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Differential equation\higher order I need help solving this task, if anyone had a similar problem it would help me.
The task is: Find the general solution of the differential equation
$$ y''-y' - 2y=e^{2x} \cos^2 x$$
For homogeneous I get:
$$y_H=c_1e^{2x}+c_2e^{-x}$$
I have a problem with the particular, I tried:
$$y=(A\sin^2x+B\cos^2x)e^{2x}\\y'=(2A\sin x \cos x + 2B \sin x \cos x) e^{2x} +(A\sin^2 x + B \cos ^2 x) 2e^{2x}\\y''=2e^{2x}(2A \sin x \cos x - 2B \sin x \cos x + A\sin^2 x +B \cos^2 x)$$
I have no idea if this is correct?
I don't know what's next from here ..
Thanks in advance !
| Your homogeneous solution is correct.
Note that $\cos^2(x)=\frac{1}{2}(1+\cos(2x))$, so the particular solution should be of the form $$y_{p}(x)=Axe^{2x}+Be^{2x}\cos(2x)+Ce^{2x}\sin(2x).$$
Substituting we find that $A=\frac{1}{6}$, $B=-\frac{1}{26}$ and $C=\frac{3}{52}$.
$$y'_p=2C\mathrm{e}^{2x}\sin\left(2x\right)-2B\mathrm{e}^{2x}\sin\left(2x\right)+2C\mathrm{e}^{2x}\cos\left(2x\right)+2B\mathrm{e}^{2x}\cos\left(2x\right)+2Ax\mathrm{e}^{2x}+A\mathrm{e}^{2x}$$
and $$y''_p=\mathrm{e}^{2x}\left(-2\left(2C+2B\right)\sin\left(2x\right)+2\left(2C-2B\right)\cos\left(2x\right)+2A\right)$$
$$+2\mathrm{e}^{2x}\left(\left(2C-2B\right)\sin\left(2x\right)+\left(2C+2B\right)\cos\left(2x\right)+2Ax+A\right)$$
So we have $$y''_p-y'_p-2y_p$$
$$=e^{2x}\big[\cos(2x)(4C-4B+4C+4B-2C-2B-2B)+\sin(2x)(-4C-4B+4C-4B-2C+2B-2C)+2A+4Ax+2A-2Ax-A-2Ax\big]$$
$$=e^{2x}\big[\cos(2x)(6C-4B)+\sin(2x)(-4C-6B)+3A\big]$$
$$=\frac{e^{2x}}{2}(1+\cos(2x)).$$
So we have $3A=\frac{1}{2}\implies A=\frac{1}{6}$, $6C-4B=\frac{1}{2}$ and $-4C-6B=0$ which give $C=\frac{3}{52}$ and $B=-\frac{1}{26}$.
| {
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Show that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$ I cam across a question in The Book of Proof that states:
Show that $$\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$$
Which I found the answer to be:
Assume n ≥ 3. Then:
$$\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2} = \binom{n - 2}{3} + \binom{n-2}{2}+ \binom{n - 1}{2} = \binom{2}{2} + \binom{3}{2}+...+\binom{n-1}{2}$$
I have just learned about the binomial theorem, and I am not sure how we got to that answer.
I understand that $\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2}$ because that is the sum of the two previous rows in the Pascal triangle. But I can't find a simple explanation for the steps that follow or for why we assumed $n \geq 3$
| $$S=\sum_{k=2}^{n-1} {k \choose 2} =\text{co-efficient of}~ x^2~ \text{in}\sum_{k=0}^{n-1}(1+x)^k= \text{co-efficient of} ~x^2~ \text{in} \frac{(1+x)^n-1}{1+x-1}. $$
$$\implies S= \text{co-efficient of} ~x^3~ \text{in} [(1+x)^n-1]={n \choose 3}$$
| {
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Given $S$ is maximum, find $\dfrac{h}{x}$ A question is given as such: Given $h=\dfrac{L(R+\sqrt{R^2-x^2})}{2R}$ and $S=\dfrac{4}{3}xh$, find $\dfrac{h}{x}$ in terms of $L$ and $R$ when $S$ is maximum.
By differentiation, I got that $R\sqrt{R^2-x^2}+R^2-2x^2=0$, which gives that $\dfrac{h}{x}=\dfrac{Lx}{R^2}$. On the other hand I can also get that $2h=\dfrac{L}{R}\dfrac{x^2}{\sqrt{R^2-x^2}}$, but neither of them gives anything close to $\dfrac{h}{x}$.
The final answer is $\dfrac{h}{x}=\dfrac{\sqrt{3}L}{2R}$.
| If you know that $\frac{h}{x} = \frac{Lx}{R^2}$ and $2h = \frac{L}{R} \frac{x^2}{\sqrt{R^2-x^2}}$, you can solve for $x$ in terms of $L, R$. Multiplying the first equation by $x$ and dividing the second by $2$ yields $$\frac{Lx^2}{R^2} = h = \frac{L}{2R} \frac{x^2}{\sqrt{R^2-x^2}}$$
Multiplying both sides by $\frac{R}{Lx^2}$ makes it $$\frac{1}{R} = \frac{1}{2\sqrt{R^2-x^2}}$$
Then this means that $R^2-x^2 = \frac{R^2}{4} \to x^2 = \frac{3R^2}{4}$. Take the square root to get that $x = \frac{R\sqrt{3}}{2}$.
Using $\frac{h}{x} = \frac{Lx}{R^2}$, you can plug in $x = \frac{R\sqrt{3}}{2}$ to get $$\frac{h}{x} = \frac{L\frac{R\sqrt{3}}{2}}{R^2} = \frac{L\sqrt{3}}{2R}$$
| {
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If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .
If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .
What I Tried :- First I thought for a moment and found out that I can write this :
$$\frac{(a-c)(b-d)}{(b-c)(d-a)} = \frac{(a-c)(b-d)}{(a-b)(c-d)} * \frac{2016}{2017}$$
But how is it going to help?
Then I thought maybe cross-multiply everything and try to factor it? I get :-
$$2017(a-b)(c-d) = 2016(b-c)(d-a)$$
$$\rightarrow 2017(a-b)(c-d) - 2016(b-c)(d-a) = 0$$
$$\rightarrow 2017(ac - ad - bc + bd) - 2016(bd - ab - cd - ac) = 0$$
$$\rightarrow 2017ac + bd + 2016ab + 2016cd + 2016ac - 2017ad - 2017bc - 2016bd = 0$$
I am hopeless at this point, how do you even factorise this?
Can anyone help?
| Let
$$t=\frac{(a-b)(c-d)}{(b-c)(d-a)}=\frac{ac+bd-ad-bc}{ac+bd-ab-cd}$$
then
$$t-1=\frac{ab+cd-ad-bc}{ac+bd-ab-cd}=\frac{(a-c)(b-d)}{(b-c)(d-a)}$$
so
$$\frac{(a-c)(b-d)}{(a-b)(c-d)}=\frac{t-1}{t}.$$
| {
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Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.
Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$
attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks
| One acxtually gets $$\sum_{k=1}^{n} (-1)^k (2k-1)^2=\frac{1}{2}[1-(-1)^n+ 4(-1)^n n^2]$$
| {
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Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
| It's $$x^2(x-4)-4(x-4)=0$$ or
$$(x-4)(x^2-4)=0.$$
Can you end it now?
| {
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Proof $f(x)>x$ for $x>0$ I'm a little stuck at proving that for these function:
$$\ f(x) = \frac{x}{\sqrt{1+x^2}} + \sqrt{1+x^2}\cdot\ln(1+x^2) $$
$f(x)>x$ for every $x>0$. Another question: is there any $a$ that $f(x)>x^2$ for every $x>a$?
My steps
I have computed the derivative:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) $$
then compared to zero:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) > 0 $$
and calculated that:
$$ 4>x^2(1+x^2)(x-\ln(1+x^2))^2 $$
Each of the components is $>0$ I can see from the primary function that the root is 0, but I cannot write it in more precise way than the equation above. I am thinking of Mean Value Theroem, but I cannot see, how to apply it in this case. I was considering something like:
$$ f'(c) = \frac{f(x) - f(0)}{x} = \frac{f(x)}{x} > 0 $$
Since the function gives positive values for $x>0$, but it has leaded me to nowhere. I would really appreciate your help.
EDIT
As Andrei suggested, I have corrected the derivative:
$$ \frac{x\ln(x^2+1)}{\sqrt{x^2+1}} + \frac{2x}{\sqrt{x^2+1}} + \frac{1}{(x^2+1)^{3/2}} $$
Then calculated:
$$ 1>(x^2+1)(\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
Since
$$ \sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 \\
x^2+1<4x^2+4x^2\ln(x^2+1)+x^2\ln(x^2+1)^2 $$
The positive and negative multiplication is smaller than 1, thus the inequality is true.
However, what about $f(x)>x^2$? Do I have to calculate the second derivative?
I have continued Andrei method and for $f'(x)-2x>0$ obtained:
$$ 1>(x^2+1)(2x\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
$$ 2x\sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 $$
And reduced to:
$$ x^4<\frac{3}{2}\ln(x^2+1) /^e(\cdot) \\
e^{x^4} < (x^2+1)\sqrt{x^2+1} $$
The order of growth indicates that this is false (only even power of x) and there is no such a that fulfil these requirements.
Is it correct?
| For $x>0$
$$\frac{x}{\sqrt{x^2+1}}+\sqrt{x^2+1} \ln \left(x^2+1\right)>\frac{x}{\sqrt{x^2+1}}+\sqrt{x^2+1}>x;\forall x\in\mathbb{R}$$
Furthermore $f(x)>x^2$ only for $0<x<2.51$ (approximated), so there is no $a$ such that $f(x)>x^2$ for all $x>a$.
Hope this helps
| {
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Find the value of $\frac{1}{(pq)^2-10r^2+300}+\frac{1}{(qr)^2-10p^2+300}+\frac{1}{(pr)^2-10q^2+300}$
If $p+q+r=6,pq+pr+qr=8,pqr=2$, what is the value of:
$$\frac{1}{(pq)^2-10r^2+300}+\frac{1}{(qr)^2-10p^2+300}+\frac{1}{(pr)^2-10q^2+300}$$
I tried to change $\frac{1}{(pq)^2-10r^2+300}$ to $\frac{1}{\frac{4}{r^2}-10r^2+300}$ but I don't know how to solve it.
Please provide a hint.
| $p$, $q$ and $r$ are roots of the equation:
$$x^3-6x^2+8x-2=0,$$ which by your work gives:
$$\sum_{cyc}\frac{1}{p^2q^2-10r^2+300}=\sum_{cyc}\frac{r^2}{4-10r^4+300r^2}=$$
$$=\sum_{cyc}\frac{r^2}{2(r^3-6r^2+8r)-10r^4+300r^2}=\sum_{cyc}\frac{r}{16+288r+2r^2-10r^3}=$$
$$=\sum_{cyc}\frac{r}{8(r^3-6r^2+8r)+288r+2r^2-10r^3}=-\frac{1}{2}\sum_{cyc}\frac{1}{r^2+23r-176}.$$
Can you end it now?
For example, I got
$$\prod_{cyc}(r^2+23r-176)=(pqr)^2+pqr(23(pq+pr+qr)+881(p+q+r)+24311)+$$
$$+30976(p+q+r)^2-176(pq+pr+qr)^2+712448(p+q+r)-$$
$$-4048(p+q+r)(pq+pr+qr)-155056(pq+pr+qr)-5451776=$$
$$=-1448162.$$
| {
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Proving a divisibility given a recurrence relation I am given the sequence $a_0 = 0$, $a_1 = 1$, and for all $n \geq 2$, $a_n = a_{n-1} + 2a_{n-2}$. I must prove that if $3 \mid n$, then $3 \mid a_n$. By logic, if $3 \nmid n$, then the conclusion $3 \mid a_n$ is automatically true. So, I should only prove that $3 \mid a_n$ is true for multiples of 3, right? So what I did was suppose $3 \mid n$ and try to prove that $3 \mid a_{3n}$. I will argue by induction on $n$.
*
*Base Case. Suppose $n = 0$. Then, $3 \mid 0 = a_0$. So, the base case holds.
*Induction Hypothesis Suppose $3 \mid a_{3k}$ for some integer $k \geq 0$. We must show that $3 \mid a_{3k+1}$.
I know that $a_{3k+1} = a_{3k} + 2a_{3k-1}$, which implies $a_{3k+1} = 3m + 2a_{3k-1}$ for some integer $m$. I don't know how to continue the proof from here. Is it even correct so far? Please help!
| First strategy:
Let examine the three first terms $\begin{cases}a_0\equiv 0\pmod 3\\a_1\equiv 1\pmod 3\\
a_2= 1+2\times 0\equiv 1\pmod 3\end{cases}$
Now notice that $a_3\equiv 1+2\times 1\equiv 0\pmod 3$ like $a_0$
So the aim it to show by induction that the sequence of remainders is
$0,1,1,0,2,2,0,1,1,0,2,2,0,1,1\cdots$
Thus set as base hypothesis either $$P(n): a_{3n}\equiv 0\pmod 3,a_{3n+1}\equiv 1\pmod 3,a_{3n+2}\equiv 1\pmod 3$$
$$Q(n): a_{3n}\equiv 0\pmod 3,a_{3n+1}\equiv 2\pmod 3,a_{3n+2}\equiv 2\pmod 3$$
And go on to show that $a_{3n+3},a_{3n+4},a_{3n+5}$ follow the $0,2,2$ pattern if $P(n)$ is assumed and $0,1,1$ pattern if $Q(n)$ is assumed.
Second strategy:
$\begin{cases}a_{n+3}=a_{n+2}+2a_{n+1}\\a_{n+2}=a_{n+1}+2a_{n}\end{cases}\ $ now go on and replace $a_{n+2}$
Thus $\ a_{n+3}=3a_{n+1}+2a_n\ $ and it is immediate that if $3$ divides $a_n$ then $3$ also divides $a_{n+3}$
(because if $a_n=3k$ then $a_{n+3}=3(a_{n+1}+2k)$)
| {
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What's wrong with my even Fibonacci generating function? I'm trying to find the closed form for the generating function of an even Fibonacci series $F_{2n}$, but I'm not getting the right answer. My idea was to use the even Fibonacci series to find the odd Fibonacci series, then combine them. i.e. if
\begin{align}
f(x) = f_0 + f_2x^2 + f_4x^4 + \dots
\end{align}
then
\begin{align}
x^2f(x) = f_0x^2 + f_2x^4 + \dots
\end{align}
so subtracting one from the other,
\begin{align}
(1 - x^2)f(x) &= f_0 + (f_2 - f_0)x^2 + (f_4 - f_2)x^4 + \dots \\
&= f_0 + f_1x^2 + f_3x^4 + \dots \\
\implies \frac{(1-x^2)f(x) - f_0}{x} &= f_1x + f_3x^3 + \dots
\end{align}
Then I add odd and even parts together (and putting $f_0 = 1$),
\begin{align}
\frac{(1-x^2)f(x) - 1}{x} + f(x) = f_0 + f_1x + f_2x^2 + f_3x^3 + \dots = \frac{x}{1-x-x^2}
\end{align}
Then when I solve for $f(x)$ I get
\begin{align}
f(x) = \frac{1-x}{(1-x-x^2)(1+x-x^2)}
\end{align}
But when I chuck this in Mathematica it gives me $1-x+3x^2-3x^3+8x^4-8x^5+21x^6-21x^7 + ...$. What went wrong?
| What Mathematica gives you indicates that you're close. The desired terms are repeated, and those repeats are because of the $1-x$ in the numerator, so the correct answer is
$$\frac{x^2}{(1-x-x^2)(1+x-x^2)}=\frac{x^2}{1-3x^2+x^4}$$
where the $x^2$ shifts the series so $F_0=0$ is the $x^0$ coefficient.
| {
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Why am I not able to find the intersection of circles $(x-2)^2+(y-5)^2=5$ and $(x-1)^2+(y+3)^2=50$ algebraically? I'm answering a question on finding the intersection of two circles. I am confused why my algebraic method is not leading me to the correct answers.
The two circles have equations:
$$(x-2)^2+(y-5)^2=5 \quad\text{and}\quad (x-1)^2+(y+3)^2=50$$
First, I expanded both and got:
$$\begin{align}
x^2-4x+4+y^2-10y+25&=\phantom{1}5 \\
x^2-2x+1+y^2+\phantom{1}6y+\phantom{1}9&=50
\end{align}$$
I then multiplied the first equation by -1, and added it to the second equation to get:
$x=32-8y$
I know that everything up to here is correct, because as far as I'm aware, simultaneously solving them should tell me equation of the line that goes through their intersection points. I graphed the line and the two circles on Desmos and this confirmed the line was right. It also clearly shows one of the solutions as (0,4)
So now, surely just substitute the equation $x=32-8y$ into one of the circles to find the intersections?
I substituted it into the first one:
$(32-8y)^2-4(32-8y)+4+y^2-10y+25=5$
Which simplified to:
$65y^2-494y+920=0$
But solving this quadratic doesn't give a y-coordinate of 4 as a solution. It seems to give totally random solutions.
Where have I gone wrong?
| You simplified the quadratic wrong:\begin{eqnarray}
(32 - 8y)^2 - 4(32-8y) + 4 + y^2 - 10y + 25 &=& 5\\
1024 - 512 y + 64 y^2 - 128 + 32 y + 4 + y^2 - 10y + 25&=&5\\
(64+1) y^2 + (-512 + 32 - 10) y + (1024-128+4+25-5) &=& 0\\
65 y^2 - 490 y + 920 &=&0
\end{eqnarray}
The $y$ coefficient should be $-490$ not $-494$.
| {
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Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition I got a question regarding my answer of proving limit using epsilon-delta, here's the question
Prove $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$
Here's the answer I've come up so far
let $f(x) = \frac{x+1}{x-2} + x$
by algebra manipulation, we get
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1| $ $=|\frac{x^2 - 1}{x-2}|$ $=|\frac{(x-1)(x+1)}{x-2}|$
let $|x-1| < 1$, by triangle inequality we get $|x| < 2$, then
$|x + 1| < 3$ and $|x - 2| < 1$
now, using the definiton of limit,
for every $\epsilon > 0$, there exist $\delta = min\{1, \frac{\epsilon}{3}\}$ such that
if $0 < |x - 1| < \delta$ then,
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1|$ $=|\frac{(x-1)(x+1)}{x-2}|$ $=|\frac{1 \cdot 3}{1}|$ $< \epsilon$
Is this correct? honestly I'm not sure on getting the upper bound of $|x-2|$, so I used the assumption of $|x-1| < 1$
Any tips would help, thanks beforehand.
| $f(x) = \frac{x+1}{x-2}+x=\frac{x^2 +1-x}{x-2}$
We can prove that :
$|f(x) - l|<\delta $ $ \Leftarrow $ $ |x-a|<\alpha $
$\alpha , \delta > 0$
$|f(x) - l|= |\frac{x^2 +1-x}{x-2}+1|=|\frac{(x+1)(x-1)}{x-2}|$
$|f(x) - l|<\delta$ $\Rightarrow $ $|\frac{(x+1)(x-1)}{x-2}|<\delta$
$\Rightarrow $ $|x-1| |\frac{(x+1)}{x-2}|<\delta$
Suppose $x\in [\frac{1}{2}, \frac{3}{2}] $
$\Rightarrow $
$\frac{3}{2}\leq x+1\leq\frac{5}{2}$
and:
$\frac{-3}{2} \leq x-2\leq\frac{-1}{2}$
$\Rightarrow$ $\frac{2}{3}\leq \frac{-1}{x-2}\leq2$
$\Rightarrow $ $|\frac{x+1}{x-2}|\leq5$
$\Rightarrow $ $|x-1| |\frac{(x+1)}{x-2}|\leq5|x-1|$
we know that :
$|x-1| |\frac{x+1}{x-2}|<\delta$
So:
$5|x-1|<\delta$
$\Rightarrow $ $|x-1|<\frac{\delta}{5} $
We put $\alpha=\frac{\delta} {5}$
Finally :
After the definition of limite we proved $\lim_{x\to 1} f(x) =-1$
| {
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"source": "stackexchange",
"question_score": "6",
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If $x,y, z>0$ and $x+y+z=1$ then prove that $\frac 1x+\frac 1y+\frac 1z≥9$? Can some please help me out with this question?
I tried it in this way
Note that $a,b>0$ then $a≥b\iff\frac 1a≤\frac1b$ .
