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Finding the recursive formula for calculating the solution of the linear equation system $A$x = b. Let $K$ be a field and n $\in$ $\mathbb{N}$. It's given that $A$ = $
\begin{bmatrix}
0 & p_{1} & 0 & 0 & \dots & 0 \\
q_{2} & 0 & p_{2} & 0 & \dots & 0 \\
0 & q_{3} & 0 & p_{3} & \dots & 0 \\
\vdots & & \ddots & \ddots & \ddots & \vdots \\
0 & \dots & 0 & q_{n-1} & 0 & p_{n-1} \\
0 & \dots & 0 & 0 & q_{n} & p_{n}
\end{bmatrix}
$ $\in$ $K^{n,n}$,
$b$ = $
\begin{bmatrix}
b_{1} \\
\vdots \\
b_{n} \\
\end{bmatrix}
$ $\in$ $K^{n,1}$ with $p_{i}$,$q_{i}$ $\neq$ 0 for all i = 1, $\dots$ ,n.
Find a recursive formula for calculating the solution of the linear equation system $A$x = b.
My ideas and thoughts:
I thought about writing the matrix as the following:$$
\left[\begin{array}{cccccc|c}
0 & p_{1} & 0 & 0 & \dots & 0 & b_{1} \\
q_{2}&0&p_{2}&0&\dots&0&\vdots\\
0&q_{3}&0&p_{3}&\dots&0&\vdots\\
\vdots&&\ddots&\ddots&\ddots&\vdots&\vdots \\
0 & \dots & 0 & q_{n-1}&0&p_{n-1}&\vdots \\
0&\dots&0&0&q_{n}&p_{n}&b_{n} \end{array}\right]
$$ so that \begin{equation}
=
\begin{cases}
0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1} \\
q_{2} \cdot x_{1} + 0 \cdot x_{2} + p_{2} \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{2} \\
0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3} \\
\vdots \\
0 \cdot x_{1} \dots 0 \cdot x_{n} + q_{n-1} \cdot x_{n+1} + 0 \cdot x_{n+2} + p_{n-1} \cdot x_{n+3} = b_{n-1} \\
0 \cdot x_{1} \dots 0 \cdot x_{n} + 0 \cdot x_{n+1} + q_{n} \cdot x_{n+2} + p_{n} \cdot x_{n+3} = b_{n}
\end{cases}
\end{equation} Now I could calculate the solution for each row:
(1) $0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1}$
$\Leftrightarrow$ $ p_{1} \cdot x_{2} = b_{1}$ $\quad$ $\mid$ $\div$ $p_{1}$
$\Leftrightarrow$ $x_{2}$ = $\frac{b_{1}}{p_{1}}$
(3) $ 0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot x_{2} + p_{3} \cdot x_{4} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot \frac{b_{1}}{p_{1}} + p_{3} \cdot x_{4} = b_{3}$ $\quad$ $\mid$ $- q_{3} \cdot \frac{b_{1}}{p_{1}}$
$\Leftrightarrow$ $p_{3} \cdot x_{4} = b_{3} - q_{3} \cdot \frac{b_{1}}{p_{1}}$$\quad$ $\mid$ $\div p_{3}$
$\Leftrightarrow$ $x_{4}$ = $\frac{b_{3}-q_{3} \cdot \frac{b_{1}}{p_{1}}}{p_{3}}$
$\Leftrightarrow$ $\frac{p_{1} \cdot b_{3}-q_{3} \cdot b_{1}}{p_{1} \cdot p_{3}}$
And so on $\dots$
I don't know how to continue at this point to find a recursive formula for calculating the solution of the linear equation system. My ideas did not really lead me anywhere (I don't even think that they are correct) and im not sure if I even understand the matrix correctly. So how should I start instead?
Any hints guiding me to the right direction I much appreciate.
|
So,
a) solve the 1st as $x_2=b_1/p_1$;
b) solve the 2nd in terms of $x_3=(b_2-q_2x_1)/p_2$, the $x_1$ is unknown: keep it as such and continue;
c) all the following $x_{2k}$ will be a function of the precedent even-index and are therefore determined;
d) all the following $x_{2k+1}$ will be a linear function of the precedent odd-indexed and thus a linear function of $x_1$;
e) continue till the last two rows, which you are going to solve both in terms of $x_n$: one will refer to $x_{n-2}$ and one to $x_{n-1}$,
thus one will contain the unknown and the other not, and you can solve for both $x_1$ and $p_n$;
f) replace the found value of $x_1$ in the precedent results depending on it.
|
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|
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?
|
You write the equation in the form
$$
[(9+2)^k-9^k]-[(7+1)^k-7^k]-[(5+1)^k-5^k]=0.
$$
If $k\ge 0$, then you can use that
$$
(9+2)^k-9^k=\int_9^{9+2}kt^{k-1}dt\ge 2k9^{k-1},\\
-[(7+1)^k-7^k]=-\int_7^{7+1}kt^{k-1}dt\ge -k8^{k-1},\\
-[(5+1)^k-5^k]\ge -k6^{k-1}.
$$
Your equation becomes
$$
0\ge k\cdot\left(2\cdot 9^{k-1}-8^{k-1}-6^{k-1}\right).
$$
If we discard the $k=0$ solution, it suffices to examine the inequality
\begin{align}
6^{k-1}+8^{k-1}&\ge 2\cdot 9^{k-1},\qquad k=1,2,3,4,...
\end{align}
This is equivalent to the inequality
$$
6^\ell+8^\ell\ge 2\cdot 9^\ell,\qquad \ell=0,1,2,3,...,
$$
if $k=\ell+1$. Observe that
$$
9^\ell +9^\ell\ge 6^\ell+8^\ell\ge 2\cdot 9^\ell
$$
since $\ell\ge 0$. This means equality holds, and the only way that can be true is if $\ell=0$. (to see this step, rewrite as $0<9^\ell-6^\ell=-(9^\ell-8^\ell)<0$.)
|
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|
Determining in which quadrant an angle terminates. Let $A$ be in the fourth quadrant and let $\sec(A) = \dfrac{13}{5}$, Let $B$ be in the third quadrant and let $\csc(B)=\dfrac{-5}{3}$. Find $\sin(A + B)$ and determine in which quadrant $A + B$ terminates.
$\sec(A) = \dfrac{13}{5} \Rightarrow \cos(A) = \dfrac{5}{13}$
$\csc(B) = \dfrac{-5}{3} \Rightarrow \sin(B) = \dfrac{-3}{5}$
$\sin(A) = \dfrac{-12}{13}$ and $\cos(B) = \dfrac{-4}{5}$
$\sin(A + B) = \sin(A) \cos(B) + \sin(B) \cos(A)$
after performing the substitutions, I arrive at $\sin(A + B) = \dfrac{33}{65}$
$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
= $\dfrac{5}{13}(\dfrac{-4}{5}) - \dfrac{-12}{13}(\dfrac{-3}{5}) = \dfrac{-56}{65}$
|
Hint: Your calculations above are correct (I cleaned them up a little bit). If sine is positive but cosine is negative, what quadrant are you in? Does this give a hint to $A + B$?
|
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|
How to calculate $\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$ How to calculate $$\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$$
The limit equals a but any hints for the method?
|
Let us consider $$A=(x+a)^{1+\frac1x}\qquad \text {and} \qquad B=(x) ^{1+\frac{1}{x+a}}$$
$$A=(x+a)^{1+\frac1x}\implies \log(A)=\left({1+\frac1x}\right)\log(x+a)$$ Using Taylor we then have
$$\log(A)=\log \left({x}\right)+\frac{a+\log
\left({x}\right)}{x}+\frac{a-\frac{a^2}{2}}{x^2}+O\left(\frac{1}{x^3
}\right)$$
$$A=e^{\log(A)}=x+a+\log \left({x}\right)+\frac{2
\left(a-\frac{a^2}{2}\right)+\left(a+\log \left({x}\right)\right)^2}{2
x}+O\left(\frac{1}{x^2}\right)$$
$$B=(x) ^{1+\frac{1}{x+a}}\implies \log(B)=\left({1+\frac{1}{x+a}}\right)\log(x)$$
$$\log(B)=\log \left({x}\right)+\frac{\log \left({x}\right)}{x}-\frac{a \log
\left({x}\right)}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$B=e^{\log(B)}=x+\log \left({x}\right)+\frac{-2 a \log \left({x}\right)+\log
^2\left({x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$A-B=a\left(1+\frac{1+2 \log
\left({x}\right)}{x}\right)+O\left(\frac{1}{x^2}\right)$$
Trying with $x=10^6$ and $a=\pi$, the exact value would be $3.1416826014$ while the approximation leads to $3.1416826006$.
|
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|
How do you evaluate limit of $\frac{\sqrt{1+x^2} - \sqrt {1+x}}{\sqrt {1+x^3} - \sqrt {{1+x}}}$ when $x$ tends to $0$? I tried rationalization method and got as $\frac{x^2-x}{\sqrt {1+x^3} - \sqrt {1+x} ({\sqrt{1+x^2} + \sqrt {1+x}})}$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.
|
Using the same rationalization method, when $x\not\in\{0,1\}$,
$$
\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}
=\frac1{x+1}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\int \frac{\cos^2 x}{1 + \sin^2 x}dx$ My approach was to use a double-arc formula for cosine and sine, but I did not succeed.
$\sin^2 x = \frac{1 - \cos(2x)}{2}$ and $\cos^2x = \frac{1 + \cos(2x)}{2}$
|
Note that we can write
$$\begin{align}
\frac{\cos^2(x)}{1+\sin^2(x)}&=\frac{1+\cos(2x)}{3-\cos(2x)}\\\\
&=\frac{4-(3-\cos(2x))}{3-\cos(2x)}\\\\
&=-1+\frac{4}{3-\cos(2x)}
\end{align}$$
Enforce the substitution $y=\tan(x)$, so that $\cos(2x)=\frac{1-y^2}{1+y^2}$ and $dx=\frac1{1+y^2}\,dy$. Proceeding reveals
$$\begin{align}
\int\frac{4}{3-\cos(2x)}\,dx&=\int \frac{2}{1+2y^2}\,dy\\\\
&=\sqrt2\arctan(\sqrt2\,y)+C\\\\
&=\sqrt{2}\arctan(\sqrt2 \tan(x))+C
\end{align}$$
Putting it all together yield
$$\int \frac{\cos^2(x)}{1+\sin^2(x)}\,dx=-x+\sqrt{2}\arctan(\sqrt2 \tan(x))+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.
|
This fails for $a=1,$ $b=2,$ $c=3,$ & $d=4$. The left-hand side becomes $\frac{275}{504}=0.5456349...$ and the right-hand side is $\frac{16}{10}=1.6$
|
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|
Showing that if $p_1 + \cdots p_n = 1$ then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \right)^2 \ge n^3+2n+\dfrac 1n$? This problem is from the book "Cauchy-Schwarz Masterclass":
Show that if $p_1 + \cdots p_n = 1$ with each $p_i$ positive, then $\displaystyle \sum_{k=1}^n \left(p_k + \dfrac {1}{p_k} \right)^2 \ge n^3+2n+\dfrac 1n$
I expanded the LHS and arrived at
$$(p_1^2 + \cdots + p_n^2) + \left(\dfrac {1}{p_1^2} + \cdots + \dfrac {1}{p_n^2}\right) \ge n^3 + \dfrac 1n$$
I was able to show that $p_1^2 + \cdots + p_n^2 \ge \dfrac 1n$ by applying C-S to the sequences $(p_1, ..., p_n)$ and $(p_1, ..., p_n)$. I also think that $\dfrac {1}{p_1^2} + \cdots + \dfrac {1}{p_n^2} \ge n^3$ is true (because equality holds for $p_i = \dfrac 1n$ and I checked numerically for some values) but I am not able to prove it.
At this point in the book we only proed that classic $C-S$ inequality and the $C-S$ for inner product space, so I am hoping there is a solution only with these tools.
|
This is a proof that only uses Cauchy Schwarz, the second part can be simplified if you are allowed to use AM $\ge$ HM.
