped! :)
| {
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Elementary manipulation with elements of group Let $(G,*)$ be a group with identity $e$ , let $a,b∈G$ such that $a*b^3*a^{-1}=b^2$ and $b^{-1}*a^2*b=a^3$ , then how do we prove that $a=b=e$ ?
| Given $$ab^3a^{-1}=b^2$$ thus we get $$ab^6a^{-1}=b^4\;,\;\;ab^9a^{-1}=b^6$$
Replaceing $b^6$ in $ab^6a^{-1}=b^4$, we get $a^2b^9(a^{-1})^2=b^4$. Thus $a^2b^{18}(a^{-1})^2=b^8$. Again $ab^3a^{-1}=b^2$ gives $ab^{27}a^{-1}=b^{18}$. Replacing $b^{18}$, we get $a^3b^{27}(a^{-1})^3=b^8$. But $b^{-1}a^2b=a^3$. Thus $(b^{-1}a^2b)b^{27}(b^{-1}a^2b)^{-1}=b^8$. This gives $a^2b^{27}(a^2)^{-1}=b^8$. Also $a^2b^9(a^{-1})^2=b^4$ and $a^2b^{18}(a^{-1})^2=b^8$ give $a^2b^{27}(a^2)^{-1}=b^{12}$. Thus $b^8=b^{12}, i.e. b^{4}=e$. Thus $ab^6a^{-1}=e$, and thus $b^6=e$. so, $b^2=e$ and similarly it can be shown that $b^3=e$. So, $b=e$. Now showing $a=e$ is easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it up into two parts; proving that the $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $9$, and that it's divisible by $11$. Unfortunately, I factor it out and get $10(4\cdot10^{2n} + 9\cdot10^{2n-1} + 5)-45$, which is equal to $10(99k)-45$, but of $99$'s divisors, only $3$ divides $45$.
Is there a more ingenious way to factor this equation so that all of my problems are solved?
| Assuming that $4\times10^{2n}+9\times10^{2n-1}+5=99k$,
$$
4\times10^{2(n+1)}+9\times10^{2(n+1)-1}+5=4\times10^{2n+2}+9\times10^{2n+1}+5=100\times(4\times10^{2n})+100\times(9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1}+5-5)+5=100\times(4\times10^{2n}+9\times10^{2n-1}+5)-500+5=100\times99k-495=100\times99k-5\times99=(100k-5)\times99.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implicit differentiation of a lemniscate at a point: So here's the problem:
Find the slope of the tangent line of :
$2(x^2 +y^2)^2 = 25(x^2 - y^2)$ at the point (3,1)
Cool:
So here's what I did:
Simplification step:
$2(x^2 +y^2)^2 = 25(x^2 - y^2) \Rightarrow 2(x^2+y^2)^2 = 25x^2 - 25y^2$
Differentiate both sides:
$4(x^2 +y^2)(2x(x') + 2y(y')) = 50x(x') - 50y(y')$
Simplify:
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$(4x^2 + 4y^2)(2x + 2y(y'))= 50x - 50y(y')$
$8x^3 + 8x^2y(y') + 8xy^2 + 8y^3(y')= 50x - 50y(y')$
$8x^2y(y') + 8y^3(y') + 50y(y') = 50x - 8x^3 + 8xy^2$
$(y') (8x^2y + 8y^3 + 50y) = 50x - 8x^3 + 8xy^2$
$y' = \frac{50x - 8x^3 + 8xy^2}{8x^2y + 8y^3 + 50y}$
Cool, got y'...
Now, should I take the original equation and solve for y and substitute? I'm not even going to try putting down my steps on here.
The prompt is to find m=?
Someone please help?
| The answers above were wrong since the equation is wrong. It should be the following -
y1= 8(x^2+y^2)-50)x/-(8(x^2+y^2)+50)y
Then you get the slope as -9/13.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Pattern matching puzzles Problems: I wish to have an equation for below data:
$x = 2, y = 5$
$x = 2, y = 6 $
$x = 2, y = 7 $
$x = 2, y = 8 $
$x = 1, y = 9$
$x = 0, y = 10$
And Will you show me the steps to acquire that equation?
| First sqap the values for $x$ and $y$. Then apply the Lagrange interpolation. The first 4 points lie on the the line $y=2$ Then do the Lagrange interpolation explained here:
After the application, the first 5 points then lie on the function
$$y = 2 - \frac{(x - 5) (x - 6) (x - 7) (x - 8)}{24}$$
Do this method again and the final function should be:
$$y = 2 - \frac{(x - 5) (x - 6) (x - 7) (x - 8)}{24} + \frac{(x - 5) (x - 6) (x - 7) (x - 8) (x - 9)}{40}$$
To get the function for your initial question, just swap $x$ and $y$. So the final function is:
$$x = 2 - \frac{(y - 5) (y - 6) (y - 7) (y - 8)}{24} + \frac{(y - 5) (y - 6) (y - 7) (y - 8) (y - 9)}{40}$$
Or in expanded form:
$$x = -446+\frac{4894}{15} y - \frac{1123}{12}y^2+\frac{317}{24}y^3-\frac{11}{12}y^4 +\frac{1}{40}y^5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
cubic equations which have exactly one real root Question is to check :
For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .
the way in which i have proceeded is :
let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$
i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$
But, $a^3+a+c=0$. So, $-a^3-a=c$.
so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$
i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$
Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 <0$ for any real number $a$
Thus, quadratic has no real root and so is the cubic $x^3+x+c$
I would like to know if this justification is sufficient and if this can be generalized.
I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)
can this be generalized to any odd degree polynomial (at least for some special cases)
| It is a bit unclear what you want to generalise. $x^3+x+c$ is a strictly increasing function of $x$ - a strictly increasing (or decreasing) polynomial of odd order with real coefficients will always have a single real root.
Or you may be saying that the polynomial can be written as $p(x)=(x-a)q(x)$ where $q(x)$ is a polynomial of even degree which has no real roots. Any odd degree polynomial with a single simple zero (ie not a multiple zero) can be written in this way. However this form of factorisation does not imply that $p(x)$ is strictly increasing, simply that it has just one real zero.
For example take $p(x)=x\left((x-3)^2+1\right)$ with $a=0$
We have $p(0)=0, P(1)=5, p(2)=4, p(3)=3, p(4)=8$, so $p(x)$ is not increasing, while $q(x)$, which is a sum of squares, one of which is non-zero, has no real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
Find hypotenuse given acute angle bisectors
In a right triangle $ABC$ (right-angled at $B$), $D$ and $E$ are points of $\overline{AB}$ and $\overline{BC}$ respectively such that $\overline{CD}$ and $\overline{AE}$ are the angle bisectors of the acute angles of the triangle.
Given that $AE=9$ and $CD=8\sqrt{2}$, find the length of the hypotenuse $\overline{AC}$.
| Let $\alpha = \angle BAE$. Let $\beta = \angle BCD$. What we want is $AC = \sqrt{AB^2 + BC^2}$. In other words, we want to solve for $\alpha, \beta, AB, BC$ such that
$$AB = 9 \cos\alpha \quad\quad BC = 8\sqrt{2} \cos\beta \quad\quad \alpha + \beta = \frac{\pi}{4} \quad\quad \frac{BC}{AB} = \tan 2\alpha$$
From the last equation we have
$$\begin{align*}
\frac{8\sqrt{2}\cos\beta}{9\cos\alpha} &= \tan 2\alpha \\
\frac{8\sqrt{2}}{9} \cdot \frac{\cos(\pi/4 - \alpha)}{\cos\alpha} &= \tan 2\alpha\\
\frac{8\sqrt{2}}{9} \cdot \frac{(\cos\alpha + \sin\alpha)/\sqrt{2}}{\cos\alpha} &= \tan2\alpha \\
\frac{8}{9}(1+\tan\alpha) &= \frac{2\tan\alpha}{1 - \tan^2 \alpha} \\
4(1 -\tan^2\alpha + \tan\alpha - \tan^3 \alpha) &= 9\tan\alpha \\
4\tan^3 \alpha + 4\tan^2 \alpha + 5\tan\alpha - 4 &= 0
\end{align*}$$
At this point we have a cubic equation to solve. But worry not: using the rational root theorem we can search for rational roots relatively easily. It turns out that $\tan\alpha = 1/2$ is a root. Is this the only solution we want? Let's factorise the equation:
$$(2\tan\alpha - 1)(2\tan^2 \alpha + 3\tan\alpha + 4) = 0$$
The discriminant of the quadratic factor is $3^2 - 4\cdot 2 \cdot 4 < 0$. Indeed, $\tan\alpha = 1/2$ is the only real root.
We find $AB = 18/\sqrt{5}$, $BC = 24/\sqrt{5}$. So $AC = 6\sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Logarithm problem : Prove that $log_{3^2} \frac{1}{2} > 0$ Logarithm problem :
Prove that $log_{3^2} \frac{1}{2} > 0$
My approach :
$log_{3^2} \frac{1}{2} > 0$
$\Rightarrow \frac{1}{2} log_3 \frac{1}{2} >0$
$\Rightarrow \frac{1}{2} [ log_3 1 -log_3 2] >0 $
$\Rightarrow \frac{1}{2} [ 0 - log_3 2] >0 $
$\Rightarrow -\frac{1}{2} [ log_3 2] > 0 $ { which is false}
Please suggest... thanks...
| To evaluate $log_{3^2}\frac{1}{2}$ you solve $(3^2)^x=\frac{1}{2} \rightarrow xlog(3^2)=log(\frac{1}{2}) \rightarrow x =\frac{log(\frac{1}{2})}{log(3^2)}$, but $log(\frac{1}{2}) = -ln(2)$. Hence, $log_{3^2}\frac{1}{2}$ is not greater than 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that:$ \sum\limits_{n=1}^{p} \left\lfloor \frac{n(n+1)}{p} \right\rfloor= \frac{2p^2+3p+7}{6} $ where $p$ is a prime and $p \equiv 7 \mod{8}$ Proof that:$ \sum\limits_{n=1}^{p} \left\lfloor \frac{n(n+1)}{p} \right\rfloor= \frac{2p^2+3p+7}{6} $
where $p$ is a prime number such that $p \equiv 7 \mod{8}$
I tried to separate the sum into parts but it does not seems to go anywhere. I also tried to make a substitutions for $p$ ,but, I don't think it is entriely correct to call $p=7+8t$. Any ideas?
| $$\sum_{i=1}^p \frac{n(n+1)}p = \frac1 p \frac {p(p+1)(p+2)}3 = \frac{p^2+3p+2}3$$, and
$$\frac{p^2+3p+2}3 - \frac{2p^2+3p+7}6 = \frac{p-1}2$$
So you are asking to prove that $$\sum_{n=1}^p \frac{n(n+1)}p - \lfloor\frac{n(n+1)}p\rfloor = \frac{p-1}2$$.
The term being summed is $\dfrac 1 p$ times the residue of $n(n+1)$ modulo $p$.
So this becomes showing $\displaystyle \sum_{n \in \Bbb F_p} (n(n+1) \pmod p) = p(p-1)/2$
Let $f(x)$ be the number of solutions to $n(n+1)=x$ in $\Bbb F_p$.
$n(n+1) = x \iff n^2 + n = x \iff (2n+1)^2 = 4x+1$, hence $\displaystyle f(x) = 1 + \binom{4x+1}p$,
and the sum becomes $\displaystyle \sum_{x=0}^{p-1} x f(x) = \sum x + \sum x \binom{4x+1}p$.