So it is sufficient to prove $\frac {1}{1/x+1/y+1/z}\leq\frac 19$ .
|
If $x,y,z>0$ and $x+y+z=1$, then prove that
\begin{align}\frac1x+\frac1y+\frac1z\ge9 \tag{1}\label{1}. \end{align}
Alternatively, using Ravi substitution,
where the triplet $a, b, c$ represents the sides of a valid triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$,
\begin{align}
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{2}\label{2}
,
\end{align}
\eqref{1} becomes
\begin{align}
\frac{3\rho^2-2(a+c+b)\,\rho+(ab+bc+ac)}
{\rho^3-(a+b+c)\rho^2+(ab+bc+ac)\,\rho-abc}
&\ge9
\tag{3}\label{3}
,\\
\frac{3\rho^2-2\,(2\rho)\,\rho+(\rho^2+r^2+4rR)}
{\rho^3-(2\,\rho)\rho^2+(\rho^2+r^2+4rR)\,\rho-4\rho r R}
-9
&\ge0
\tag{4}\label{4}
,
\end{align}
\begin{align}
\frac{4R+r-9\rho r}{\rho r}
&\ge0
\tag{5}\label{5}
.
\end{align}
Since $\rho=x+y+z=1$,
\eqref{5} simplifies to
\begin{align}
R-2r&\ge0
\tag{6}\label{6}
,
\end{align}
which always holds for a valid triangle.
The equality corresponds only to the equilateral triangle,
$a=b=c$, that is, $x=y=z=\tfrac13$
in the original \eqref{1}.
| {
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"answer_count": 3,
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} |
Matrix $A$ such that $A^{3} - A^{2} = 0$ and its eigenvalues We have the matrices $A \in M_{n \times n}$ and $X \in M_{n \times 1}$ such that $A \cdot X = \lambda \cdot X$ for any scalar $\lambda$. Let $q(x) = \alpha_{n} \cdot x^{r} + \alpha_{n-1} \cdot x^{r-1} + \cdots + \alpha_{1} \cdot x + \alpha$ be a polynomial equation and $q(A) = \alpha_{n} \cdot A^{r} + \alpha_{n-1} \cdot A^{r-1} + \cdots + \alpha_{1} \cdot A + \alpha \cdot I_{n}$.
In the book I'm studying there is a property of eigenvalues that goes like this: If $\lambda$ is an eigenvalue of $A$ and $q(A) = 0$ then $q(\lambda) = 0$.
If we know that $A$ is a matrix such that $A^{3} - A^{2} = 0$ and we have a polynomial equation $q(x) = x^{3} - x^{2}$ then we know that $q(A) = 0$. Shouldn't this tell us that the only possible eigenvalues of $A$ for any eigenvector are either $0 \lor 1$?
Because the only solutions for $q(\lambda) = 0$ are $\lambda = 0 \lor \lambda = 1$.
| if the minimal polynomial is $x^2(x-1)$ the Jordan form can be of this type:
$$
\left(
\begin{array}{ccc|cc|cc|ccc}
1&0&0&0&0&0&0&0&0&0 \\
0&1&0&0&0&0&0&0&0&0 \\
0&0&1&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&1&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&0&0&1&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\ \hline
0&0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0&0 \\
\end{array}
\right)
$$
The degree of $k$ each factor $(x-\lambda)^k$ in the minimal polynomial is the size of the largest Jordan block with that eigenvalue. Here, we are allowed one or more 2 by 2 Jordan blocks of Dietrich's type (in comment above).
| {
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"answer_count": 2,
"answer_id": 1
} |
Nested radicals like Ramanujan's infinite radicals $$\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+...}}}}}=?$$
This like the
My question is about "how to start this problem ?".I've been thinking for over two hours but get stuck at the end. I tried by a calculator to approximate the value, I find out if I define a sequence like below,It seems $a_n\to 2$ $$a_1=\sqrt1\\
a_2=\sqrt{1+\sqrt{5}}\sim 1.798907\\
a_3=\sqrt{1+\sqrt{5+\sqrt{11}}}\sim 1.97075\\
a_4=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19}}}}\sim 1.99661\\
a_5=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29}}}}}\sim 1.99967\\\vdots \\a_n \text{tends to } 2$$
But I have no idea to solve it analytically. Can someone help me? or get me the clue?
| $$2=\sqrt{1+\sqrt9}=\sqrt{1+\sqrt{5+\sqrt{16}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{25}}}}=$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}}=...$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+2)^2}}}}}=$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+1)^2+(n+1)-1+\sqrt{(n+3)^2}}}}}}=...$$
| {
"language": "en",
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"answer_id": 0
} |
Property of positive matrix Let $A$ be a $C^*$-algebra and let $a = \begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2}& \dots & a_{nn}\end{pmatrix} \in M_n(A)$ be a positive matrix.
Is it true that if $\lambda_1, \dots, \lambda_n \in \mathbb{C}$, then
$$\begin{pmatrix}\overline{\lambda}_1 & \overline{\lambda}_2 & \dots & \overline{\lambda}_n\end{pmatrix}\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2}& \dots & a_{nn}\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{pmatrix}$$ is a positive element of $A$?
I tried using the characterisation $a= x^*x$ but the computation became quite ugly so I was wondering if there is a conceptual easy way to see this.
| If $a$ is a positive element of $M_n(A)$, then there is a $b \in M_n(A)$ with $a = b^*b$. It follows that
$$
\begin{pmatrix}\overline{\lambda}_1 & \overline{\lambda}_2 & \dots & \overline{\lambda}_n\end{pmatrix}\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2}& \dots & a_{nn}\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{pmatrix} =
\left[b\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{pmatrix} \right]^*\left[b\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{pmatrix}\right].
$$
Now, write
$$
b \pmatrix{\lambda_1\\ \vdots \\ \lambda_n} = \pmatrix{c_1\\ \vdots \\ c_n}
$$
for some $c_1,\dots,c_n \in A$. From the above, we have
$$
\begin{pmatrix}\overline{\lambda}_1 & \overline{\lambda}_2 & \dots & \overline{\lambda}_n\end{pmatrix}\begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n}\\ a_{21} & a_{22} & \dots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2}& \dots & a_{nn}\end{pmatrix}\begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\end{pmatrix} =
\pmatrix{c_1 \\ \vdots \\ c_n}^* \pmatrix{c_1 \\ \vdots \\ c_n} =
\sum_{i=1}^n c_i^*c_i.
$$
Now, it suffices to note that the sum of positive elements is positive.
| {
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"answer_count": 1,
"answer_id": 0
} |
Multiplicative inverse of complex numbers proof I recently attempted to show that the multiplicative inverse for complex numbers exists and expressed it in complex form, as follows:
Suppose $z = a + bi$ is a non-zero complex number. Show that $z$ has a multiplicative inverse and express it in the form $c + di$.
Let $z^{-1}$ denote the multiplicative inverse of Z. Then,
$$z^{-1}z = 1 = zz^{-1}$$
$$\implies z^{-1}(a+bi) = 1 = (a+bi)z^{-1}$$
So,
$$z^{-1} = \frac{1}{a+bi}$$
Multiplying the numerator and denominator by the conjugate:
$$z^{-1} = \frac{a-bi}{a^2 + b^2}$$
$$z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$$
Thus, for all non-zero complex numbers $z$, there exists a multiplicative inverse, $z^{-1}$, where $z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$
QED.
However, I was told that this proof is circular because I assumed that the inverse exists. How can I rectify this?
Responses are much appreciated.
| What you have done is useful but it is not a proof. Now just verify that $$(a+bi)(\frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2}))$$ $$=(\frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2}))(a+bi)=1$$ by direct calculation.
| {
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"answer_count": 3,
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Use the division algorithm to establish that, The cube of any integer is either of the form $9k ,9k + 1, 9k + 8$.
Use the division algorithm to establish that, The cube of any integer is either of the form $9k ,9k + 1, 9k + 8$.
Let $a$ is an integer, write $a = 9k + r, 0 \le r < 9$ hence $r = \{0,1,2,3,4,5,6,7,8\}$ then $$a^3 =(9k + r)^3 = 9(9k^3 + 3kr(9k +r )) + r^3, 0 \le r³ < 9$$
when $r=0 \to r^3 = 0$
when $r=1 \to r^3 = 1$
when $r=2 \to r^3 = 8$
when $r=3 \to r^3 = 27$
Since $0\le r^3 < 9$ above $8$ values cannot be accepted. Hence $r^3\in{0,1,8}$. Hence $a^3$ can express in $a^3 = 9k, a^3 = 9k + 1, a^3 = 9k + 8$ forms. Therefore cube of any integer is form $9k, 9k + 1,$ or $9k + 8$.
Is this correct? Are there other solutions?
| This is a supplementary answer:
You want to be a little careful for why $27$ can be disregarded. Note that if $a=9k+27$ then $a=9(k+3)+0$. So what's really happening is that remainder $27$ can be rewritten to be remainder $0$. For example, consider what happens if you do $6$ instead of $9$.
Write $a=6k+r$ such that $0\leq r <6$. Then $a^3=6k'+r^3$ for some appropriate $k'$.
If $r=0$ then $r^3=0$.
If $r=1$ then $r^3=1$.
If $r=2$ then $r^3=8$.
If $r=3$ then $r^3=27$.
If $r=4$ then $r^3=64$.
If $r=5$ then $r^3=125$.
If I stopped after the first two and said that values above $5$ can be disregarded, then I might incorrectly conclude that the cube of any number can be written as $6k$ or $6k+1$ for some $k$. But in fact $8$ corresponds to remainder $2$ (mod $6$), $27$ to remainder $3$ (mod $6$), and $125$ to remainder $5$ (mod $6$).
So in fact cubes are represented by $6k$, $6k+1$, $6k+2$, $6k+3$, and $6k+5$.
| {
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"answer_id": 1
} |
Evaluate the limit $\lim_{x\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$ Evaluate the limit:
$$\lim_{n\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$$
Using the fact that ${(1 + x)^{1/2} \approx 1 + x/2}$ for "small" x,
I have that $\sqrt{n+1}\approx\sqrt{n}(\frac{1}{2n}+1)$ then $n\rightarrow \infty$. However, following this procedure I end up with the following limit: $\lim_{n\rightarrow \infty}2n^2=\infty$, but the answer is $\frac{1}{2}$. I would be thankful for any help.
| $$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})=$$
$$=\sqrt{n^3}\left(\frac{2}{\sqrt{n+2}+\sqrt{n}}-\frac{2}{\sqrt{n+3}+\sqrt{n+1}}\right)=$$
$$=\frac{2\sqrt{n^3}\left(\sqrt{n+3}-\sqrt{n+2}+\sqrt{n+1}-\sqrt{n}\right)}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+3})}=$$
$$=\tfrac{2\sqrt{n^3}\left(\frac{1}{\sqrt{n+3}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+3})}=\tfrac{2\left(\frac{1}{\sqrt{1+\frac{3}{n}}+\sqrt{1+\frac{2}{n}}}+\frac{1}{\sqrt{1+\frac{1}{n}}+1}\right)}{(\sqrt{1+\frac{2}{n}}+1)(\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{3}{n}})}\rightarrow\frac{2\left(\frac{1}{2}+\frac{1}{2}\right)}{2\cdot2}=\frac{1}{2}.$$
| {
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"source": "stackexchange",
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Finding $\frac{d^2y}{dx^2}$ for a curve given by $(x,y)=(t^2+2t,3t^4+4t^3)$. Why does my method yield the incorrect answer? The other day I saw what was, seemingly, a fairly simple question.
A curve in the $xy$-plane is given parametrically by the equations:
$$\begin{align}
x &=\phantom{3}t^2+2t \\
y &=3t^4+4t^3
\end{align}$$
for all $t>0$. Find the value of $\frac{d^2y}{dx^2}$ at $(8,80)$.
My method to obtain the correct answer is as follows:
$$\frac{dx}{dt}=2t+2 \tag1$$
$$(\frac{dx}{dt})^2=\frac{dx^2}{dt^2}=4t^2+8t+4 \tag2$$
$$\frac{dy}{dt}=12t^3+12t^2\implies\frac{d^2y}{dt^2}=36t^2+24t \tag3$$
$$\frac{\frac{d^2y}{dt^2}}{\frac{dx^2}{dt^2}}=\frac{d^2y}{dt^2}\frac{dt^2}{dx^2}=\frac{d^2y}{dx^2}=\frac{36t^2+24t}{4t^2+8t+4}\tag4$$
$(8,80)$ occurs when $t=2$. Plugging this value of $t$ into the equation above yields an answer of $\frac{16}{3}$. The answer, however, should be $4$.
Where did I go wrong and why, exactly, did this particular method fail to give the correct answer?
| As mentioned in the comments, you cannot divide the individual second derivatives to get the answer. The correct way of doing this is as follows:
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot\frac{dt}{dx}$$
we have
$$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{12t^2(t+1)}{2(t+1)} = 6t^2$$
and
$$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2(t+1)}$$
which means that the final answer would be
$$\frac{d^2y}{dx^2} = \frac{3t^2}{t+1}$$
substituting $t=2$ yields $4$, which is the correct answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Functional Equation (similar to Cauchy's): $ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) $
Solve the following functional equation: $ k \in \mathbb R $, $ f : \mathbb R \to \mathbb R $ and
$$ f ( x + y ) = f ( x ) f ( y ) + k x y ( x + y ) \text , \forall x , y \in \mathbb R \text . $$
I have only got $ f ( 0 ) = 1 $, by letting $ f ( 0 ) = a $ and solving for $ a $.
If I use this, everything cancels out and leaves me with $ f ( x ) = f ( x ) $ which is fruitless.
| Sketch:
Assume $k \ne 0$.
Note that
\begin{align}
f(x)f(-x)= f(0) = 1
\end{align}
then it follows $f$ is never zero and
\begin{align}
f(x) = \frac{1}{f(-x)}.
\end{align}
Also, we have that
\begin{align}
f(2x) = f(x)^2+2kx^3
\end{align}
which means
\begin{align}
1=&\ f(2x)f(-2x) \\
=&\ (f(x)^2+2kx^3)(f(-x)^2-2kx^3) \\
=&\ (f(x)^2+2kx^3)(\frac{1}{f(x)^2}-2kx^3)\\
=&\ 1+\frac{2kx^3}{f(x)^2}-2kx^3f(x)^2-4k^2x^6
\end{align}
Hence
\begin{align}
&0=2kx^3f(x)^4+4k^2x^6f(x)^2-2kx^3\\
&\implies f(x)^4+2kx^3f(x)^2-1=0
\end{align}
So, we have that
\begin{align}
f(x)^2=\frac{-2kx^3\pm \sqrt{4k^2x^6+4}}{2}= -kx^3\pm \sqrt{k^2x^6+1}
\end{align}
which means
\begin{align}
f(x) = \sqrt{\sqrt{k^2x^6+1}-kx^3}.
\end{align}
However, by plugging $f$ back into the functional equation, we immediately arrive at a contradiction. Hence $f(x)^4+2kx^3f(x)^2-1\ne 0$, which means $k = 0$. This reduces the functional equation to $f(x+y) = f(x)f(y)$, which means $f(x) = e^{cx}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$? We are given the following $n\times n$ matrix, whose terms are:
*
*$a+b$ for the terms in the diagonal ($i=j$).
*$a$ for the rest of the elements.
I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?
| \begin{align*}
& \begin{vmatrix}
a + b & a & a & \cdots & a \\
a & a + b & a & \cdots & a \\
a & a & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a & a & a & \cdots &a + b
\end{vmatrix} \\
= & \begin{vmatrix}
na + b & na + b & na + b & \cdots & na + b \\
a & a + b & a & \cdots & a \\
a & a & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a & a & a & \cdots &a + b
\end{vmatrix} \\
= & (na + b)\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
a & a + b & a & \cdots & a \\
a & a & a + b & \cdots & a\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a & a & a & \cdots &a + b
\end{vmatrix} \\
= & (na + b)\begin{vmatrix}
1 & 1 & 1 & \cdots & 1 \\
0 & b & 0 & \cdots & 0 \\
0 & 0 & b & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & b
\end{vmatrix} \\
=& (na + b)b^{n - 1}
\end{align*}
Can you figure out what operations I used in each step?
| {
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"answer_count": 4,
"answer_id": 3
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Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$ Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$
I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:
$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$
I haven't been taught that as equal to $tan x$.
| By Tangent half-angle formulas with $t=\tan x$
*
*$\cos 2x=\frac{1-t^2}{1+t^2}$
*$\sin 2x=\frac{2t}{1+t^2}$
we have
$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\frac{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}=\frac{1+t^2-1+t^2+2t}{1+t^2+1-t^2+2t}=\frac{2t+2t^2}{2+2t}=t$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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All solutions $(x, y, z) \in (\mathbb{N},\mathbb{N},\mathbb{N})$ for $3x^2 + 6y^2 = z^2$ My attempt:
$z^2 = 3(x^2 + 2y^2)$
so $3|z^2$ and thus $3|z$
letting $z = 3k$, we reduce the equation to $x^2+2y^2=3k^2$
Obviously $(n,n,n) $ represents infinitely many solutions, but not all since $(5,1,3)$ is a solution for example.
Also, if $x$ and $k$ are even then $4|2y^2$, then $y$ is even
If $x$ and $k$ are odd, then $x^2$ and $k^2$ are congruent to 1 mod 8 so $2y^2$ is congruent to 2 mod 8 so y is also odd.Therefore, it's sufficient to find the odd solutions.
But I don't know how to proceed from here.
Any hints on how to parametrize the odd solutions or a suitable mod $n$ to consider to simplify it further would be appreciated.
| There is a systematic way to solve such equation: the idea is to parametrize the ellipse $C: 3X^2+6Y^2=1$ by a family of lines through a fixed rational point. (in other words, an ellipse is a rational curve).
Here $A=({1\over 3}, {1\over 3})$ is a rational point on this ellipse. Note that a line through $A$ has equation $L_t:(X-1/3)=t(Y-1/3)$ must cut the ellipse in exactly one otherpoint which is a rational fonction of $t.
It is preferable to write $C= (3.(X-1/3)^2+6.(Y-1/3)^2+6(X-1/3)+12(Y-1/3)=0$
Let $U=X-1/3, V=Y-1/3$
The intersection of $L_t\cap C$ is $U=tV, 3U^2+6V^2+6U+12V=0$
Or $3t^2V^2+6V^2+6tV+12V=0$
If we exclude the point A $(V=0)$ we get; $V(3t^2+6)=-6T-12$, $V=-{t+4\over t^2+3}, U=t V$
Now $t={a\over b}\in \bf Q$, ${x\over z}=U+1/3=-t{t+4\over t^2+3}+1/3$ and ${y\over z}=V+1/3=-{t+4\over t^2+3}$
${x\over z}=-{a^2+4ab\over a^2+3b^2}+1/3={-2a^2+3b^2-12ab\over a^2+3b^2}$
${y\over z}=-{a/b+4\over (a/b)^2+3}+1/3=$
$-{ab+4b^2\over a^2+3b^2}+1/3=$ $a^2-3ab-9b^2\over 3(a^2+b^2)$
So $x=-2a^2+3b^2-12ab$, $y=a^2-3ab-9b^2$, $z= 3(a^2+b^2)$ is a solution if $(a,b,c)\in \bf Z$.
Conversely, for every solution, the point $ P=(x/z, y/z)$ is on the ellipse, so $(x,y,z)$ is proportional to the triple $(-2a^2+3b^2-12ab , a^2-3ab-9b^2,3(a^2+b^2))$, for $(a,b)$ the slope of the line $(A,P)$.
Note that a different choice for $A$ yields a different parametrization of the set of solution...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$(a^4+b^2) \leq (a^2+b^2)^2$ for all $a, b \in \mathbb{R}_+$? Let $a, b \in \mathbb{R}_+$.
Question. Is valid that
$$(a^4+b^2) \leq (a^2+b^2)^2?\tag{1}$$
I tried what follows:
$$(a^2+b^2)^2=a^4+2a^2b^2+b^4 \geq a^4+2a^2b^2.$$
For this, can I to conclude that $(1)$ holds?
| If $b\ne 0$ then $b^2>0$ so $(b^2C\le b^2D\iff C\le D)$ for any $C,D.$
So with $C=1$ and $D=2a^2+b^2,$ if $b\ne 0$ then $$a^4+b^2\le (a^2+b^2)^2 \iff a^4+b^2\le a^4+2a^2b^2+b^4 \iff$$ $$\iff b^2\le 2a^2b^2+b^4\iff$$ $$\iff b^2(1)\le b^2(2a^2+b^2)\iff$$ $$\iff 1\le 2a^2+b^2.$$ The last inequality above cannot be true for every $a,b>0.$ E.g. if $a=b=1/1000000000000.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence and the limit of $x_n=(1+\frac{1}{2})(1+\frac{1}{2^2})...(1+\frac{1}{2^{2^n}})=\prod_{i=0}^n (1+\frac{1}{2^{2^i}})$ for $n \ge 0$ It is easy to see that the sequence is bounded below by $1$ and increasing as we multiply the terms by a number larger than $1$ each time.
I was also able to show that $x_n \le a_n$ for all $n \ge 0$ where $a_n=\sum_{i=0}^n \frac{1}{2^i}$. Since we know $(a_n)_n$ is bounded above by $2$, so do $(x_n)_n$ and now we know it also converges.
However, I couldn't find its limit. Can you help me on finding its limit?