Apply CS to $n$ copies of $1$ and $\displaystyle\;p_k+\frac{1}{p_k}$, we obtain
$$n \sum_{k=1}^n \left(p_k + \frac{1}{p_k}\right)^2
= \left( \sum_{k=1}^n 1^2\right)\sum_{k=1}^n\left(p_k + \frac{1}{p_k}\right)^2
\ge \left(\sum_{k=1}^n p_k + \frac{1}{p_k}\right)^2
= \left(1 + \sum_{k=1}^n \frac{1}{p_k}\right)^2$$
Apply CS again to $\sqrt{p_k}$ and $\displaystyle\;\frac{1}{\sqrt{p_k}}$, we obtain
$$\sum_{k=1}^n \frac{1}{p_k} = \sum_{k=1}^n \sqrt{p_k}^2 \sum_{k=1}^n \frac{1}{\sqrt{p_k}^2}
\ge \left(\sum_{k=1}^n \frac{\sqrt{p_k}}{\sqrt{p_k}}\right)^2 = n^2$$
Combine these two inequalities, we obtain
$$\sum_{k=1}^n \left(p_k + \frac{1}{p_k}\right)^2
\ge \frac1n \left(1 + n^2\right)^2 = n^3 + 2n + \frac1n$$
|
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|
Finding all measurable functions maximizing an expression given integral conditions I'd like to know how to answer the question located in this comprehensive exam from the 1990s. The question is:
Find the maximum value of $\int_{-1}^{1} x^3 g(x) dx$ for measurable functions g(x) satisfying
$$ \int_{-1}^{1} g(x) dx = \int_{-1}^{1} x g(x) dx = \int_{-1}^{1} x^2 g(x) dx = 0 . $$
and $\int_{-1}^{1} |g(x)|^2 dx = 1$.
In principle I think I could reduce to the case where $g$ is continuous, and then even smooth/polynomial, and try to formulate things as some kind of optimization problem, but I suspect there's some sort of trick I'm not aware of.
|
We have four conditions. Define
$$
g(x) = a_0+a_1 x+ a_2 x^2+ a_3 x^3
$$
and then calculate
$$
\int_{-1}^1 g(x)dx = 2\left(a_0+\frac 13 a_2\right)=0\\
\int_{-1}^1 g(x)x dx = 2\left(\frac 13 a_1+\frac 15 a_3\right) = 0\\
\int_{-1}^1 g(x)x^2 dx =2\left(\frac 13 a_0 +\frac 15 a_2\right) = 0\\
\int_{-1}^1 |g(x)|^2 dx = 2 a_0^2+\frac{4 a_0 a_2}{3}+\frac{2 a_1^2}{3}+\frac{4 a_1a_3}{5}+\frac{2 a_2^2}{5}+\frac{2
a_3^2}{7}=1
$$
and solving we have
$$
\left[
\begin{array}{cccc}
a_0 & a_1 & a_2 & a_3 \\
0 & -\frac{3 \sqrt{\frac{7}{2}}}{2} & 0 & \frac{5 \sqrt{\frac{7}{2}}}{2} \\
0 & \frac{3 \sqrt{\frac{7}{2}}}{2} & 0 & -\frac{5 \sqrt{\frac{7}{2}}}{2} \\
\end{array}
\right]
$$
and the maximum is $\frac{5}{\sqrt{14}}-\frac{3 \sqrt{\frac{7}{2}}}{5}$
|
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|
Find the values of $a$ and $b$ such that the following matrices are similar
Find the values of $a$ and $b$ such that the following matrices are similar.
$$A=\begin{bmatrix} -2& 0 & 0 \\ 2 & a & 2 \\ 3& 1 & 1 \end{bmatrix}, \qquad B=\begin{bmatrix} -1& 0 & 0 \\ 0 & 2 & 0 \\ 0& 0 & b \end{bmatrix}$$
I know that similar matrices must have the same ranks. Hence, I can take that value of $a$ as any real number and $b \neq 0$. Is my answer is true or false? Any hints or solutions will be appreciated. Thank you.
|
Calculate the characteristic polynomials:
$$\det (A - \lambda I) = (-2-\lambda) \begin{vmatrix}a-2 & 2 \\ 1 & 1-\lambda\end{vmatrix} = (-2-\lambda)(\lambda^2 - \lambda(a+1) + (a-2)) = (-2 - \lambda)\left(\frac{1+a-\sqrt{a^2-2a+9}}2 - \lambda\right)\left(\frac{1+a+\sqrt{a^2-2a+9}}2 - \lambda\right)$$
$$\det (B - \lambda I) = (-1-\lambda)(2 - \lambda)(b - \lambda)$$
If $A$ and $B$ are similar, then the characteristic polynomials are equal so uniqueness of factorization in $\mathbb{C}[\lambda]$ gives $b = -2$ and
$$\left\{\frac{1+a-\sqrt{a^2-2a+9}}2, \frac{1+a+\sqrt{a^2-2a+9}}2\right\} = \{-1, 2\}$$
for which the only solution is $a = 0$.
On the other hand, if $a = 0, b= 2$ then both $A$ and $B$ diagonalize with distinct eigenvalues $\{-1, 2, -2\}$ so they are similar.
|
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|
If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ If $ab \mid c(c^2-c+1)$ and $c^2+1 \mid a+b$ then prove that $\{a, b\}=\{c, c^2-c+1 \}$ (equal sets), where $a$, $b$, and $c$ are positive integers.
This is math contest problem (I don't know the source). I was struggling to solve this, but I cannot find a way to make it easier. Can you help me?
All I did was like this:
$$a+b=d(c^2+1)=d(c^2-c+1+c) \\ c(a+b)=dc(c^2-c+1)+dc^2 \\ c(a+b)=deab+dc^2=d(eab+c^2)$$
I suppose this is a quadratic in $c$ and determine its discriminant. But it doesn't make the problem any easier!
|
Suppose we have:
$m^2k^2(c^2+1)^2-4mc(c^2-c+1)=t^2$
$(m^2k^2(c^2+1)^2-t)(m^2k^2(c^2+1)^2+t)=4mc(c^2-c+1)$
We may have a system of equations as follows:
$m^2k^2(c^2+1)^2-t=4mc$
$m^2k^2(c^2+1)^2+t=c^2-c+1$
Summing two equations we get:
$2m^2k^2(c^2+1)^2=4mc+c^2-c+1$
$k^2=\frac{2mc+(c^2-c+1)/2}{2m^2(c^2+1)^2}$
Now problem reduces to : what is condition for $2mc+(c^2-c+1)/2$ to be a perfect square.The discriminant of quadratic relation $2mc+(c^2-c+1)/2≥0$ will give a range of value for m, which is finally the condition for initial relation to be a perfect square.
|
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|
Integration of $\frac{1}{x^2-a^2}$ by trigonometric substitution?
$$\int \frac{1}{x^2-a^2}dx$$
Now, I know this can be done by splitting the function into two integrable functions,
$\displaystyle\dfrac{1}{2a}\int \bigg(\dfrac{1}{x-a} - \dfrac{1}{x+a}\bigg)dx$
And then doing the usual stuff.
My question is, how can we do this by using trigonometric substitution?
The only thing that gets in my mind is $x=a\sec\theta$, but then got stuck on proceeding further.
Any help would be appreciated.
|
You might use $x=a\sec\theta$: $dx=a\sec\theta\tan\theta\,d\theta$ and the integral becomes
$$
\int\frac{1}{a^2(\sec^2\theta-1)}a\sec\theta\tan\theta\,d\theta
=
\frac{1}{a}\int\frac{\cos^2\theta}{\sin^2\theta}\frac{1}{\cos\theta}\frac{\sin\theta}{\cos\theta}\,d\theta
=
\frac{1}{a}\int\frac{1}{\sin\theta}\,d\theta
$$
and this is a known fellow: set $\theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
$$
\int\frac{1}{\sin u\cos u}\,du=
\int\frac{1}{\cos^2u+\sin^2u}{\sin u\cos u}\,du=
\int\frac{\cos u}{\sin u}\,du+\int\frac{\sin u}{\cos u}\,du
$$
You thus get
$$
\log\lvert\sin u\rvert-\log\lvert\cos u\rvert+c=
\log\left\lvert\tan\frac{\theta}{2}\right\rvert+c
$$
Now back substitute.
Oh, well, but if we do $x=a\cos t$, instead?
$$
\int\frac{1}{a^2(\cos^2t-1)}(-a\sin t)\,dt=
\frac{1}{a}\int\frac{1}{\sin t}\,dt
$$
Speedier, isn't it? Anyway, not easier than observing that
$$
\frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2866731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Arithmetic series problem Given $\left\{a_n\right\}$ arithmetic progression, $a_1=2$, $a_{n+1}=a_n+2n$ $\left(n\:\ge \:1\right)$. $a_{50}=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_{50}=2+d\left(n-1\right)$$
$$a_{50}=2+2\left(n^2-n\right)$$
$$a_{50}=2+2\cdot 2450$$
$$a_{50}=4902$$
But this is wrong. Answers:
$$A=2452,\:B=2450,\:C=2552,\:D=2500$$
|
Alternatively, note: $a_{n+1}-a_n=2n$. So:
$$\begin{align} a_2-a_1&=2\cdot 1\\
a_3-a_2&=2\cdot 2\\
a_4-a_3&=2\cdot 3\\
&\vdots \\
a_{50}-a_{49}&=2\cdot 49 \end{align}$$
Summing all (midterms telescope):
$$a_{50}-a_1=2(1+2+3+\cdots +49) \Rightarrow \\
a_{50}=a_1+2\cdot \frac{1+49}{2}\cdot 49=2+2450=2452.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2868149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Convergence of the series $\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$.
Check whether the series
$$\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$$
is convergent.
I tried to sandwich the function by $\dfrac{1}{1+x^2}$ and $\dfrac{x}{1+x^2}$ , but this did not help at all.
Any other way of approaching?
|
When $x\in [0,1], \frac {\sqrt x}{2} \le\frac {\sqrt x}{1+x^2} \le \sqrt x$
Which means that $\frac 12\int_0^\frac1n \sqrt x\ dx \le \int_0^\frac1n \frac {\sqrt x}{1+x^2} \ dx \le \int_0^\frac1n \sqrt x \ dx$
if $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \sqrt x \ dx $ converges then $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac{\sqrt x}{1+x^2} \ dx $ converges.
and if $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \sqrt x \ dx $ diverges then $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac 12\sqrt x \ dx $ diverges and $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac{\sqrt x}{1+x^2} \ dx $ diverges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2869442",
"timestamp": "2023-03-29T00:00:00",
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|
The limits of integration for a paraboloid and plane.
Consider the volume, $V$, enclosed by the paraboloid
$$z=x^2+y^2$$
and the plane $2x+z=3$. Determine the limits for the enclosed volume. Do not attempt to evaluate the integral.
Hello, I’m confused on how to get limits for this integral, I drew it and I think $z$ limits are from the paraboloid equation to $z=3-2x$, but after that I’m stumped.