The first sum is $p(p-1)/2$, so we are left with showing that the second sum is zero.
Let us do a last rearrangement by setting $y = 1+4x$ and writing the sum as $\displaystyle \sum x(y) \binom y p $, where $x(y) = \frac 14 (y-1 + k(y)p)$ and $k(y)$ is the remainder of $y-1$ mod $4$.
Let $\displaystyle S_i = \sum_{y \equiv i \pmod 4} \binom y p$.
Since $-1$ is not a square, $S_0 = - S_3$ and $S_1 = - S_2$.
Since $2$ is a square, $S_1 + S_3 = S_2 + S_3$ (and $S_0 + S_2 = S_0 + S_1$) hence $S_1 = S_2 = 0$.
We can rewrite the sum into $$\frac 1 4 \left(\sum y\binom y p + (3p-1)S_0 - S_1 + (p-1)S_2 + (2p-1)S_3\right) = \frac 1 4 \left(\sum y\binom y p + p(S_0 + S_2)\right)$$
Since $(-1)$ is not a square, $$\sum y \binom y p = \sum_0^{(p-1)/2} (2y - p) \binom y p$$
By this question, this is $$-p \sum_0^{(p-1)/2} \binom y p = -p \sum_0^{(p-1)/2} \binom {2y} p = -p(S_0 + S_2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$ Is it possible to evaluate the following integral in a closed form?
$$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx,$$
where $\phi$ is the golden ratio:
$$\phi=\frac{1+\sqrt{5}}{2}.$$
| Since $\frac1\phi=\phi-1$, we get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\int_0^\infty\frac{x^\phi\arctan(x)}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{1}\\
&=\int_0^\infty\frac{x^\phi(\frac\pi2-\arctan(x))}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{2}
\end{align}
$$
Average $(1)$ and $(2)$ to get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\frac\pi4\int_0^\infty\frac{x^\phi}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{3}\\
&=\frac\pi{4\phi}\int_0^\infty\frac{x}{\left(x+1\right)^2}\frac{\mathrm{d}x}{x}\tag{4}\\
&=\frac\pi{4\phi}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\frac1\phi=\phi-1$
$(2)$: Substitute $x\mapsto\frac1x$
$(3)$: Average $(1)$ and $(2)$
$(4)$: Substitute $x\mapsto x^{1/\phi}$
$(5)$: $\int_0^\infty\frac{\mathrm{d}x}{(x+1)^2}=\left[-\frac1{x+1}\right]_0^\infty=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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Writing u as a linear combination of the vectors in S. Write vector
u = $$\left[\begin{array}{ccc|c}2 \\10 \\1\end{array}\right]$$
as a linear combination of the vectors in S. Use elementary row operations on an augmented matrix to find the necessary coefficients.
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v2\left[\begin{matrix}5\\4\\1\end{matrix}\right]
$ }. If it is not possible, explain why?
This is what i have so far:
S = {
$v1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , v2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
v3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
v4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
$ }
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]
=\left[\begin{matrix}0\\0\\0\end{matrix}\right]$
$c1$$\left[\begin{matrix}1\\2\\2\end{matrix}\right] , c2\left[\begin{matrix}4\\2\\1\end{matrix}\right],
c3\left[\begin{matrix}5\\4\\1\end{matrix}\right].
c4\left[\begin{matrix}2\\10\\1\end{matrix}\right]$
$
\begin{bmatrix}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{bmatrix}
$
Now i don't know how to do this. Help will greatly be appreciated.
Thanks
| You're heading in the right direction, you've found the augmented matrix:
$$
\left[\begin{array}{ccc|c}
1 & 4 & 5 & 2\\
2 & 2 & 4 & 10\\
2 & 1 & 1 & 1\\
\end{array}\right].
$$
The solutions $(x,y,z)$ to this system of linear equations are precisely the values for which $$x\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}+y\begin{bmatrix}4 \\ 2 \\ 1 \end{bmatrix}+z\begin{bmatrix}5 \\ 4 \\ 1 \end{bmatrix}=\begin{bmatrix}2 \\ 10\\ 1 \end{bmatrix}.$$
So, we solve the above system of linear equations using elementary row operations on the augmented matrix, to reduce it to row echelon form. Then, if the system of equations turns out to be consistent (i.e., if there is a solution), we use back substitution to find all the solutions $(x,y,z)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fibonacci Sequence Exercise I need some help checking the following solution.
The Fib sequence is defined by $a_1 = 1, a_2 = 1$ and for all $n\geq 2$, $a_{n+1} = a_n + a_{n-1}$. Thus, the sequence begins:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...
Prove that for all $n\geq 1$, $a_n <\left(\frac{5}{3}\right)^n$.
So far I have:
by induction
$a_{n+1} = \left(\frac{5}{3}\right)^n+1 = a_n+a_{n-1} < \frac{5}{3}\cdot\left(\frac{5}{3}\right)^n $ from the definition and factoring.
$\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n > a_n+a_{n-1} $
$\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n > a_n\cdot \frac{5}{3}$ from above
$a_n\cdot \frac{5}{3} > a_n + a_{n-1}$
This is where I get stuck, is this a complete solution? Or is there further computation needed?
| We need Strong induction here
Let $a_n<\left(\frac53\right)^n$ for integer $1\le n\le m$
$\implies \displaystyle a_{m+1}=a_m+a_{m-1}<\left(\frac53\right)^m+\left(\frac53\right)^{m-1}=\left(\frac53\right)^{m-1}\left(\frac53+1\right)$
which will be $<\left(\frac53\right)^{m+1}$ if $\displaystyle \left(\frac53+1\right)<\left(\frac53\right)^2\iff 8\cdot3^2<3\cdot5^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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finding factors for gcd To compute $gcd(25, 11)$, Euclid's algorithm would proceed as follows:
$$\underline{25} = 2 \cdot \underline{11}+3$$
$$\underline{11} = 3 \cdot \underline{3}+2$$
$$\underline{3} = 1 \cdot \underline{2}+1$$
$$\underline{2} = 2 \cdot \underline{1}+0$$
Thus $gcd(25,11)$ = $gcd(11,3)$ = $gcd(3,2)$ = $gcd(2,1)$ = $gcd(1,0)=1$
*THIS PART I DO NOT UNDERSTAND CAN SOMEONE GO OVER IT PLEASE *
To find $x$ and $y$ such that $25x+11y=1$, we start by expressing 1 in terms of the last pair (1,0). Then work backwards and express it in terms of (2,1) , (3,1), (11,3), and finally (25,11). The first step is:
$$1= 1-0$$
to rewrite this in terms of (2,1), we use the substitution $0 = 2-2 \cdot 1$ from the last line of the gcd calculation to get:**
$$1 = \underline{1}-(\underline{2} -2 \cdot\underline{1}) = -1 \cdot \underline{2} + 3 \cdot \underline{1}$$**
$$$$
how did they get the last statement of the expression i understand that if you evalute it it will equal 1 but to me it seems like they pulled number out of thin air , every time i try to do it, i end up simplifying and i dont get quiet what they got? how do i come up with this statement: $-1 \cdot \underline{2} + 3 \cdot \underline{1}$ from the previosu statement: $\underline{1}-(\underline{2} -2 \cdot\underline{1})$
(they keep going):
The second last line of the gcd calculation tells us that $1=3-1 \cdot 2$. Substituting:
$$ 1 = -1 \cdot \underline{2} + 3( \underline{3} - 1 \cdot \underline{2} = 3 \cdot \underline{3} - 4 \cdot \underline{2} ) $$
continuing in this same way with substitutions $2=11-3 \cdot3 $ and $3 = 25-2 \cdot 11$ gives:
$$1= 3 \cdot \underline{3} - 4(\underline{1} - 3 \cdot \underline{3}) = -4 \cdot \underline{11} + 15 \cdot \underline{3} = -4 \cdot \underline{11} + 15 (\underline{25} - 2 \cdot {11}) = 15 \cdot \underline{25} - 34 \cdot \underline{11}$$
We're done: $15 \cdot 25 - 34 \cdot 11 =1$, so $x=15$ and $y=-34$
| Perhaps it would actually be easier to see what’s going on in a symbolic example. Suppose that we applied the algorithm to $m$ and $n$ and get:
$$\begin{align*}
n&=q_0m+r_0\tag{1}\\
m&=q_1r_0+r_1\tag{2}\\
r_0&=q_2r_1+r_2\tag{3}\\
r_1&=q_3r_2+0\tag{4}\\
\end{align*}$$
We know then that $\gcd(n,m)=\gcd(m,r_0)=\gcd(r_0,r_1)=\gcd(r_1,r_2)=\gcd(r_2,0)=r_2$. We start by expressing $r_2$ as an integer combination of that last pair, $r_2$ and $0$:
$$r_2=1\cdot r_2-1\cdot 0\;.\tag{5}$$
We can solve $(4)$ for $0$ in terms of $r_1$ and $r_2$ to get $$0=1\cdot r_1-q_3\cdot r_2$$ and substitute this into $(5)$ to eliminate the $0$:
$$\begin{align*}
r_2&=1\cdot r_2-1\cdot(1\cdot r_1-q_3\cdot r_2)\\
&=-1\cdot r_1+(1+q_3)\cdot r_2\;,\tag{6}
\end{align*}$$
expressing $r_2$ as an integer combination of $r_1$ and $r_2$. We can now solve $(3)$ for $r_2$ in terms of $r_0$ and $r_1$ to get
$$r_2=1\cdot r_0-q_2\cdot r_1$$
and substitute this into the righthand side of $(6)$ to eliminate the $r_2$ on the righthand side:
$$\begin{align*}
r_2&=-1\cdot r_1+(1+q_3)\cdot(1\cdot r_0-q_2\cdot r_1)\\
&=(1+q_3)\cdot r_0-(1+(1+q_3)q_2)\cdot r_1\\
&=(1+q_3)\cdot r_0-(1+q_2+q_3q_2)\cdot r_1\;,\tag{7}
\end{align*}$$
expressing $r_2$ as an integer combination of $r_0$ and $r_1$. Keep going in the same way. The next step is to eliminate $r_1$ from the righthand side of $(7)$ by solving $(2)$ for $r_1$ to get $r_1=1\cdot m-q_1\cdot r_0$ and substituting into $(7)$ to get
$$\begin{align*}
r_2&=(1+q_3)\cdot r_0-(1+q_2+q_3q_2)\cdot(1\cdot m-q_1\cdot r_0)\\
&=-(1+q_2+q_3q_2)\cdot m+\big((1+q_3)+(1+q_2+q_3q_2)q_1\big)\cdot r_0\\
&=-(1+q_2+q_3q_2)\cdot m+(1+q_3+q_1+q_2q_1+q_3q_2q_1)\cdot r_0\;.\tag{8}
\end{align*}$$
Finally, solve $(1)$ for $r_0$ to get $r_0=1\cdot n-q_0\cdot m$ and substitute that into the righthand side of $(8)$ to eliminate $r_0$ from that side of the equation:
$$\begin{align*}
r_2&=-(1+q_2+q_3q_2)\cdot m+(1+q_3+q_1+q_2q_1+q_3q_2q_1)\cdot(1\cdot n-q_0\cdot m)\\
&=(1+q_3+q_1+q_2q_1+q_3q_2q_1)\cdot n\\
&\qquad-(1+q_2+q_3q_2+q_0+q_3q_0+q_2q_1q_0+q_3q_2q_1q_0)\cdot m\;,
\end{align*}$$
finally expressing the gcd $r_2$ as an integer combination of the original integers $n$ and $m$.