Thanks for your effort and time in advance.
| $$x_n=\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\\=\frac{\left(1-\frac 1 2\right)\left[\left(1+\frac 1 2\right)\left(1+\frac 1 {2^2}\right)\cdots\left(1+\frac 1 {2^{2^n}}\right)\right]}{\left(1-\frac 1 2\right)}\\=\frac{\left(1-\frac 1 {2^{2^{n+1}}}\right)}{\left(1-\frac 1 2\right)}.$$
Thus, $$\lim x_n=\frac 1{1/2}=2.$$
See $x_n=1+\frac 1 2+\frac 1 {2^2}+\frac 1 {2^3}+\frac 1 {2^4}+\cdots+\frac 1 {2^{1+2+\cdots +2^{2^n}}}>\sum_{i=0}^n \frac 1 {2^i}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a$ is sufficiently large as compared with $b,$ and $\sqrt \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the value of $k?$
If $a$ is sufficiently large as compared with $b,$ and $\sqrt
\frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the
value of $k?$
Query
Can we find it using particular value?,like putting $a=4,b=2,$ then we get
$k=4\sqrt2+4\sqrt \frac{2}{3}-8$
This problem is given in Multiple Choice Question exercise,the possible answers given are
A.$\frac{2}{3}$
B.$\frac{3}{4}$
C.$\frac{4}{5}$
D.$\frac{5}{6}$
None of the answer mathches with mine....
Please give some hint to determine value of $k?$
| Suppose that for small $\epsilon$:
$$(1+\epsilon)^{-1/2}=1+\binom{-1/2}{1}\epsilon+\binom{-1/2}{2}{\epsilon^2}+O(\epsilon^3)=1-\frac12\epsilon+\frac{3}{8}\epsilon^2+O(\epsilon^3)$$
If you put $-\epsilon$ instead of $\epsilon$ into the last expression you get:
$$(1-\epsilon)^{-1/2}=1+\frac12\epsilon+\frac{3}{8}e^2+O(\epsilon^3)$$
It means that:
$$(1+\epsilon)^{-1/2}+(1-\epsilon)^{-1/2}=2+\frac34\epsilon^2+O(\epsilon^3)$$
If you put $\epsilon=b/a$, you are done. Obviously $k=\frac34$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good? I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:
$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$
so, the solutions are:
$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$
I am not sure if that should be done like I did it.
| I would rather mark (easier to find a mistake) $$z+{1\over z} = c$$ where $c$ is real and now solve $$z^2-zc+1=0$$ It discriminant is $c^2-4$ so $$z_{1,2}= {c\pm \sqrt{c^2-4}\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Study the convergence of $ \sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$ I have to study the convergence of the following series
$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$
What I tried was to compare the following terms:
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}} = \frac{1}{n^{5 / 6}}$$
We know that
$$\frac{1}{n^{5/6}} \rightarrow 0 \text{, as } n \rightarrow \infty$$
so using the First Comparison Test I concluded that the series
$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$
is convergent. However, I think I got something wrong. The First Comparison Test tells us that if the series is convergent, then the limit of the term that makes the series is $0$, not the other way around. So I think my argument is invalid. How should I approach this?
| Your inequality
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}} = \frac{1}{n^{5 / 6}}$$
is fine but since $\sum \frac{1}{n^{5 / 6}}$ diverges it is inconclusive.
As an alternative we can use limit comparison test with $\sum \frac{1}{n^{5 / 6}}$ to conclude for divergence or by direct comparison we have
$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \ge \frac{\sqrt{n}}{n \sqrt[3]{n}+\sqrt n} = \frac{1}{n^{5 / 6}+1} \ge \frac{1}{n^{5 / 6}+n^{5 / 6}}=\frac12\frac{1}{n^{5 / 6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Differentiate logarithmically the expressions for $\sin{\theta}$ and $\cos{\theta}$ in factors and deduce the sums to infinity. (i)$$\frac{1}{ \theta^2 - \pi^2} + \frac{1}{ \theta^2- 2^2\pi^2} + \frac{1}{ \theta- 3^2\pi^2 }... $$
(ii)$$\frac{1}{1^2 + x^2} + \frac{1}{2^2 +x^2} + \frac{1}{3^2 + x^2}... $$
These are the problems given in "Differential Calculus for Beginners"" by Joseph Edwards. I am beginner and have no idea how to approach such problems. The phrase 'differentiate logarithmically' is confusing me.
| The phrase "the expressions for $\sin\theta$ and $\cos\theta$ in factors" seems to refer to the infinite product representations of those functions.
For (i):
\begin{align*}
\sin\theta &=
\theta \; \prod_{n=1}^{\infty}\left(1 - \frac{\theta^2}{n^2 \pi^2}\right)\\
\frac{d}{d\theta}\ln(\sin\theta) &=
\frac{d}{d\theta}\left[\ln\theta \;+\; \sum_{n=1}^{\infty}\ln\left(1-\frac{\theta^2}{n^2\pi^2}\right)\right]\\
&=
\frac{1}{\theta} \;+\; 2\theta \sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2}
\end{align*}
From there you can solve for the sum in question, to yield:
$$
\sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2} \;=\;
\frac{1}{2\theta}\left(\cot\theta \;-\; \frac{1}{\theta}\right)
$$
For (ii), the simplest thing to do is to use the above result with $\theta\rightarrow i \pi x$.
| {
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Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$ Month ago I encounter a nice result numerically checked by Mathematica
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2
$$
where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.
At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like
$$
\int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi}
$$
just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.
So I question it here for some further discussion, and thanks in advance for any suggestion.
| Here is a solution based on @G Cab's remarkable observation.
Reduction. Let $I$ denote the integral, and apply the substitution $(\theta,\varphi)\mapsto(2\theta,2\varphi)$ to write
$$ I = 4 \int_{0}^{\pi/4} \int_{0}^{\pi/4} \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \, \mathrm{d}\theta\mathrm{d}\varphi. $$
If we define the $1$-form $\omega$ by
$$ \omega = \left( \log\frac{\cot\theta}{\cot\varphi} \right) \biggl( \frac{\mathrm{d}( (\theta-\varphi)^2 )}{2\sin^2(\theta-\varphi)} - \frac{\mathrm{d}( (\theta+\varphi)^2 )}{2\sin^2(\theta+\varphi)} \biggr), $$
then $\omega$ is smooth in the region $(0, \pi/2)^2$ via continuation along the diagonal $\varphi = \theta$. Moreover, @G Cab's computation for the differential reduces to1)
$$ \mathrm{d}\omega
= 4 \left( \frac{2\theta\cot(2\theta) - 2\varphi \cot(2\varphi)}{\cos(2\theta)-\cos(2\varphi)} \right) \, \mathrm{d}\theta \wedge \mathrm{d}\varphi. $$
Now let $\epsilon > 0$ be small and consider the triangular region
$$ \mathcal{T}_{\epsilon} = \bigl\{ (\theta, \varphi) : \epsilon \leq \varphi \leq \theta \leq \tfrac{\pi}{4} \bigr\}. $$
Its boundary $\partial\mathcal{T}_{\epsilon}$ consists of three oriented line segments
Then by the symmetry and the Stokes' theorem,
\begin{align*}
I
&= 2 \lim_{\epsilon \to 0^+} \int_{\mathcal{T}_{\epsilon}} \mathrm{d}\omega
= 2 \lim_{\epsilon \to 0^+} \int_{\partial \mathcal{T}_{\epsilon}} \omega
= 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}),
\end{align*}
where $\rightarrow$, $\uparrow$, and $\swarrow$ denote the oriented bottom/height/hypothenus of $\partial\mathcal{T}_{\epsilon}$, respectively, and
$$ I_{\bullet} = \lim_{\epsilon \to 0^+} \int_{\bullet} \omega $$
for each $\bullet \in \{ \rightarrow, \uparrow, \swarrow \}$.
Integral Computations. We will evaluate each $I_{\bullet}$:
1. It is not hard to check that $\omega = 0$ along $\varphi = \theta$, and so, we get $I_{\swarrow} = 0$.
2. For $I_{\uparrow}$, we note that
\begin{align*}
I_{\uparrow}
&= \int_{\varphi = 0}^{\varphi = \pi/4} \omega|_{\theta=\pi/4} \\
&= \int_{\varphi = 0}^{\varphi = \pi/4} \left( - \log\cot\varphi \right) \biggl( \frac{\mathrm{d}( (\pi/4-\varphi)^2 )}{2\sin^2(\pi/4-\varphi)} - \frac{\mathrm{d}( (\pi/4+\varphi)^2 )}{2\sin^2(\pi/4+\varphi)} \biggr) \\
&= 2 \int_{0}^{\pi/4} \left( \log\cot\varphi \right) \biggl( \frac{\pi/4-\varphi}{1-\sin(2\varphi)} + \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \biggr) \, \mathrm{d}\varphi \\
&= 2 \int_{-\pi/4}^{\pi/4} \frac{\pi/4+\varphi}{1+\sin(2\varphi)} \log \left| \cot\varphi \right| \, \mathrm{d}\varphi \\
&= \frac{1}{2} \int_{0}^{\pi} \frac{x}{1-\cos x} \log \left| \cot \left(\frac{x}{2} - \frac{\pi}{4}\right) \right| \, \mathrm{d}x,
\end{align*}
where the last step is the result of the substitution $\varphi=\tfrac{x}{2}-\tfrac{\pi}{4}$. Then by noting that
$$ \log \left| \cot \left(\frac{\pi}{4} - \frac{x}{2} \right) \right|
= \operatorname{Re}\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)
= i (\arctan(e^{-ix}) - \arctan(e^{ix})), $$
we get
\begin{align*}
I_{\uparrow}
&= \frac{1}{2}\int_{0}^{\pi} \frac{ix}{1-\cos x} (\arctan(e^{-ix}) - \arctan(e^{ix})) \, \mathrm{d}x \\
&= \frac{1}{2}\int_{-\pi}^{\pi} \frac{ix}{1-\cos x} \left(\frac{\pi}{4} - \arctan(e^{ix})\right) \, \mathrm{d}x \\
&= i \int_{\mathcal{C}} \frac{(\log z )(\frac{\pi}{4} - \arctan z)}{(z-1)^2} \, \mathrm{d}z,
\end{align*}
where the substitution $z = e^{ix}$ is applied to the last step and $\mathcal{C}$ is the path parametrized by $e^{ix}$ for $-\pi < x < \pi$. Now deforming the contour $\mathcal{C}$ so as to enclose the principal branch cut for $\log$,
\begin{align*}
I_{\uparrow}
&= 2\pi \int_{0}^{1} \frac{\frac{\pi}{4} + \arctan t}{(t+1)^2} \, \mathrm{d}t
= 2\pi \int_{0}^{1} \frac{1}{(t^2+1)(t+1)} \, \mathrm{d}t
= \frac{\pi^2}{4} + \frac{\pi}{2} \log 2.
\end{align*}
3. For the integral along the bottom,
\begin{align*}
\int_{\rightarrow} \omega
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \omega|_{\varphi=\epsilon} \\
&= \int_{\theta = \epsilon}^{\theta = \pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\mathrm{d}( (\theta-\epsilon)^2 )}{2\sin^2(\theta-\epsilon)} - \frac{\mathrm{d}( (\theta+\epsilon)^2 )}{2\sin^2(\theta+\epsilon)} \biggr) \\
&= \int_{\epsilon}^{\pi/4} \log\left( \frac{\cot\theta}{\cot\epsilon} \right) \biggl( \frac{\theta-\epsilon}{\sin^2(\theta-\epsilon)} - \frac{\theta+\epsilon}{2\sin^2(\theta+\epsilon)} \biggr) \, \mathrm{d}\theta.
\end{align*}
Substituting $\theta = \epsilon x$,
\begin{align*}
\int_{\rightarrow} \omega
&= \int_{1}^{\pi/4\epsilon} \log\left( \frac{\cot(\epsilon x)}{\cot\epsilon} \right) \biggl( \frac{\epsilon^2 (x-1)}{\sin^2(\epsilon(x-1))} - \frac{\epsilon^2 (x+1)}{2\sin^2(\epsilon(x+1))} \biggr) \, \mathrm{d}x.
\end{align*}
It is not hard (although not entirely trivial) to show that the above integral converges to
\begin{align*}
I_{\rightarrow}
= \lim_{\epsilon \to 0^+} \int_{\rightarrow} \omega
= \int_{1}^{\infty} (-\log x) \biggl( \frac{1}{x-1} - \frac{1}{x+1} \biggr) \, \mathrm{d}x
= -\frac{\pi^2}{4}.
\end{align*}
Conclusion. Combining all these observations altogether, we have
$$ \bbox[10px,#ffd]{ I = 2 (I_{\rightarrow} + I_{\uparrow} + I_{\swarrow}) = \pi \log 2 } $$
as desired.
${}^{1)}$ We include the sketch of computation for $\mathrm{d}\omega$, which is essentially @G Cab's computation in backward direction. Substituting $u=\theta-\phi$ and $v=\theta+\phi$,
$$ \omega = \left( \log \left( \frac{\sin v - \sin u}{\sin v + \sin u} \right) \right)\biggl( \frac{u \, \mathrm{d}u}{\sin^2 u} - \frac{v \, \mathrm{d}v}{\sin^2 v} \biggr) . $$
Taking $\mathrm{d}$,
\begin{align*}
\mathrm{d}\omega
&= \mathrm{d}\left( \frac{u}{\sin^2 u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}u \\
&\qquad - \mathrm{d}\left( \frac{v}{\sin^2 v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \wedge \mathrm{d}v \\
&= \frac{u}{\sin^2 u} \left( \frac{\partial}{\partial v} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}v \wedge \mathrm{d}u\\
&\qquad - \frac{v}{\sin^2 v} \left( \frac{\partial}{\partial u} \log \left( \frac{\sin v - \sin u}{\sin u + \sin v} \right) \right) \mathrm{d}u \wedge \mathrm{d}v \\
&= 4 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) \, \mathrm{d}u \wedge \mathrm{d}v.
\end{align*}
By noting that $\mathrm{d}u \wedge \mathrm{d}v = 2 \, \mathrm{d}\theta \wedge \mathrm{d} \varphi$, the question boils down to showing
$$ 2 \left( \frac{v \cos u \csc v - u \cos v \csc u}{\cos(2u) - \cos (2v)} \right) = \frac{(v+u)\cot(v+u) - (v-u)\cot(v-u)}{\cos(v+u) - \cos(v-u)}. $$
This can be verified by expanding the right-hand side.
| {
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"question_score": "17",
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Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$ Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$ where we are using polar coordinates.
I was wondering if you guys could show me multiple ways to do this problem. I did it by using implicit differentiation, but whenever I ask questions here I get a bunch of brilliant solutions so I thought I'd ask. Thanks!
| $$r^2=\sin(2\theta)$$
$$r^2=2\sin\theta\cos\theta\to r^4=2(r\sin\theta)(r\cos\theta)$$
as $r\cos\theta=x;\;r\sin\theta=y$ we get the cartesian equation
$$(x^2+y^2)^2=2xy\to f(x,y)=(x^2+y^2)^2-2xy=0$$
I use the formula for implicit diferentiation
$$\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}$$
So we have
$$\frac{dy}{dx}=\frac{2 y-4 x \left(x^2+y^2\right)}{4 y \left(x^2+y^2\right)-2 x}$$
Horizontal tangents are where derivative is zero.
$$2 y-4 x \left(x^2+y^2\right)=0\to x^2+y^2=\frac{y}{2x}$$
Plug in the equation of the curve
$$\left(\frac{y}{2x}\right)^2=2xy\to y=0;\;y=8x^3$$
The first solution is then $(0,0)$
Plugging $y=8x^3$ in the equation of the curve we get
$$(x^2+64x^6)^2=16x^4\to x^4 \left(64 x^4-3\right) \left(64 x^4+5\right)=0$$
$x=\pm\sqrt[4]{\frac{3}{64}}\approx \pm 0.4653$ and $y\approx \pm 0.806$.
The points where the tangents are vertical are the symmetric wrt line $y=x$ bisector of the first quadrant.
$$
.
$$
| {
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"url": "https://math.stackexchange.com/questions/3896924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?
| $x^2+y^2=(x+y)^2-2xy=16-4=12.$
$x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$.
Now you can use your factorization:
$x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Why $\frac{(1-x)\log(1-x)}{x\log x}$ is increasing on (0, 1)? Let $f(x)=\frac{(1-x)\log(1-x)}{x\log x}$. I find one claims this function is increasing on $(0,1)$. While I find the numerator of $f'(x)$ is
$$-x\log x -\log x\log(1-x)-(1-x)\log(1-x).$$
I am wondering why this is non-negative?
| (Not a complete answer but hopefully close enough)
You want to show that
$$ x \log(x)+ (1-x)\log(1-x) \le - \log(x) \log(1-x)$$
Divide on both sides by $\log(x) \log(1-x)$ and this is
$$ \frac{x}{\log(1-x)} + \frac{1-x}{\log(x)} \le -1 $$
This is of the form
$$ f(x) + f(1-x) \le c$$
where $f(x) = \frac{x}{\log(1-x)}$. I don't know if there is a nice way to show this.
But the not-so-nice way is to Taylor expand
$$ \frac{x}{\log(1-x)} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dotsm $$
$$ \frac{1-x}{\log(x)} = -1 + \frac{1-x}{2} + \frac{(1-x)^2}{12} + \frac{(1-x)^3}{24} + \dotsm $$
Now you add them up, you see that the RHS becomes
$$ -2 + \frac{1}{2} + \frac{x^2 + (1-x)^2}{12} + \frac{x^3 + (1-x)^3}{24} + \dotsm $$
$$ \le -2 + 1 + \frac{1}{12} + \frac{1}{24} + \dotsm $$
But then showing that the last sum is $\le -1$, seems clear to me but maybe you might want more concrete justification...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of the improper integral $\int_0^2\frac x{(16-x^4)^{1/3}}\,\mathrm dx$ Need your help in the checking convergence of this improper integral :$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}$$I tried my luck and somehow was able to establish a relationship between $(1-t^4)$ and $t^4$ i.e $$(1-t^4)\geq t^4$$
$$\frac{1}{1-t^4} \leq \frac{1}{t^4}$$
$$\Rightarrow \frac{1}{(1-t^4)\frac{1}{3}} \leq \frac{1}{t^\frac{4}{3}}$$ $$\Rightarrow \frac{t}{(1-t^4)\frac{1}{3}} \leq \frac{t}{t^\frac{4}{3}}$$
for some $$t\in\left[0,\frac{1}{2^\frac{1}{4}}\right]$$ It can be shown that$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}=2^\frac{2}{3}\int_0^1\frac{tdt}{(1-t^4)^\frac{1}3} $$
Therefore ,
$$F(x) =\int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{(1-t^4)^\frac{1}3} \leq \int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{t^\frac{4}{3}} = g(x)$$ For this particular interval I was able to prove its convergence as $g(x)$ converges but I am unable to comment on the remaining interval. Please help and also provide some insight which might help me in future. Thanks in advance...
| The problem with your approach, as you observed, is that the inequality $1-t^4\ge t^4$ doesn't hold near $t=1$, which is where the impropriety in the integral occurs. Instead it's better to argue along the following lines:
For $0\le x\le2$, we have
$$16-x^4=(2-x)(8+4x+2x^2+x^3)\ge8(2-x)$$
and thus
$${x\over(16-x^4)^{1/3}}\le{2\over2(2-x)^{1/3}}=(2-x)^{-1/3}$$
so that
$$\int_0^2{x\over(16-x^4)^{1/3}}\,dx\le\int_0^2(2-x)^{-1/3}\,dx=-{3\over2}(2-x)^{2/3}\big|_0^2={3\over2^{1/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational? $$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.
I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
Sorry if this is a stupid question.
| Note that
$$
\sqrt[3]{2+\sqrt{5}}=\frac{1}{2}\sqrt[3]{16+8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3+3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2+(\sqrt{5})^3}=\frac{1+\sqrt{5}}{2}.
$$
Similarly,
$$
\sqrt[3]{2-\sqrt{5}}=\frac{1}{2}\sqrt[3]{16-8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3-3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2-(\sqrt{5})^3}=\frac{1-\sqrt{5}}{2}.
$$
Thus,
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1,
$$
as desired.
As for the second question, we can use the @player3236's approach. Let
$$
a=\sqrt[3]{x+\sqrt{y}},~b=\sqrt[3]{x-\sqrt{y}}.
$$
Then, for $c=a+b$ we have
$$
c^3=2x+3c\sqrt[3]{x^2-y},
$$
so $c$ is a root of the polynomial $t^3-3\sqrt[3]{x^2-y}\cdot t-2x$. If $c=a+b=1$, then
$$
1-3\sqrt[3]{x^2-y}-2x=0,
$$
or
$$
\sqrt[3]{y-x^2}=\frac{2x-1}{3},
$$
or
$$
y=\frac{1}{27}(27x^2+(2x-1)^3).
$$
We've obtained some necessary condition. However, I'm not sure whether this condition is sufficient or not (I guess that proof should be similar to case $x=2$, $y=5$, which was discussed above).