Thanks!
|
The two surfaces intersect when $$z=x^2+y^2=3-2x \Rightarrow x^2+2x+y^2 = 3$$
Now $$x^2+2x+y^2 = 3 \Rightarrow \left(x+1\right)^2+y^2=4$$
Therefore in cartesian coordinates:
$$V= \int{\int{\int dz}dy}dx=\int_{x=-3}^{x=1}{\int_{y=-\sqrt{4-(x+1)^2}}^{y=\sqrt{4-(x+1)^2}}{\int_{z=x^2+y^2}^{z=3-2x}1dz}dy}dx$$
In Cylindrical coordinates, starting from:
$$V= \int{\int{\int_{z=x^2+y^2}^{z=3-2x}1dz}}dA=\int{\int{\left[4-\left(x+1\right)^2-y^2\right]}}dA$$
Then using $r=x^2+y^2$, $x=r\cos\theta$ and $y=r\sin\theta$:
$$V=\int_{\theta=0}^{\theta=2\pi}{\int_{r=0}^{r=2}{\left[ {4-\left(r\cos\theta+1\right)^2-\left( {r\sin\theta} \right)^2} \right]}}rdrd\theta$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to integrate this indefinite integral? $$\int\frac{x^2\sec^2x}{(x \tan{x}+1)^{2}}\,\mathrm{d}x$$
I tried the online available calculators but they cannot calculate the answer or provide the solution.
|
This is basically working backwards, like @Arthur suggests:
\begin{align}
\frac{x^2\sec^2 x}{(x\tan x+1)^2} &= \frac{x^2}{\left(x\frac{\sin x}{\cos x} + 1\right)^2\cos^2 x}\\
&= \frac{x^2}{(x\sin x + \cos x)^2}\\
&= \frac{x^2(\sin^2 x + \cos ^2 x) - x\cos x\sin x + x \cos x\sin x}{(x\sin x + \cos x)^2}\\
&= \frac{x\sin x(x \sin x + \cos x) - (\sin x - x\cos x)x\cos x}{(x\sin x + \cos x)^2}\\
&= \frac{(\sin x - x\cos x)'(x \sin x + \cos x) - (\sin x - x\cos x)(x\sin x + \cos x)'}{(x\sin x + \cos x)^2}\\
&= \left(\frac{\sin x - x\cos x}{x \sin x + \cos x}\right)'
\end{align}
so $$\int \frac{x^2\sec^2 x}{(x\tan x+1)^2} \,dx = \frac{\sin x - x\cos x}{x \sin x + \cos x}$$
|
{
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|
Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin(\sin12)+\arccos(\cos12)$ for every $x \in \mathbb {R}$ Find the least value of $n\in N$ for which $(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12})$ for every $x \in \mathbb {R}$
$(n-2)x^2+8x+n+4>\arcsin({\sin12})+\arccos({\cos12}) $
$(n-2)x^2+8x+n+4>4\pi-12+4\pi-12$
$(n-2)x^2+8x+n+4>8\pi-24$
I don't know how to solve further. Please help
|
Consider $f(x)=\arcsin(\sin x)+\arccos(\cos x)$; then
$$
f'(x)=\frac{1}{\sqrt{1-\sin^2x}}\cos x+\frac{1}{\sqrt{1-\cos^2x}}\sin x=
\frac{\cos x}{\lvert\cos x\rvert}+\frac{\sin x}{\lvert\sin x\rvert}
$$
Note that $7\pi/2<12<4\pi$; one inequality is obvious; the other one is equivalent to $\pi<24/7$, but we know from Archimedes that $\pi<22/7$.
For $x\in(7\pi/2,4\pi)$ we have $\cos x>0$ and $\sin x<0$; thus $f$ is constant over $(7\pi/2,4\pi)$. Since
$$
f(4\pi-\pi/4)=0
$$
and $4\pi-\pi/4\in(7\pi/2,4\pi)$, we conclude that $f(12)=0$.
For $n\le2$, the polynomial $p(x)=(n-2)x^2+8x+n+4$ takes on negative values (it's a concave parabola for $n<1$ and a line for $n=2$).
Assuming $n>2$, you need that the discriminant of $p$ is negative:
$$
64-4(n-2)(n+4)<0
$$
that is
$$
n^2+2n-24>0
$$
so $n>4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$ Solve the equation:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$
Things I have done so far:
$$\sqrt[3]{x^{2}+4}=\sqrt{x-1}+2x-3$$ Deducting 2 from both sides of equation
$$\Leftrightarrow (\sqrt[3]{x^2+4}-2)=(\sqrt{x-1}-1)+2x-4$$
$$\Leftrightarrow \frac{x^2+4-8}{\sqrt[3]{(x^2+4)^2}+2\sqrt[3]{x^2+4}+4}=\frac{x-2}{\sqrt{x-1}+1}+2(x-2)$$
$$\Leftrightarrow (x-2)(\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2)=0$$
+)$$x-2=0\Leftrightarrow x=2$$
+)$$\frac{x+2}{\sqrt[3]{(x^{2}+4)^2}+2\sqrt[3]{x^{2}+4}+4}-\frac{1}{\sqrt{x-1}+1}-2=0(*)$$
I don't know how to solve the equation (*).
By the way, I think there must be "smart" way to solve this equation.
|
Actually, there is only one solution in this equation.
let $f(x)=(x^2+4)^{1/3}-\sqrt{x-1}-2x-3(x\geq 1)$,then $f'(x)=\frac{2x}{3(x^2+4)^{\frac{2}{3}}}-\frac{1}{2\sqrt{x-1}}-2 \leq \frac{2x}{3(x^2+4)^{\frac{2}{3}}} -2\leq \frac{2x}{3(x^2)^{\frac{2}{3}}} -2=\frac{2}{3x^{\frac{1}{3}}}-2 \leq -\frac{4}{3}$.
So $f(x)$ is strictly decreasing function and this equation has only one solution.
|
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|
Prove that inequality $1+\frac{1}{2\sqrt{2}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$ Let $n$ is a natural number. Prove that inequality $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+...+\frac{1}{n\sqrt{n}}<2\sqrt{2}$$
My try: $$\frac{1}{n\sqrt{n}}=\frac{\sqrt{n}}{n^2}<\frac{\sqrt{n}}{n\left(n-1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n-1}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{n-1}}\right)<\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}$$
Please check my solution for me and give me some idea.
|
Your sum can be written as $\sum_{k=1}^n\frac{1}{k^{3/2}}=1+2^{-3/2}+\sum_{k=3}^n k^{-3/2}$. Since $f(x)=x^{-3/2}$ is decreasing, we can bound the sum from above by an integral, that is,
$$
\sum_{k=3}^n\frac{1}{k^{3/2}}\leq\int_2^n x^{-3/2}\,dx.
$$
Note specifically the lower bound of the integral.
Evaluating the integral on the right yields
$$
\int_2^n x^{-3/2}\,dx=-2x^{-1/2}\bigg|_{x=2}^{x=n}=\frac{2}{\sqrt2}-\frac{2}{\sqrt n}.
$$
Thus, we have
$$
\sum_{k=1}^n\leq1+2^{-3/2}+\frac{2}{\sqrt2}-\frac2{\sqrt n}\leq\frac{2^{3/2}+1+4}{2^{3/2}}.
$$
Clearly $2^{3/2}=2\sqrt2=\sqrt8\leq\sqrt9=3$, so that we have
$$
\frac{2^{3/2}+1+4}{2^{3/2}}\leq\frac{8}{2^{3/2}}=2\sqrt2.
$$
Putting it all together, then, we have
$$
\sum_{k=1}^nk^{-3/2}\leq2\sqrt2.
$$
|
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|
Evaluate the Sum $S=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots \infty$ Evaluate the Sum
$$S=\frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}+\cdots \infty$$
My try: We have the $n$ th term as
$$T_n=\frac{1.3.5. \cdots (2n-1)}{4.6.8 \cdots (2n+2)}$$ $\implies$
$$T_n=\frac{1.3.5. \cdots (2n-1)}{2^n \times (n+1)!}$$
$$T_n=\frac{(2n)!}{4^n \times n! \times (n+1)!}$$
$$T_n=\frac{\binom{2n}{n-1}}{n \times 4^n}$$
Any clue here?
|
Hint: See Catalan Number with $x=\dfrac14$.
|
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|
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdot$ $1$ Tail $+$ $8$ Heads $\cdot$ $2$ tails $+$ $7$ Heads $\cdot$ $3$ tails.
The answer is $176$ though.
|
the combinations notation ${10\choose 7} = \frac {10!}{7!3!}$ accounts for the fact that if you have 7 heads you also have 3 non-heads. ${10\choose 7}{10\choose 3}$ is effectively squaring the value that you need for that term. Similar for the other terms.
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x}{x}=1$$
which means that the function grows linearly.
$$f(x)-mx=\sqrt{x^2+3x}-x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}-1\right)\sim x\left(\frac {1}{2}\cdot\frac{3}{x}\right)=\frac{3}{2}$$
The oblique asymptote is $y=x+\frac 3 2$ which is correct. For $x\rightarrow-\infty$ we have:
$$\frac{f(x)}{x}=\frac{|x|}{x}=\frac{-x}{x}=-1$$
This means that
$$f(x)-mx=\sqrt{x^2+3x}+x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty$$
Which is not what my textbook reports ($-\frac{3}{2}$). Any hints on what I did wrong to find the $q$ for $x\rightarrow-\infty$?
|
The problem is in
$$
f(x)-mx=\sqrt{x^2+3x}+x\underset{\begin{array}{c} \uparrow \\ \text{problem}\end{array}}{=}x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty
$$
which should be
$$
f(x)-mx=\sqrt{x^2+3x}+x=x\left(-\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)
$$
because we are near $-\infty$, so $\sqrt{x^2}=-x$.
In order to avoid such common mistakes, I suggest to change $x=-t$, when the limit for $x\to-\infty$ is involved:
$$
\lim_{x\to-\infty}\bigl(\sqrt{x^2+3x}+x\bigr)=
\lim_{t\to\infty}\bigl(\sqrt{t^2-3t}-t\bigr)=
\lim_{t\to\infty}\frac{-3t}{\sqrt{t^2-3t}+t}
$$
From here it should be clear.
Another strategy, in this case, is to set $x=-1/t$, that transforms the limit into
$$
\lim_{t\to0^+}\left(\sqrt{\frac{1}{t^2}-\frac{3}{t}}-\frac{1}{t}\right)=
\lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t}
$$
which is a simple derivative: if $f(t)=\sqrt{1-3t}$, then $f'(t)=\frac{-3}{2\sqrt{1-3t}}$ and
$$
\lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t}=f'(0)=-\frac{3}{2}
$$
|
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|
If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$
Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions, find the possible values of $n$.
I tried taking $2x$ as $a$, and then $2y$ as $b$, and then finding the possibilities. However, I am not sure about the approach I've used.
A little hint would be appreciated.
$x$, $ y$ and $z$ belong to whole numbers.
|
Assume $x,y,z$ are required to be positive integers.
Consider two cases . . .
Case $(1)$:$\;n$ is even.
From the equation $2x+2y+z=n$, it follows that $z$ is also even.
Writing $z=2w$, and $n=2m$, for some positive integers $w,m$, we get
\begin{align*}
&2x+2y+z=n\\[4pt]
\iff\;&2x+2y+2w=2m\\[4pt]
\iff\;&x+y+w=m\\[4pt]
\end{align*}
By the Stars-and-Bars formula, the equation
$$x+y+w=m$$
has exactly ${\large{\binom{m-1}{2}}}$ positive integer solution triples $(x,y,w)$, for any given positive integer value of $m$, hence
\begin{align*}
&{\small{\binom{m-1}{2}}}=28\\[4pt]
\implies\;\;&{\small{\frac{(m-1)(m-2)}{2}}}=28\\[4pt]
\implies\;\;&m^2-3m+2=56\\[4pt]
\implies\;\;&m^2-3m-54=0\\[4pt]
\implies\;\;&(m-9)(m+6)=0\\[4pt]
\implies\;\;&m=9\\[4pt]
\implies\;\;&n=18\\[4pt]
\end{align*}
Case $(2)$:$\;n$ is odd.
From the equation $2x+2y+z=n$, it follows that $z$ is also odd.
Writing $z=2w-1$, and $n=2m-1$, for some positive integers $w,m$, we get
\begin{align*}
&2x+2y+z=n\\[4pt]
\iff\;&2x+2y+(2w-1)=2m-1\\[4pt]
\iff\;&x+y+w=m\\[4pt]
\end{align*}
As for case $(1)$, since $m$ is a positive integer such that the equation
$$
x+y+w=m
\qquad\qquad
$$
has exactly $28$ positive integer solution triples $(x,y,w)$, it follows that $m=9$, hence $n=2m-1=17$.