It’s just repeated substitution to eliminate the remainders $r_k$ until only $n$ and $m$ are left. (And it’s much easier when you’re working with numbers.)
| {
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Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\tag1\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1
$$
My Attempt:
Let $2^x = a$ and $3^x = b$ . Then $(1)$ becomes
$$
\sqrt{a^2-a\cdot b+b^2}+\sqrt{b^2-b+1}+\sqrt{a^2-a+1} = a+b+1
$$
How can I complete the solution from this point?
| Hint:
$$a^2+b^2\ge 2ab$$
$$\implies 4a^2+4b^2-4ab\ge a^2+b^2+2ab$$
$$\implies \sqrt{a^2+b^2-ab}\ge \frac{a+b}{2}$$
With equality iff $a=b$
You see where this is going?
| {
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"source": "stackexchange",
"question_score": "7",
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How to solve the inequality $\frac {5x+1}{4x-1}\geq1$ Please help me solve the following inequality.
\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}
I have tried the following method but it is wrong. Why?
\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}
Thank you for your attention
| As Carlos Eugenio Thompson Pinzón has commented, we need $\displaystyle4x-1\ne0$
Method $1:$
If $\displaystyle4x-1\ne0, (4x-1)^2>0$
Multiplying either sides of $\displaystyle\frac{5x+1}{4x-1}\ge1$ by $(4x-1)^2$
we get $\displaystyle(5x+1)(4x-1)\ge(4x-1)^2$
$\displaystyle\implies (4x-1)\{5x+1-(4x-1)\}\ge0 $
$\displaystyle\implies \left(x-\frac14\right)\{x-(-2)\}\ge0$
We know if $(x-a)(x-b)\ge0$ where $a<b$ either $x\le a$ or $x\ge b$
Method $2:$
As we know $a\ge b\implies \begin{cases} ac\ge bc &\mbox{if } c> 0 \\
ac\le bc &\mbox{if } c< 0 \end{cases}$
If $\displaystyle 4x-1>0\iff x>\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\ge 4x-1\iff x\ge-2$
$\displaystyle\implies x>\frac14$ is one of the solutions
If $4x-1<0\iff x<\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\le 4x-1\iff x\le-2$
$\displaystyle\implies x\le -2$ is the other solution
| {
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} |
Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.
| Without words:
$$\begin{align}
\left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1.
\end{align}$$
| {
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If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:
$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$
for some integers $m, n$
$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$
$$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$
Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.
For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.
$$\implies (p + q)^2 = \frac{m^2}{n^2}$$
$$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$
Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.
Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.
$$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$
is rational as well.
So for all $i$,
$$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$
and
$$a^{\frac{i}{3}}b^{\frac{i}{3}}$$
are rational.
I know I'm really close to the answer, but it somehow just keeps slipping through my fingers.
| As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so
$$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$
Combining this with $p+q\in \Bbb Q$, we are done.
| {
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"url": "https://math.stackexchange.com/questions/530778",
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"source": "stackexchange",
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How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.
My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't
| IN $\bigtriangleup$ABC we take AB=c;AC=b &BC=a
*
*NOW $a^2$ = $6^2$+$7^2$ -2.6.7.cos$\angle$BPC
*NOW cos$\theta$ $\geq$ -1$\longrightarrow$ -cos$\theta$$\leq$1. from this we get $a^2$$\leq$13
*$c^2$= $10^2$+$6^2$ -2.10.6.cos$\angle$APB.similarly from this we get [applying the inequality] c$\leq$16
*now the area of $\bigtriangleup$ABC=$\frac{1}{2}$acsin60=$\sqrt[2]{3}$.$\frac{1}{4}$.ac$\leq$$\sqrt[2]{3}$.$\frac{1}{4}$.16. 13=52$\sqrt[2]{3}$
-
| {
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Calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using pen and paper How can you calculate $\sum\limits_{n=1}^\infty (n-1)/10^n$ using nothing more than a pen and pencil? Simply typing this in any symbolic calculator will give us $1/81$. I could also possibly find this formula if I was actually looking at given numbers but I have never tried working backwards. By backwards, I mean to be given the summation formula and determine the convergent limit for it (if it exists).
So supposing that we did not know the limit for this summation formula was $1/81$ and that we do not have any software for assistance, how do we calculate this summation formula without having to take this summation to infinity?
| We know that
$$
\begin{align}
A&=\hphantom{1+}\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\
10A&=1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots\\[4pt]
9A&=1
\end{align}
$$
So $\frac1{10}+\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\dots=A=\frac19$.
Likewise,
$$
\begin{align}
\color{#C00000}{\frac1{10}}A&=\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^2}}A&=\hphantom{\frac1{10^2}+}\frac1{10^3}+\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^3}}A&=\hphantom{\frac1{10^2}+\frac1{10^3}+}\frac1{10^4}+\frac1{10^5}+\dots\\
\color{#C00000}{\frac1{10^4}}A&=\hphantom{\frac1{10^2}+\frac1{10^3}+\frac1{10^4}+}\frac1{10^5}+\dots\\
&\vdots\quad\text{summing the equations above}\\
\color{#C00000}{A}\hphantom{A}A&=\frac1{10^2}+\frac2{10^3}+\frac3{10^4}+\frac4{10^5}+\dots
\end{align}
$$
So $\frac1{10^2}+\frac2{10^3}+\frac3{10^4}+\frac4{10^5}+\dots=A^2=\frac1{81}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding minimum $\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2}$ I would appreciate if somebody could help me with the following problem
Q. Finding minimum $f(x)$
$$f(x)=\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2}$$
| Hint: Since
$$f(x)=\sqrt{x^2-3x+3}+\sqrt{x^2-3x+2} = \sqrt{\left(x-\dfrac32\right)^2 + \dfrac34} + \sqrt{\left(x-\dfrac32\right)^2 - \dfrac14},$$
we have $\left(x-\dfrac32\right)^2 + \dfrac34 \ge \dfrac14+\dfrac34 = 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?
| Suppose that $4k+1 = a^2$ and $9k+1 = b^2$, where $a, b$ are positive integers.
Then, we get that $ 9a^2 - 4b^2 = 5 $.
Since we have $(3a-2b)(3a+2b) = 5$, and both terms are integers,$3a+2b > 0$, $3a+2b>3a-2b$, so we must have $3a+2b = 5, 3a-2b = 1$. This gives us $a=1, b= 1$ and hence $k=0$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find series $\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt[m]{n}]}}{n^a}$?
let $m$ is give a positive integers,
Determine for which values of $a$,the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}$$ converges
where $[x]$ is the largest integer not greater than $x$.
and I have see this not hard problem :show that
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges,
solution: note
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}=\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}\right)$$
and we have
let $$a_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}$$
$$0<a_{n}<\dfrac{2n+1}{n^2}<\dfrac{3}{n}\longrightarrow 0$$
and
$$a_{n}-a_{n+1}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}+\dfrac{1}{n^2+1}-\dfrac{1}{(n+1)^2+1}+\cdots+\dfrac{1}{n^2+2n}-\dfrac{1}{(n+1)^2+2n}-\left(\dfrac{1}{(n+1)^2+2n+1}+\dfrac{1}{(n+1)^2+2(n+1)}\right)$$
so
$$a_{n}-a_{n+1}=(2n+1)\left(\dfrac{1}{n^2(n+1)^2}+\cdots+\dfrac{1}{(n^2+2n)(n^2+4n+1)}\right)-\dfrac{1}{n^2+4n+2}-\dfrac{1}{n^2+4n+3}$$
so
$$a_{n}-a_{n+1}>\dfrac{(2n+1)^2}{(n^2+2n)(n^2+4n+1)}-\dfrac{2}{n^2+4n+1}=\dfrac{2n^2+1}{(n^2+2n)(n^2+4n+1)}>0$$
so
$$\sum_{n=1}^{\infty}(-1)^na_{n}=\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges
But for my problem,I can't
My try:
$$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}=\sum_{n=1}^{\infty}(-1)^n\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$
let
$$b_{n}=\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$
then how prove
$$b_{n}\longrightarrow 0$$and $$b_{n}>b_{n+1}$$
| Let's start with the observation that if $a > 1$, then the sum converges absolutely, and if $a \leqslant 0$, the terms of the sum don't converge to $0$, hence the sum cannot converge. So in the following, we always assume $0 < a$.
The case $m = 1$ is immediate by Leibniz' criterion, $\sum\limits_{n=1}^\infty \dfrac{(-1)^n}{n^a}$ converges if and only if $a > 0$.
Generally, as you found, we must consider the segments
$$b_n = \sum_{k=n^m}^{(n+1)^m-1} \frac{1}{k^a}.$$
We do need that $b_n \to 0$, but we don't necessarily need that $b_n > b_{n+1}$. If we can write $b_n = c_n + d_n$, where $c_n \downarrow 0$, and $\sum_n d_n$ converges absolutely, that suffices to show the (conditional) convergence of the original sum. Now,
$$b_n = \underbrace{\int_{n^m}^{(n+1)^m} \frac{dt}{t^a}}_{c_n} + \underbrace{\sum_{k=n^m}^{(n+1)^m-1} \int_{k}^{k+1}\left(\frac{1}{k^a} - \frac{1}{t^a}\right)\,dt}_{d_n}.$$
We have $\frac{1}{k^a} - \frac{1}{t^a} = \frac{a(t-a)}{\xi^{a+1}}$ for some $\xi \in [k,t]$ by the mean value theorem, so
$$0 < \int_k^{k+1} \frac{1}{k^a} - \frac{1}{t^a}\,dt < \frac{a}{2k^{a+1}},$$
whence $$\sum_{n=1}^\infty \lvert d_n\rvert < \frac{a}{2}\sum_{k=1}^\infty \frac{1}{k^{a+1}} < \infty.$$
It remains to consider $c_n$. For $a = 1$, we get
$$c_n = \int_{n^m}^{(n+1)^m} \frac{dt}{t} = \log \left((n+1)^m\right) - \log \left(n^m\right) = m\left( \log (n+1)-\log n\right) = m\log\left(1+\frac1n\right).$$
It's evident that $c_n \to 0$ and $c_n > c_{n+1}$ in this case.
For $0 < a < 1$, we get
$$c_n = \int_{n^m}^{(n+1)^m} \frac{dt}{t^a} \overset{t=u^m}{=} \int_n^{n+1} \frac{mu^{m-1}}{u^{ma}}\,du = m\int_n^{n+1} u^{m(1-a)-1}\,du,$$
and we see that $c_n \to 0$ if and only if $m(1-a)-1 < 0 \iff a > \dfrac{m-1}{m}$. $c_n$ is easily seen to be monotonic (constant for $m(1-a)-1 = 0$, increasing for $m(1-a)-1 > 0$, and decreasing in the convergent case).
Wrapping up:
$$\sum_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt[m]{n}\rfloor}}{n^a}$$
converges if and only if $a > \dfrac{m-1}{m}$.
| {
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$\int_{2}^{\infty}\frac{dx}{x^2-1}$ converge or diverge How can I find whether the integral converge or diverge.
I did the following.