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $ 2^{16^x} = 16^{2^x} $. Find $x$ such that $ 2^{16^x} = 16^{2^x}.$
I am a bit confused, what will happen when we expand $16^{2^x}$, will we get $4^{2^{2^x}}$ or $4^{2^{x+1}}$ ?
| Alternative approach
$$16^{2^x} = [2^4]^{2^x} = 2^{4 \times 2^x}.$$
Since this is equal to $2^{16^x}$ you have
$$16^x = 4 \times 2^x \implies 8^x = 4 \implies x = (2/3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$ without using calculus? Hi mathematics stack exchange, what is the minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$? I know how to solve this problem using calculus, you take a derivative, but I am wondering if there is an elementary method to find the minimum using precalculus methods.
| Let $y = x^2$, notice
$$\begin{align}
x^2 + (y-3)^2 & = x^2 + (x^2-3)^2 = x^4 - 5x^2+9\\
(x-3)^2+(y+1)^2 & = (x-3)^2 + (x^2 + 1)^2 = x^4 - 3x^2 -6x + 10\end{align}$$
The problem at hand can be rephrased as:
Given $A = (0,3)$, $B = (3,-1)$ and $P = (x,y)$ be a point on the parabola $y = x^2$. What is the minimum value of $AP + PB$?
If one make a plot of the parabola $y = x^2$, one will notice the parabola intersect with the line segment $AB$, this means the minimum value of $AP + PB$ is $$AB = \sqrt{(0-3)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
| The equation resolves to $(2^k+x)(2^k-x) \mid (2^k+1)(2^k-1)$. Since we have the first factor larger than the rest, we would look for a common factor in $(x- 1)$ and $(2^k-1)$ or $(x+1)$ and $2^k+1$,
We could suppose that the first is a product of say $(ab)(cd)$, and that the second is a product of $(ac)(bd)e$, where $ac$ divides $2^k+1$ and $bd$ divides $2^k-1$. The common divisor between the first two and the last two factors, is $x+1$, and between the second and third, and the first and fourth, $x-1$.
But this common factor must also divide $2^k+1$ and $2^k-1$, and so must divide $2$.
So there is no other number which divides pairs of divisors (ie $2^k \pm 1$ and $2^k\pm x$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903856",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "14",
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Continuity of the function $f(x,y)=\frac{\cos(x+y)+\cos(x-y) -2}{x^2 +y^2 }$
Find the natural domain of the function
$$f(x,y)=\frac{\cos(x+y)+\cos(x-y) -2}{x^2 +y^2 }$$
find the points where it's continuous and for each accumulation point of it's natural domain, determine whether or not the function can be expanded to a continuous one.
My attempt:
I think the natural domain is $\mathscr D_f=\Bbb R^2\setminus\{0,0\}$ and that the function is continuous at every $x\in\mathscr D_f$. The denominator is always positive.
I also think $f(x,y)$ does have a limit as $(x,y)\to (0,0)$.
I was looking at the Taylor expansion of cosine:
$\cos x = 1 -\frac12x^2 +\frac{1}{24} x^4 + \cdots$
so
$\begin{aligned}\cos(x+y)&=1-\frac12(x+y)^2+\frac1{24}(x+y)^4 + \cdots\\\cos(x-y)&=1-\frac12(x-y)^2+\frac1{24}(x-y)^4 + \cdots\end{aligned}$
The point is that $(x+y)^2 +(x-y)^2 =2x^2 +2y^2,$ so
$\cos(x+y)+\cos(x-y)-2=-(x^2 +y^2) +\frac1{12} (x^4 +6x^2 y^2 +y^4 )+\cdots$
which shows that
$\lim\limits_{(x,y)\to(0,0)} f(x,y) = -1$
so there is then a removable discontinuity at $(0,0)$ and therefore, we can extend $f(x,y)$ there.
Is this correct? If so, is there anything I could improve or any easier approach?
Thank you very much!
| Corrrect. Another way: $\cos(x\pm y)=\cos x \cos y \mp \sin x \sin y$.
So $\cos(x+y)+\cos(x-y)=2\cos x \cos y.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$
I've calculated several values of this function, and it seems to have a limit of $1$.
| $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$
$$\therefore \ (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
$$ = \left(2 \left(x^{1/2} - \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} -...\right) + x^{1/2} \sin \frac{1}{x}\right)^x$$
$$ = \left(x^{1/2} \left(2 + \sin \frac{1}{x} - \frac{2x^{3/2}}{3!} + \frac{2x^{5/2}}{5!} -...\right) \right)^x$$
$$ = \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x.$$
Now $ \left(x^{1/2}\right)^x = \left(x^{x}\right)^{1/2} = (e^{x \ln x})^{1/2} = e^{\frac{1}{2} x \ln x},\ $ and $\ \lim\limits_{x\rightarrow 0^{+}}e^{\frac{1}{2}x\ln(x)}=e^{\lim\limits_{x\rightarrow 0^{+}}\frac{1}{2}x\ln(x)}=1. $
Lastly, for $ \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x,\ $ make the substitution $u = \frac{1}{x}$ to get:
$\lim\limits_{u\rightarrow \infty} \left( 2 + \sin u - \frac{2}{3!u} + \frac{2}{5!u^2} - ...\right)^{\frac{1}{u}}$, and the bracket oscillates between $1$ and $3$ for large $u$, so this limit is $1$. Thus:
$$ \lim\limits_{x\rightarrow 0^{+}} \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x = 1 \times 1 = 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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proof with induction $2^n + 4 > n^2 + 2n$ prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here
| It is true that for $n\geq5$
$$2^n+4>n^2+2n$$
Assume that for $n=k$
$$2^k+4>k^2+2k$$
Let's show for $n=k+1$
$$2^{k+1}+4>(k+1)^2+2(k+1)$$
Now
$$2(2^k+4)>2(k^2+2k)$$
$$2^{k+1}+4>2k^2+2k-4$$
$$2^{k+1}+4>2k^2+2k-4>(k+1)^2+2(k+1)$$
$$2k^2+2k-4>k^2+4k+3$$
$$k^2-2k-7>0$$
for $k\geq5$
$$(k-1)^2-8>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$ For $x,y,z \in [2,\infty)$, prove that
$\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$
I tried to group the terms and prove that $\frac{xy+z}{x+yz} - \frac{y}{3}\leq \frac{1}{3}$
| using the lemma:if $a,b\ge 2$ then $ab\ge a+b$
$$\sum_{cyc} \frac{xy+z}{x+\color{red}{yz}}-\sum \frac{x}{3}$$ $$\le \sum_{cyc} \frac{xy+z}{x+\color{red}{y+z}}-\sum\frac{x}{3}=1-\dfrac{1}{6}\Big( \dfrac{{(x-y)}^2+{(y-z)}^2+{(z-x)}^2}{x+y+z}\Big)\le 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that $(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$ If $a,b,c \in \mathbb{R}$ and $a+b+c = 1$, prove that
$(2a+b)(2b+c)(2c+a)+(1+a+2b)(1+b+2c)(1+c+2a) \leq 9$
| Homogenize the inequality by writing (replace "1"s):
$$
(2a+b)(2b+c)(2c+a)+(a+b+c+a+2b)(a+b+c+b+2c)(a+b+c+c+2a) \leq 9(a+b+c)^3
$$
Evaluating the terms gives the following equivalent inequality to be shown
$$
a^3 + b^3 + c^3 - 3 abc \ge 0
$$
Now since the variables are not restricted to nonnegative ones, this must be proved "by hand" (not by AM-GM). Observe the identity, which can be shown by multiplying everything out:
$$
a^3 + b^3 + c^3 - 3 abc = \frac12 (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)
$$
Since $a+b+c=1$, this gives the required
$$
a^3 + b^3 + c^3 - 3 abc = \frac12 ((a-b)^2 + (b-c)^2 + (c-a)^2) \ge 0
$$
and equality only holds for $a=b=c=\frac13$.
$\qquad \Box$
| {
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How to prove that $a^3 + b^3 \geq a^2b + ab^2$? Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--
For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$
I have no idea how to begin but this is what I did.
By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)
Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$
Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.
(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)
| You can't apply AM-GM, as we need that $a,b\ge 0$ and not $a+b>0$.
Let's use the $a+b>0$ condition. Dividing by $a+b>0$ gives that it's enough to prove
$a^2-ab+b^2 \ge ab$
$\iff (a-b)^2 \ge 0$
The reason we divide is that we remember that $a^3+b^3=(a+b)(a^2-ab+b^2)$ since $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+...+b^{n-1})$ for odd positive $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose $F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y}F(y)$ for all $x ,y \in \mathbb{R}_+$. Is $F$ convex? Let $F : \mathbb{R}_+ \to \mathbb{R}$ be a function which is controlled in the following way: $$F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y} F(y).$$ Is this function convex?
This estimate is satisfied by convex functions, i.e., $F(x) = e^x$ and $F(x)=x^2$: \begin{eqnarray*}
(x+y) e^{x+y} &=& xe^{x+y} + y e^{x+y} \ \geq \ x e^x + ye^y, \\
(x+y)(x+y)^2 & =& x \cdot x^2 + y \cdot y^2 + (\text{positive cross terms})
\end{eqnarray*}
It might be worth noting that if $t = \frac{x}{x+y} \in (0,1]$, then the defining estimate reads: $$F\left( \frac{x}{t} \right) \geq t F(x) + (1-t) F \left( \frac{x}{t} -1 \right).$$
| $f$ does not have to be convex.
Try for example $f(x)=-x^{-½}$, for $x>0$, which is concave and satisfies
$$
f(x+y)>\frac{x}{x+y}f(y)+\frac{x}{x+y}f(y)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating the integral is not giving the same simplified function I calculated this integral below :
$$
\int_{-\pi/2}^{\pi/2} \cos(x)\cos(nx) dx
$$
And I got this result :
$\dfrac{\left(\left(n-1\right)\sin\left(\frac{{\pi}n+{\pi}}{2}\right)+\left(n+1\right)\sin\left(\frac{{\pi}n-{\pi}}{2}\right)\right)}{n^2-1}$
I had a check at the end at the solution :
$-\dfrac{2\cos\left(\frac{{\pi}n}{2}\right)}{n^2-1}$
So apparently there is some kind of simplification to do but I don't actually know how .
| Observe that
$$\sin\left(\frac{\pi n+ \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)+ \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = \cos\left(\frac{\pi n}{2}\right)$$
$$\sin\left(\frac{\pi n- \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)- \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = -\cos\left(\frac{\pi n}{2}\right)$$
when $n\in\mathbb{N}$.
Then,
\begin{equation*}
\begin{split}
(n-1)\sin\left(\frac{\pi n+ \pi}{2}\right) & + (n+1)\sin\left(\frac{\pi n- \pi}{2}\right) = (n-1)\cos\left(\frac{\pi n}{2}\right) - (n+1)\cos\left(\frac{\pi n}{2}\right) \\
& = -2\cos\left(\frac{\pi n}{2}\right)
\end{split}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x) d \theta}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} } x \in \Bbb{R}$ Compute
$$ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x)}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} }
\,\,d \theta\,,\quad x \in \Bbb{R}$$
The denominator makes it a bit annoying. We can evaluate it as
$ [(\cos \theta - x)^2 + \sin^2 \theta = (1 - 2 \cos \theta x + x^2)^{1.5}$ but it doesn't make it any easier. We can also expand $(1 + x^2 - 2 \cos \theta x )^{1.5}$ using $(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$ but it's still difficult.
The denominator also has some geometric meaning (cubed distance from $(x,0)$ to $(\cos \theta, \sin \theta)$ ) but does it help?
| Consider the sequence of substitutions:
$$u=\cos\theta-x\implies\mathrm du=-\sin\theta\,\mathrm d\theta$$
$$\implies f(x)=-\int_{1-x}^{-1-x}\frac u{(u^2+1-(u+x)^2)^{\frac32}}\,\mathrm du=\int_{-1-x}^{1-x}\frac u{(1-2ux-x^2)^{\frac32}}\,\mathrm du$$
Then
$$v=1-2ux-x^2\implies\mathrm dv=-2\,\mathrm du$$
$$\implies f(x)=\frac1{4x}\int_{x^2+2x+1}^{-3x^2+2x+1}\frac{v+x^2-1}{v^{\frac32}}\,\mathrm dv$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $x^4-3x^2+18$ irreducible over the $3$-adics? Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?
It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.
| If it's reducible, it has a factor of degree $1$ or $2$. I'll leave it up to you to
show there is no factor of degree $1$ (i.e. no root in $\mathbb Q_3$). So suppose
$x^2 + \alpha x + \beta$ divides $x^4 - 3 x^2 + 18$. The remainder on this division
is $$(-\alpha^3 + 2 \alpha \beta + 3 \alpha) x - \alpha^2 \beta + \beta^2 + 3 \beta + 18$$
If $\alpha = 0$ we'd have $\beta^2 + 3 \beta + 18 = 0$, which has no root in $\mathbb Q_3$,
so $\alpha \ne 0$ and $\beta = (\alpha^2-3)/2$. Substituting in to the remainder,
this gives us $$-\frac{\alpha^4}{4} + \frac{3}{2} \alpha^2 + \frac{63}{4} = 0$$
which has no solution in $\mathbb Q_3$.
EDIT: Here's one way to see $\beta^2 + 3 \beta + 18=0$ has no solution in $\mathbb Q_3$. If $\beta$ has $3$-adic order $m$, then $\beta^2$ has order $2m$ and $3\beta$ has order $m+1$, while of course $18$ has order $2$.
So the only possibility is $m=1$.
Thus we have the $3$-adic expansion $\beta = \beta_1 3^{1} + \beta_2 3^{2} + \ldots$, with $\beta_i \in \{0,1,2\}$. But these three possibilities make $\beta^2 + 3\beta + 18 =
2 \cdot 3^2 + \ldots, 1 \cdot 3^2 + \ldots, 2 \cdot 3^2 + \ldots$ respectively, and not $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| $6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
$6\sin{x}\cos{x} -9\sin^2{x} - \cos^2{x} -2\cos{x} + 6\sin{x} - 1=0$
$-(3 \sin x - \cos x)^2 + 2(3 \sin x - \cos x) - 1= 0$
$(3 \sin x - \cos x)^2 - 2(3 \sin x - \cos x) + 1= 0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What kind of mathematical rule was broken in here? I just wanted to write the iterated version of:
$$ a^2 - b^2 = (a+b)(a-b)$$
As:
$$ n^x - 1 = (n^{ \sqrt{x}} +1) \color{red}{(n^{ \sqrt{x}} -1)} = \\
(n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1) \color{red}{(n^{ \sqrt{\sqrt{x}}} -1)} = \\
(n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) \color{red}{(n^{ \sqrt{\sqrt{\sqrt{x}}}} - 1)} = \dots
$$
And so:
$$n^x - 1 = (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) ... \Rightarrow \\n^x - 1 = \prod_{k=1}^{\infty} (n^{x^{\left (\frac{1}{2k}\right)}} + 1)$$
But when plugging $n=1$ we get:
$$ 1^x - 1 \equiv 0 = \prod_{k=1}^{\infty} (1^{x^{\left (\frac{1}{2k}\right)}} + 1) \equiv \prod_{k=1}^{\infty} (1+1) = \prod_{k=1}^{\infty} 2 = \infty$$
Where did I go wrong? I think that I "omitted" the $n^{x^{(\frac{1}{2k})}} - 1$ part, but isn't it valid when we talk about infinite series? I would appreciate if you could clear things up.
Thank you!
| Should be instead $$n^x-1 = (n^{x/2}-1)(n^{x/2}+1)$$,
as in general,
$$n^x-1 \not = (n^{\sqrt{x}}-1)(n^{\sqrt{x}}+1).$$
This was your error.
| {
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"url": "https://math.stackexchange.com/questions/3924358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Decompose $\frac{1}{a \sin x +b}$ into sum of trigonometric functions I am wondering if it is possible to write
$$f(x)=\frac{1}{a \sin x +b}$$
as a sum of functions that don't have trigonometric functions in the denominator? Or just pure trigonometric functions, without a constant added. So maybe something like this:
$$\frac{1}{a \sin x +b}=A\sin^3 x \cos^2x + B\sin x \cos ^4 x + C\frac{1}{\cos x}+ D\frac{\cos^2 x }{\sin x}+\dots $$
The reason I'm asking is that I need to integrate products of $f(x)$ with powers $\cos^n x$, and it would be really neat to get rid of the constant in the denominator, as integrals of the form $\int\frac{\cos^n x}{a \sin x +b}dx$ are extremely hard to solve, whereas integral tables contain solutions to $\int \sin^p x \cos^q x dx$ and similar integrals.
One potential starting point could be
$$\frac{1}{a \sin x +b}=\frac{a \sin x -b}{(a \sin x +b)(a \sin x -b)}=\frac{a \sin x -b}{(a^2 \sin^2 x -b^2)}=...$$
with the goal to eventually make the replacement $\sin^2 x +\cos^2 x =1$. Can such a method be formalized? Or is the decomposition I have in mind provable not possible? Thanks for your help!
| Assume $b>a$ such that $b+a\sin(x)\ne0$. Then, per the Fourier series
$$\frac{\sqrt{b^2-a^2}}{b+a\sin(x)}=1+2\sum_{n=1}^{\infty}\left(\frac{\sqrt{b^2-a^2}-b}{a}\right)^n[\sin\frac{n\pi}2\sin{(nx )}+ \cos\frac{n\pi}2\cos{(nx )} ]$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this series converge I need to check by Cauchy's convergence test if this series converges:
$$a_{n}=\frac{\sin(5)}{1*2} - \frac{\sin(5^2)}{2*3}+\ldots+(-1)^{n+1}\frac{\sin(5^n)}{n(n+1)}$$
I started to value
$$|a_{n+p} - a_{n}| < \epsilon$$
But don't know how to continue from:
$$\Biggl|(-1)^{n+2}*\frac{\sin(5^{n+1})}{(n+1)(n+2)}+\ldots+(-1)^{n+p+1}\frac{\sin(5^{n+p})}{(n+p)(n+p+1)}\Biggr|=\ldots$$
| Let's the call the expression you have obtained as $S$. Applying triangle inequality to $S$ and using the fact that $|\sin(x)|\leq 1$ we get,
\begin{align*}
S \leq \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)}+ \cdots + \frac{1}{(n+p)(n+p+1)}
\end{align*}
By using the well known identity$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, the above expression simplifies to
\begin{align*}
\frac{1}{n+1} - \frac{1}{n+2}+\cdots+\frac{1}{n+p}-\frac{1}{n+p+1} = \frac{p}{(n+1)(n+p+1)}.
\end{align*}
Now choose an $n$ such that $\frac{1}{n+1}<\frac{\epsilon}{2} $. This implies
\begin{align*}
&&0 < \frac{1}{n+p+1}\leq\frac{1}{n+1}<\frac{\epsilon}{2}\\
\implies && S\leq \frac{p}{(n+p+1)(n+1)}<\epsilon
\end{align*}
Hence the above series is convergent
| {
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"timestamp": "2023-03-29T00:00:00",
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How can we find the derivative of a circle if a circle is not a function? If we consider the equation of a circle, $x^2+y^2=r^2$, then I understand that $dy/dx$ can be computed in the following way via implicit differentiation:
\begin{align}
2x + 2y\frac{dy}{dx} &= 0 \\
\frac{dy}{dx} &= -\frac{2x}{2y} = -\frac{x}{y} \, .
\end{align}
Although I feel comfortable deriving this result, I don't really understand how I should interpret it. On an intuitive level, the formula $dy/dx = -x/y$ seems to suggest that the gradient of the tangent to any given point $(x,y)$ is $-x/y$. However, since the curve $x^2+y^2=r^2$ fails the vertical line test, it doesn't look like it is even a function. Usually, $dy/dx$ can be thought of as a shorthand for
$$
\lim_{h \to 0}\frac{y(x+h)-y(x)}{h} \, .
$$
However, in this case each $x$-value maps to two $y$-values, and so the limit definition doesn't seem to apply here. So what does $dy/dx$ actually represent in this context?
| You can still think of it in terms of the slope of a tangent line, and even in terms of a limit. However, as you've pointed out, $x^2 + y^2 = r^2$ isn't a function because it fails the vertical line test. However, by the Implicit Function Theorem we can consider $F(x,y) = x^2 + y^2 - r^2$, and for any $(x_{0},y_{0})$ where $\frac{\partial F}{\partial y}\ne 0$ then there exists some neighborhood around the point $(x_{0},y_{0})$ for which we can express $F(x,y) = 0$ as some function $y = f(x)$. Note that in this case, $$\frac{\partial F}{\partial y} = 2y,$$ which is zero whenever $y = 0$, so at the points $(r,0)$ and $(-r,0)$. On the circle. Anywhere else we can define the curve by either $$y = \sqrt{r^{2} - x^{2}}$$ or $$y = -\sqrt{r^{2} - x^{2}}$$ and these functions are differentiable so long as $y\ne 0$. To verify that the derivative via the limit definition matches that obtained by implicit differentiation, we can compute as follows for the positive semicircle:
\begin{align}
y' &= \lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h}\\
&=\lim_{h\to 0}\frac{\sqrt{r^{2} - (x+h)^{2}} - \sqrt{r^2 - x^2}}{h} \cdot \frac{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=\lim_{h\to 0}\frac{r^{2} - (x+h)^{2} - r^{2} - x^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2xh -h^{2}}{h(\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2})}\\
&=\lim_{h\to 0}\frac{-2x - h}{\sqrt{r^{2} - (x+h)^{2}} + \sqrt{r^2 - x^2}}\\
&=-\frac{2x}{2\sqrt{r^{2} - x^{2}}}\\
&=-\frac{x}{y}.