Note:$\;$If $x,y,z$ are only required to be nonnegative integers, then using arguments analogous to the ones given above, we get $n=12$ or $n=11$.
|
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|
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
|
Abbreviating $\sin\theta$ by $s$ and $\cos\theta$ by $s$ we have on base of $c^2+s^2=1$:$$1+2sc=c^2+2sc+s^2=(c+s)^2$$ so that:$$(1+2sc)(c-s)=(c+s)^2(c-s)=(c^2-s^2)(c+s)$$
Dividing boths sides by $c$ and abbreviating $\tan(\theta)$ by $t$ we get on base of $t=s/c$:$$(1+2sc)(1-t)=(c^2-s^2)(1+t)$$
Now divide both sides by $(1-t)(c^2-s^2)$ to achieve:$$\frac{1+2sc}{c^2-s^2}=\frac{1+t}{1-t}$$
|
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|
Value of a triangle inscribed in a rectangle I was doing a little trigonometry challenge given by my professor, which is not anything especially complicated. However, it is completely based on getting the area of the green triangle in relation to the side of the square and the triangle, equal to $x$.
Reference image:
I've already got a bit of data from some trig. relations. What doesn't seem to be doable is finding a relation between $x$ and the sides $L_1$ and $L_2$ or isolating the sine of $\beta$ from the area of that triangle, in relation to $x$ as well.
Anyway, the following is the data gathered so far.
*
*Sine of $\alpha$ (both $\alpha_1$ and $\alpha_2$):
$$ \begin{aligned}
h_1^2 &= x^2 + \frac{x^2}{4} = \frac{5x^2}{4} \\
h_1 &= \frac{x \sqrt{5}}{2}
\end{aligned} \\
\begin{aligned}
\sin {\alpha} &= \frac{x}{h_1} = \frac{x}{\frac{x \sqrt{5}}{2}} = \frac{2x}{x \sqrt 5} = \frac{2}{\sqrt 5} = \frac{2 \sqrt 5}{5}
\end{aligned} $$
*Relation between sides $L_1$ and $L_2$, as well as sine of $\beta$ (through the law of sines):
$$ \begin{aligned}
\frac{\sin \alpha}{L_2} &= \frac{\sin \beta}{\frac{x}{2}} = \frac{\sin 30^\circ}{L_1} \\
\frac{2 \sqrt 5}{5L_2} &= \frac{2 \sin \beta}{x} = \frac{1}{2L_1}
\end{aligned} \\
\begin{aligned}
L_1 &= \frac{L_2 \sqrt 5}{4} = \frac{x}{4 \sin \beta} \\
L_2 &= \frac{4L_1 \sqrt 5}{5} = \frac{x \sqrt 5}{5 \sin \beta}
\end{aligned} $$
*Relation between $h_2$ and sides $L_1$ and $L_2$:
$$ \begin{aligned}
\sin \alpha = \frac{2 \sqrt 5}{5} &= \frac{h_2}{L_1} \\
\sin 30^\circ = \frac{1}{2} &= \frac{h_2}{L_2}
\end{aligned} \\
\begin{aligned}
h_2 &= \frac{L_2}{2} = \frac{2L_1 \sqrt 5}{5} = L_1 \sin \alpha
\end{aligned} $$
*Calculating the area with the data so far (1) (using sine of $\beta$):
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot L_1 \sin \alpha}{2} \\
&= \frac{x}{4} \cdot L_1 \sin \alpha \\
&= \frac{x}{4} \cdot \frac{x}{4 \sin \beta} \sin \alpha \\
&= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta}
\end{aligned} $$
*Calculating the area with the data so far (2) (using the sides' values)
$$ \begin{aligned}
A &= \frac{\frac{x}{2} \cdot \frac{2L_1 \sqrt 5}{5}}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_1}{2} \\
&= \frac{x \sqrt 5}{5} \cdot \frac{L_2 \sqrt 5}{8} \\
&= x \cdot \frac{L_2}{8}
\end{aligned} $$
Additionally, I've tried assuming (for the sake of an approximation) that $\beta$ is a right angle, since it is roughly equal to 86 degrees, and applied a theorem I've noticed in right triangles while meddling with this challenge.
*
*Given a triangle $\Delta ABC$, where $a = BC$, $b = CA$ and $c = AB$, the height $h$ of the triangle, perpendicular to the base (assumed to be AB), is equal to the product of the product of the cathetuss divided by the hypotenuse. That is,
$$ \begin{aligned}
h = \frac{a \cdot b}{c}.
\end{aligned} $$
*Moreover, with a substitution of these values, it is possible to get the value of the sides from $x$, as well as confirm the angle of $\beta$.
$$ \begin{aligned}
h_2 &= \frac{2L_1L_2}{x} \\
&= \frac{L_2}{2} = L_1 \sin \alpha
\end{aligned} \\
L_1 = \frac{x}{4},\ L_2 = \frac{x \sqrt 5}{5} \\
x = 4L_1 = L_2 \sqrt 5 \\
\sin \beta = \sin 90^\circ = 1
$$
*And finally, calculate an estimated approximation for the area, either with the value of $\beta$ or the value of the sides in respect to $x$.
$$ \begin{align}
A &= \frac{x^2}{16} \cdot \frac{\sin \alpha}{\sin \beta} \\
&= \frac{x^2}{16} \cdot \frac{2 \sqrt 5}{5} \\
&= \frac{2x^2 \sqrt 5}{80} = \frac{x^2 \sqrt 5}{40} \\ \newline
A &= x \cdot \frac{L_2}{2} \\
&= x \cdot \frac{ \frac{x \sqrt 5}{5} }{8} \\
&= x \cdot \frac{x \sqrt 5}{40} = \frac{x^2 \sqrt 5}{40} \\
\end{align} $$
How one would come to derive $x$ into the sides' values or the sine of $\beta$, in order to get the area only through $x$? What sort of hindsight is required to do so, even?
|
I am not sure that this is a solution you want, since it uses vectors and not trigonometry. I'll put it here anyway since it is a simpler approach and gives the right answer. I will use $l$ instead of $x$ for the length of the sides of the square and triangle to avoid confusion with the variable $x$.
The idea is to find the coordinates of the vertices of the triangle, so that we can find two vectors forming the sides of the triangle. The determinant of these vectors give twice of the area of the triangle. Firstly, the line from the top left vertex of the square to the bottom right foot of the triangle is given by $y=l-\frac{1}{2}x$. The right edge of the square and left edge of the triangle are given by $x=l$ and $y=\sqrt{3}x-l\sqrt{3}$. Therefore, the vertices of the triangle are given by $\left(l,\frac{1}{2}l\right)$, $\bigg(l\left(1+\frac{1}{2\sqrt{3}+1}\right),\frac{1}{2}l\left(1-\frac{1}{2\sqrt{3}+1}\right)\bigg)$ and of course $(l,0)$.
The vectors we want are therefore $\left(0,\frac{1}{2}l\right)$ and $\bigg(l\left(\frac{1}{2\sqrt{3}+1}\right),\frac{1}{2}l\left(1-\frac{1}{2\sqrt{3}+1}\right)\bigg)$. Taking the absolute value of their determinant gives
$$ \left\vert -\frac{1}{2}l^2\left(\frac{1}{2\sqrt{3}+1}\right) \right\vert = \frac{l^2}{2(2\sqrt{3}+1)}. $$
This is the area of the paralleogram with the two vectors as its sides, which is twice of the area of the triangle, which we want. Therefore, halving this expression gives
$$ \mathrm{Area}=\frac{l^2}{4(2\sqrt{3}+1)}. $$
|
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|
Find all $x,\ y,\ z$ that satisfy: $(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$
Find all integers $x,\ y,\ z$ that satisfy:
$$(x+y-z)^{5}+(x+z-y)^{5}+(z+y-x)^{5}=0$$
I guessed that, considering the symmetry and the similar case $\sum x_{i}^{2n}=0$, that these three expressions must be equal to zero, which leads to $x=y=z=0$. This leaves two other cases:
i. $x+y-z<0,\ x+z-y<0,\ z+y-x>0$
ii. $x+y-z<0,\ x+z-y>0,\ z+y-x>0$
But this didn't help much. My other attempt was considering this equation mod $2x,\ 2y,\ 2z$.
Note. This problem states that $x,y,z$ are integers, but if you have solved the case that $x,\ y,\ z$ are real, please share your solution.
|
Since $x, y, z$ are integers so are $a=x+y-z, b=x+z-y, c=z+y-x$.
We know from Fermat's last theorem that $a^5+b^5+c^5=0$ is impossible for non-zero integers $a, b, c$.
This means that $x+y-z, x+z-y$ or $z+y-x$ must be zero.
If $x+y-z=0$ then your equation gives $x+z-y=x-y-z$ which means that $z=0$ and $x=-y$.
If $x+z-y=0$ then your equation gives $x+y-z=x-y-z$ which means that $y=0$ and $x=-z$.
If $z+y-x=0$ then your equation gives $x+y-z=y-x-z$ which means that $x=0$ and $y=-z$.
So, a general solution is $(x, y, z)=(-a, a, 0)$ or $(x, y, z)=(-a, 0, a)$ or $(x, y, z)=(0, -a, a)$ for any integer $a$.
If you want real solutions, then there are many of them. Simply choose $y=z=1$ and solve for $x$. You can find that $(x, y, z)=(-\frac{2}{\sqrt[5]{2}-1}, 1, 1)$
|
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|
For $\triangle ABC$ with inradius $r$, if $\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$, then which angle is $90^\circ$?
For $\triangle ABC$ with inradius $r$, if
$$\frac{c + a}{b} +\frac{c + b}{a} = \frac{c}{r}$$ then which angle is $90^\circ$?
My try :
I am stuck here I think my process is wrong please help me in this.
|
The equality on the $3^{rd}$ line can be written as:
$$r = \frac{abc}{a^2+b^2+(a+b)c}$$
The symmetry in $\,a,b\,$ suggests that the right angle is $\,C\,$, in which case $\,a^2+b^2=c^2\,$, then:
$$\require{cancel}
r = \frac{abc}{c^2+(a+b)c} = \frac{ab\cancel{c}}{(a+b+c)\cancel{c}} =\frac{ab}{a+b+c} = \frac{\cancel{2}S}{\cancel{2}p} = \dfrac{S}{p} \tag{1}
$$
In any triangle the inradius $\,r = \dfrac{S}{p}\,$ where $\,S\,$ is the area and $\,p = \dfrac{a+b+c}{2}\,$ is the semi-perimeter. For a right triangle $S = \dfrac{1}{2}ab\,$, so $(1)$ holds true, therefore the right angle is indeed $\,C\,$.
|
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|
On Lame's Theorem I was trying to study Lame's theorem but I'm a little bit stuck on one specific step... I write here the theorem and the proof:
Lame's theorem:
using the Euclidean algorithm to find the greatest common divisor of two positive integers has number of divisions less than or equal five times the number of decimal digits in the minimum of the two integers.
Proof:
Let $a$ and $b$ be two positive integers where $a > b$.
Applying the Euclidean algorithm to find the greatest common divisor of two integers with $a = r_{0}$ and $b = r_{1}$, we get
$r_{0} = r_{1}q_{1}+r_{2}$, $0≤r_{2}<r_{1}, $
$r_{1} = r_{2}q_{2}+r_{3}$, $0≤r_{3}<r_{2}, $
. . .
$r_{n-2} = r_{n-1}q_{n-1}+r_{n}$, $0≤r_{n}<r_{n-1}, $
$r_{n-1} = r_{n}q_{n}$
Notice that each of the quotients $q_{1}, q_{2}, ..., q_{n-1} $ are all greater than $1$ and $q_{n} ≥ 2$ and this is because $r_{n} < r_{n-1}.$
Thus we have
$r_{n} ≥ 1=f_{2}$,
$r_{n-1} ≥ 2r_{n} ≥ 2f_{2} = f_{3}$,
$r_{n-2} ≥ r_{n-1} + r_{n} ≥ f_{3} + f_{2} = f_{4}$,
$r_{n-3} ≥ r_{n-2} + r_{n-1} ≥ f_{4} + f_{3} = f_{5}$,
...
$r_{2} ≥ r_{3} + r_{4} ≥ f_{n-1} + f_{n-2} = f_{n}$,
$b = r_{1} ≥ r_{2} + r_{3} ≥ f_{n} + f_{n-1} = f_{n+1}$.
Thus notice that $b≥f_{n+1}$.