$\int_{2}^{\infty}\frac{dx}{x^2-1}$
I did the following
$\frac{a}{x-1}+\frac{b}{x+1}$
$A(x+1)+B(x-1)=1$
$t\rightarrow\infty$
$\int_{2}^{t}\frac{1/2}{x-1}-\frac{1/2}{x+1}$
$\frac{1}{2}[\ln(x-1)-\ln(x+1)$
$\ln(t-1)-ln(t+1)-\ln(2-1)-\ln(2+1)$
Then I got
$\infty-\infty-ln(1)-ln(2)$
Since I got an intermediate form I did the hospital rule.
$\frac{\ln(t+1)}{1/\ln (t-1)}$
then I took the derivative
$\frac{\frac{1}{t+1}}{\frac{1}{1/t-1}}$
Thus I got
$\frac{1}{(t-1)(t+1)}$ as t approach infity zero
thus it is convergent at at $ln(1)-ln(3)$
| One can prove convergence, without evaluating, by noting that for $x\ge 2$, we have $x^2-1\gt x^2-\frac{x^2}{2}=\frac{x^2}{2}$. Thus
$$0\lt \frac{1}{x^2-1}\lt \frac{2}{x^2}.$$
It is a standard fact that $\int_2^\infty \frac{1}{x^2}\,dx$ converges, so our integral does.
| {
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What are the generators of $\mathbb{Z}_9^*$? I am to understand that $\mathbb{Z}_9^*$ is cyclic because $9=3^2$, where $3^2$ is of the form $p^{\alpha}$, with $p$ an odd prime... but I can't find any generators for the set...
$|\mathbb{Z}_9^*|=6$, and $\mathbb{Z}_9^*=\{\overline{1},\overline{2},\overline{4},\overline{5},\overline{7},\overline{8}\}$
Using a table:
\begin{array}{c|cccccc}
\cdot & 1 & 2 & 4 & 5 & 7 & 8 \\\hline
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
2 & 2 & 4 & 7 & 5 & 2 & 4 \\
4 & 4 & 7 & 4 & 7 & 4 & 7 \\
5 & 5 & 7 & 4 & 2 & 1 & 5 \\
7 & 7 & 4 & 7 & 4 & 7 & 4 \\
8 & 8 & 1 & 8 & 1 & 8 & 1 \\
\end{array}
I'm not sure if I'm misunderstanding the definition of a generator, but since I'm assuming that $\mathbb{Z}_9^*$ is cyclic, it should have $\varphi(\varphi(9))=2$ generators... but I can't find a single one.
| When you are looking for generators, a trick to reduce the work is to note that, once you know that $a$ is not a generator, then neither is any power $a^i$. So in your case you'd start by testing $2$ rather than $2^2=4$ or $2^3=8$.
A good way to check $a$ is a generator module ${\mathbb Z}_{p^k}^\times$ is to check whether $a^{\phi(p^k) / q} \not\equiv 1(\bmod. p^k)$ for each prime $q$ dividing $\phi(p^k)$. In thi scase, you can check $2$ is a generator.
More generally, $2$ is a generator of ${\mathbb Z}_{3^k}^{\times}$ for each $k=1,2,3\ldots$. The proof goes as follows:
Lemma: $2^{2\times 3^{k-1}} = 3^k \times q + 1$ for some $q\in {\mathbb Z}$ with $3 \not | q$.
Proof: For $k=1$ this is trivial, indeed $2^2 = 3\times 1 + 1$ and clearly $3\not | 1$.
Assume the result to hold for some $k$, now for $k+1$ we note that:
\begin{align} 2^{2^\times 3^k} = \left(2^{2\times 3^{k-1}} \right)^3 = (3^k \times q + 1)^3 &= 3^{3k} q^3 + 3 \times 3^{2k} q^2 + 3 \times 3^k q + 1 \end{align}
which is of the form $3^{k+1} q^\prime + 1$ where $q^\prime = 3^{2k-1} q^3 + 3^{k}q^2 + q$, now since $3 \not | q$ and $k\geq 1$, we get that $3\not | q^\prime$, finishing the induction. $\square$
Now, with this lemma, note that if the order of $2$ modulo $3^k$ were not $\phi(3^k) = 2 \times 3^{k-1}$, then it'd either divide $2\times 3^{k-2}$ or $3^{k-1}$ (by taking out one prime factor from each, this is the "least we can take out"). The latter is impossible since $2^{3^{k-1}} \equiv -1 (\bmod. 3)$ and so $2^{3^{k-1}} \equiv 1 (\bmod. 3^k)$ is impossible since it'd imply $2^{3^{k-1}} \equiv 1 (\bmod. 3)$.
So, suppose the order of $2$ were $2\times 3^{k-2}$, we showed a moment ago that $2^{2\times 3^{k-2}} = 3^{k-1} \times q + 1$ where $3\not | q$, so $2^{2\times 3^{k-2}} \equiv 1(\bmod. 3^{k})$ is impossible since it would imply $3^{k-1} \times q \equiv 0(\bmod. 3^k)$ yet $3\not | q$.
| {
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$x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.
Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Suppose $x^2+xy+y^2=a^2$ and $x^2-xy+y^2=b^2$. Then $2xy=a^2-b^2=(a+b)(a-b)$, so one of $x$ and $y$ is even, but then I am stuck.
| The question amounts to showing that the elliptic curve
$Y^2 = X^2+X+1$, $Z^2 = X^2-X+1$
has no rational points other than those with $X=0$ and $X=\infty$.
That can be done by elementary means but requires a Fermat-style descent.
According to Dickson's History of the Theory of Numbers
(Volume II, Chapter XVI, bottom of page 481, reference to footnote 119 on p.480),
the result was proved in 1876 by Genocchi.
[In modern language, a Weierstrass form for this curve is
$$y^2 = x^3+x^2-24x+36 = (x-2)(x-3)(x+6),$$
with conductor $48$ and thus in the "Antwerp Tables"
(see curve 48C),
where we find that it has Mordell-Weil group
$({\bf Z}/2{\bf Z}) \oplus ({\bf Z}/4{\bf Z})$.]
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I determine this function? If i know that$$ f(3)-f(-1/2) = 7$$ and $$ f(2)-f(1/2) = 3 $$ and $$f(3/2)-f(-2) =7$$ How can I determine the function?
| $$
f(x) = \left\{
\begin{array}{ll}
110004 & \quad x = -2 \\
68 & \quad x = -\frac{1}{2} \\
4 & \quad x = \frac{1}{2} \\
110011 & \quad x = \frac{3}{2} \\
7 & \quad x = 2 \\
75 & \quad x = 3 \\
0 & \quad x\neq-2,-\frac{1}{2},\frac{1}{2},\frac{3}{2},2,3
\end{array}
\right.
$$
| {
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Solving trigonometry equation Please help me understand how to solve this for $0\leq x\leq360 $
I seem to have a problem with equations with powers.
$$3\sin^2 x-3\cos^2x+\cos x-1=0 $$
thinking that I would start by simplifying:
$$3 (\sin^2 x- \cos^2x)+\cos x - 1=0 $$
How I wish the equation in the bracket was in form $\sin^2 x+ \cos^2x$ which is equal to 1.
I also tried to substitute $\sin^2 x=1- \cos^2x$
| $3\sin^2{x} - 3\cos^2{x} + \cos{x} - 1 = 0$
$6\cos^2{x} - \cos{x} - 2 = 0$
$\cos{x} = -\frac{1}{2}\:\:\:\cos{x} = \frac{2}{3}$
| {
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $2+\sqrt{2}$ ... how is it ?
| $2ab = (a+b)^2 - (a^2+b^2)$, hence we need to find the minimum value of $a + b + \frac{2}{(a+b)^2 - (a^2+b^2)}$ $ = a + b + \frac{2}{(a+b)^2 - 1}$.
Let us find the maximum value (k) of $a+b$ given $a^2+b^2=1$. Thinking graphically, when $a+b = k$ is tangent to the circle, its gradient is $-1$ and hence the gradient of the normal is $1$, or $a=b$. Hence $k = \frac{1}{\sqrt2} + \frac{1}{\sqrt2} = \sqrt{2}$, which gives $\sqrt{2} + \sqrt{2} + \frac{2}{2-1} = 2 + \sqrt{2}$ as the answer.
This relies on the fact that $f(x) = x + \frac{2}{x^2 - 1}$ is decreasing for $x \in (1, \sqrt{2}]$. Partial fractions gives $f(x) = x + \frac{1}{x-1} - \frac{1}{x+1}$ and taking the derivative shows that $f'(x) < 0$ in the given interval. There might be a way without calculus.
| {
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Finding the limit of $(\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$ Any Ideas/Hints?
$$\lim\limits_{n \to \infty } (\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$$
| $$\lim_{n\to\infty}{\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot 1$$
$$=\lim_{n\to\infty}\left({\sqrt[3]{n^3+n^2}}-\sqrt[3]{n^3+1}\right)\cdot \frac{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}$$
$$=\lim_{n\to\infty}\frac{{\sqrt[3]{n^3+n^2}^3}-\sqrt[3]{n^3+1}^3}{{{\sqrt[3]{(n^3+n^2)^2}}++\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}}=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{(n^3+n^2)^2}}+\sqrt[3]{(n^3+n^2)(n^3+1)}+\sqrt[3]{(n^3+1)^2}}}$$
$$=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{n^6+2n^5+n^4}}+\sqrt[3]{n^6+n^5+n^3+n^2}+\sqrt[3]{n^6+2n^3+1}}}$$
$$=\lim_{n\to\infty}\frac{n^2-1}{{{\sqrt[3]{n^6(1+2\frac{1}{n}+\frac{1}{n^2})}}+\sqrt[3]{n^6(1+\frac{1}{n}+\frac{1}{n^3}+\frac{1}{n^4})}+\sqrt[3]{n^6(1+2\frac{1}{n^3}+\frac{1}{n^6})}}}$$
$$=\lim_{n\to\infty}\frac{n^2-1}{\sqrt[3]n^6+\sqrt[3]n^6+\sqrt[3]n^6}=\lim_{n\to\infty}\frac{n^2-1}{n^2+n^2+n^2}=\frac{1}{3}$$
| {
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Find all positive integers m, n, p such that $(m+n)(mn+1)=2^p$ Find all positive integers m, n, p such that
$$(m+n)(mn+1)=2^p$$
Please give me some hints
Thanks
| We have $m+n=2^a$ and $mn+1=2^b$ with $a+b=p$. First note that both $m$ and $n$ need to be odd.
First case: Suppose $m=1$ or $n=1$, then the other is equal to $2^{p/2}-1$ and this is a solution for even $p$.
Now suppose $m>1$ so $b>a>1$. Adding the two equations above we get
$(m+1)(n+1)=mn+1+m+n=2^a(2^{b-a}+1)$. Let $m+1=2^xw$ and $n+1=2^zy$ with $w,y$ odd, $x,z>0$ and $x+z=a$. Then $2^a=2^xw+2^zy-2$, so one of $x$ or $z$ needs to be $1$ (if both were $>1$ the RHS would be $\equiv 2\mod 4$.
Case $x=1$: We have $x+z=a$, so $z=a-1$. The equation for $2^a$ becomes
$$
2^a=2w+2^{a-1}y-2\geq 2^{a-1}y.
$$
This implies $y=1$, since it is odd. We get $m-1=2w-2=2^{a-1}$ and $n+1=2^z=2^{a-1}$ and indeed this gives a solution.