\end{align}
Where the last equality comes from the equation $y = \sqrt{r^2 - x^2}.$ We could do the same for the negative semicircle.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the SVD of a matrix with one zero eigenvalue? I am trying to find the SVD of the matrix \begin{pmatrix}1&2\\3&6\end{pmatrix}
$A^TA=\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}10&20\\20&40\end{pmatrix}$
$\lambda_1=50$
$\lambda_2=0$
$\lambda_1>\lambda_2>=0$
$\Sigma=\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}$
When $\lambda_1=50$:
$v_1=\begin{pmatrix}1\\2\end{pmatrix}$
Normalize:
$u_1=\begin{pmatrix}\frac{1}{\sqrt{5}}\\\frac{2}{\sqrt{5}}\end{pmatrix}$
When $\lambda_1=0$:
$v_2=\begin{pmatrix}-2\\1\end{pmatrix}$
Normalize:
$u_2=\begin{pmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{pmatrix}$
$V=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{-2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$V^T=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$A=U\Sigma V^T$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}u_1&u_2\\u_3&u_4\end{pmatrix}\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}5\sqrt{2}u_1&0\\5\sqrt{2}u_3&0\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}\sqrt{10}u_1&2\sqrt{10}u_1\\\sqrt{10}u_3&2\sqrt{10}u_3\end{pmatrix}$
$u_1=\frac{1}{\sqrt{10}}$
$u_3=\frac{3}{\sqrt{10}}$
How can I find $u_2$ and $u_4$?
| You have correctly noticed that the second column of $U$ doesn't really matter in terms of satisfying $A=U\Sigma V^\top$. In order to be a proper SVD, you just need to find a vector orthonormal to the first column of $U$ to make $U$ an orthogonal matrix (but again, this is not necessary for the equality $A = U \Sigma V^\top$ to hold).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find, in radians the general solution of cos 3x = sin 5x I am studying maths as a hobby. I have come across this problem:
Find a general solution for the equation cos 3x = sin 5x
I have said, $\sin 5x = \cos(\frac{\pi}{2} - 5x)$
so
$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$
When I add $(\frac{\pi}{2} - 5x)$ to $2n\pi$ I get the answer $x = \frac{\pi}{16}(4n +1)$, which the book says is correct.
But when I subtract I get a different answer to the book. My working is as follows:
$3x = 2n\pi - \frac{\pi}{2} + 5x$
$2x = \frac{\pi}{2} - 2n\pi$
$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$
but my text book says the answer is $\frac{\pi}{4}(4n + 1)$
Is the book wrong?
| $$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x $$
$$3x=\frac{\pi}{2}-5x+2k\pi$$
$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$
or
$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$
$$x=\frac{\pi}{4}-k\pi$$
$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$
where $k\in Z$
writing $-k\pi$ or $k\pi$ does not change the solution set. Because $-k$ is the opposite of $k$ in integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Power of a point tangents Two circles $A$ and $B$ with centers $P$ and $Q$, respectively, are externally tangent to each other. The power of point $P$ with respect to circle $B$ is $8$. The power of point $Q$ with respect to circle $A$ is $15$. What is the ratio of the areas of circle $A$ and circle $B$?
How do I go about solving this problem? Is there perhaps a formula for ratios of areas of circles given different powers of points?
| Power of a point $X$ wrt circle $(O,r)$ is $OX^2-r^2$
Given is $$(a+b)^2-b^2=8 \quad , \quad (a+b)^2-a^2=15$$
Define $x:= a/b$
$$\dfrac{(a+b)^2-a^2}{(a+b)^2-b^2}=\dfrac{15}{8}$$
$$\Rightarrow \dfrac{(x+1)^2-x^2}{(x+1)^2-1^2}=\dfrac{15}{8}$$
$$\Rightarrow x=\dfrac{2}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$. Let $f$ be differentiable function and $y=-2x+1$ be tangent line to graph of $y=f(x)$ at the point $x=2$.
How can I find the equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.
I created a function $f (x)$ myself, but it didn't work. Any ideas?
| Find $f(2)$ and $f'(2)$ as follows $$\begin{align} y \quad &= \quad f(2) + f'(2)(x-2) \\&=\quad f'(2)x + f(2) - 2f'(2) \\&=\quad -2x + 1 \end{align}$$
Since the coefficients should be equal, we get
$$\begin{cases} f'(2) = -2 \\ f(2) = -3 \end{cases}$$
Now, take $g = f^2$. To find the tangent equation, we will need the derivative of $g$ at $2$. $$g'(2) = [f(x)^2]'\bigg|_2 = 2f(2)f'(2) = 2(-3)(-2) = 12$$
Hence, the required tangent is
$$\begin{align} y \quad &= \quad g(2) + g'(2)(x - 2) \\&=\quad (-3)^2 + 12(x-2) \tag{$g(2) = f(2)^2$} \\&=\quad 12x - 15 \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$ A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$.
Here's the link:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$
The proof given there left some details for me, in particular I have to prove that
$$
\sum_{k = 1}^{n}{1 \over k}
\sin\left(k\pi\frac{2m+1}{n+1}\right) >
\sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right)
$$
for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$
where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$.
Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?
| Remark: One year ago, I analyzed the monotonicity of the function values at all local minimizers,
when I tried to answer Inequality $\sum\limits_{1\le k\le n}\frac{\sin kx}{k}\ge 0$ (Fejer-Jackson).
Here, I give the analysis of the monotonicity of the function values at all local maximizers.
Problem: Given positive integer $n\ge 3$, all the local maximizer of $f(x) = \sum_{k=1}^n \frac{\sin kx}{k}$ on $(0, \pi)$ are given by $x_m = \frac{2m+1}{n+1}\pi, m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor$.
Prove that $f(x_m) > f(x_{m+1})$ for all $m=0, 1, \cdots, \lfloor \frac{n-1}{2}\rfloor - 1$.
Proof: Since $f(x) = \int_0^x f'(t) \mathrm{d} t + f(0)$, we have
\begin{align}
f(x_{m+1}) - f(x_m) &= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+3}{n+1}\pi} f'(t) \mathrm{d} t
= \int_{\frac{2m+1}{n+1}\pi}^{\frac{2m+2}{n+1}\pi} \Big( f'(t) + f'(t + \tfrac{\pi}{n+1})\Big) \mathrm{d} t. \tag{1}
\end{align}
Since $f'(x) = \sum_{k=1}^n \cos kx = \frac{\sin \frac{nx}{2} \cos \frac{(n+1)x}{2}}{\sin \frac{x}{2}}$, we have,
for all $t\in [\frac{2m+1}{n+1}\pi, \frac{2m+2}{n+1}\pi]$,
\begin{align*}
&f'(t) + f'(t + \tfrac{\pi}{n+1})\\
=\ & \frac{\sin \frac{nt}{2} \cos \frac{(n+1)t}{2}}{\sin \frac{t}{2}} + \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi)
\cos (\frac{(n+1)t}{2} + \frac{\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\
=\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\
& - \frac{\sin (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi)\sin (\frac{(n+1)t}{2} + \frac{\pi}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{2}\\
=\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin \frac{t}{2}} \\
& - \frac{\cos (\frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2})\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})}\\
\le\ & - \frac{\sin (\frac{nt}{2} - m\pi)\sin (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\
& - \frac{\cos (\frac{nt}{2} - m\pi )\cos (\frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2})}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \tag{3}\\
\le\ & - \frac{\cos \frac{\pi - t}{2}}{\sin (\frac{t}{2} + \frac{\pi}{2n+2})} \\
\le\ & 0
\end{align*}
where in (2) we have used
$\sin(\alpha - m\pi) \sin (\beta - \frac{(2m+1)\pi}{2}) = -\sin \alpha \cos \beta$,
and in (3) we have used
\begin{align}
&0 < \frac{nt}{2} + \frac{n}{2n+2}\pi - m\pi - \frac{\pi}{2} < \frac{nt}{2} - m\pi < \pi, \\
&0 \le \frac{(n+1)t}{2} - \frac{(2m+1)\pi}{2} \le \frac{\pi}{2},\\
&0 < \frac{t}{2} < \frac{t}{2} + \frac{\pi}{2n+2} < \frac{\pi}{2}.
\end{align}
Also, note that $f'(\frac{2m+1}{n+1}\pi) + f'(\frac{2m+1}{n+1}\pi + \tfrac{\pi}{n+1}) = -1 < 0$.
From (1), we have $f(x_{m+1}) < f(x_m)$. We are done.
| {
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"question_score": "5",
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Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$ Question:
Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$
My process:
I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.
So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.
Here is the expression I got:
$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$
Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.
And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?
| $g(x)=(x-a)(x-b)$ where $a=-1+\sqrt 3 i, b=-1-\sqrt 3 i$ are cubic roots of 8.
For $n=1, 2, \ldots, 6, a_n-a$ are the roots of $h_1 (x) = f(x+a)$, $a_n-b$ are the roots of $h_2(x)=f(x+b)$.
Recall that $a^3=b^3=8$, then the desired product is the product of $h_1(x)$ and $h_2(x)$'s constant terms
$$
h_1(0) \cdot h_2(0) = (a^6-2a^3-8)\cdot (b^6-2b^3-8) = (8^2-2 \cdot 8 - 8)^2 = 40^2=1600.
$$
| {
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} |
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$
I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$
From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$
Hence $1+9abc\ge 4(ab+bc+ac)$.
This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?
| My second solution:
By using Schur's inequality and the identity
$$a^3 + b^3 + c^3 + \frac{15}{4} abc =
\frac{3}{4}[a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)] + \frac{1}{4}(a+b+c)^3,$$
we have
$$a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$
If $d \ge \frac{15}{4}$, we have
$$a^3 + b^3 + c^3 + abcd \ge a^3 + b^3 + c^3 + \frac{15}{4} abc \ge \frac{1}{4}.$$
If $d < \frac{15}{4}$, we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= a^3 + b^3 + c^3 + \tfrac{15}{4} abc - (\tfrac{15}{4} - d)abc \\
&\ge \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{a+b+c}{3})^3\\
&= \frac{1}{4} - (\tfrac{15}{4} - d)(\tfrac{1}{3})^3\\
&= \frac{1}{9} + \frac{d}{27}.
\end{align}
We are done.
| {
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} |
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?
| Bioche's rules say you should set $\;u=\tan \frac t2,\enspace \mathrm du=\frac12(1+u^2)\,\mathrm dt$, whence the integral of a rational function
$$\int\frac{2(1-u^2)\,\mathrm du}{(1+u^2)(u^2+2u+3)},$$
which is easily calculated using partial fractions decomposition.
Some more details: using the half angle formulæ, we get
\begin{align}
\frac{\cos t}{2 + \sin t + \cos t}\,\mathrm dt&=\frac{\cfrac{1-u^2}{1+u^2}}{2+\cfrac{2u}{1+u^2}+\cfrac{1-u^2}{1+u^2}}\,\frac{2\,\mathrm du}{1+u^2} \\
&=\frac{2(1-u^2)}{(1+u^2)\bigl(2(1+u^2)+2u+1-u^2\bigr)} \\
&=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}. \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Cyclic inequality with a+b+c=3 Let $a;b;c$ be non-negative real numbers such that $a+b+c=3$
Prove:
$(a+b)(b+c)(c+a) \geq (a+bc)(b+ca)(c+ab)$
Can someone prove that problem without using $pqr$ or $uvw$ method? Thanks so much.
| Using the AM-GM inequality, we have
$$(a + bc)(b + ca) \leqslant \left[\frac{(a + bc)+(b + ca)}{2}\right]^2 = \frac{(a + b)^2(c + 1)^2}{4} .$$
Therefore
$$(a+bc)(b+ca)(c+ab) \leqslant \frac{\displaystyle \prod (a+b) \prod (a+1)}{8}.$$
It remains to prove that
$$(a+1)(b+1)(c+1) \leqslant 8.$$
Which is true because
$$(a+1)(b+1)(c+1) \leqslant \left(\frac{a+1+b+1+c+1}{3}\right)^3=8.$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
For which a the matrix is positive-definite We have $A=\begin{pmatrix}a&2\\ 2&a+1\end{pmatrix}$
And I want to know for which a the matrix is positive-definite.
So our requirement is $det(A)\ge0$
$(a*(a+1))-4\ge0$
$a^2+a-4\ge0$
I understand that in the points $\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$ the function are equal to zero.
Is it true to say that because the function is equal to zero in the points so $ ∀a>\frac{-1+\sqrt{17}}{2},\frac{-1-\sqrt{17}}{2}$
$a^2+a-4\ge0$?
| You can prove a matrix is positive definite by checking if the determinant of all of its leading minors are greater then 0.
$$\begin{vmatrix} a\end{vmatrix}>0 \iff a>0 \space\space\space(1)$$
$$\begin{vmatrix}
a & 2 \\
2 & a+1
\end{vmatrix} = a^2+a-4$$
$$a^2+a-4>0 \iff a>\frac{-1+\sqrt{17}}{2} \space\space\space OR \space\space\space\space a<\frac{-1-\sqrt{17}}{2}\space\space\space\space\space(2)$$
From $(1)$ and $(2)$ we can get that A is positive definite for any $a>\frac{-1+\sqrt{17}}{2}$
| {
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Find point farthest on $(x-5)^2+(y+1)^2+(z-7)^2=69$ from point $(2, 3, -6)$ So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.
I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.
| The line connecting $P(2,3,-6)$ and the farthest point $M(a,b,c)$ is parallel to $M$’s normal surface vector $(a-5,b+1,c-7)$, i.e.
$$\frac{a-2}{a-5}=\frac{b-3}{b+1}=\frac{c+6}{c-7}$$
Substitute into $(a-5)^2+(b+1)^2+(c-7)^2=69$ to obtain the point
$$M\left(5+3\sqrt{\frac{69}{194}}, -1-4\sqrt{\frac{69}{194}}, 7+13\sqrt{\frac{69}{194}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949141",
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"source": "stackexchange",
"question_score": "1",
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How to prove this identity: $4\cdot(\frac12)!=\pi$ Progress: I recently found this identity on an YouTube Video: $$4\cdot\left(\frac{1}{2}\right)!=\pi$$
But it didn't provide any kind of rigorous proof or simply a 'proof' in that regard. It only showed some Calculus identities to give some insight. Also importantly this is the only identity or equation I know that exactly equals to '$\pi$'. The part which is very interesting is $(\frac{1}{2})!$, the factorial of a fraction not an integer.
So if anyone can give a rigorous proof and even elementary proof or share more heuristics and insights, it would be useful. I'll really appreciate your efforts, thanks.
| Note that gamma function is defined as$$\Gamma{(n)}=\int x^{n-1}e^{-x}dx$$
for $n>0$, $n\in\mathbb{R}$.
Gamma function is generalization of factorial function.
$$\Gamma (n)=(n-1)!.$$
Gamma function has property
$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$
and
$$\Gamma(n+1)=n\Gamma(n).$$
So,
\begin{align}
4\left(\dfrac{1}{2}\right)!&=4\left(\dfrac{3}{2}-1\right)!\\
&=4\Gamma\left(\dfrac{3}{2}\right)\\
&=4\Gamma\left(1+\dfrac{1}{2}\right)\\
&=4\cdot \dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\\
&=2\sqrt{\pi}.
\end{align}
I guess you want to prove
$$4\left(\left(\dfrac{1}{2}\right)!\right)^2=\pi.$$
\begin{align}
4\left(\left(\dfrac{1}{2}\right)!\right)^2&=4\left(\left(\dfrac{3}{2}-1\right)!\right)^2\\
&=4\left(\Gamma\left(\dfrac{3}{2}\right)\right)^2\\
&=4\left(\Gamma\left(1+\dfrac{1}{2}\right)\right)^2\\ &=4\cdot
\left(\dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\right)^2\\ &=4\cdot
\left(\dfrac{1}{2}\sqrt{\pi}\right)^2\\ &=4\cdot \dfrac{1}{4}\pi\\
&=\pi. \end{align}
| {
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"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Analyzing the Coefficient of a generating function and its asymptotic The generating function
$\sum_{n\geq 0} D(n) x^n = \frac{1}{\sqrt{1-6x+x^2}}$ is the gf of the Delannoy number.
See last paragraph in https://en.wikipedia.org/wiki/Delannoy_number
In this link, they mentioned that the coefficient of this gf is $D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}$.
My question is how can this coefficient be calculated/extracted from its gf.And how they conclude that it behaves asymptotically as
${\displaystyle D(n)={\frac {c\,\alpha ^{n}}{\sqrt {n}}}\,(1+O(n^{-1}))}$
where
$\alpha =3+2{\sqrt {2}}\approx 5.828$
and
$c=(4\pi (3{\sqrt {2}}-4))^{-1/2}\approx 0.5727$.
How to calculate/extract $D(n)$ from its gf and how to analyze it asymptotically then?
Is that possible to do that with a direct calculations?
Edit
I tried to find the asymptotic of the codfficidnt in the above gf this way:
The gf has two poles: $r_1=3+\sqrt{8}$, $r_2=3-\sqrt{8}$. So the smallest/dominant pole is $r_2=3-\sqrt{8}$.
We can write the gf as:
$A(x)=1/ \sqrt{r_2-x} * 1/ \sqrt{r_1-x}$
Then A(x) behaves as:
$1/ \sqrt {r_1-r_2} * 1/ \sqrt{r_2-x}=
1/ \sqrt {2 * 8^{1/4}} * \sum_{n\geq 0} {-1/2 \choose n} (-1)^n (r_1)^{n+1}x^n$
Here I used the expansion's formula of $(1+x)^a$ for $a=-1/2$ and $x=-x/r_2$.
So the coefficient behaves asymptotically as:
$1/ \sqrt 2 * 1/ 8^{1/4} * (3+\sqrt 8)^{n+1/2} {-1/2 \choose n} (-1)^n$.
We can simplify it more, by using the asymptotic of ${-1/2 \choose n}$.
Is that approach fine?
| For the asymptotics we use the Wilf text, Theorem 5.3.1 (page 179)
(Darboux) as suggested in the comments. We seek the asymptotics of
$$[z^n] \frac{1}{\sqrt{1-6z+z^2}}
= [z^n] \frac{1}{\sqrt{(z-(3+2\sqrt{2}))(z-(3-2\sqrt{2}))}}
\\ = \frac{1}{(3-2\sqrt{2})^n} (3-2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{(z-(3+2\sqrt{2}))(z-(3-2\sqrt{2}))}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{((3-2\sqrt{2})z-(3+2\sqrt{2}))
((3-2\sqrt{2})z-(3-2\sqrt{2}))}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{((17-12\sqrt{2})z-1)
(z-1)}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{(1-(17-12\sqrt{2})z)
(1-z)}}.$$
Now here we have one as the dominant singularity on the circle of
convergence and the theorem applies, taking the parameter $\beta =
-1/2.$ Expanding the term containing the subdominant singularity around
one we get for the first (constant) term
$$\frac{1}{\sqrt{1-(17-12\sqrt{2})\times 1}}
= \frac{1}{\sqrt{12\sqrt{2}-16}}
= \frac{1}{2\sqrt{3\sqrt{2}-4}}.$$
This gives the asymptotic
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2\sqrt{3\sqrt{2}-4}}
(3+2\sqrt{2})^n
{n-1/2\choose n}.}$$
The Wilf text gives $O(n^{-3/2})$ for the error in this approximation,
in sync with Wikipedia.
Wilf quotes on the same page an asymptotic for the remaining binomial
coefficient namely
$${n-\alpha-1\choose n} \sim \frac{n^{-\alpha-1}}{\Gamma(-\alpha)}$$
with $\alpha$ not a nonnegative integer. In the present case we have
$\alpha = -1/2$ so we obtain
$$\frac{1}{\sqrt{n}} \frac{1}{\Gamma(1/2)} =
\frac{1}{\sqrt{n}} \frac{1}{\sqrt{\pi}}.$$
This gives the form from the Wikipedia entry which is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2\sqrt{\pi(3\sqrt{2}-4)}}
\frac{1}{\sqrt{n}}
(3+2\sqrt{2})^n.}$$
| {
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If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.