By a Lemma I don't report here, we have $f_{n+1}>α^{n−1}$ for $n>2$.
As a result, we have $b > α^{n−1}$.
Now notice since $\log_{10} \alpha > \frac{1}{5}$
we see that
$\log_{10}b > (n − 1)/5$.
Thus we have
$(n - 1)< 5 log_{10}b$
Now let $b$ has $k$ decimal digits. As a result, we have $b < 10^k$ and thus $log_{10}b < k$. Hence we conclude that $(n − 1) < 5k$.
Since $k$ is an integer, we conclude that $n ≤ 5k$.
What I really don't understand is just this line:
Now notice since $\log_{10} \alpha > \frac{1}{5}$,
My question is: why $\frac{1}{5}$ has been chosen?
Where does it come from?
Has it been chosen because the theorem says:
[...]less than or equal five times [...]
?
Thank you
|
Let's chase through the rest of the proof without using the line $\log_{10}\alpha>1/5$. So instead of writing $\log_{10}b>(n-1)/5$, we'll just write $$\log_{10}b>(n-1)\log_{10}\alpha.$$ Continuing through the rest of the algebra, we end up with $$n-1<k/\log_{10}\alpha.$$
So, this actually gives a more precise conclusion to the theorem: the number of steps $n$ is less than $1$ plus the number of decimal digits $k$ times the constant $C=1/\log_{10}\alpha$. If we now observe that $\log_{10}\alpha>1/5$ so $C<5$, we can conclude that $n$ is at most the number of decimal digits times $5$, which is the originally stated (but weaker) conclusion. It is natural to choose $5$ here since it is the smallest integer greater than $C$.
|
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|
How to calculate the limit $\lim\limits_{n\to\infty} \int\limits_0^1 \frac{n(2nx^{n-1}-(1+x))}{2(1+x)}\,dx$? How to calculate the limit $\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$?
I have to calculate the limit when solving
Find $a,b$ for $\displaystyle\int_0^1 \dfrac{x^{n-1}}{x+1} dx=\dfrac{a}{n}+\dfrac{b}{n^2}+o(\dfrac{1}{n^2}) (n\to\infty)$
First I calculated that
$\lim\limits_{n\to\infty} \displaystyle\int_0^1 \dfrac{nx^{n-1}}{x+1} dx=\dfrac{1}{2}$, thus $a=\dfrac{1}{2}$, then
$2b=\lim\limits_{n\to\infty}\displaystyle\int_0^1\dfrac{n(2nx^{n-1}-(1+x))}{2(1+x)} dx$.
However, I cannot find a good way to calculate it.
|
\begin{align*}
2b &= \lim_n n \int_0^1 \frac{nx^{n-1}} {x+1} \mathrm dx - \frac n 2 \\
&= \lim_n n \int_0^1 \frac {\mathrm d (x^n)} {x+1} - \frac n2\\
&= \lim_n n \left. \frac {x^n} {1+x}\right|_0^1 + n \int_0^1 \frac {x^n \mathrm dx} {(1+x)^2} - \frac n 2\\
&= \lim_n \frac n {n+1} \cdot \left.\frac {x^{n+1}}{(x+1)^2}\right|_0^1 + \frac {2n}{n+1}\int_0^1 \frac {x^{n+1}}{(x+1)^3} \mathrm d x\\
&= \frac 14 + 2\lim_n \int_0^1 \frac {x^{n+1} \mathrm dx} {(x+1)^3}\\
&= \frac 14,
\end{align*}
where
$$
0 \gets \frac 18 \int_0^1 x^{n+1} \mathrm dx \leqslant \int_0^1 \frac {x^{n+1} \mathrm dx} {(1+x)^3} \leqslant \int_0^1 x^{n+1} \mathrm d x \to 0.
$$
UPDATE
The limit you gave at the very first is actually $b$, not $2b$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is the function is differentiable at 0? Is the function given by
$ f(x) = \begin{cases} \frac{1}{x \log2} - \frac{1}{2^x-1},\text{if} \ x\neq 0 ,\\\frac{ 1}{2} \ \text {if} \ x=0 \end{cases}$.
is differentiable at zero ?
My Attempts:
$f'(0) = \frac{f(x) - f(0)}{x-0} =\frac{\frac{1}{x \log2} - \frac{1}{2^x-1}- \frac{1}{2} }{x} $
after that i can not able to proceed further
Pliz help me
Any hints/solution will be appreciated
thanks in advance
|
$$\frac1{2^x-1}=\frac1{\exp(x\ln2)-1}=\frac1{x\ln 2+x^2(\ln 2)^2/2+O(x^3)}
=\frac1{x\ln 2}\left(1-\frac{\ln 2}2+O(x)\right)$$
etc.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $ a^2+b^2+c^2=1$ implies $ab+bc+ca \in [-\frac12,1]$? If $a,b,c$ belong to set of real numbers and $ a^2+b^2+c^2=1$ then prove that $ab+bc+ca$ belongs to $[-\frac12,1]$
I have tried AM>GM>HM(progressions mean inequality) but I am unable to do anything.
I have even attempted to assume a,b and c to be sides of an triangle and using $|a-b|<|c|$ and squaring but still I don't get the right-hand limit right.
Please tell me how to prove it.
|
Let's denote $S = ab+bc+ca$.
For the lower bound:
$$(a+b+c)^2 \geq 0 \Leftrightarrow a^2+b^2+c^2 + 2S = 1 + 2S \geq 0 \Leftrightarrow \boxed{S \geq -\frac{1}{2}}$$
For the upper bound we use Cauchy-Schwarz:
$$1+2S = (a+b+c)^2 = (1\cdot a + 1\cdot b+1\cdot c)^2 \leq 3\cdot (a^2+b^2+c^2) = 3 $$$$\Leftrightarrow 1+2S \leq 3 \Leftrightarrow \boxed{S\leq 1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I show that $\cos^4x=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x )+\frac{3}{8}$ I know how to prove that
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
by substituting $\cos(2x)$ with $2\cos^2x-1$ according to the double angle identity
$$\cos(2x)=2\cos^2x-1$$
However, how do I do that for $\cos^4x$?
Do I do it by writing $\cos^4x$ as $$\cos^2(x)\cdot \cos^2(x)$$ and thus get it by squaring the LHS of $$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
Im not sure how to proceed. Any ideas?
|
Since
$$\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)$$
we have
$$\cos^2(2x)=\frac{1}{2}+\frac{1}{2}\cos(4x)$$
and thus, if we square first equation we get
$$\cos^4x=\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\cos ^2(2x)=$$
$$ =\frac{1}{4}+{1\over 2}\cos(2x)+{1\over 4}\Big{(}\frac{1}{2}+\frac{1}{2}\cos(4x)\Big{)}$$
$$ =\frac{3}{8}+{1\over 2}\cos(2x)+\frac{1}{8}\cos(4x)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A sequence of full rank matrices $A_i \in \mathbb{R}^{m\times n}$ such that $A_i \rightarrow A$ in $\|\cdot\|_2$ $\textbf{Question:}$
Prove that for every $A\in \mathbb{R}^{m\times n}$, $\exists$ a sequence of full rank matrices $A_i \in \mathbb{R}^{m\times n}$ such that $A_i \rightarrow A$ in $\|\cdot\|_2$.
$\textbf{Answer Given:}$
If $A$ is of full rank, then $A_i = A$ $\forall i$.
If A is not of full rank, then $A$ has the form
$$A= U
\begin{bmatrix}
\sigma_1 & 0 & \dots & \dots& 0\\
0 & \sigma_2 & 0 &\dots & 0 \\
\vdots & & \ddots & & \vdots \\
0 & \dots & 0 & \sigma_r & 0 \\
0 & \dots & \dots & \dots & \mathcal{O}
\end{bmatrix}
V^T
$$
where $\mathcal{O}$ is a $(m-r)\times (n-r)$ matrix of zeros.
Then for $m=n$,
$$A_i= U
\begin{bmatrix}
\sigma_1 & 0 & \dots & \dots& 0 & \dots & 0\\
0 & \sigma_2 & 0 &\dots & 0 & & \vdots\\
\vdots & & \ddots & & \vdots & &\vdots\\
0 & \dots & 0 & \sigma_r & 0 & & \vdots\\
0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots\\
\vdots & & & & & \ddots & \vdots\\
0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i}
\end{bmatrix}
V^T
$$
and for $m > n$,
$$A_i= U
\begin{bmatrix}
\sigma_1 & 0 & \dots & \dots& 0 & \dots & 0\\
0 & \sigma_2 & 0 &\dots & 0 & & \vdots\\
\vdots & & \ddots & & \vdots & &\vdots\\
0 & \dots & 0 & \sigma_r & 0 & & \vdots\\
0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots\\
\vdots & & & & & \ddots & \vdots\\
0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i} \\
0 & \dots & \dots & \dots & \dots & \dots & 0\\
\vdots & & & & & & \vdots\\
0 & \dots &\dots & \dots & \dots & \dots & 0
\end{bmatrix}
V^T
$$
and for $m<n$,
$$A_i= U
\begin{bmatrix}
\sigma_1 & 0 & \dots & \dots& 0 & \dots & 0 & 0 & \dots & 0\\
0 & \sigma_2 & 0 &\dots & 0 & & \vdots & \vdots & &\vdots\\
\vdots & & \ddots & & \vdots & &\vdots & \vdots & &\vdots\\
0 & \dots & 0 & \sigma_r & 0 & & \vdots & \vdots & &\vdots\\
0 & \dots & \dots & \dots & \frac{1}{i} & & \vdots & \vdots & &\vdots\\
\vdots & & & & & \ddots & \vdots & \vdots & &\vdots\\
0 & \dots & \dots & \dots & \dots & \dots & \frac{1}{i} & 0 & \dots & 0
\end{bmatrix}
V^T
$$
Thus $\operatorname{rank}(A_i) = \min\{m,n\}$ and each $A_i$ is of full rank, which gives us
$$ \|A-A_i \|_2 =
\begin{vmatrix} \begin{vmatrix} \begin{bmatrix}
0 & \dots & \dots & \dots & \dots & 0\\
\vdots & \ddots & & & & \vdots\\
0 & & 0 \\
0 & & 0 & \frac{1}{i} & \dots & 0 \\
\vdots & & & & \ddots & \vdots \\
0 &\dots &\dots &\dots & 0 & \frac{1}{i}
\end{bmatrix}\end{vmatrix}\end{vmatrix}_2
= \sigma_{i+1} = \frac{1}{i} \rightarrow 0 \text{ as } i \rightarrow \infty.
$$
by the best rank-$k$ approximation theorem with respect to $\| \cdot \|_2$.
$\textbf{My Question (the only part that I do not understand):}$
How were the $\frac{1}{i}$ diagonals derived?
|
They were by choice, with the goal of obtain $\|A-A_i\|_2=\frac1{i}$, in fact, you can replace $\frac1{i}$ by another positive function of $i$ that goes to $0$ as $i \to \infty$.
|
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|
Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$
Then, we have:
$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.
Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.
Then, I used the following substitution:
$c + d = t$ and $a + b = 4 - t$.
We assume wlog that $a \leq b \leq c \leq d$.
Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.
Then, what we have to prove is:
$9t^2(4 - t)^2 \geq 256$.
We can rewrite this as:
$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or
$(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,
at which point I got stuck.
|
We need to prove that
$$\sum_{cyc}\ln(a^2+3)\geq4\ln4$$ or
$$\sum_{cyc}\left(\ln(a^2+3)-\ln4-\frac{1}{2}(a-1)\right)\geq0.$$
Now, let $f(x)=\ln(x^2+3)-\ln4-\frac{1}{2}(x-1).$
Thus, $$f'(x)=\frac{(x-1)(3-x)}{2(x^2+3)},$$ which says that $f$ decreases on $[3,+\infty)$ and since $f(3)>0$, there is unique $x_0>3$, for which $f(x_0)=0$.
And indeed, $x_0=4.586...$ and since $f(1)=0$, our inequality is proven for $\max\{a,b,c,d\}\leq x_0$.
Let $a>x_0$.