Case $z=1$: The same just with $m$ and $n$ swapped.
So the only solutions are $m=1$ and $y=2^{p/2}-1$ and
$(m,n)=(2^k\pm 1,2^k\mp 1)$
| {
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binomial theorem with conditional probability There are four men in a room, 1 pair of brothers, and 2 unrelated men. The
probability that any man has blood-group X is
1
4
. The probability that if one
brother has blood-group X, the other brother also has X is
3
4
, otherwise the
blood-groups are independent. Find the probability that exactly 2 men in the
room have blood-group X.
| $$\frac{3}{4}\cdot\left({3 \choose 2}\cdot\left(\frac{1}{4}\right)^2\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{3}{4}\cdot \left({2\choose 1}\frac{1}{4}\cdot\frac{3}{4}\right)+\frac{1}{4}\cdot\left(\frac{1}{4}\right)^2\right)=\frac{23}{128}$$
Chance of 1$^{\mathrm{st}}$ brother having X and not having X on the right and left, respectively.
| {
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How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + 2y^2 + 4y + 2. $$
I've been thinking about maybe somehow making $c$ equal $2y^2 + 4y + 2$?
The answer from the answerbook is: $$ x^2 - 3x + 2y^2 + 4y + 2 = \left( x - \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 - \dfrac {9}{4}. $$
| Note that the expression can be written as $f(x)+g(y)$, where $f(x) = x^2-3x+2 $ and $g(y) = 2y(y+2)$.
The two parts are independent of each other, so we can minimize them separately.
Setting $f'(x) = 0$ gives $2x-3 = 0$, so we see that $x = \frac{3}{2}$ is the minimizer. Noting that $(x-\frac{3}{2})^2 = x^2-3x+\frac{9}{4}$, we see that we can write $f(x) = (x-\frac{3}{2})^2 - \frac{1}{4}$, and that since the square term is always non-negative, we have $f(x) \ge - \frac{1}{4}$, and $f(\frac{3}{2}) = - \frac{1}{4}$.
Similarly, setting $g'(y) = 0$ gives $y=-1$, and repeating the above, we note that $(y+1)^2 = y^2+2y+1$, so we can write $g(y) = 2(y+1)^2 -2$, and so $g(y) \ge -2$ and $g(-1) = -2$.
Putting these together, we have
$f(x)+g(y) = (x-\frac{3}{2})^2 - \frac{1}{4} + 2(y+1)^2 -2 =
(x-\frac{3}{2})^2 + 2(y+1)^2 - \frac{9}{4}$.
Then $f(x)+g(y) \ge - \frac{9}{4}$, and $f(\frac{3}{2}) + g(-1) = - \frac{9}{4}$.
| {
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If : $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1\dots$ Problem :
If $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1$
Then find the value of $\sin^2\alpha + \sin^2\beta +\sin^2\gamma$
Please suggest how to proceed in such problem.
| HINT:
If $\sin^2\alpha=x$ etc,
$\displaystyle\tan^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}=\frac x{1-x}$
Just simplify to find $x+y+z=1$ assuming $(1-x)(1-y)(1-z)\ne0$ i.e, $\tan\alpha$ etc. are finite
| {
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How to solve the equation $x^2 + 4 |x| - 4 = 0$ How do I get the value of x from the equation that is provided
| Putting $x=a+ib$ where $a,b$ are real
we get $\sqrt{a^2+b^2}=4-(a+ib)^2=4+b^2-a^2-2abi$
Equating the imaginary parts $ b=0$
Now for real $x,$
$$|x|= \begin{cases} +x &\mbox{if } x\ge0 \\
-x & \mbox{if } x<0 \end{cases}$$
| {
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Limit of $x_1=1, x_{n+1}=x_n+\frac1{x_n^2}$
Given that $x_1 = 1$ and $x_{n+1}=x_{n}+ 1/x_{n}^{2}$. Find the limit of the sequence.
Let $ c $ be the limit of the sequence, then $ c=c+\frac{1}{c^2} $, that means $ \frac{1}{c^2}=0 $. That can't be like that.What is wrong???
| Notice $$x_{n+1}^3 = x_n^3 (1 + \frac{1}{x_n^3})^3 = x_n^3 + 3 + \frac{3}{x_n^3} + \frac{1}{x_n^6}$$
We have $x_{n+1}^3 - x_n^3 > 3$ for all $n$ and hence we can bound $x_n$ from below
$$x_n^3 \ge x_1^3 + 3(n-1) = 3n-2 \quad\implies\quad x_n \ge \sqrt[3]{3n-2}$$
As a result, the sequence $x_n$ diverges.
| {
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verify this limit of radical function $$\lim_{x \to -\infty} \frac{\sqrt[3]{x}- \sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}} $$
$$= \lim_{x \to -\infty} \frac{x^{1/3}-x^{1/5}}{4x^{1/3}+x^{1/5}} $$
divide numerator and denominator by highest power of $x$ in the denominator, $x^{1/3},$ to get:
$$= \lim_{x \to -\infty} \frac{1 - x^{-2/15}}{4+x^{-2/15}}$$
Which gives an answer of $\frac{1}{4}.$ Is this correct? Did I show work correctly?
| Your work is correct
$$\lim_{x\to-\infty}\frac{\sqrt[3]{x}-\sqrt[5]{x}}{4\sqrt[3]{x}+\sqrt[5]{x}}=$$
$$\lim_{x\to-\infty}\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{4\sqrt[15]{x^5}+\sqrt[15]{x^3}}=$$
$$\lim_{x\to-\infty}\frac{\frac{\sqrt[15]{x^5}-\sqrt[15]{x^3}}{\sqrt[15]{x^5}}}{4\frac{\sqrt[15]{x^5}+\sqrt[15]{x^3}}{\sqrt[15]{x^5}}}=$$
$$\lim_{x\to-\infty}\frac{1-\sqrt[15]{1/x^2}}{4(1+\sqrt[15]{1/x^2})}=1/4$$
| {
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How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem find this minimum,then
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^2}\Longrightarrow f'(1)=-\dfrac{1}{2}$$
so
$$a=-\dfrac{1}{2},b=1$$
and
$$\dfrac{1}{1+x^2}+\dfrac{1}{2}x-1=\dfrac{x(x-1)^2}{(x^2+1)^2}\ge 0$$
then
$$\dfrac{1}{1+x^2}\ge-\dfrac{1}{2}x+1$$
so
$$\sum_{cyc}\dfrac{1}{1+a^2}\ge\sum_{cyc}-\dfrac{1}{2}a+1=-\dfrac{1}{2}(a+b+c)+3=\dfrac{3}{2}$$
But for maximum,I can't prove it.Thank you
| By symmetry, it just suffices to look at the case $0 \le a \le b \le c$ where $a + b + c = 3$.
Let $f(x) = \frac{1}{1+x^2}$ and $F(a,b,c) = f(a)+f(b)+f(c)$. Notice
$$\frac{d^2}{dx^2} f(x) = \frac{2(3x^2-1)}{(1+x^2)^3} \longrightarrow
\begin{cases} > 0,&\text{ for } x > \frac{1}{\sqrt{3}}\\< 0,&\text{ for } x <\frac{1}{\sqrt{3}}\end{cases}$$
The function $f(x)$ is concave on $[0,\frac{1}{\sqrt{3}}]$ and convex on $[\frac{1}{\sqrt{3}},3]$.
Since $a + b + c = 3$, we know $c \ge 1$. This implies $c \in (\frac{1}{\sqrt{3}}, 3]$ where $f(x)$ is convex. If $b$ also belongs to this interval, then we can increase the value of $F(a,b,c)$ by decreasing $b$ and increasing $c$ symmetrically w.r.t their mean $\frac{3-a}{2}$ until either $b$ hits $\frac{1}{\sqrt{3}}$ or $c$ hits $3$. It is easy to see $b$ will hit $\frac{1}{\sqrt{3}}$ first. From this, we can conclude when $b > \frac{1}{\sqrt{3}}$,
$$F(a,b,c) \le F(a, \frac{1}{\sqrt{3}}, 3-a-\frac{1}{\sqrt{3}})$$
If $a > \frac{1}{\sqrt{3}}$, the new $b$ will be smaller than $a$. We can flip the roles of $a$ and $b$ and repeat above trick. At the end, we will find a pair of $a', b'
\in [0, \frac{1}{\sqrt{3}}]$ such that
$$F(a,b,c) \le F(a',b',c')\quad\text{ where }\;\;c' = 3 - a'-b'$$
Since $f(x)$ is concave there, we have
$$F(a',b',c') \le F(\frac{a'+b'}{2},\frac{a'+b'}{2},c')$$
This means in order to find the optimal $(a,b,c)$ which maximize $F(a,b,c)$, one only need to look at those points of the form
$$(a,b,c) = (x,x,3-2x)\quad\text{ where } x \in [0,\frac{1}{\sqrt{3}}]$$
For such points, $F(a,b,c)$ reduces to
$$F(x,x,3-2x) = \frac{2}{1+x^2} + \frac{1}{1 + (3-2x)^2}$$
Let us call this function as $G(x)$. Notice
$$\frac{dG(x)}{dx} = \frac{4(3-2x)}{(1+(3-2x)^2)^2} - \frac{4x}{(1+x^2)^2}$$
We have
$$G'(0) = \frac{3}{25}> 0\quad\text{ and }\quad G'(\frac{1}{\sqrt{3}}) = \frac{1944\sqrt{3}-3645}{1588\sqrt{3}-2448} \sim -0.91866538126469 < 0$$
So the optimal $(a,b,c) = (x,x,3-2x) $ occurs for a $x \in (0,\frac{1}{\sqrt{3}})$.
We can determine the corresponding value of $x$ by solving
$$G'(x) = 0\quad\iff\quad 6x^4-27x^3+49x^2-33x+1 = 0$$
The polynomial in R.H.S has 2 real roots, only one of it $\sim 0.031776261136413$ falls inside the interval $[0,\frac{1}{\sqrt{3}}]$. So the maximum value of $F(a,b,c)$ is about
$$\max F(a,b,c) \sim G(0.031776261136413) \sim 2.101903255548146 < \frac{11}{5}$$
| {
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problem with comparing inequality Hello I just wondering if it's possible to prove this inequality: there are positive, various $a,b,c$ and
$ \frac{3a-b}{3} \ge x \ge \frac{3(a^2-b^2)}{3a+b}$
$ \frac{3b-c}{3} \ge y \ge \frac{3(b^2-c^2)}{3b+c}$
$ \frac{3c-a}{3} \ge z \ge \frac{3(c^2-a^2)}{3c+a}$
I want to prove that $x+y \ge z$ when we add we get
$\frac{3a+2b-c}{3} \ge x+y \ge \frac {3(a-b)(a+b)}{3 a+b}+\frac {3(b-c)(b+c)}{3 b+c} $ but further I don't know how to compare it with $z$
| Hint:
To prove that $x+y\ge z$ we need to prove that minimum value of $x$ plus minimum value of $y$ is $\ge$ maximum value of $z$.
So you have:
$x=\frac{3(a^2-b^2)}{3a+b}; y=\frac{3(b^2-c^2)}{3b+c}; z=\frac{3c-a}{3}$
Just prove that $\frac{3(a^2-b^2)}{3a+b}+\frac{3(b^2-c^2)}{3b+c}\ge\frac{3c-a}{3}$
| {
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Summation formula for $x^2+x$ Since I learned easier ways of calculating summations I've been curious as to how I could find formulas for as many equations as possible. I came across the equation $x^2+x$, I've spent quite some time on this problem and could not find a solution. If someone has maybe already done this or have any suggestions on how I could get the formula that would be greatly appreciated.