If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.
What I Tried: I found $1446921630 = 2 * 3^5 * 5 * 7 * 11^2 * 19 * 37$ , but that did not really give me useful information.
A little information I got is that $m$ and $n$ each will have a factor of $11$ , and at-least $1$ factor of $3$, as both will be divisible by $3$ and $11$ .
I could have assumed the numbers to be of the form $10x + y$ or something, but I can't as I don't know how many digits both $m$ and $n$ will have, and that will more like Trial and Error.
Another thing is that among $m$ and $n$, one will be even and one will be odd. The odd number will start with an even digit which is not $0$, and the even number cannot end with $0$. Also $5$ should divide the odd number and it will end with $5$. That is all I could conclude.
Can anyone help me?
| Just another approach...
$1$st, estimate the number of digits in $m,n$:
$$
\sqrt{1446921630} \approx 38038,
$$
so we expect that $m,n$ are of the form
$$
m = 10^4 a+10^3b+10^2c+10d+e, \\
n = 10^4 e+10^3d+10^2c+10b+a, \\
$$
where $a,b,c,d,e$ are just digits ($a,e \ne 0$).
$2$nd,
$$
m n = 10^8ae+10(\ldots) + ae,
$$
so:
a) $ae\le 14$;
b) $ae$ has last digit $0$.
From here we have only one pair for $a\le e$:
$$a=2, \;e=5.$$
$3$rd,
now $mn$ has the form
$$
mn = 10\cdot 10^8 + 10^7(2d+5b) + 10^2(...) + 10(2d+5b) + 2\cdot 5.
$$
and we have estimation:
$$
mn > 10\cdot 10^8+10^7(2d+5b).
$$
$$
mn < (2\cdot 10^4 + 10^3(b+1))(5\cdot 10^4+10^3(d+1)) \le 10\cdot 10^8 + 10^7(2d+5b)+10^6\cdot 100.
$$
from here we conclude:
a) $34.69 <2d+5b < 44.69$
b) last digit of $2d+5b$ is $2$.
Definitely $2d+5b = 42$.
Therefore, only $2$ possible cases work for $b,d$:
$b=6, d=6$;
$b=8, d=1$.
$4$th: so we have $2$ candidates:
a) $m = 26?65$, $n = 56?62$,
b) $m = 28?15$, $n = 51?82$.
case a) is too big: even if $?=0$, we'll have $mn>14.6\times 10^8$.
case b) leads to $m=28215, n=51282$.
| {
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Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different Find number of skew-symmetric matrices of order $3\times 3$ in which all non-diagonal elements are different and belong to the set $\{-9,-8,-7,...,7,8,9\}$
My Attempt:
I did a simple calculation and obtained $$\binom{9}{3}\times(3!)\times 2^3=4032$$
But answer given is $$\frac{4032}{6}=672$$
Why has it been divided by $6$
| As suggested by @aarbee in comment section i am providing a detailed solution to given problem.
Definition: $\;(i)\;$Since it is skew symmetric matrix so all the elements in diagonal will be zero.
$\;(ii)\;$ $a_{ij}=−a_{ji}\;∀\;i≠j$ for skew symmetric matrix.
Explanation of $9\choose 3$
First choose $3$ elements either from set $\{−9,−8,−7,−6,−5,−4,−3,−2,−1\}$ or $\{1,2,3,4,5,6,7,8,9\}$ because $a_{ij}=−a_{ji}\;∀\;i≠j$ for skew symmetric matrix. So number of ways to choose $3$ elements out of $9$ is ${9\choose 3}$
and these three entries can be arranged in three place say $a_{12},a_{13},a_{23}$ in $3!$ ways
We can't do ${18 \choose 3}\cdot3!$ because in this way we can get $3$ elements $1,-1,\;2$ which won't form skew-symmetric matrix because in this case $3$ elements in upper diagonal or lower diagonal will $1 , -1, \; 2$.
And $2^3$ because we can permute $a_{12}$ and $a_{21},\;$ $\;a_{13}$ and $a_{31},\;$ $\;a_{23}$ and $a_{32}$.
So total number of ways are ${9\choose 3}\cdot3!\cdot2^3$
Detailed explanation of $2^3$: Let say selected elements are $1,2,3$
Here are those $8$ matrcies whose elmenets are $1,2,3$
$\begin{bmatrix}
0 & 1 & 2\\
-1 & 0 & 3\\ -2 & -3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & -1 & 2\\
1 & 0 & 3\\ -2 & -3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & 1 & -2\\
-1 & 0 & 3\\ 2 & -3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & 1 & 2\\
-1 & 0 & -3\\ -2 & 3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & -1 & -2\\
1 & 0 & 3\\ 2 & -3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & -1 & 2\\
1 & 0 & -3\\ -2 & 3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & 1 & -2\\
-1 & 0 & -3\\ 2 & 3 & 0\\
\end{bmatrix},\;$ $\begin{bmatrix}
0 & -1 & -2\\
1 & 0 & -3\\ 2 & 3 & 0\\
\end{bmatrix},\;$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $
I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.
I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.
| Let $u=\sqrt2x$, $y=v$, $z=w$. Then, it is equivalent to find the maximum $k^2=u^2+v^2+w^2$ subject to
$$\frac u{10\sqrt2} + \frac v{10} + \frac w{10} =1\tag1$$
Observe that $k$ is the distance from the origin to the plane (1) with the axis-intercepts $10\sqrt2,10,10$. In the first quadrant, (1) is a tetrahedron with its furthest corner at the $x$-intercept $10\sqrt2$. Thus, the maximum is $k^2_{max}=(10\sqrt2)^2 =200$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Inverse of Multivariable Functions on Manifolds Consider the unit circle, $S=${$(x,y) | x^2 + y^2 =1$} with the stereographic charts $(U_N,\phi_N)$, $(U_S,\phi_S)$
i.e. $U_N=S$ \ {$(0,1)$}, $\phi_N((x,y))=\frac{x}{1-y}$
$U_S=S$ \ {$(0,-1)$}, $\phi_S((x,y))=\frac{x}{1+y}$
How would I find ${\phi_N}^{-1}$ and ${\phi_S}^{-1}$?
These functions are bijective on their domains ($U_N$ and $U_S$ respectively) so their inverses exist.
For ${\phi_N}^{-1}$, I have considered a point $z$ in $\phi_N(U_N)$ and set $\frac{xy}{1-y}=z$. I believe I need to write both x and y in terms of $z$ to get something like ${\phi_N}^{-1}(z)=(f(z), g(z))$ where I need to find $f(z)$ and $g(z)$. And then similarly for $\phi_S$.
How would I do this? Thank you for any help!
| Let $\phi_N((x,y))=z=\frac{x}{1-y}$
Then $x=z(1-y)$
Substituting this into $x^2+y^2=1$ we have $z^2{(1-y)}^2+y^2=1$
Then, solving this for $y$, (I used the quadratic formula) $$(1+z^2)y^2-2z^2y+z^2-1=0$$
$$y=\frac{2z^2\pm\sqrt{4z^4-4(1+z^2)(z^2-1)}}{2(1+z^2)}$$
$$y=\frac{2z^2\pm2\sqrt{z^4-(z^4-1)}}{2(1+z^2)}$$
$$y=\frac{z^2\pm1}{1+z^2}$$
Now, if we had $y=\frac{z^2+1}{z^2+1}=1$, then this is not actually in the domain of $\phi_N$, since $(x,1)\notin$ $U_N$ for any x.
So we must have $$y=\frac{z^2-1}{z^2+1}$$
Then, from $x=z(1-y)$, we can just substitute our expression for y in to get
$$x=\frac{2z}{z^2+1}$$
Thus, $${\phi_N}^{-1}(z)=(\frac{2z}{z^2+1}, \frac{z^2-1}{z^2+1})$$
Similar procedure for finding ${\phi_S}^{-1}$
| {
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"source": "stackexchange",
"question_score": "2",
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| Another proof inspired by marty cohen.
By Power Mean Inequality,
$$\left(\frac{x^3+y^3}{2}\right)^{1/3} \ge \left( \frac{x^2+y^2}{2}\right)^{1/2}=2^{-\frac 12}\\
\implies x^3+y^3\ge 2\cdot 2^{-\frac 32}=\frac{1}{\sqrt 2} = \frac{x^2+y^2}{\sqrt 2}\ge \sqrt 2 xy.\blacksquare$$
| {
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"source": "stackexchange",
"question_score": "2",
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Indefinite Integral of $\dfrac {x^{p-1}}{x^{2m} - a^{2m}}$ I am looking at a bunch of related integrals found in Spiegel's "Mathematical Handbook of Formulas and Tables", (Schaum, 1968), item $14.336$.
It's a complicated and messy old beast:
For $m \in \mathbb Z$ such that $m \ge 1$:
$\displaystyle \int \dfrac {x^{p-1} \, \mathrm d x}{x^{2m} - a^{2m}} = \frac 1 {2 m a^{2 m - p}} \sum_{k=1}^m \cos \frac {k p \pi} m \ln \left({x^2 - 2 a x \cos \frac {k \pi} m + a^2}\right)$
$\displaystyle - \frac 1 {m a^{2 m - p}} \sum_{k=1}^m \sin \frac {k p \pi} m \arctan \left({\frac {x - a \cos \frac {k \pi} m}{a \sin \frac {k \pi} m}}\right)$
$\displaystyle + \frac 1 {2 m a^{2 m - p}} (\ln (x - a) + (-1)^p \ln (x + a)) + C$
for all $0 < p \le 2 m$.
(Note that he glosses over the sign of $x$, presenting the expression for positive $x$ only -- I will explore the negative $x$ case later.)
I thought of $2$ different approaches:
*
*Factorise the denominator using the standard result:
$\displaystyle x^{2 n} - y^{2 n} = (x - y) (x + y) \prod_{k \mathop = 1}^{n - 1} \left({x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}\right)$
and then explore the possibility of doing a partial fraction expansion. However, this did not work out so well, as I was then unable to see how to simplify the expression on the top of the resulting terms in all but the $x - a$ and $x + a$ denominators, so I didn't proceed further with that.
*Prove it by induction on $m$, which first concerns establishing the base case:
$\displaystyle \int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\dfrac {x - a} {x + a}}\right) + C$ (that is: $m = 1$, $p = 1$)
$\displaystyle \int \frac {x \, \mathrm d x} {x^2 - a^2} = \ln \left({x^2 - a^2}\right) + C$ (that is: $m = 1$, $p = 2$)
But when I put $m = 1$ into the above expression, I get:
$\dfrac 1 {2 a^{2 - p} } ( (-1)^p \ln (x^2 + 2 a x + a^2) ) + \dfrac 1 {2 a^{2 - p} } (\ln (x - a) + (-1)^p \ln (x + a) )$
But that pesky term on the left, in $(-1)^p \ln (x^2 + 2 a x + a^2)$ is clearly incorrect.
This arises from the $\cos \dfrac {k p \pi} m \ln \left({x^2 - 2 a x \cos \dfrac {k \pi} m + a^2}\right)$ expression in the given solution.
But for integer $p$ and $m = 1$, that $\cos \dfrac {k p \pi} m$ term, which I would hope to vanish, defiantly remains in place.
It doesn't help much changing it to $\cos \dfrac {k p \pi} {2 m}$, because this time it does not vanish for $p = 2$.
So either Spiegel has reported this result wrong, or I'm misunderstanding something.
Should that cosine in fact be a sine? Then it would vanish away like I want it to.
For $m = 2$ and $p = 1$ you get this:
$\dfrac 1 {4 a^3} \ln \left( {\dfrac {x - a} {x + a} }\right) - \dfrac 1 {2 a^3} \arctan \dfrac x a$
which is consistent with the result quoted. I have gone a little further, trying $p = 2, 3, 4$, but now it is looking as though the cosine term is completely erroneous and perhaps should not be there at all, because any contribution it makes is consistently absent.
I have gone no further with higher $m$ values.
| There is a typo in the formula
\begin{align}
I(m,p) =\int \dfrac {x^{p-1} \, \mathrm d x}{x^{2m} - a^{2m}}
& = \frac 1 {2 m a^{2 m - p}} \sum_{k=1}^{m-1} \cos \frac {k p \pi} m \ln \left({x^2 - 2 a x \cos \frac {k \pi} m + a^2}\right)\\
& \>\>\>\>\>- \frac 1 {m a^{2 m - p}} \sum_{k=1}^{m-1}\sin \frac {k p \pi} m \arctan \left({\frac {x - a \cos \frac {k \pi} m}{a \sin \frac {k \pi} m}}\right)\\
& \>\>\>\>\>+ \frac 1 {2 m a^{2 m - p}} (\ln (x - a) + (-1)^p \ln (x + a))
\end{align}
where the summation should run from $1$ to $m-1$, instead of $m$. Then, the followings check out
$$I(1,1)=\int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\dfrac {x - a} {x + a}}\right)$$
$$I(1,2)=\int \frac {x \, \mathrm d x} {x^2 - a^2} = \ln \left({x^2 - a^2}\right)$$
$$I(2,1)=\int \frac {\, \mathrm d x} {x^4 - a^4}
= \dfrac 1 {4 a^3} \ln \left( {\dfrac {x - a} {x + a} }\right) - \dfrac 1 {2 a^3} \arctan \dfrac x a$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the probability of the outcome which has highest chance of occurring. A biased die has numbers $1, 2, 3, 4, 5, 6$. The probability of obtaining one of the numbers is greater
than ${1\over 6}$, whereas the probability of obtaining a number opposite to it is less than ${1\over 6}$ . The remaining four numbers each have a probability of ${1\over 6}$ of being obtained. Given that any two opposite faces add up to $7$. When two such dice are rolled, probability of obtaining a sum $7$ is ${47\over 288}$. If the number which has the highest probability of being obtained has probability ${p\over q}$, for ($p, q$) = $1$, find $p + q$.
Well I know that $P(E)$ = ${No.\;of\;Favourable\;Outcomes\over Total\;Outcomes}$ but I am confused as to how to find the Probability in this case. Because the formula which I wrote, I guess, is applicable only when all the possible outcomes have equal chances of occurring which is not happening in this case.
| You may obtain the sum $7$ with $1+6,2+5,3+4,4+3,5+2,6+1$. Among these sums, two are biased and the other $4$ have probability $\frac{1}{6}\times\frac{1}{6}$. The other two pairs have faces with probabilities $\frac16+x$ and $\frac16-x$ for some positive $x$.
Hence, the probability of obtaining the sum $7$ is
$$\frac{4}{36}+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}$$
$$\frac{1}{36}-x^2=\frac12(\frac{47}{288}-\frac{4}{36})=\frac{5}{192}$$
$$x^2=\frac{1}{36}-\frac{5}{192}=\frac{1}{576}=(\frac{1}{24})^2$$
Hence, the highest probability is $\frac16+\frac1{24}=\frac{5}{24}$ and $p+q=5+24=29$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital? $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$
$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:
$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$ I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?
| fast solution,as another idea$$\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+0n^2+n+1}-\sqrt{n^2-n+2}}=\\
\lim_{n \to \infty}\frac{1}{\sqrt[3]{(n+0)^3-0+0n^2+n+1}-\sqrt{(n-\frac12)^2-\frac14+2}}=\\
\lim_{n \to \infty}\frac{1}{n-(n-\frac12)}=\\2$$
REMARK:
$$n^3+an^2+bn+c=\\(n+\frac a3)^3-(3n^2.\frac a3+3n(\frac a3)^2+(\frac a3)^3)+bn+c\\(n\to \infty) \implies n^3+an^2+bn+c\sim (n+\frac a3)^3 $$ so $$n^3+n+1=(n+0)^3-0^3+n+1$$ also for $n^2+an+b=(n+\frac a2)^2-(\frac a2)^2+c$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\sqrt{a-\sqrt{a+x}} = x$ Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$
My approach:
Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.
| $$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
Consider it in terms of a quadratic in $a$
$$a^2+a(-2x^2-1)+x^4-x=0$$
Now we see that $x^4-x=x(x^3-1)=x(x-1)(x^2+x+1)=(x^2-x)(x^2+x+1)$. And $(x^2+x+1)+(x^2-x)=2x^2+1$. By Vieta, its roots are $x^2-x$ and $x^2+x+1$. You could also do it using quadratic formula.
| {
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Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.
What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.
This is where I am stuck, how should I proceed next?
Can anyone help me?
| I'm leaving this here, but it's wrong. I added a correct answer below.
\begin{align}
\lfloor 3x + 1 \rfloor &= 2x - \frac{1}{2} \\
\lfloor 3x \rfloor + 1 &= 2x - \frac{1}{2} \\
\lfloor 3x \rfloor &= 2x - \frac 32
\end{align}
So, for some integer, $n$,
\begin{array}{ccccc}
&&\lfloor 3x \rfloor = n \\
\hline
n &\le &3x &< &n+1 \\
\frac n3 &\le &x &< &\frac n3 + \frac 13 \\
\hline
\frac {2n}3 &\le &2x &< &\frac {2n}3 + \frac 23 \\
\frac {2n}3 - \frac 32 &\le &2x - \frac 32 &< &\frac {2n}3 - \frac 56 \\
\frac {2n}3 - \frac 32 &\le &n &< &\frac {2n}3 - \frac 56 \\
-\frac 32 &\le &\frac n3 &< &-\frac 56 \\
-\frac 92 &\le & n &< &-\frac 52 \\
-4 &\le & n &\le &-3 \\
\hline
-4 &\le & \lfloor 3x \rfloor &\le &-3 \\
-4 &\le & 2x - \frac 12 &\le &-3
\end{array}
Hence $x \in \left\{ -\dfrac 74, -\dfrac 54 \right\}$
The sum of the roots is $-3$.
This is a correct answer.
Let $x = n + \epsilon$ where $n \in \mathbb R$ and $\epsilon \in [0,1)$.
\begin{align}
\lfloor 3x \rfloor &= 2x - \frac 32 \\
3n + \lfloor 3\epsilon \rfloor &= 2n + 2\epsilon - \frac 32 \\
6n + 2\lfloor 3\epsilon \rfloor &= 4n + 4\epsilon - 3 \\
2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
\end{align}
Since the LHS is an integer, then so too must $4\epsilon$ be an integer.
It follows that $4\epsilon \in \{0,1,2,3\}$
If $\epsilon = 0$.
\begin{align}
2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
2n + 3 &= 0 \\
&\text{No solution.}
\end{align}
If $\epsilon = \dfrac 14$.
\begin{align}
2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
2n + 3 &= 1 \\
n &= -1 \\
x &= -\dfrac 34
\end{align}
If $\epsilon = \dfrac 12$.
\begin{align}
2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
2n + 5 &= 2 \\
&\text{No solution.}
\end{align}
If $\epsilon = \dfrac 34$.
\begin{align}
2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
2n + 7 &= 3 \\
n &= -2 \\
x &= -\dfrac 54
\end{align}
| {
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"question_score": "2",
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Show $ (2 - b^2c^2) (1 + b^2) \ge (b^2 + c^2)(b + c - 3)^2 $ in some range. Show $ F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0$ in the following two ranges of $(b,c)$:
*
*$b \in [0 \quad 1]$ and $0 \le c\le b$,
*$b \in [1 \quad 1.2]$ and $0 \le c\le 3-2b$.
Equality occurs only at $(b,c)= (1,0)$.
This can be solved by treating $F$ as a polynomial in one of the variables (the other a parameter), and then using Budan's theorem. This has been performed here, but this is a very lengthy business. Does someone have a more direct method? Thanks in advance!
| Proof of (1): Clearly, if $b = 0$, the inequality is true. Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.
If $b\in (0, 1]$ and $c\in (0, b]$, let $b = \frac{1}{1+s}, c = b \frac{1}{1+t}$ for $s, t \ge 0$.
The inequality becomes
$$\frac{1}{(1+s)^6(1+t)^4}[f(s,t) + 14s^4 - 3s^2 + 2s] \ge 0$$
where $f(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.
We are done.
$\phantom{2}$
Proof of (2): If $b = 1$, the inequality becomes $c(1-c)(c^2 - 3c + 4) \ge 0$ which is true.
Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.
If $b\in (1, 6/5]$ and $c\in (0, 3-2b]$, let $b = 1 + \frac{1}{5}\cdot \frac{1}{1+s}$
and $c = (3-2b)\frac{1}{1+t}$ for $s, t\ge 0$. The inequality becomes
$$\frac{1}{15625(1+s)^6(1+t)^4}g(s, t) \ge 0$$
where $g(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.