Thus, $$\prod_{cyc}(a^2+3)\geq(x_0^2+3)\cdot3^3>256$$ and we are done!
|
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|
Show that the equation represents an ellipsoid. Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.
Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
Common factor $\left(\dfrac32\right)^{\frac12}$ not sure how to find length and position half-axes??
|
We have $$\frac32 = (x+y)^2 + (y+z)^2 + (x+z)^2 = 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz + 2xy$$
so set $A =\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$ and the equation can be rewritten as $$\left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \frac32$$
Now diagonalize $A$:
$$A= P^TDP = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
-\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\
0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}^T \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4\end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\
-\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\
0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\\end{bmatrix}$$
so we have
$$\frac32 = \left\langle A\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle P^TDP\begin{bmatrix} x \\ y \\ z\end{bmatrix},\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle = \left\langle DP\begin{bmatrix} x \\ y \\ z\end{bmatrix},P\begin{bmatrix} x \\ y \\ z\end{bmatrix}\right\rangle$$
Therefore we can define new coordinates $$\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix} = P\begin{bmatrix} x \\ y \\ z\end{bmatrix} =
\begin{bmatrix}
\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{3}} \\
\frac{z}{\sqrt{3}}-\frac{x}{\sqrt{2}} \\
\frac{z}{\sqrt{3}}-\frac{y}{\sqrt{2}} \\
\end{bmatrix}$$
so we get $$\frac32 = \left\langle D\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix},\begin{bmatrix} x' \\ y' \\ z'\end{bmatrix}\right\rangle = x'^2 + y'^2 + 4z'^2$$
or $$\left(\frac{x'}{\sqrt{\frac32}}\right)^2 + \left(\frac{y'}{\sqrt{\frac32}}\right)^2 + \left(\frac{z'}{\frac12\sqrt{\frac32}}\right)^2 = 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial Fraction Decomposition of quadratic factor I'm trying to break up the following equation into partial fractions:
$$\frac{1}{(x^2-1)^2}=\frac{1}{(x+1)^2(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
I'm ending up with two equations and four unknowns though:
$$A+C=0 \\ -A + B + C+D = 0$$
(these are the $x$ and constant coefficient variables).
How do I break this up completely?
|
We have
$$\frac{1}{(x^2-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$$
$$=\frac{A(x+1)(x^2-1)+B(x+1)^2+C(x-1)(x^2-1)+D(x-1)^2}{(x^2-1)^2}=$$
$$=\frac{A(x^3+x^2-x-1)+B(x^2+2x+1)+C(x^3-x^2-x+1)+D(x^2-2x+1)}{(x^2-1)^2}=$$
that is
*
*$A+C=0 \implies A=-C$
*$A+B-C+D=0\implies 2A+B+D=0$
*$-A+2B-C-2D=0\implies B=D \implies A=-B$
*$-A+B+C+D=1\implies -2A+2B=1 \implies A=-\frac14$
and then $A=-B=-C=-D=-\frac14$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If x is a number between 0 and 1, what number results from dropping all odd powers from the binary representation of x? Let $x$ be a number in the interval $(0,1).$ $x$ has a representation in base 2, using the finite one if the sequence terminates. Clearly we can drop either all the even powers or the odd powers in the reprentation of $x$ in base 2 and thus we can write $x = A+B,$ where numbers $A$ and $B$ are also numbers in the interval $(0,1)$ and $A$ can be represented as a sum of the reciprocals of even powers of $2$ and $B$ can be represented as a sum of the reciprocals of odd powers of $2$. I wish to solve $A$ and $B$.
For example, $\frac{1}{7}=0.001001001001001..._2.$ Dropping off even powers results in $1.000001000001..._2×2^{-6} = \frac{1}{63}$ and for the odd powers we get $0.001000001000001..._2 = \frac{8}{63}.$ Thus $A = \frac{1}{63}$ and $ B = \frac{8}{63}.$ Note that $63 = 9\times7.$
Being able to write $A$ using only even powers of $2$ is the same as being able to write $A$ in base 4 using only 0s and 1s and being able to write $B$ using only odd powers of $2$ is the same as being able to write $B$ in base 4 using only 0s and 2s. Indeed, from previous example, $\frac{1}{63} = 0.001001001001001..._4$ and $\frac{8}{63} = 0.02002002002002..._4.$ Thus we can also try to solve $A$ and $B$ such that $x = A + B$ where $A$ has a representation in base 4 using only 0s and 1s and $B$ using only 0s and 2s.
The $A$ and $B$ seem pretty difficult to calculate generally. We had that $\frac{1}{7} = \frac{1}{63} + \frac{8}{63}$ with $63 = 9\times 7$. For the next prime we have $\frac{1}{11} = \frac{85}{1023} + \frac{8}{1023},$ as $1023 = 93\times11.$
Finding either $A$ or $B$ is enough to get both, so: Is there a general method for finding $A$?
|
One way to represent the result of such dropping seems to be with the series:
$$\sum_{n=0}^\infty \frac{\lfloor4^{n+1}x\rfloor-\lfloor4^nx\rfloor}{4^{2n+2}}$$
But this, admittedly, is a rather cheap trick
|
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|
Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges.
Now
$$ S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)$$
but I don't know how to go on with this. Now $ \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0$ but the right answer should be $1$.
|
The partial sum must be:
$$S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\\
\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...\color{red}{\frac1N-\frac{1}{N+1}}= \\
\left( \color{blue}{\frac{1}{1}}-\require{cancel}\cancel{\frac{1}{2}} \right)+\left( \cancel{\frac{1}{2}}-\cancel{\frac{1}{3}} \right) + \;...+ \left( \cancel{\frac{1}{N}}-\color{blue}{\frac{1}{N+1}} \right)=\\
\color{blue}1-\color{blue}{\frac{1}{N+1}}$$
Hence:
$$\lim_{N\to\infty} S_N=\lim_{N\to\infty} \left(1-\frac1{N+1}\right)=1.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the Matrix X and Y $\text { Find the matrix } X \text { and } Y , \text { if } X + Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] \text { and } X - Y = \left[ \begin{array} { c c } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$
Adding 1 and 2
$x + y +x-y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] + \left[ \begin{array} { l l } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$
$2 x = \frac { 1 } { 2 } \left[ \begin{array} { l l } { 8 } & { 8 } \\ { 0 } & { 8 } \end{array} \right]$
$x = \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$
to Find Y
$\left[ \begin{array} { l l } { 1 } & { 4 } \\ { 0 } & { 4 } \end{array} \right] + y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right]$
$y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] - \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$
$y = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 0 } & { 3 } \end{array} \right]$
so $x = \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$ and $y = \left[ \begin{array} { l l } { 1 } & { 2 } \\ { 0 } & { 3 } \end{array} \right]$ is this correct or not ?
|
You have the right ideas but I see multiple mistakes.
$$X+Y+X-Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] + \left[ \begin{array} { l l } { 3 } & { 6 } \\ { 0 } & { - 1 } \end{array} \right]$$
$$2 X = \color{red}{1} \cdot \left[ \begin{array} { l l } { 8 } & { 8 } \\ { 0 } & { 8 } \end{array} \right]$$
$$X = \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$$
to Find $Y$
$$\begin{bmatrix} {\color{red}{4}} & { 4 } \\ { 0 } & { 4 } \end{bmatrix} + y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right]$$
$$Y = \left[ \begin{array} { l l } { 5 } & { 2 } \\ { 0 } & { 9 } \end{array} \right] - \left[ \begin{array} { l l } { 4 } & { 4 } \\ { 0 } & { 4 } \end{array} \right]$$
$$Y = \left[ \begin{array} { l l } { 1 } & { \color{red}-2 } \\ { 0 } & { \color{red}5 } \end{array} \right]$$
|
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|
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
|
By l'Hopital we have
$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1} {\sin^2 x} =\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{\sin 2x-2x}{2x\sin^2 x+x^2\sin 2x }$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{2\cos 2x-2}{2\sin^2 x+2x\sin 2x+2x\sin 2x +2x^2\cos 2x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-4\sin 2x}{2\sin 2 x+8x\cos 2x+4 \sin 2x+4x\cos 2x-4x^2\sin 2x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-8\cos 2x}{12\cos 2 x+8\cos 2x-16x \sin 2x-8x\sin 2x+4\cos 2x-8x\sin 2x-8x^2\cos2x}$$
$$=\lim_{x \to 0}\frac{-8\cos 2x}{24\cos 2 x-32x \sin 2x-8x^2\cos2x} =\frac{-8}{24-0-0}=-\frac13$$
|
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|
Find sum of all integer solution of $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ Given $\frac{2}{a} + \frac{3}{b} + \frac{4}{c} + \frac{5}{d} = \frac{1}{14}$ , which $a,b,c ,d$ are an positive integer , all solution of $a,b,c,d$ are members of Set "S" ,then find sum all of members in set "S"
I don't know how to start to solve this problem but I've got 2 solution that is $\frac{2}{52} + \frac{3}{598} + \frac{4}{437} + \frac{5}{266} = \frac{1}{14}$ and $\frac{2}{224} + \frac{3}{304} + \frac{4}{437} + \frac{5}{115} = \frac{1}{14}$
question 2 If problem say that $a,b,c,d$ are an all integer , they will have negative integer to this solution ? please give me hint or theorem relevant. Thank you in advance .
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Hint: For first question you can use following algorithm; it is a bit long operation.
$2/a+3/b+4/c+5/d=1/14$
we rewrite the equation as:
$$\frac{1}{\frac{a}{28}}+\frac{1}{\frac{b}{42}}+\frac{1}{\frac{c}{56}}+\frac{1}{\frac{d}{70}}=1$$
Now we solve equation:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1$$
Number of terms is 4 ,so:
$\frac{1}{x}≥\frac{1}{4}$ ⇒ $x≤4$, let $x=2$ then we have:
$\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1-\frac{1}{2}=\frac{1}{2}$
$\frac{1}{y}<\frac{1}{2}$⇒ $y>2$, meanwhile $\frac{3}{y}≥\frac{1}{2}$ so $y≤6$ so $3≤y≤6$; let $y=3$ then we have:
$\frac{1}{z}+\frac{1}{t}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$
$\frac{1}{z}<\frac{1}{6}$⇒ $z>6$. Meanwhile, $\frac{3}{z}≥\frac{1}{6}$⇒$z<18$
, so $6<z<18$ , let $z=7$ then we must have:
$\frac{1}{t}=\frac{1}{6}-\frac{1}{7}=\frac{1}{42}$⇒ $t=42$ .
$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$
So following numbers can be a set of solutions to your equation:
$a=2\times 28=56$
$b=3\times 42=126$
$c=7\times 56=392$
$d=42\times 70=2940$
x can be $2, 3, 4$, each of these numbers leads to one set of solutions. y can be $3, 4, 5, 6 $ , z can be $ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18$ among which certain values give integer solutions for t.
Note that equation of parameters x, y, x and t is symmetric for initial equation i.e each term of this equation can be equal to any term of initial equation.
|
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|
Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational. Can someone please check my proof?
Let $n \in \mathbb{N}$ such that n is a non-perfect square. Show that $\sqrt{n}$ is irrational.
Let's prove it by contradiction, that is, suppose there are $n,a,b \in \mathbb{N}$ such that $\frac{a}{b}=\sqrt{n}$, n is a non-perfect square and there are no common factors for a and b, that is, they are coprime. n is either even or odd. Suppose n is even, that means $n=2k$ for some $k \in \mathbb{N}$. Therefore:
$$
\begin{align*}
\frac{a^2}{b^2}&=2k\\
a^2 &= 2k\cdot b^2 = 2(k\cdot b^2)
\end{align*}
$$
Which means that $a^2$ is even and therefore a is even and that means that there is some $j\in \mathbb{N}$ such that $a=2j$. Plugging in that value for a:
$$
\begin{align*}
\frac{2j}{b^2}&=2k\\
4j^2 &= 2k\cdot b^2 \rightarrow b^2 = 2\big(\frac{2j^2}{k}\big)
\end{align*}
$$
Using the same reasoning b is also even. That contradicts the fact that a and b are coprime and therefore $\sqrt{n}$ is irrational.