Example of another summation with a equation:
$\sum\limits_{i=1}^n$ = $x^2$
Equation to solve this is $\frac{n(n+1)(2n+1)}{6}$
| Note that
$$k^2+k=\frac{1}{3}\left((k+1)^3-k^3-1\right).$$
Thus our sum $\sum_{k=1}^n (k^2+k)$ is equal to
$$\frac{1}{3}\left((2^3-1^3-1)+(3^3-2^3-1)+(4^3-3^3-1)+(5^3-4^3-1)+\cdots +((n+1)^3-n^3-1)\right).$$
Observe the nice almost total cancellations. We end up with
$$\frac{1}{3}\left((n+1)^3-1^3-n\right).$$
Remarks: $1.$ Since we know $\sum_1^n k$, this gives a way to derive the formula for $\sum_1^n k^2$.
$2.$ The sums $\sum k(k+1)$, $\sum k(k+1)(k+2)$, $\sum k(k+1)(k+2)(k+3)$ and so on are nice, much nicer than $\sum k^2$, $\sum k^3$, $\sum k^4$ and so on.
| {
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Find $ S_{20} $ if $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n Define $ a_{n} $ is a sequence and all terms of $ a_{n} $ are positive. $ S_{n} $ is the summation of the first n terms. If $ \frac{ a_{n} +2}{2} = \sqrt{2S_{n}} $ for all integer n, then find $ S_{20} $
| $S_n=a_1+a_2+\cdots+a_n=S_{n-1}+a_n=S_{n-1}+2(\sqrt{2S_n}-1)\Rightarrow$ $S_{19}+2(\sqrt{2S_{20}}-1)=S_{20}$. Also we have $S_{20}-S_{19}=a_{20}$. Adding these two we get
$2\sqrt{2S_{20}}=a_{20}\Rightarrow S_{20}=\frac{a_{20}^2}{8}$.
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"url": "https://math.stackexchange.com/questions/568785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.
My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases}
$$
So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?
| We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation.
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd then, $$x^2+y^2+z^2\equiv(0+0+1)\mod4 ; 2xyz \equiv 0 \mod4$$ So, the only possibility is that all are even.
Let x = 2X , y = 2Y ,z = 2Z. Then, $$4X^2 +4Y^2 +4Z^2 = 16XYZ $$ $$\implies X^2 + Y^2 + Z^2 = 4XYZ$$ The same argument goes show that X,Y,Z are even. The process can be continued indefinitely.
This is possible only when x = y = z = 0. So ,there exist no other solution other than this.
| {
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"source": "stackexchange",
"question_score": "13",
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Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
| Actually you have shown that there is a number $b$ that if you divide it by $15$ you will get $16^n-1$. But this does not mean that $16^n-1$ is divisible by $15$. For example $23=\frac{345}{15}$ but it does not mean that $23$ is divisible by $15$ . To show that $a$ is divisible by $b$ you have to show that $a=kb$ ($k$ is an integer, just like $a,b$)not $a=\frac kb$. Adi Dani Hinted for a solution by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating the integral $\int_0^{\frac{\pi}{2}}\log\left(\frac{1+a\cos(x)}{1-a\cos(x)}\right)\frac{1}{\cos(x)}dx$ How can I evaluate the following integral?
$$
\int_0^{\pi/2}
\log\left(\frac{1 + a\cos\left(x\right)}{1 - a\cos\left(x\right)}\right)\,
\frac{1}{\cos\left(x\right)}\,{\rm d}x\,,
\qquad\left\vert\,a\,\right\vert \le 1$$
I tried differentiating under the integral with respect to the parameter $a$, and I also tried expanding the log term in a Taylor series and then switching the order of integration and summation. I ran into difficulties with both approaches.
| $$\begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1+ b\cos x}\right)\frac{1}{\cos x}dx &= \int_{0}^{\pi/2} \int_{b}^{a} \frac{1}{1+t \cos x} \ dt \ dx \\ &= \int_{b}^{a} \int_{0}^{\pi/2}\frac{1}{1+t \cos x} \ dx \ dt \end{align}$$
Let $ \displaystyle u = \tan \frac{x}{2}$.
$$\begin{align} &= \int_{b}^{a} \int_{0}^{1} \frac{1}{1+ t \left(\frac{1-u^{2}}{1+u^{2}} \right)} \frac{2}{1+u^{2}} \ du \ dt \\ &= 2 \int_{b}^{a} \int_{0}^{1} \frac{1}{1+t} \frac{1}{1+ \frac{1-t}{1+t} u^{2}} du \ dt \end{align}$$
Let $\displaystyle w = \sqrt{\frac{1-t}{1+t}} u $.
$$ \begin{align} &= 2 \int_{b}^{a} \int_{0}^\sqrt{\frac{1-t}{1+t}} \frac{1}{\sqrt{1-t^{2}}} \frac{1}{1+w^{2}} \ dw \ dt \\ &= 2 \int_{b}^{a} \frac{1}{\sqrt{1-t^{2}}} \arctan \sqrt{\frac{1-t}{1+t}}\ dt \\ &= \int_{b}^{a} \frac{\arccos t}{\sqrt{1-t^{2}}} \ dt \\ &= \frac{1}{2} \Big(\arccos^{2} (b)- \arccos^{2} (a)\Big) \end{align}$$
Then
$$ \begin{align} \int_0^{\frac{\pi}{2}}\log\left(\frac{1+a \cos x}{1-a \cos x}\right)\frac{1}{\cos x}dx &= \frac{1}{2} \Big(\arccos^{2} (-a)- \arccos^{2} (a)\Big) \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} - \arcsin (-a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big[ \Big(\frac{\pi}{2} + \arcsin (a)\Big)^{2} - \Big(\frac{\pi}{2} - \arcsin (a)\Big)^{2} \Big] \\ &= \frac{1}{2} \Big(2 \pi \arcsin a\Big) = \pi \arcsin a \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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show continuity of $\frac{xy}{\sqrt{x^2+y^2}}$ How to show this function's continuity?
My book's 'Hint' says $|xy| \leq \frac12(x^2+y^2)$ can be used.
$ f(x,y) = \left\{
\begin{array}{l l}
\frac{xy}{\sqrt{x^2+y^2}} & \quad , \quad(x,y)\neq(0,0)\\
0 & \quad , \quad(x,y)=(0,0)
\end{array} \right.$
| Here is another way to prove the continuity of $f(x,y)$ at $(0,0)$.
$$\left|\frac{xy}{\sqrt{x^2+y^2}}-0\right|={{|x||y|}\over{\sqrt{x^2+y^2}}} <{{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\over{\sqrt{x^2+y^2}}}=\sqrt{x^2+y^2} < \varepsilon \\\,\,(\text{where}\,\,\, \varepsilon \,\text{is a preassigned positive number})$$ if $x^2+y^2 < \delta^2,$ where $\delta =\varepsilon$. So,given any $\varepsilon >0, \exists \delta >0 $ such that $$|f(x,y)-f(0,0)|<\varepsilon \forall (x,y) \in x^2+y^2< \delta^2$$ and so $f(x,y)$ is continuous at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\right)^2}dx = \int_{0}^{\pi}\frac{1+\tan^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
$\displaystyle I = \int_{0}^{\pi}\frac{\sec^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
Now Let $\tan \frac{x}{2} = t$ and $\sec^2 \frac{x}{2}dx = 2dt$
$\displaystyle I = 2\int_{0}^{\infty}\frac{1}{(9+t^2)^2dt}$
Now I did not understand how can i solve after that
Help Required
Thanks
| HINT
Let $$u = \tan(\frac{x}{2})$$
Using the trig. identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$ (1)
$$u = \tan(\frac{x}{2}) = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
Squaring both sides $$u^2 \cdot \cos^2(\frac{x}{2}) = \sin^2(\frac{x}{2})$$
Using the trig. identity $\sin^2(x) + cos^2(x) = 1$ (2) This becomes
$$u^2 \cdot \cos^2(\frac{x}{2}) = 1 - \cos^2(\frac{x}{2})$$
$$u^2 \cdot \cos^2(\frac{x}{2}) + \cos^2(\frac{x}{2}) = 1 \implies \cos^2(\frac{x}{2}) \cdot (u^2 + 1) = 1 \implies \cos^2(\frac{x}{2}) = \frac{1}{(u^2 + 1)}$$
Using trig identity (2) again becomes
$$\sin^2(\frac{x}{2}) = 1 - \frac{1}{(u^2 + 1)} = \frac{u^2}{(u^2 + 1)}$$
Then
$$\cos(x) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = \frac{1}{(u^2 + 1)} - \frac{u^2}{(u^2 + 1)} = \frac{1 - u^2}{u^2 + 1}$$
Back to the original statement, $$u = \tan(\frac{x}{2})$$ inverse $\tan$ both sides to get
$$\tan^{-1}u = \frac{x}{2}$$
Then differentiate both sides and using trig. identity $\frac{d}{dx} \tan^{-1}(x) = \frac{1}{x^2 + 1}$
$$dx = \frac{2 \cdot du}{u^2 + 1}$$
Substitute these back into the integral and solve.
| {
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"url": "https://math.stackexchange.com/questions/574099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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why $x^3=11^3$ mod $5083$ has only one solution? Why $x^3=11^3$ mod $5083$ has only one solution? (The only answer is $x=11$)
I know it has at most 3 roots but how to find that there isn't another answer?
| As $5083=13\cdot 17\cdot 23$
We have $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{13}\ \ \ \ (1)$
As all primes have primitive roots, taking discrete logarithm wrt some primitive root $g$ of $13,$(say $2$)
$3$ind$\displaystyle_g \left(\frac x{11}\right)\equiv0\pmod {12}\ \ \ \ (2)$
Using Linear Congruence Theorem, $(2)$ hence $(1)$ have $(3,12)=3$ solutions namely, $4k\pmod{13}$ where $k=0,1,2$ of $(2)$
$\implies 2^{4k}\pmod{13},k=0,1,2$ of $(1)$
Similarly, $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{17}$ and $\displaystyle\left(\frac x{11}\right)^3\equiv1\pmod{23}$ should have unqiue solutions namely $1$
Using Chinese Remainder Theorem, we shall get three in-congruent solutions $\pmod{13\cdot 17\cdot 23}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Triangle inequality problem. Prove that from sections $x,y,z$ where
$x= \sqrt[3]{(a-b)^2(a+b)}, y=\sqrt[3]{(b-c)^2(b+c)}, z=\sqrt[3]{(c-a)^2(c+a)}$ and $a,b,c>0$, $a\neq\ b \neq c$
it is possible to construct a triangle.