We are done.
| {
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} |
Prove this inequality (please using AM-GM or Cauchy-Schwarz if possible) Let $a ; b ; c > 0$ such that $a+b+c=3$
Prove: $P=\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{6abc}{ab+bc+ca} \geq 5$
I tried:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=\dfrac{a^2c+b^2a+c^2b}{abc}=\dfrac{(a^2c+b^2a+c^2b)(c+a+b)}{3abc} \geq \dfrac{(ab+bc+ca)^2}{3abc}$
Then: $P \geq \dfrac{(ab+bc+ca)^2}{3abc}+\dfrac{6abc}{ab+bc+ca}$
But I think it's a wrong way, cuz $\dfrac{(ab+bc+ca)^2}{3abc}$ seems too weak.
| Suppose $c = \min \{a,b,c\},$ we have
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} -3 = \frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac},$$
$$(a+b+c)(ab+bc+ca)-9abc = 2c(a-b)^2+(a+b)(a-c)(b-c).$$
We write the inequality as
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{18abc}{(a+b+c)(ab+bc+ca)} \geqslant 5.$$
equivalent to
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} - 3 \geqslant 2\left[1 - \dfrac{9abc}{(a+b+c)(ab+bc+ca)}\right],$$
or
$$\frac{(a-b)^2}{ab} + \frac{(a-c)(b-c)}{ac} \geqslant \frac{4c(a-b)^2}{(a+b+c)(ab+bc+ca)}+\frac{2(a+b)(a-c)(b-c)}{(a+b+c)(ab+bc+ca)},$$
or
$$\left[\frac{1}{ab} - \frac{4c}{(a+b+c)(ab+bc+ca)}\right](a-b)^2 + \left[\frac{1}{ac} -\frac{2(a+b)}{(a+b+c)(ab+bc+ca)}\right](a-c)(b-c) \geqslant 0.$$
But
$$(a+b+c)(ab+bc+ca) \geqslant 9abc \geqslant 4abc,$$
and
$$\begin{aligned}(a+b+c)(ab+bc+ca) - 2ca(a+b) & \geqslant (a+b)(ab+bc+ca) - 2ca(a+b) \\&= (a+b)[a(b-c)+bc] \geqslant 0.\end{aligned}$$
The proof is completed.
| {
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$\lim_{ x \to 0 } \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$ I have to evaluate this limit:
$$\lim_{ x \to 0 } \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$$
I have an indeterminated form $[\frac{0}{0}]$.
So I applied the Hopital rule
$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \sim $$
$$-e^{\frac{\log(1+\arctan x)}{x}} \cdot \frac{\frac{x}{(1+\arctan x)(1+x^2)}-\log(1+\arctan x)}{x^2} \sim$$
$$ -e \cdot \frac{1}{x^2} \cdot \left(\frac{x}{(1+x)(1+x^2)}-x\right)
\sim +e $$
According to my book and wolfram alpha the final result should be $\frac{e}{2}$ and I don't know where I made mistakes.
| You can avoid the exponentials by a simple trick using the standard limit $\lim_{y\to 0}\frac{e^y-1}y=1$.
\begin{eqnarray*}\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}
& = & e\cdot \frac{\left(1-e^{\frac{\ln(1+\arctan x)-x}{x}}\right)}{x}\\
& \stackrel{y=\frac{\ln(1+\arctan x)-x}{x}}{=} & e\cdot \underbrace{\frac{1-e^y}y}_{\stackrel{x\to 0}{\longrightarrow}-1}\cdot \frac{\ln(1+\arctan x)-x}{x^2}
\end{eqnarray*}
Here I use that $\lim_{x\to 0}\frac{\ln(1+\arctan x)-x}{x} =0$, which can quickly be shown by L'Hospital.
Now, we only have to deal with
\begin{eqnarray*}\frac{\ln(1+\arctan x)-x}{x^2}
& \stackrel{L'Hosp.}{\sim} & \underbrace{\frac 1{(1+x^2)(1+\arctan x)}}_{\stackrel{x\to 0}{\longrightarrow}1}\cdot \frac{1-(1+\arctan x)(1+x^2)}{2x}
\end{eqnarray*}
Finally,
\begin{eqnarray*}\frac{1-(1+\arctan x)(1+x^2)}{2x}
& \stackrel{L'Hosp.}{\sim} & \frac{-(1+\arctan x)2x-(1+x^2)\frac 1{1+x^2}}{2}\\
& \stackrel{x\to 0}{\longrightarrow} & -\frac 12\\
\end{eqnarray*}
All together
$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \stackrel{x\to 0}{\longrightarrow} e\cdot (-1) \cdot \left(-\frac 12\right) = \frac e2 $$
| {
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Is the conclusion wrong in this proof of Case 2 of FLT for n=3? The classic proofs of some cases of Fermat´s last theorem are separated into two parts. These proofs use the reduced Fermat equation $z^n=x^n+y^n$. The positive integers $x,y,z $ are coprime to each other. In Case$1$ they are coprime to the positive integer n, but in Case $2$, one of $x, y, z$ is divisible by $n$.
This attempt to prove Case $2$ for $n=3$ is inspired by the Theorems on FLT by the Norwegian mathematician Niels Henrik Abel. The proof is one of contradiction. The crucial point is the conclusion in the end of the proof.
Assume that (I) $z^3=x^3+y^3$ is true, when the positive integers $x,y,z $ are coprime to each other. Assume $z$ is the one which is divisible by $3$. Factorize (I) $z^3=x^3+y^3$ $\implies $ (II) $z^3=(x+y)((x+y)^2-3xy)$. The left side has $3$ factors $=3$. $ (x+y)^2-3xy$ has only $1$ factor $=3$ since $3xy$ has only $1$ factor $=3$ $\implies$ $x+y $ has $2$ factors $=3$.
Also, $x+y$ divides $z^3$ $\implies$ $x+y$ is coprime to $x$ and to $y$ $\implies$
If $x+y$ is not divisible by $3$, $x+y$ is coprime to $ (x+y)^2-3xy$.
Then $3$ is the only possible common factor between $x+y$ and $ (x+y)^2-3xy$.
$z^3=x^3+y^3$ $\iff $ $y^3=z^3-x^3$ $\iff $ $x^3=z^3-y^3$.
Factorizing of $y^3=z^3-x^3$ and $x^3=z^3-y^3$ $\implies$
$y^3=(z-x)((z-x)^2+3xz)$ and $x^3=(z-y)((z-y)^2+3yz)$.
Since $x$ and $y$ are not divisible by $3$ , the factors of these factorizations are coprimes and therefore cubes.
Let $z-x=b^3$ and $z-y=d^3$. $b,d$ are positive coprime integers. Then
$ b^3+d^3=(z-x) + (z-y) =2z- (x+y)$ $\implies$ (III) $ b^3+d^3=2z- (x+y) $.
The right side has $1$ factor $=3$ $\implies$ the left side is divisible by $3$.
Factorizing of the left side $\implies$ $ b^3+d^3=(b+d)((b+d)^2-3bd)$. Then,
If one factor in the left side is divisible by $3$, the other factor is too $\implies$
The left side has $2$ factors $=3$ since $x+y$ has 2 factors $=3$ but $z$ only $1$ factor. That is a contradiction. $z^3\neq x^3+y^3$.
If $z$ has $2$ factors $=3$ $\implies$ $x+y $ in (II) $z^3=(x+y)((x+y)^2-3xy)$
has $5$ factors $=3$. That does not give a contradiction in (III) $ b^3+d^3=2z- (x+y)$.
But a contradiction is already reached for $z$ divisible by $1$ factor $=3$.
It is shown that $z^3\neq x^3+y^3$ $\iff $ $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $1$ factor $=3$. Then $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $2$ factors$=3$. If left side does not equal right side when both sides of the equation is divided by $3$ factors $=3$, division of both sides by more factors $=3$ does not make them equal.
In the same way, inequality is shown when $x$ or $y$ divisible by $3$.
Case $2$ of FLT for n=3 is true.
| The proof is not correct. In line -10 you have used the factorisation of $b^3+d^3$ to prove that this expression is divisible by $9$. You then correctly deduce that $z$ is divisible by $9$.
You then claim that this is a contradiction but it isn't a contradiction, it just means that $9|z$ and $3^5|x+y$.
FLT for $n=3$ can be proved by elementary means - see for example Fermat's Affirmative Questions, The Mathematical Gazette, Vol 96, Number 535, March 2012.
The sort of equations you were using were used very successfully and with great ingenuity by Sophie Germain, for quite general values of $n$, but only for Case 1 of FLT.
| {
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Proving $abcd+3\geq a+b+c+d$
If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that $$abcd+3\ge a+b+c+d$$
The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$ upto permutation .(I don't know if there are further equality cases.)
I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $a$ it suffices to show the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $a=\sqrt{3},b=c=d=0$)
P.S ;I am not aware of using Lagranges multipliers
Its from here
Update :An answer has been posted in the link above (AOPS) similar to Dr Mathva's answer
| Due to symmetry, assume wlog that $a\geqslant b\geqslant c\geqslant d\geqslant 0$. Since I will use this fact later, I am proving it before the cases:
Lemma: We have that $2\geqslant a+d$.
In fact, this follows from Cauchy-Schwarz: $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d$$
We will now consider some cases depending on the position of the number $1$:
*
*Case 1: $1\geqslant a\geqslant b\geqslant c\geqslant d\geqslant 0$. Observe that one may rewrite the inequality as $$3+abcd-(a+b+c+d)=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd) $$ Which is clearly positive.
*Case 2: $a\geqslant 1\geqslant b\geqslant c\geqslant d\geqslant 0$. If $ab\geqslant 1$, we might proceed as in the next case. So we will work with $ab<1$. Thus \begin{align*}3+abcd-(a+b+c+d)&=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd)\\&\geqslant \underbrace{(1-a)}_{<0}(1-b)+(1-c)(1-d)\\&\geqslant (1-a)(1-c)+(1-c)(1-d)\\&=(1-c)(2-(a+d))\geqslant 0 \end{align*} Where the last inequality follows from the lemma.
*Case 3: $a\geqslant b\geqslant 1\geqslant c\geqslant d\geqslant 0$. This implies that \begin{align*}6+2abcd-2(a+b+c+d)&=a^2+b^2+c^2+d^2+3+2abcd-2(a+b+c+d)\\
&=(a-1)^2+(b-1)^2+(c+d-1)^2+2cd(ab-1)\geqslant 0\end{align*} Or, equivalently $3+abcd\geqslant a+b+c+d$.
*Case 4: $a\geqslant b\geqslant c\geqslant 1\geqslant d\geqslant 0$. As @dezdichado noticed, this case is straightforward, since it forces directly $a=b=c=1, d=0$ due to the constraint $a^2+b^2+c^2+d^2=3$.
*Case 5: $a\geqslant b\geqslant c\geqslant d\geqslant 1$. Clearly impossible, since this would yield $a^2+b^2+c^2+d^2\geqslant 4>3$.
Done!
| {
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Matrices which determinant is obvious. At least in my current Linear Algebra course, exercises concerning the determinant of a matrix of dimension $n\geq 5$ more often than not require one to notice some property of the matrix so that its determinant becomes obvious and one needs not compute it in the usual way. I'm trying to find a simple way to realize that the determinant of
$$\begin{pmatrix} 0 & -1 & -1 & -1 & -1
\\
1 & 0 & -1 & -1 & -1
\\
1 & 1 & a & -1 & -1
\\
1 & 1 & 1 & 0 & -1
\\
1 & 1 & 1 & 1 & 0\end{pmatrix}$$
is equal to $a$, and I would appreciate anyone who points me in the right direction.
Also, since these are the types of questions that might go to my (as well as other students') final, I'm open to users giving other examples of matrices which determinants are obvious without the need to compute them, here are a few:
$$\det \begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
1 & 1 & -4 & 1 & 1
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4\end{pmatrix}
=
\det\begin{pmatrix}
-4 & 1 & 1 & 1 & 1
\\
1 & 1 & 1 & -4 & 1
\\
0 & 0 & 0 & 0 & 0
\\
1 & -4 & 1 & 1 & 1
\\
1 & 1 & 1 & 1 & -4
\end{pmatrix}=0
$$
by adding rows $1,2,4,5$ to row $3$.
$$\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
2 & 3 & 4 & 5 & 6
\\
3 & 4 & 5 & 6 & 7
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}
=\det \begin{pmatrix}
1 & 2 & 3 & 4 & 5
\\
1 & 1 & 1 & 1 & 1
\\
2 & 2 & 2 & 2 & 2
\\
4 & 5 & 6 & 7 & 8
\\
5 & 6 & 7 & 8 & 9\end{pmatrix}=0$$
by subtracting row $1$ from row $2$ & $3$.
$$\det \begin{pmatrix}
0 & -a & -b & -d & -g
\\
a & 0 & -c & -e & -h
\\
b & c & 0 & -f & -i
\\
d & e & f & 0 & -j
\\
g & h & i & j & 0\end{pmatrix}=0$$
since $(-1)^5\det (A)=\det(A^T)=\det(A)$. All of these can be generalized to other dimensions, with the exception of the last one for which the dimension of the matrix must be odd.
| Denote your determinant by $f(a)$ so we get a function $f : \Bbb{R} \to M_5(\Bbb{R})$. Notice that $f(0)$ is a skew-symmetric $5 \times 5$ matrix and hence has determinant $0$ so $f(0)=0$.
The derivative of a determinant of a matrix function can be calculated by differentiating every column of the matrix and summing them up:
$$f'(a) = \begin{vmatrix} 0 & -1 & 0 & -1 & -1\\1 & 0 & 0 & -1 & -1\\1 & 1 & 1 & -1 & -1\\1 & 1 & 0 & 0 & -1\\1 & 1 & 0 & 1 & 0\end{vmatrix} = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\1 & 1 & 0 & -1\\1 & 1 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0 \end{vmatrix}.$$
by adding the second and subtracting the first row from the third and fourth. The latter matrix is block upper-triangular so
$$f'(a) = \begin{vmatrix} 0 & -1 & -1 & -1\\1 & 0 & -1 & -1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix}\cdot \begin{vmatrix} 0 & -1 \\ 1 & 0\end{vmatrix} = (-1)\cdot(-1)=1.$$
Therefore $f(a)= a+C$ for some constant $C \in \Bbb{R}$. From $f(0)=0$ we get $C=0$ and hence $f(a)=a$.
| {
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How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?
| What you suggest basically seems to work - I'm going to change the leading value into $n$ rather than $n-1$ for ease of reading:
First, $2^1 = 1+1$ and $2^2 = 1+1+2 = 1+2^0+2^1$
Then using induction, assuming that the summation holds for lower $n$, we can use your observation $2^n = 2^{n-1} + 2^{n-1}$ and change the first $2^{n-1}$ into the summation form, giving $2^{n} = 1+1 + 2 + \cdots + 2^{n-2} + 2^{n-1}$ as required.
| {
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Checking an introductory Linear Algebra Exercise regarding linear applications. My final is few days away and I'm doing the list of exercises recommended by the teacher. Not all exercises have solutions however, and the teacher isn't answering my questions, so I just want to check whether I did the following exercise correctly:
Let $f:\mathbb{R}^3\rightarrow \mathbb{R}^3 : (2x_1-x_2+x_3,x_2+x_3,x_1+x_2+2x_3)$.
*
*Prove $f$ is a linear application.
*Find the matrix $A$ that corresponds with the linear applicaiton $f$.
*Find a basis for $\text{Im}(f)$ and $\text{Ker}(f)$.
*
*Proving that $f(c\textbf{v})=cf(\textbf{v})$ and that $f(\textbf{v}+\textbf{u})=f(\textbf{v})+f(\textbf{u})$ is easy enough.
*$$A=\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$
*This is the bit I'm more concerned about. Vectors of $\text{Im}(f)$ are precisely those of the form
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\x_2 \\ x_3\end{pmatrix}$$
we can apply elementary column operations since these don't affect the column space ($\text{Im}(f)$),
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \end{pmatrix} \sim \begin{pmatrix} 0 & -1 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 0 \end{pmatrix}$$
Since the first two columns are clearly linearly independent, and vectors of $\text{Im}(f)$ are preciesly those of the form
$$x_1\begin{pmatrix} 0 \\2 \\ 3\end{pmatrix}+x_2\begin{pmatrix} -1 \\1 \\ 1\end{pmatrix}$$
for $x_1,x_2\in \mathbb{R}$, we must have that $\{ (0, 2, 3),(-1, 1, 1)\}$ is a basis of $\text{Im}(f)$. Solutions of the equation $$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\textbf{v}=\textbf{0}$$
are not changed by multiplying on the left both sides by elementary matrices.
$$\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & -1 & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -3 & -3 \end{pmatrix}\sim\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$
Thus $\text{Ker}(f)$ is precisely the set of vectors $v$ such that $x_1=x_2=-x_3$, or $$v=t\begin{pmatrix} 1 \\ 1\\ -1\end{pmatrix}$$ for $t\in \mathbb{R}$. Therefore $\{ (1,1,-1)\}$ is a basis of $\text{Ker}(f)$.
| That all looks fine to me (except that for item 1, you'd get no credit unless you actually wrote out the proof, of course).
An alternative to your solution for 1 is to do item 2 first; then you have
$$
f(v) = Av
$$
for all $v \in \Bbb R^3$. So
$$
f(v+w) = A(v + w) = Av + Aw = f(v)+ f(w)
$$
by the distributive law for matrix multiplication, and a similar thing works for scalar multiples.
| {
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Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$ Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:
Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
&\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}
Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$
Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$
So $x = 13$ or $x = 11$.
Is it correct that way?
| As other comments and solutions have noted, you need to consider that 35 is composite. But nearly all of your work is correct and completing the square is a great strategy for arbitrary quadratic congruences.
$$(3x-4)(3x+2)\equiv0\pmod{35}$$
This is correct. But then you need to go from there to
$$(3x-4)(3x+2)\equiv0\pmod 5 \qquad\text{ and }\qquad (3x-4)(3x+2)\equiv0\pmod 7$$
we'll tackle those one at a time.
$$3x-4\equiv0\pmod 5\qquad\text{ or }\qquad 3x+2\equiv0\pmod 5\\
3x\equiv4\pmod 5\qquad\text{ or }\qquad 3x\equiv3\pmod 5\\
x\equiv3\pmod 5\qquad\text{ or }\qquad x\equiv1\pmod 5\\
$$
$$3x-4\equiv0\pmod 7\qquad\text{ or }\qquad 3x+2\equiv0\pmod 7\\
3x\equiv4\pmod 7\qquad\text{ or }\qquad 3x\equiv5\pmod 7\\
x\equiv6\pmod 7\qquad\text{ or }\qquad x\equiv4\pmod 7\\
$$
Pairing those solutions together with the CRT (Chinese Remainder Theorem) gives you the four solutions others have mentioned.
| {
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Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?
update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} + 1)^2=9 \implies
(\frac{y^2}{4} + 1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
| When you are searching for intersection of two figures the intersection dot must satisfy parabola equation which indicates that
$$ y^2 = 4x - 4$$
Which means that
$$ 4x - 4 \geq 0$$
$$ x \geq 1 $$
The last inequality will exclude $x = -3$ solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find $a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$ A sequence $\left\{a_n\right\}$ is defined as $a_n=a_{n-1}+2a_{n-2}-a_{n-3}$ and $a_1=a_2=\frac{a_3}{3}=1$
Find the value of $$a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$$
I actually tried this using difference equation method.Let the solution be of the form $a_n=\lambda^n$
$$\lambda^n=\lambda^{n-1}+2\lambda^{n-2}-\lambda^{n-3}$$ which gives the cubic equation $\lambda^3-\lambda^2-2\lambda+1=0$. But i am not able to find the roots manually.
| Let $S=a_1+\dfrac{a_2}{2}+\dfrac{a_3}{2^2}+\dfrac{a_4}{2^3}+\ldots $
Then $\dfrac{S}{2} = \dfrac{a_1}{2}+\dfrac{a_2}{2^2}+\dfrac{a_3}{2^3}+ \ldots$
Subtracting we get
$\dfrac{S}{2} = a_1 +\dfrac{a_2-a_1}{2}+\dfrac{a_3-a_2}{2^2}+\dfrac{a_4-a_3}{2^3}+ \ldots$
Now $a_4-a_3 = 2a_2-a_1, a_5-a_4=2a_3-a_2$ etc.
So $\dfrac{S}{2} = 1 +\dfrac{1-1}{2}+\dfrac{3-1}{2^2}+\dfrac{2a_2-a_1}{2^3}+ \dfrac{2a_3-a_2}{2^4}+\ldots $
$=1+\dfrac{1}{2}-\dfrac{1}{8}+3\left(\dfrac{a_2}{2^4}+\dfrac{a_3}{2^5}+\ldots \right)$
$=\dfrac{11}{8}+\dfrac{3}{8} (S-1) =1+\dfrac{3S}{8}$
$ \Rightarrow \dfrac{S}{8} = 1 $ so that $S=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Limit to $e^2$. I have to show
$$
\lim_{x \to \infty} {\left(\frac{x^2 + 1}{1 - x^2}\right)}^{x^2} = e^2,
$$
but I don't get the trick to see it, I suppose I can use something like
$$
{\left(\lim_{x \to \infty} {\left(1 + \frac1x\right)}^x\right)}^2 = e^2,
$$
but I do not see how I can apply to the problem. On the other hand,
$$
\frac{x^2 + 1}{1 - x^2} = 1 + \frac{2 x^2}{1 - x^2}.
$$
Any hint?
| If the limit of interest is $\lim_{x\to\infty}\left(\frac{x^2+1}{x^2-1}\right)^{x^2}$, then we can write
$$\begin{align}
\left(\frac{x^2+1}{x^2-1}\right)^{x^2}&=\left(\frac{x^2-1+2}{x^2-1}\right)^{x^2}\\\\
&=\left(\left(1+\frac{2}{x^2-1}\right)^{x^2-1}\right)^{x^2/(x^2-1)}
\end{align}$$
Now, use $\lim_{y\to\infty}\left(1+\frac{t}{y}\right)^y=e^t$ with $y=x^2-1$. Can you wrap this up now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Let $A$ be a real $5 \times 5$ matrix satisfying $A^3-4A^2+5A-2I=O$. Is the matrix $A^4+A^3-3A^2-3A$ invertible?