Now suppose n is odd, that means that $n=2k+1$ for some $k\in \mathbb{N}$. Therefore:
$$
\begin{align*}
\frac{a^2}{b^2}&=2k+1\\
a^2 &= (2k+1)\cdot b^2 \rightarrow (2k+1)|a^2
\end{align*}
$$
Since $(2k+1)|a^2$ we can conclude that $a^2 = q(2k+1)$. Plugging in that value for a:
$$
\begin{align*}
\frac{q(2k+1)}{b^2}&=2k+1\\
q(2k+1) &= (2k+1)\cdot b^2 \rightarrow q=b^2
\end{align*}
$$
That way q is a common factor for a and b, which contradicts the fact that they are coprime, hence $\sqrt{n}$ is irrational.
Is my reasoning correct? I've seen easier and shorter proofs, by I tried to generalize the reasoning that I used to prove that $\sqrt{2}$ and $\sqrt{3}$ are irrational...
Thank you for reading it!
|
As the comments suggest, you overanalyze. If $\frac{a}{b}=\sqrt{n}$ then $\frac{a^2}{b^2}=n$ and $a^2=nb^2$. Thus, as with the cases of $2$ and $3$, $n\mid a^2$. Now, for every prime factor $p_i$ of $n$, if $p_i\mid a^2$, then $p_i\mid a$. Thus $(p_i)^2\mid a^2$. Remove that $p_i$ from each occurrence of $n$ and $a$, and you are left with the fact that $\frac{n}{p_i}\mid \frac{a^2}{(p_i)^2}$. Repeat for every other prime factor of $n$ to arrive at the fact if $n\mid a^2$ then $n\mid a$. So $a=kn,a^2=k^2n^2$, and from here the proof is the same as for $2$ and $3$ in that you can show that $n$ must be a factor of $b$.
|
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|
Find the three roots of the equation $\left(z+1\right)^3=-1$
Give your answers in the form $x+ iy$, where $x\in \mathbb{R}\:and\:y\in \mathbb{R}$.
I suspect I'll have to use de Moivre's theorem to solve this, but I don't know how to factor in the $+1$, because increasing the real part of a complex number by one doesn't necessarily increase its modulus by one.
I know that one of the roots will be $-2$, by inspection, but that's about it.
|
Consider instead
$$c^3 = -1$$
where $c = z + 1$. Denoting in geometric form $c = re^{i\theta}$ and $-1 = e^{i \pi}$, we get
$$r^3 e^{i3\theta} = e^{i\pi + 2k\pi}, \qquad k \in \lbrace 0,1,2 \rbrace$$
We get $r = 1$ and three different angles
\begin{align}
\theta_1 &= \frac{\pi}{3} = \\
\theta_2 &= \frac{\pi}{3} + \frac{2\pi}{3} = \pi\\
\theta_3 &= \frac{\pi}{3} + \frac{4\pi}{3} = \frac{5\pi}{3}\\
\end{align}
So we get $c_1 = e^{i\frac{\pi}{3}}$, $c_2 = e^{i\pi}$ and $c_3 = e^{i\frac{5\pi}{3}}$. The corresponding roots are
\begin{align}
z_1 &= -1 + e^{i\frac{\pi}{3}} = (-1 + \cos(\frac{\pi}{3})) + i\sin(\frac{\pi}{3}) \\
z_2 &= -1 + e^{i\pi} = (-1 + \cos(\pi)) + i\sin(\pi)\\
z_3 &= -1 + e^{i\frac{5\pi}{3}}= (-1 + \cos(\frac{5\pi}{3})) + i\sin(\frac{5\pi}{3})\\
\end{align}
|
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|
Functions satisfying $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\}.$ I have used this identity: if $g(x)=1/(1-x),$ then
$$g^{-1}(x)=1-\frac{1}{x},$$
to get all functions satisfying: $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\},$ but I didn't get a general form of its solution. My question here is:
Is there any simple method to solve the titled functional equation?
|
We will prove that $$f(x)=x-\frac{1}{2(1-x)}$$ when
$$f(x)+f\left(\frac{1}{1-x}\right)=x$$
At first we get $$f\left(\frac{1}{1-x}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$$ and we get
$$f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}$$
Now we get
$$f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-\frac{x-1}{x}}\right)=\frac{x-1}{x}$$
so
$$f\left(\frac{x-1}{x}\right)+f(x)=\frac{x-1}{x}$$
and we obtain
using that
$$f\left(\frac{1}{1-x}\right)=x-f(x)$$
$$x-f(x)+\frac{x-1}{x}-f(x)=\frac{1}{1-x}$$
so $$2f(x)=x+\frac{x-1}{x}-\frac{1}{1-x}$$
and $$f(x)=\frac{-x^3+x-1}{x}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2916771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Finding the inverse of a matrix using elementary matrixes This should be fairly simple as I know matrix A can be found by $$A = E_1^{-1} E_2^{-1}...E_k^{-1}$$
So it should go that $$A^-$$ can be found by $$E_kE_(k-1)...E_1$$
But for some reason my numbers aren't matching up with the book.
I have a matrix $$\begin{bmatrix}1&0&-1\\0 &6&-1\\0&0&4\end{bmatrix}$$
I find that I can get an Identity Matrix from this matrix by doing (1/6)R2 -> R2, (1/4)R3 -> R3, 1/6R3 + R2 -> R2, R3 + R1 -> R1. From there I can find the inverse of the elementary matrices no problem but for some reason my normal E does not multiply into the inverse. Did I do something wrong in my steps?
It should be noted I need to solve the matrix using elementary matrices.
|
If $E_1\cdots E_n A = I$, then $A^{-1} = E_1\cdots E_n I$.
So you can transform $A$ into $I$ and simultaneously transform $I$ to $A^{-1}$ using the same transformations:
$$
\left[\begin{array}{rrr|rrr}
1&0&-1 & 1 & 0 & 0\\0 &6&-1& 0 & 1 & 0\\0&0&4& 0 & 0& 1
\end{array}\right]
\sim \left[\begin{array}{rrr|rrr}
1&0&-1 & 1 & 0 & 0\\0 &1&-\frac16& 0 & \frac16 & 0\\0&0&1& 0 & 0& \frac14
\end{array}\right]\sim \left[\begin{array}{rrr|rrr}
1&0&0 & 1 & 0 & \frac14\\0 &1&0& 0 & \frac16 & \frac1{24}\\0&0&1& 0 & 0& \frac14
\end{array}\right]$$
so $A^{-1} = \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix}$.
Using elementary matrices with this notation, our transformations are $$L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)A = I$$
so \begin{align}
A^{-1} &= L_{3,2}\left(\frac16\right)L_{3,1}(1)D_3\left(\frac14\right)D_2\left(\frac16\right)\\
&= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & \frac16 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac14 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & \frac16 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac14 \\ 0 & \frac16 & \frac1{24} \\ 0 & 0 & \frac14\end{bmatrix}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2917756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
"Binomial coefficients" generalized via polynomial iteration This is a question I will answer myself immediately by repeating one of my old AoPS posts, since the latter post no longer renders on AoPS.
Convention. In the following, whenever $A$ is a commutative ring with $1$, and $f$ and $g$ are two polynomials over $A$ in the variable $x$, then we denote by $f\left[g\right]$ the evaluation of the polynomial $f$ at $g$. This evaluation is defined as the image of $f$ under the $A$-algebra homomorphism $A\left[x\right] \to A\left[x\right]$ which maps $x$ to $g$ (this homomorphism is unique, due to the universal property of the polynomial ring). Equivalently, this evaluation is $\sum_{i\geq 0} f_i g^i$, where $f_0, f_1, f_2, \ldots$ are the coefficients of the polynomial $f$ before $x^0, x^1, x^2, \ldots$, respectively.
Note that the evaluation $f\left[g\right]$ is frequently denoted by $f\left(g\right)$ in literature, but this $f\left(g\right)$ notation is slightly ambiguous, because $f\left(g\right)$ can also mean the product of the polynomials $f$ and $g$ (in particular, this often happens when $g$ is a complicated sum, so that the parentheses around $g$ are required), and this is an entirely different thing. Thus we are always going to denote the evaluation by $f\left[g\right]$, and never by $f\left(g\right)$.
Also note that every polynomial $f\in A\left[x\right]$ satisfies $f\left[x\right]=f$ and $x\left[f\right]=f$.
Also, I let $\mathbb{N} = \left\{0,1,2,\ldots\right\}$.
Theorem 1. Let $A$ be a commutative ring with $1$, and let $f\in A\left[x\right]$ be a polynomial over $A$ in the variable $x$. Let us define a polynomial $f_n\in A\left[x\right]$ for every $n\in\mathbb{N}$. Namely, we define $f_n$ by induction over $n$: We start with $f_0=x$, and continue with the recursive equation $f_{n+1}=f\left[f_n\right]$ for every $n\in\mathbb N$. [Note that I hesitate to denote $f_n$ by $f^n$ since $f^n$ already stands for "$n$-th power of $f$ with respect to multiplication of polynomials", and that's something different from $f_n$.] Then, for any nonnegative integers $n$ and $k$, we have
\begin{equation}
\prod_{i=1}^n\left(f_i-x\right) \mid \prod_{i=k+1}^{k+n}\left(f_i-x\right)
\end{equation}
in the ring $A\left[x\right]$.
Theorem 2. Let $A$, $f$ and $f_n$ be as in Theorem 1. Then, for any two coprime positive integers $m$ and $n$, we have
\begin{equation}
\left(f_m - x\right) \left(f_n - x\right) \mid \left(f_{mn} - x\right) \left(f - x\right)
\end{equation}
in the ring $A\left[x\right]$.
The question is to prove these two theorems.
Theorem 1 has been posted by gammaduc at https://artofproblemsolving.com/community/c7h412796 ; Theorem 2 has been posted by gammaduc at https://artofproblemsolving.com/community/c7h412797 .
|
Proof of Theorem 2
The proof of Theorem 2 I shall give below is merely a generalization of GreenKeeper's proof at AoPS, with all uses of specific properties of $A$ excised.
If $a_1, a_2, \ldots, a_m$ are some elements of a commutative ring $R$, then we let $\left< a_1, a_2, \ldots, a_m \right>$ denote the ideal of $R$ generated by $a_1, a_2, \ldots, a_m$. (This is commonly denoted by $\left(a_1, a_2, \ldots, a_m\right)$, but I want a less ambiguous notation.)
Lemma 3. Let $R$ be a commutative ring. Let $a, b, c, u, v, p$ be six elements of $R$ such that $a \mid p$ and $b \mid p$ and $c = au+bv$. Then, $ab \mid cp$.
Proof of Lemma 3. We have $a \mid p$; thus, $p = ad$ for some $d \in R$. Consider this $d$.
We have $b \mid p$; thus, $p = be$ for some $e \in R$. Consider this $e$.
Now, from $c = au+bv$, we obtain $cp = \left(au+bv\right)p = au\underbrace{p}_{=be} + bv\underbrace{p}_{=ad} = aube + bvad = ab\left(ue+vd\right)$. Hence, $ab \mid cp$. This proves Lemma 3. $\blacksquare$
Lemma 4. Let $A$, $f$ and $f_n$ be as in Theorem 1. Then, for any nonnegative integers $a$ and $b$ such that $a \geq b$, we have
\begin{equation}
f_{a-b}-x\mid f_a-f_b \qquad \text{ in } A\left[x\right] .
\end{equation}
Proof of Lemma 4. Lemma 4 is precisely the relation \eqref{darij-proof.3} from the proof of Theorem 1; thus, we have already shown it. $\blacksquare$
Lemma 5. Let $A$, $f$ and $f_n$ be as in Theorem 1. Let $a$ and $b$ be nonnegative integers such that $a \geq b$. Then, $\left< f_a - x, f_b - x \right> = \left< f_{a-b} - x, f_b - x \right>$ as ideals of $A\left[x\right]$.