I started the limitation $x,y,z$ from cauchy inequality, but I am not sure if inequality I need to prove is $x+y \ge z \ge |x-y|$ and how to prove it.
| $x^3+y^3-z^3+3xyz=(x+y-z)(x^2+y^2+z^2-xy+yz+xz)$
sinc $(x^2+y^2+z^2-xy+yz+xz) >0 \implies x^3+y^3-z^3+3xyz>0 \iff x+y-z>0$
$x^3+y^3-z^3+3xyz>0 \iff -(a-b)(b-c)(a+c+2b)+3xyz>0 \iff 27(a-b)^2(b-c)^2(c-a)^2(a+b)(b+c)(a+c)> (a-b)^3(b-c)^3(a+c+2b)^3 \iff 27(c-a)^2(a+b)(b+c)(a+c) >(a-b)(b-c)(a+c+2b)^3$
if$(a-b)(b-c)<0$, then it is proved.
in case $(a-b)(b-c)>0$, note $a,c$ is symmetry, WOLG,let $c$ is min {$a,b,c$},$a=c+u,b=c+v$
$ 27(c-a)^2(a+b)(b+c)(a+c) -(a-b)(b-c)(a+c+2b)^3=(32v^2-32uv+108u^2)c^3+(48v^3-24uv^2+84u^2v+108u^3)c^2+(24v^4+9u^2v^2+75u^3v+27u^4)c+4v^5+2uv^4-3u^2v^3+11u^3v^2+13u^4v $
$32v^2-32uv+108u^2>0$
$48v^3-24uv^2+84u^2v+108u^3>0$
$4v^5+2uv^4-3u^2v^3+11u^3v^2+13u^4v>0$
$\implies27(c-a)^2(a+b)(b+c)(a+c) -(a-b)(b-c)(a+c+2b)^3>0$
with same method,we have $y+z>x,x+z>y$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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FoxTrot Bill Amend Problems So I found this on the Wolfram website today:
So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:
*
*$ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $
*$ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $
*$ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $
*$ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$
Ideas
*
*The degree of the numerator is $\frac{3}{2}$. The degree of the
denominator is $\frac{3}{2}$. The expression is of an indeterminate
form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.
*No ideas, really.
*Derivatives are always pretty easy, although this one is a bit
bashy. Basically bash Product and Chain Rules, etc.
*Integration by Parts bash? The $\mathrm{d}\theta$ part is very
trivial. I mean for the $\rho$ and $\phi$.
| My take on $\#3$: $$ \begin {align*} \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] &= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot (n+1) \ln u - \dfrac {u^{n+1}}{(n+1)^2} \right] \\&= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1} \ln u}{n+1} - \dfrac {1}{(n+1)^2} \cdot u^{n+1} \right] \\&= \dfrac {1}{n+1} \cdot \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ u^{n+1} \ln u - \dfrac {u^{n+1}}{n+1} \right] \\&= \dfrac {1}{n+1} \cdot \left[ \dfrac {\mathrm{d}}{\mathrm{d}u} \left( u^{n+1} \ln u \right) - \dfrac {\mathrm{d}}{\mathrm{d}u} \left( \dfrac {u^{n+1}}{n+1} \right) \right] \\&= \dfrac {1}{n+1} \cdot \left[ \dfrac {u^{n+1}}{u} + (n+1) \cdot u^n \ln u - u^n \right] \\&= \dfrac {1}{n+1} \cdot \left[ {u^n} + (n+1) \cdot u^n \ln u {-u^n} \right] \\&= \boxed {u^n \ln u}. \end {align*} $$Beautifully done intentionally?
| {
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"timestamp": "2023-03-29T00:00:00",
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Expressing $\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$ in closed form I want to express $$\sum_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.
I know that $$\pi z\cot(\pi z)=1+2z^2\sum_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.
| First rewrite it as:
$$\frac{1}{z^3} + \sum_{n=1}^\infty \left(\frac{1}{z^3-n^3} + \frac{1}{z^3+n^3}\right) = \frac{1}{z^3} +\sum_1^\infty \frac{2z^3}{z^6-n^6}$$
so we only need to find:
$$\sum_{n=1}^\infty \frac{1}{z^6-n^6}$$
Now, if $\omega$ is a primitive cube root of unity, there are functions $a(u),b(u),c(u)$ such that for real $u\neq v$:
$$\frac{1}{u^3-v^3} = \frac{a(u)}{u-v} +\frac{b(u)}{\omega u -v} + \frac{c(u)}{\omega^2u-v}$$
Setting $u=z^2$ and $v=n^2$, and finding a solution $\zeta$ so that $\zeta^2=\omega$, then we have:
$$\frac{1}{z^6-n^6} = \frac{a(z^2)}{z^2 - n^2}+\frac{b\left((\zeta z)^2\right)}{(\zeta z)^2-n^2} + \frac{c\left((\zeta^2 z)^2\right)}{(\zeta^2 z)^2 -n^2}$$
So we can use your first result to find:
$$\sum_{n=1}^\infty \frac{1}{z^6-n^6} = a(z^2)f(z) + b(\zeta^2z^2)f(\zeta z) + c(\zeta^4 z^2)f(\zeta^2 z)$$
where $$f(z)=\sum_{n=1}^\infty \frac{1}{z^2-n^2} = \frac{\pi z\cot(\pi z)-1}{2z^2}$$
That seems a bit messy, and requires jumping into complex numbers...
| {
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"timestamp": "2023-03-29T00:00:00",
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All roots of the quartic equation $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real Problem
Prove that all roots of $a x^4 + b x^3 + x^2 + x + 1 = 0$ cannot be real. Here $a,b \in \mathbb R$, and $a \neq 0$.
Source
This is one of the previous year problem of Regional Math Olympiad (India). I had a hard time solving it, so thought I'd better ask here.
Observations
*
*Some real roots are possible: when $a<0$, the equation has two of them.
*If one more coefficient was allowed to be arbitrary: $a x^4 + b x^3 + cx^2 + x + 1 = 0$, then the roots could be all real, since every quartic can be brought into such form by scaling
| Let $f(x) = x^4 + x^3 + x^2 + bx + a$. Then the quartic
$ax^4 + bx^3 + x^2 + x + 1$ is $x^4 f(1/x)$, and has four real roots
iff $f$ does. But then the same is true of
$f(x-\frac14) = x^4 + \frac58 x^2 + Bx + A$ for some $B$ and $A$
(we don't need the formula).
If this is $(x-a)(x-b)(x-c)(x-d)$ for some $a,b,c,d$
then $a^2+b^2+c^2+d^2 = -2\frac58 < 0$, so $a,b,c,d$ cannot all be real,
QED.
| {
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Proving that $(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$, where $a$, $b$, $c$ are the sides, and $\Delta$ the area, of a triangle Let $a$, $b$, $c$ be the sides of $\triangle ABC$.
Prove $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$
where $\Delta$ is the area of the triangle.
(Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$$ I incorporated the correction into the title. —@Blue)
*
*Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$
*I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.
*Then $\angle AFB = \angle AFC = \angle BFC=120^\circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.
*Observe that
$$xy+yz+zx=\frac{4\Delta}{\sqrt{3}}$$
*So, we have a big inequality to prove:
$$\left(x^2+xy+y^2\right)\left(y^2+yz+z^2\right)\left(z^2+zx+x^2\right)\geq xy+yz+zx$$ but I can't prove it.
Thanks for help
| I think you mean to the following:
$$(abc)^{\frac{2}{3}}\geq\frac{4\Delta}{\sqrt3},$$ which is
$$16\Delta^2\leq3\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}(2a^2b^2-a^4)\leq\sqrt[3]{a^4b^4c^4}$$ or
$$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)\geq0,$$ which is true by Schur and AM-GM:
$$\sum_{cyc}\left(a^4-2a^2b^2+\sqrt[3]{a^4b^4c^4}\right)=\sum_{cyc}\left(\sqrt[3]{a^{12}}+\sqrt[3]{a^4b^4c^4}-2a^2b^2\right)\geq$$
$$\geq\sum_{cyc}\left(\sqrt[3]{a^9b^3}+\sqrt[3]{a^9c^3}-2a^2b^2\right)=\sum_{cyc}(a^3b+ab^3-2a^2b^2)\geq0.$$
| {
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Limit at infinity involving $e$ I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.
| $$
\begin{align}
\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^{\frac{x^3+8}{x}}
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{x^2+8/x}\\
&=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)\frac{x^2+8/x}{5x^2+x+8/x}}\\
&=\left(\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)}\right)^{\lim\limits_{x\to\infty}\frac{x^2+8/x}{5x^2+x+8/x}}\\[6pt]
&=e^{1/5}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
| We count the ways in which Team A can win the series, and double the result. To count the ways A can win the series, we make a list like yours.
A wins in $4$: There is $1$ way this can happen.
A wins in $5$: A has to win $3$ of the first $4$, and then win. There are $\binom{4}{3}$ ways this can happen.
A wins in $6$: A has to win $3$ of the first $5$, then win. There are $\binom{5}{3}$ ways this can happen.
A wins in $7$: A has to win $3$ of the first $6$, then win. There are $\binom{6}{3}$ ways this can happen.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
| $\dfrac{1}{3}+\dfrac{1}{4}\dfrac{1}{2!}+\dfrac{1}{5}\dfrac{1}{3!}+\dots\\=\displaystyle\sum_3^\infty\dfrac{1}{n}\times\dfrac{1}{(n-2)!}\\=\displaystyle\sum_3^\infty\dfrac{n-1}{n!}\\=\displaystyle\sum_3^\infty\left[\dfrac{1}{(n-1)!}-\dfrac{1}{n!}\right]\\=\dfrac{1}{2!}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
| $$
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}=
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x}
$$
Apply l'Hôpital or Taylor expansion (better) to the first fraction.
For the Taylor expansion, it's easier to do it in pieces:
$$
\sin x = x-\frac{x^3}{6}+o(x^4),\quad
\cos x = 1-\frac{x^2}{2}+o(x^4)
$$
so
$$
\sin x-x\cos x=x-\frac{x^3}{6}-x+\frac{x^3}{2}+o(x^4)=\frac{x^3}{3}+o(x^4)
$$
while
$$
\sin x+x\cos x=x-\frac{x^3}{6}+x-\frac{x^3}{2}+o(x^4)=2x+o(x^2)
$$
so
$$
\lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}=
\lim_{x\to 0}\frac{\frac{1}{3}x^3+o(x^4)}{x^3}\frac{2x+o(x^2)}{x}=
\lim_{x\to 0}\left(\frac{1}{3}+o(x)\right)(2+o(x))=\frac{2}{3}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} +\cdots+{1 \over {\sqrt n }}\right)$ $$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }}+\cdots+{1 \over {\sqrt n }}\right)$$
( Without use of integrals ).
I was able to squeeze it from the bottom to ${\lim }=1$, but that's not good enough.
I'd be glad for help.
| $$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{{\sqrt{n} + \sqrt{n}}} < \frac{1}{2\sqrt{n}}$$
add inequalities from 1 to n,
$$\sqrt{n+1} - 1 > \frac{1}{2}\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$
also for lower bound note that,
$$\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \left( \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \cdots+ \frac{1}{\sqrt{n}}\right)= \frac{n}{\sqrt{n}}$$
so we have $$2(\sqrt{n+1} - 1) > \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \sqrt{n}$$
Now sandwich theorem tells us that limit is 2
EDIT:
$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}} < \frac{1}{\sqrt{n}+\sqrt{n+1}} < \frac{1}{\sqrt{n}+\sqrt{n}}$$
$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}}< \sqrt{n+1} - \sqrt{n} < \frac{1}{\sqrt{n}+\sqrt{n}} $$
add from 1 to n and let $$S = \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$,
then,
$$\frac{1}{2}\left( S + \frac{1}{\sqrt{n+1}} -1\right) < \sqrt{n+1} -1 < \frac{1}{2}S$$
$$2(\sqrt{n+1} -1) <S<2(\sqrt{n+1} -1) +1 -\frac{1}{\sqrt{n+1}}$$
| {
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Exponentials in complex numbers If $\displaystyle z-\frac1z=i$, then find $\displaystyle z^{2014}+\frac{1}{z^{2014}}$.