Consider the polynomial $t^3-4t^2+5t-2=(t-1)^2(t-2)$. The minimal polynomial of $A$ divides $(t-1)^2(t-2)$. Let $p(t)$ be the minimal polynomial of $A$.
If $p(x)=t-1$ or $t-2$, then $A$ is a scalar multiple of the identity and $A^4+A^3-3A^2-3A=-4I \text{ or }6I$, so $A^4+A^3-3A^2-3A$ is invertible.
If $p(t)=(t-1)^2(t-2)$ or $(t-1)(t-2)$, then $A$ is similar to a upper triangular matrix $J$ with diagonal entries $1,2$. Then $A^4+A^3-3A^2-3A=P(J^4+J^3-3J^2-3J)P^{-1}$ and $\det(A^4+A^3-3A^2-3A)=\det(J^4+J^3-3J^2-3J)$. Since $J^4+J^3-3J^2-3J$ is upper triangular and the diagonal entries do not vanish, $\det(J^4+J^3-3J^2-3J) \ne 0$.
If $p(t)=(t-1)^2$, then by the same reasoning, $A^4+A^3-3A^2-3A$ is invertible.
Is there more efficient way to solve this kind of problems or I have to discuss any possible situations every time?
| The long division of $x^4+x^3-3x^2-3x$ by $x^3-4x^2+5x-2$ leaves a remainder $12x^2-26x+10$, so your matrix to be checked for invertibility is
$$
2(6A^2-13A+5I)
$$
Suppose $v\ne0$ is in the null space of $6A^2-13A+5I$, so $6A^2v-13Av+5v=0$.
We also know that $A^3v-4A^2v+5Av-2v=0$. Multiplying the former by $A$, the latter by $6$ and subtracting, we obtain $11A^2v-25Av+12v=0$ and an elimination from the two “degree two” relations yields
$$
7Av-17v=0
$$
which means that $A$ should have $17/7$ as an eigenvalue. However, this is not possible, because an eigenvalue $\lambda$ of $A$ must satisfy $\lambda^3-4\lambda^2+5\lambda-2=0$.
The above doesn't use the Jordan canonical form. If you know it, you can prove that the eigenvalues of $p(A)$ (for $p$ a polynomial) have the form $p(\lambda)$ where $\lambda$ is an eigenvalue of $A$. Since the eigenvalues of $A$ must satisfy $(\lambda-1)^2(\lambda-2)=0$, they can only be $1$ or $2$; however
$$
x^4+x^3-3x^2-3x=x(x^2-3)(x+1)
$$
and neither $1$ nor $2$ are roots of this polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$ We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.
$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
= 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$
Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.
| A monomial giving a contribution to degree 16 must be of the form
$$x^{a_1}\cdot x^{a_2}\cdot x^{a_3}\cdot x^{a_4}$$
with $a_i <6$ and $a_1+a_2+a_3+a_4=16$.
Suppose $a_1 \geq a_2 \geq a_3 \geq a_4$, then you have only five possibilities: $(5,5,5,1)$, $(5,5,4,2)$, $(5,5,3,3)$, $(5,4,4,3)$ and $(4,4,4,4)$.
The coefficient of $x^{16}$ equals the cardinality of the set composed by these vectors and the ones obtained permuting their coordinates.
It is easy to prove that this cardinality is equal to:
$$\binom{4}{3}+2\binom{4}{2}+\binom{4}{2}+2\binom{4}{2}+1=35$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Upper bound for - $\sum_{i=1}^n i {2n \choose n+i} $. I am tryin to calculate an upper bound for the following sum:
$\sum_{i=1}^n i {2n \choose n+i} $.
I know that the limit should be $\Theta (4^n\sqrt n)$ but I am not sure how to proceed. I already tried to bound by the biggest element, but the bound was too high.
| We have that
$$S_n = \sum_{q=1}^n q {2n\choose n+q}
= \sum_{q=1}^n q [w^{n-q}] \frac{1}{(1-w)^{n+q+1}}
= [w^n] \frac{1}{(1-w)^{n+1}}
\sum_{q=1}^n q \frac{w^q}{(1-w)^q}.$$
Now here the coefficient extractor enforces the range and we get
$$[w^n] \frac{1}{(1-w)^{n+1}}
\sum_{q\ge 1} q \frac{w^q}{(1-w)^q}
= [w^n] \frac{1}{(1-w)^{n+1}}
\frac{w/(1-w)}{(1-w/(1-w))^2}
\\ = [w^{n-1}] \frac{1}{(1-w)^n} \frac{1}{(1-2w)^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^n} \frac{1}{(1-w)^n} \frac{1}{(1-2w)^2}.$$
We now apply the change of variables rule 1.8 (5) from Egorychev's
Integral representation and the Computation of Combinatorial sums
(page 16) with $A(w) = \frac{w}{(1-2w)^2}$ and $f(w) = \frac{1}{1-w}.$ We
get $h(w) = w (1-w)$ and find
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{A(w)}{f(w) h'(w)} \right]\right|_{w=g(z).}$$
with $g$ the inverse of $h.$ This becomes
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{w/(1-2w)^2}{(1-2w)/(1-w)}
\right]\right|_{w=g(z)}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\left.\left[ \frac{w(1-w)}{(1-2w)^3} \right]\right|_{w=g(z)}
= \mathrm{Res}_{z=0} \frac{1}{z^{n}}
\left.\left[ \frac{1}{(1-2w)^3} \right]\right|_{w=g(z).}$$
Observe that $(1-2w)^2 = 1 - 4w + 4w^2 = 1-4w(1-w) = 1-4z$
so this is
$$\mathrm{Res}_{z=0} \frac{1}{z^{n}}
\frac{1}{(1-4z)^{3/2}} =
4^{n-1} {n-1+1/2\choose n-1}
= 4^{n-1} {n-1/2\choose n-1}.$$
Expanding the binomial coefficient we find
$$4^{n-1} \frac{1}{(n-1)!} \prod_{q=0}^{n-2} (n-1/2-q)
= 2^{n-1} \frac{1}{(n-1)!} \prod_{q=0}^{n-2} (2n-1-2q)
\\ = 2^{n-1} \frac{1}{(n-1)!} \frac{(2n-1)!}{2^{n-1} (n-1)!}
= \frac{1}{2n} n^2 {2n\choose n}
= \frac{1}{2} n {2n\choose n}.$$
Now for the central binomial
coefficient
we have ${2n\choose n} \sim \frac{4^n}{\sqrt{\pi n}}$ so this yields
$$\bbox[5px,border:2px solid #00A000]{
S_n = \frac{1}{2} n {2n\choose n}
\sim \frac{4^n}{2\sqrt{\pi}} \sqrt{n}.}$$
If we seek $\Theta(4^n\sqrt{n})$ from first principles we use from the
same reference the upper bound ${2n\choose n} \le 4^n/\sqrt{3n+1}$ so
we get
$$S_n \le \frac{1}{2} n \frac{4^n}{\sqrt{3n+1}}
= \frac{1}{2\sqrt{3}} \sqrt{n} \frac{4^n}{\sqrt{1+1/3/n}}
\lt \frac{4^n}{2\sqrt{3}} \sqrt{n}.$$
The lower bound is done similarly and we have the claim.
| {
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"url": "https://math.stackexchange.com/questions/4007052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How can one calculate $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$? I am working through a textbook for my Fourier Series class and I came across the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$. I know from the context of the question that the answer has to be $-\frac{\pi^2}{12}$, but I am not sure how to arrive at this answer. I tried partial fraction decomposition to see if there was any telescoping behavior, but I had an issue with the $(-1)^n$ part.
| Assuming that you know $S:=\sum_{n\geq 1} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$, we get (and I'll replace $(-1)^{n}$ with $(-1)^{n-1}$ which seems that you actually intended to be positive)
$$
\frac{1}{1^2} -\frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + \cdots \\
= \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \cdots \\
- 2 \left(\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + \cdots\right) \\
= S - 2\cdot \frac{1}{2^{2}}S = \frac{1}{2}S = \frac{\pi^{2}}{12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$ I want to find the solutions of $(x+1)^{63}+(x+1)^{62}(x-1)+\cdots+(x-1)^{63}=0$.
It is not hard to see $x=0$ is a root of the equation. but I don't know how to solve this equation in general. I can see terms of the equation looks very similar to binomial expansion of $[(x+1)+(x-1)]^{63}$ except the coefficient of each term is $1$ rather than $63\choose k $ (for $k=0,1,\cdots,63$ ). is it possible to use binomial theorem to solve the equation? (or other approaches)
| $$\frac {a^n - b^n}{a-b} = a^{n-1} + a^{n-2} b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1}$$
Hence:
$$(x+1)^{63} + (x+1)^{62}(x-1) + \dots+ (x-1)^{63} = \frac {(x+1)^{64}-(x-1)^{64}}{(x+1)-(x-1)} = \frac12((x+1)^{64} - (x-1)^{64})$$
If the above expression is zero, we must have $(x+1)^{64} = (x-1)^{64}$. Hence $\dfrac {x+1}{x-1}$ must be one of the 64-th roots of unity (or if you are only considering real roots, $\pm 1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$
If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$
To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?
I tried this way as an alternative ansewr:
Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$
$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.
| Another approach: Let the roots of $$x^3+3x+5=0 ~~~~(1)$$ be $a,b,c$, then $a+b+c=0, ab+bc+ca=3, abc=-5$.
Let $$F=(a+1/a)(b+1/b)(c+1/c)=\frac{(a^2+1)(b^2+1)(c^2+1)}{abc}=\frac{y_1 y_2y_3}{abc}=\frac{y_1 y_2 y_3}{-5}$$
where $y_1=(a^2+1)$ etc. let us transform $x$ equation to $y$ equation as $y=x^2+1 \implies x= \sqrt{y-1}.$ Putting it in (1) and simplifying we get $$y^3+3y^2-29=0$$ Hence $y_1y_2y_3=29$ and finally, $$F=\frac{-29}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $ I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:
$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $$
Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2} \sim \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.
In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$
but I don't understand this solution.
Trying to solve the limit with $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot \frac{\sin (ax) a x^2 - (1-\cos (ax)) 2x}{x^4}}{2x} $$$$ = \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$
| Although a solution is provided (and was accepted) I wanted provide a solution using Maclaurin series of cos:
$$
cos(x)=1-x^2/2!-x^4/4!+x^6/6!-...
\\ cos(ax) = 1-a^2x^2/2!-a^4x^4/4!+a^6x^6/6!-...
\\ 1-cos(ax) = a^2x^2/2!+a^4x^4/4!-a^6x^6/6!+...
\\ \frac{(1-cos(ax)}{x^2} = a^2/2!+a^4x^2/4!-a^6x^4/6!+...
\\ \lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2}
$$
Since the denominator in $\lim_{x\to 0}$ goes to zero, in order for the limit exist, the numerator also should go to zero for $\lim_{x\to 0}$:
$$
\lim_{x\to 0} {\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)} = \lim_{x\to 0} {\cos \left(\pi \cdot a^2/2 \right)}
$$
for ${\cos \left(\pi \cdot a^2/2 \right)}$ to be zero $a^2$ should be $1+2k$
In this case we can apply the L'Hospital's Rule:
$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)}{x^2} = \lim_{x\to 0} \frac{-\sin \left(\pi \cdot (a^2/2!+a^4x^2/4!-a^6x^4/6!+... ) \right)(\pi (a^4(2x)/4!-4a^6x^3/6!+... ))}{2x}
= {-\sin \left(\pi \cdot a^2/2 \right)\pi a^4/4!} = \pm \pi a^4/24
$$
for which it is positive if k is even, and it is negative if k is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to calculate $\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}} \right)$? This is an admission question of Tokyo University：
For all natural number $n\geqslant2$, we always have
$$n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\geqslant\cfrac{i}{10}$$
in which $i$ is integer, please calculate the maximum 0f $i$.
I have worked out this problem, furthermore, how to calculate
$$\lim_{n\to\infty}\left(n-\sum_{k=2}^n\frac{k}{\sqrt{k^2-1}}\right)$$
is the value a transcendental number？
| First note that
$$
\frac{k}{{\sqrt {k^2 - 1} }} = 1 + \frac{1}{{(k + \sqrt {k^2 - 1} )\sqrt {k^2 - 1} }}.
$$
Thus,
$$
\mathop {\lim }\limits_{n \to + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2 - 1} }}} } \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {1 - \sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2 - 1} )\sqrt {k^2 - 1} }}} } \right).
$$
Observe that our sequence is decreasing, so we indeed need to estimate this limit efficiently. Now
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2 - 1} )\sqrt {k^2 - 1} }}} \le \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2 - 1}}} \to \frac{1}{2}\sum\limits_{k = 2}^\infty {\frac{1}{{k^2 - 1}}} = \frac{3}{8}
$$
and
$$
\sum\limits_{k = 2}^n {\frac{1}{{(k + \sqrt {k^2 - 1} )\sqrt {k^2 - 1} }}} \ge \frac{1}{2}\sum\limits_{k = 2}^n {\frac{1}{{k^2 }}} \to \frac{1}{2}\sum\limits_{k = 2}^\infty {\frac{1}{{k^2 }}} = \frac{1}{2}\left( {\frac{{\pi ^2 }}{6} - 1} \right) \ge 0.322.
$$
Consequently,
$$
0.625 \le \mathop {\lim }\limits_{n \to + \infty } \left( {n - \sum\limits_{k = 2}^n {\frac{k}{{\sqrt {k^2 - 1} }}} } \right) \le 0.678,
$$
and the answer is $i=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}$ Suppose circle $I$ is $\triangle ABC$’s incircle and $D, E,$ and $F$ are the tangent points of $BC, CA, AB$ and circle $I.$ Given that $DI$ intersects $EF$ at $K, BC = 10, AC = 8,$ and $AB = 7$ find $\frac{S_{\triangle BKC}}{S_{\triangle ABC}}.$
| Since you let me have the pleasure of solving the problem (by not sharing your solution or approach), it is only fair that I reciprocate and let you enjoy the exercise of at least some parts of the proof. Here we go:
Lemma 1 - If in $\triangle ABC$ shown below, $AX$ bisects $\widehat A$ then $\frac{BX}{CX}=\frac{AB}{AC}$ .
Proof is left as an exercise for the interested reader.
Lemma 2- As in the figure below, extend $IK$ to $J$, such that $\widehat{JEK}=\widehat{IEK}$ . Then $\triangle EJI$ and $\triangle ABC$ are similar and $\frac{JI}{BC}=\frac{EJ}{AB}=\frac{EI}{AC}$.
Proof is somewhat beautiful, therefore it is left as an exercise for the interested reader.
Now for calculations, let us refer to the radius of the in-circle as $r$. Let us also write the lengths of $BC$ , $CA$ and $AB$ as $a$ , $b$ and $c$ , respectively. We have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{S_{\triangle BIC} + S_{\triangle BIK} + S_{\triangle CIK}}{S_{\triangle ABC}} $$
$$= \frac{a}{a+b+c} + \frac{a.IK}{(a+b+c)r} \qquad (1)$$
To find $IK$ , note that in $\triangle EJI$ , $EK$ bisects $\widehat{JEI}$ . From Lemma 1 we have:
$$\frac{IK}{JK} = \frac{EI}{EJ} $$
$$\therefore \frac{IK}{IJ} = \frac{EI}{EI+EJ} \qquad (2)$$
From Lemma 2 we have:
$$\frac{EI}{EI+EJ} = \frac{b}{b+c} \qquad (3)$$
and
$$\frac{IJ}{EI} = \frac{a}{b} $$
$$\therefore IJ = \frac{a.r}{b} \qquad (4)$$
From (2) , (3) and (4) we can calculate $IK$ :
$$IK = \frac{a.r}{b+c} \qquad (5)$$
Now from (5) and (1) we can calculate the desired ratio of areas:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{a}{a+b+c} (1 + \frac{a}{b+c}) = \frac{a}{b+c}$$
For the particular values of $a=10$ , $b=8$ and $c=7$ we have:
$$\frac{S_{\triangle BKC}}{S_{\triangle ABC}} = \frac{10}{8+7} = \frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a $m$ divides the area under and above x-axis of $y=x (1 - x)$ My questions is, suppose that for specific $m$, $y=mx$ divides the area under and above $x$-axis of $y=x(1- x)$ into $2$ equal areas, so what is $m$? In an interval, I estimated fast with $m \in (\frac{1}{5}, \frac{2}{5}]$, is this right? I plot this equation on graphs, and only estimated it. I didn't calculate integral precisely.
| First find the area of $x(1-x)$ from $0$ to $1$ (which is the area above the x axis) which comes out to be $\frac{1}{6}$, now the area between the line $y=mx$ and $y=x(1-x)$ would be $\frac{1}{12}$, so $$\int_0^{1-m}\left[x(1-x)-mx\right]dx=\frac{1}{12}$$
So, $$\frac{(1-m)^2}{2}-\frac{(1-m)^3}{3}-\frac{m(1-m)^2}{2}=\frac{1}{12}$$
$$\frac{m^2-2m+1}{2}+\frac{m^3-3m^2+3m-1}{3}-\frac{m^3-2m^2+m}{2}=\frac{1}{12}$$
$$3m^2-6m+3+2m^3-6m^2+6m-2-3m^3+6m^2-3m=\frac{1}{2}$$
$$-m^3+3m^2-3m+1=\frac{1}{2}$$
$$(1-m)^3=\frac{1}{2}$$
$$m=1-\frac{1}{\sqrt[3]{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality $2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3$ for $a+b+c=1$ Let $a,b,c$ be positive numbers with $a+b+c=1$. Show that
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2) +30 abc \ge 3
$$
Equality holds for $a=b=c=\frac13$. One could try homogenizing:
$$
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4 \ge 0
$$
Expanding gives
$$
2 \sum_{cyc} a^4 - \sum_{cyc} a^3(b+c) - 2 \sum_{cyc} a^2 b^2 + 2 a b c \sum_{cyc} a \ge 0
$$
but I didn't manage to advance further.
| Method 1let $p=a+b+c,q=ab+bc+ca,r=abc$
Now by schur $$a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)\ge 0$$ $$\iff p^4-5p^2q+4q^2+6pr\ge 0$$ Now we have to prove $$4 p^4 - 18 p^2 q + 8 q^2 + 30 p r\ge 0$$ or $$4(p^4-5p^2q+4q^2+6pr)+2(p^2q+3pr-4q^2)\ge 0$$ whcih is true because $$p^2q+3pr-4q^2=a^3(b+c)+b^3(a+c)+c^3(a+b)-2a^2b^2-2b^2c^2-2c^2a^2\ge 0$$
Method 2 I see that i have clashed with an existing answer so here is a different way
let $3u=a+b+c,3v^2=ab+bc+ca,w^3=abc$ then without even exanding we see that its lenear in $w^3$ Hence by uvw method it suffices to check when
*
*$c=0$
The inequality achieves the equality
*
*$a=b$ also let $t=\frac{c}{a}$ The inequality is $$ 2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but$$4{(t-1)}^2(t+1)\ge 0$$
Method 3 This method is ugly although it can be got by hand ,i used wolfram alpha to speed up my calculation
WLOG $a\ge b\ge c$ let $$f(a,b,c)=
2(a^2+b^2+c^2)^2 + 5(a^2+b^2+c^2)(a+b+c)^2 +30 abc (a+b+c) - 3 (a+b+c)^4
$$ $$f(a,b,c)-f(\frac{a+b}{2},\frac{a+b}{2},c)=1/2 (a - b)^2 (8 a^2 + 12 a b + 8 b^2 - 5 a c - 5 b c - 6 c^2)\ge 0$$
As inequality is homogenous WLOG $c=1$ and $t=\frac{a+b}{2}$ we then have to prove $$f(t,t,1)\ge 0$$ $$\iff 2(2t^2+1)^2+5(2t^2+1)(2t+1)^2+30t^2(2t+1)-3(2t+1)^4\ge 0$$ which is nothing but $$4{(t-1)}^2(t+1)\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.