The answer should be in terms of $1, -1,\;i\;or\;-i$. I am not able to understand how to simplify the given expression so that its $2014^{th}$ power can be found out easily.
| $(1)$
We have $\displaystyle z^2+\frac1{z^2}=1$
$\displaystyle\implies z^4-z^2+1=0$
Using $\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2),$
$z^6+1=(z^2+1)(z^4-z^2+1)=0\implies z^6=-1$
$(2A)$ We have $\displaystyle z^2=iz+1$
$\displaystyle \implies z^3=z\cdot z^2=z(iz+1)$
$\displaystyle=iz^2+z$
$\displaystyle=i(iz+1)+z=i+z(i^2+1)=i$
$(2B)$ As $\displaystyle z^2-iz-1=0, z^2-iz+i^2=0$
$\displaystyle\implies z^3+i^3=(z+i)(z^2-iz+i^2)=0\implies z^3=-i^3=i$
$\displaystyle\implies z^6=-1$
Now, $2014=335\cdot6+4$
Finally $\displaystyle z^4+\frac1{z^4}=\left(z^2+\frac1{z^2}\right)^2-2\cdot z^2\cdot\frac1{z^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that we flip a coin until either it comes up tails twice or we have flipped six times. What is the expected number of times we flip the coin? So this is the answer that I got (where $N$ represents the number of coin flips after which you can stop) but can someone tell me if this looks right?
$P(N=2) = P(TT) = \dfrac{1}{4} $
$P(N=3) = P(HTT , THT) = \dfrac{2}{8} $
$P(N=4) = P(HHTT;HTHT;THHT) = \dfrac{3}{16} $
$P(N=5) = \dfrac{4}{32} $
$P(N=6) = 1 - \dfrac{1}{4} - \dfrac{2}{8} - \dfrac{3}{16} - \dfrac{4}{32} = \dfrac{3}{16}$
$E(N) = 2\left(\dfrac{1}{4}\right) + 3\left(\dfrac{2}{8}\right) + 4\left(\dfrac{3}{16}\right) + 5\left(\dfrac{4}{32}\right) + 6\left(\dfrac{3}{16}\right) = \dfrac{15}{4} = 3.75$
| What you are looking at is a right-censored negative binomial distribution. That is, let $Y = X \wedge 6 = \min\{X, 6\}$ where $X \sim {\rm NegBinomial}(r = 2, p = 1/2)$, where the particular choice of parametrization for $X$ is $$\Pr[X = k] = \binom{k-1}{k-r} (1-p)^r p^{k-r}, \quad k = r, r+1, r+2, \ldots.$$ That is, $X$ counts the total number of trials needed to observe $r = 2$ tails, where each trial is an independent Bernoulli trial with probability of observing tails equal to $p = 1/2$. Then $Y$ is $X$ right-censored at $6$. So we can calculate $$\begin{align*} {\rm E}[Y] &= {\rm E}[X \wedge 6] \\ &= \sum_{k=2}^5 k \Pr[X = k] + \sum_{k=6}^\infty 6 \Pr[X = k] \\ &= 6 - \sum_{k=2}^5 (6-k)\Pr[X = k] \\ &= 6 - \frac{9}{4} = \frac{15}{4}.\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| Let $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=x $$
Clearly, $x>0$
$$\implies x^2=7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7x$$
Now left is the proof of converge(as conversed with Abdulh Khazzak Gustav ElFakiri)
Observe that the $r$th term $T_r$ of this infinite product is $\displaystyle7^{\left(\frac1{2^r}\right)}$
using Convergence/Divergence of infinite product, $$\sum_{0\le r<\infty}\ln(T_r)=\ln 7\sum_{0\le r<\infty}\frac1{2^r}$$ which is an infinite Geometric Series with common ratio $=\frac12$ which $\in(-1,1)$, hence the later Series is convergent $\left(\text{ in fact }\displaystyle=\ln7\cdot\frac1{1-\frac12}\right)$, so will be the original infinite Product
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Product of a matrix and its Hermitian transpose? Suppose I need to find a matrix B such that
$B^H B = A$
and $A = \begin{bmatrix}4 &0& 0\\ 0 &1 &i\\ 0 &-i& 1\\\end{bmatrix}$
How do I proceed with a Product of a matrix and its Hermitian transpose ?
| This will illustrate how to handle your specific, small matrix, but the other post points to a general and practical method.
Diagonalize A. It must have non-negative eigenvalues or $B^{H}B=A$ is not possible. The eigenvalues of $A$ are $4,2,0$. Check that $A$ has a basis of eigenvectors:
$$
A\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right] = 4\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right],\;\;
A\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right] = 2\left[\begin{array}{c}0 \\ i \\ 1\end{array}\right],\;\;
A\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right] = 0\left[\begin{array}{c}0 \\ i \\ -1\end{array}\right]
$$
The eigenvectors are automatically orthogonal for distinct eigenvalues. Therefore, the normalized eigenvectors give a unitary transition matrix
$$
U = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array}\right].
$$
That is $U^{H}U=UU^{H}=I$, and
$$
U^{H}AU = \left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right],\;\;\; A = U\left[\begin{array}{ccc}4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{array}\right]U^{H} =
U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
So one matrix that will work for $B$ is
$$
B=\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
However, the better choice for $B$ may be the unique Hermitian matrix ($B=B^{H}$) with non-negative eigenvales for which $B^{2}=A$:
$$
B=U\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & \sqrt{2} & 0 \\ 0 & 0 & 0\end{array}\right]U^{H}
$$
| {
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The $\sum\limits_{n = 1}^\infty {\frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}}$ series is absolutely convergent? $$\sum\limits_{n=1}^\infty\frac{\sin(\sqrt n)}{n^{\frac{3}{2}}}$$
Let ${a_n} = \frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}$, then, using the criterion of quotient I must prove that $\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|<1$, indeed:
$$\begin{array}{l}
\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{\sin (\sqrt {n + 1} )}}{{{{(n + 1)}^{\frac{3}{2}}}}}}}{{\frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} ) \cdot {n^{\frac{3}{2}}}}}{{{{(n + 1)}^{\frac{3}{2}}} \cdot \sin (\sqrt n )}}} \right|\\
= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} ) \cdot {n^{\frac{3}{2}}}}}{{{{(n + 1)}^{\frac{3}{2}}} \cdot \sin (\sqrt n )}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right|{\left| {\frac{n}{{n + 1}}} \right|^{\frac{3}{2}}}\\
= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right| \cdot \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{n}{{n + 1}}} \right|^{\frac{3}{2}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right| \cdot (1)\\
= \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right|
\end{array}$$
then I stay stagnant.
How I can know if this limit is $ \geq1 $ or $ < 1?$
| Let us take absolute series of given series,
ie.$$\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$$.since we know $|\frac {\sin \sqrt n}{n^{3/2}}|$$\leq$ $|\frac {1}{n^{3/2}}|$.But $\sum_{n=1}^{\infty} |\frac {1}{n^{3/2}}|$ is converging.Hence by comparison theorem, $\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$ converges.Since absolute series conveges,so the given series converges...
$$ $$
Hence proved
| {
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Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar{z} = 1$.
So $\left|z^2+z+4z\bar{z}\right| = |z|\cdot \left|z+4\bar{z}+1\right| = \left|z+4\bar{z}+1\right|$
Now Let $z = x+iy$. Then $\bar{z} = x-iy$ and $|z| =1\Rightarrow x^2+y^2 = 1$
So $\left|x+iy+4x-4iy+1\right| = \left|5x+1-3iy\right| = \sqrt{(5x+1)^2+9y^2}$
So Let $f(x) = \sqrt{(5x+1)^2+9(1-x^2)} = \sqrt{25x^2+1+10x+9-9x^2}$
So $\displaystyle f(x) = \sqrt{16x^2+10x+10}=4\sqrt{x^2+\frac{5}{8}x+\frac{5}{8}}$
So $\displaystyle f(x) = 4\sqrt{\left(x+\frac{5}{16}\right)^2+\left(\frac{5}{8}-\frac{25}{256}\right)}\geq 4\sqrt{\frac{27 \times 5}{256}} = 4\times \frac{3\sqrt{15}}{16} = \frac{3\sqrt{15}}{4}$
which is occur at $\displaystyle x = -\frac{5}{16}$
(2) Now I did not understand how can i solve (II) one
Help Required
Thanks
| HINT. The second part is the triangle inequality in reverse. So instead of starting with $|a+b|$ and getting $|a+b|\leq |a|+|b|$. You have $|a|+|b|$ and want to write something like $|a+b|\leq |a|+|b|$.
| {
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much
| let
$$a=\dfrac{2(x+y)}{x},b=\dfrac{x+y}{y},x,y>0$$
then
$$C=a+b+\sqrt{a^2+b^2}=\dfrac{(x+y)(x+2y+\sqrt{4y^2+x^2})}{xy}$$
so
$$C-10=\dfrac{x^2-7xy+2y^2+(x+y)\sqrt{x^2+4y^2}}{xy}$$
and note
$$(x+y)^2(x^2+4y^2)-(x^2-7xy+2y^2)^2=4xy(2x-3y)^2\ge 0$$
so
$$a+b+\sqrt{a^2+b^2}\ge 10$$
| {
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finding minimum of function Can you please give me some hints finding minimum of this function:
$ (r-1)^2 + (\frac{s}{r} -1)^2 + (\frac{t}{s}-1)^2 + (\frac{4}{t}-1)^2$
where $ 1 \le r \le s \le t \le 4 $,
$r,s,t \in \Bbb R $
| Let us write $a = \frac{r}{1}$, $b = \frac{s}{r}$, $c = \frac{t}{s}$, and $d = \frac{4}{t}$. Then the function you want to minimise is
$$f(a,b,c,d) = (a-1)^2 + (b-1)^2 + (c-1)^2 + (d-1)^2,$$
subject to the constraints $a,b,c,d \geqslant 1$ and $p(a,b,c,d) = a\cdot b\cdot c\cdot d = 4$.
That looks like using the method of Lagrange multipliers may help a lot. So considering $\dfrac{\partial f}{\partial a}(a,b,c,d) = 2(a-1)$ and similar for $b,c,d$, and $\dfrac{\partial p}{\partial a}(a,b,c,d) = bcd = \dfrac{p(a,b,c,d)}{a} = \frac{4}{a}$ on the surface $abcd = 4$, again similar for $b,c,d$. So to have $\nabla f = \lambda\cdot \nabla p$, we must have
$$\lambda = \frac{\partial f/\partial a}{\partial p/\partial a} = \frac{a(a-1)}{2} = \frac{b(b-1)}{2} = \frac{c(c-1)}{2} = \frac{d(d-1)}{2}.$$
Thus the only critical point of $f$ on $\{abcd = 4\}$ (with positive $a,b,c,d$) is $a=b=c=d=\sqrt{2}$. It is readily checked that that point is a local minimum, and then one verifies that it is the global minimum under the constraints.
Thus the function is minimised for $r = \sqrt{2}, s = 2, t = 2\sqrt{2}$, with the minimal value $4(3-2\sqrt{2})$.
| {
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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.