Proof of Lemma 5. We have $a \geq b$, thus $0 \leq b \leq a$. Hence, $0 \leq a-b \leq a$. Therefore, $a-b$ is a nonnegative integer such that $a \geq a-b$. Hence, Lemma 4 (applied to $a-b$ instead of $b$) yields $f_{a-\left(a-b\right)} - x \mid f_a - f_{a-b}$. In view of $a-\left(a-b\right) = b$, this rewrites as $f_b - x \mid f_a - f_{a-b}$. In other words, $f_a - f_{a-b} \in \left< f_b - x \right>$.
Hence, $f_a - f_{a-b} \in \left< f_b - x \right> \subseteq \left< f_{a-b} - x, f_b - x \right>$. Thus, both elements $f_a - f_{a-b}$ and $f_{a-b} - x$ of $A\left[x\right]$ belong to the ideal $\left< f_{a-b} - x, f_b - x \right>$. Hence, their sum $\left(f_a - f_{a-b}\right) + \left(f_{a-b} - x\right)$ must belong to this ideal as well. In other words,
\begin{equation}
\left(f_a - f_{a-b}\right) + \left(f_{a-b} - x\right) \in \left< f_{a-b} - x, f_b - x \right> .
\end{equation}
In view of $\left(f_a - f_{a-b}\right) + \left(f_{a-b} - x\right) = f_a - x$, this rewrites as $f_a - x \in \left< f_{a-b} - x, f_b - x \right>$. Combining this with the obvious fact that $f_b - x \in \left< f_{a-b} - x, f_b - x \right>$, we conclude that both generators of the ideal $\left< f_a - x, f_b - x \right>$ belong to $\left< f_{a-b} - x, f_b - x \right>$. Hence,
\begin{equation}
\left< f_a - x, f_b - x \right> \subseteq \left< f_{a-b} - x, f_b - x \right> .
\label{darij-proof2.1}
\tag{11}
\end{equation}
On the other hand, $f_a - f_{a-b} \in \left< f_b - x \right> \subseteq \left< f_a - x, f_b - x \right>$. Thus, both elements $f_a - f_{a-b}$ and $f_a - x$ of $A\left[x\right]$ belong to the ideal $\left< f_a - x, f_b - x \right>$. Hence, their difference $\left(f_a - x\right) - \left(f_a - f_{a-b}\right)$ must belong to this ideal as well. In other words,
\begin{equation}
\left(f_a - x\right) - \left(f_a - f_{a-b}\right) \in \left< f_a - x, f_b - x \right> .
\end{equation}
In view of $\left(f_a - x\right) - \left(f_a - f_{a-b}\right) = f_{a-b} - x$, this rewrites as $f_{a-b} - x \in \left< f_a - x, f_b - x \right>$. Combining this with the obvious fact that $f_b - x \in \left< f_a - x, f_b - x \right>$, we conclude that both generators of the ideal $\left< f_{a-b} - x, f_b - x \right>$ belong to $\left< f_a - x, f_b - x \right>$. Hence,
\begin{equation}
\left< f_{a-b} - x, f_b - x \right> \subseteq \left< f_a - x, f_b - x \right> .
\end{equation}
Combining this with \eqref{darij-proof2.1}, we obtain $\left< f_a - x, f_b - x \right> = \left< f_{a-b} - x, f_b - x \right>$. This proves Lemma 5. $\blacksquare$
Lemma 6. Let $A$, $f$ and $f_n$ be as in Theorem 1. Let $a$ and $b$ be two nonnegative integers. Then,
\begin{equation}
\left< f_a - x, f_b - x \right> = \left< f_{\gcd\left(a,b\right)} - x \right>
\end{equation}
as ideals of $A\left[x\right]$.
Here, we follow the convention that $\gcd\left(0,0\right) = 0$. Note that
\begin{equation}
\gcd\left(a,b\right) = \gcd\left(a-b,b\right)
\qquad \text{for all integers $a$ and $b$.}
\label{darij.proof2.gcd-inva}
\tag{12}
\end{equation}
Proof of Lemma 6. We shall prove Lemma 6 by strong induction on $a+b$:
Induction step: Fix a nonnegative integer $N$. Assume (as the induction hypothesis) that Lemma 6 holds whenever $a+b < N$. We must now show that Lemma 6 holds whenever $a+b = N$.
We have assumed that Lemma 6 holds whenever $a+b < N$. In other words, if $a$ and $b$ are two nonnegative integers satisfying $a+b < N$, then
\begin{equation}
\left< f_a - x, f_b - x \right> = \left< f_{\gcd\left(a,b\right)} - x \right> .
\label{darij.proof2.l6.pf.1}
\tag{13}
\end{equation}
Now, fix two nonnegative integers $a$ and $b$ satisfying $a+b = N$. We want to prove that
\begin{equation}
\left< f_a - x, f_b - x \right> = \left< f_{\gcd\left(a,b\right)} - x \right> .
\label{darij.proof2.l6.pf.2}
\tag{14}
\end{equation}
Since our situation is symmetric in $a$ and $b$, we can WLOG assume that $a \geq b$ (since otherwise, we can just swap $a$ with $b$). Assume this.
We have $f_0 = x$ and thus $f_0 - x = 0$. Hence,
\begin{equation}
\left< f_a - x, f_0 - x \right> = \left< f_a - x, 0 \right> = \left< f_a - x \right> = \left< f_{\gcd\left(a,0\right)} - x \right>
\end{equation}
(since $a = \gcd\left(a,0\right)$). Hence, \eqref{darij.proof2.l6.pf.2} holds if $b = 0$. Thus, for the rest of the proof of \eqref{darij.proof2.l6.pf.2}, we WLOG assume that we don't have $b = 0$. Hence, $b > 0$, so that $a+b > a$. Therefore, $\left(a-b\right) + b = a < a+b = N$. Moreover, $a-b$ is a nonnegative integer (since $a \geq b$). Therefore, \eqref{darij.proof2.l6.pf.1} (applied to $a-b$ instead of $a$) yields
\begin{equation}
\left< f_{a-b} - x, f_b - x \right> = \left< f_{\gcd\left(a-b,b\right)} - x \right> .
\end{equation}
Comparing this with the equality $\left< f_a - x, f_b - x \right> = \left< f_{a-b} - x, f_b - x \right>$ (which follows from Lemma 5), we obtain
\begin{equation}
\left< f_a - x, f_b - x \right> = \left< f_{\gcd\left(a-b,b\right)} - x \right> .
\end{equation}
In view of $\gcd\left(a-b,b\right) = \gcd\left(a,b\right)$ (which follows from \eqref{darij.proof2.gcd-inva}), this rewrites as
\begin{equation}
\left< f_a - x, f_b - x \right> = \left< f_{\gcd\left(a,b\right)} - x \right> .
\end{equation}
Thus, \eqref{darij.proof2.l6.pf.2} is proven.
Now, forget that we fixed $a$ and $b$. We thus have shown that if $a$ and $b$ are two nonnegative integers satisfying $a+b = N$, then \eqref{darij.proof2.l6.pf.2} holds. In other words, Lemma 6 holds whenever $a+b = N$. This completes the induction step. Hence, Lemma 6 is proven by induction. $\blacksquare$
Lemma 7. Let $A$, $f$ and $f_n$ be as in Theorem 1. Let $p$ and $q$ be nonnegative integers such that $p \mid q$ (as integers). Then, $f_p - x \mid f_q - x$ in $A\left[x\right]$.
Proof of Lemma 7. We have $\gcd\left(p,q\right) = p$ (since $p \mid q$).
Applying Lemma 6 to $a = p$ and $b = q$, we obtain
\begin{equation}
\left< f_p - x, f_q - x \right> = \left< f_{\gcd\left(p,q\right)} - x \right> = \left< f_p - x \right>
\end{equation}
(since $\gcd\left(p,q\right) = p$). Hence,
\begin{equation}
f_q - x \in \left< f_p - x, f_q - x \right> = \left< f_p - x \right> .
\end{equation}
In other words, $f_p - x \mid f_q - x$. This proves Lemma 7. $\blacksquare$
Proof of Theorem 2. Let $m$ and $n$ be two coprime positive integers. Thus, $\gcd\left(m,n\right) = 1$ (since $m$ and $n$ are coprime). Hence, $f_{\gcd\left(m,n\right)} = f_1 = f$.
Lemma 7 (applied to $p=m$ and $q = mn$) yields $f_m - x \mid f_{mn} - x$ (since $m \mid mn$ as integers). Lemma 7 (applied to $p=n$ and $q = mn$) yields $f_n - x \mid f_{mn} - x$ (since $n \mid mn$ as integers).
But Lemma 6 (applied to $a = m$ and $b = n$) yields
\begin{equation}
\left< f_m - x, f_n - x \right> = \left< f_{\gcd\left(m,n\right)} - x \right> = \left< f - x \right>
\end{equation}
(since $f_{\gcd\left(m,n\right)} = f$).
Hence, $f - x \in \left< f - x \right> = \left< f_m - x, f_n - x \right>$. In other words, there exist two elements $u$ and $v$ of $A\left[x\right]$ such that $f - x = \left(f_m - x\right) u + \left(f_n - x \right) v$. Consider these $u$ and $v$.
Now, Lemma 3 (applied to $R = A \left[x\right]$, $a = f_m - x$, $b = f_n - x$, $c = f - x$ and $p = f_{mn} - x$) yields $\left(f_m - x\right) \left(f_n - x\right) \mid \left(f - x\right) \left(f_{mn} - x\right) = \left(f_{mn} - x\right) \left(f - x\right)$. This proves Theorem 2. $\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2917929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
I was asked to evaluate the determinant of the I was asked to evaluate the determinant of the $n$x$n$ matrix
$$ A=
\begin{bmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{bmatrix}
$$
I tried starting from a 1x1 matrix to 5x5 matrix, and saw the pattern
$$\det{A}=(x-2)^{n-1}(x+2(n-1)).$$
Now I need to prove that this is the case for any nxn matrix of type A, so I tried using induction.The base case (n=1) holds.
$$A=
\begin{bmatrix}
x
\end{bmatrix}
$$
And clearly $\det{A}=x$ using both the definition of a determinant and the explicit formula above.
So now assume it holds for any $n$, and test for $n+1$.
$$\det{A}=(x-2)^n(x+2n).$$
What do I do now?
|
First subtract the last row from the first $n-1$ rows. Then add first $n-1$ columns to the last one.
\begin{align}
\det A &= \begin{vmatrix}
x & 2 & 2 & \cdots & 2 \\
2 & x & 2 & \cdots & 2 \\
2 & 2 & x & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix} \\
&= \begin{vmatrix}
x-2 & 0 & 0 & \cdots & 2-x \\
0 & x-2 & 0 & \cdots & 2-x \\
0 & 0 & x-2 & \cdots & 2-x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x \\
\end{vmatrix} \\
&= \begin{vmatrix}
x-2 & 0 & 0 & \cdots & 0 \\
0 & x-2 & 0 & \cdots & 0 \\
0 & 0 & x-2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
2 & 2 & 2 & \cdots & x + 2(n-1)
\end{vmatrix}\\
&= (x-2)^{n-1}(x-2(n-1))
\end{align}
The resulting determinant is lower-triangular so it is equal to the product of diagonal elements.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2919579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Finding the sum of squares of the real roots
Let $r_1,r_2,r_3,\cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+\cdots+r_n^2$ is $$(A)\,3\quad(B)\,14\quad(C)\,8\quad(D)\,16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
|
\begin{align}
x^8-14x^4-8x^3-x^2+1&=0
\tag{1}\label{1}
.
\end{align}
\eqref{1} can be factored as
\begin{align}
f_1(x)f_2(x)=0
,\\
f_1(x)&=x^4+4x^2+x+1
\tag{2}\label{2}
,\\
f_2(x)&=x^4-4x^2-x+1
\tag{3}\label{3}
.
\end{align}
\begin{align}
f_1(x)&=x^4+3x^2+(x^2+x+\frac14)-\tfrac14+1
\\
&=x^4+3x^2+\tfrac14(2x+1)^2+\tfrac34
>0\quad \forall x\in\mathbb{R}
.
\end{align}
We can also found that
factor $f_2(x)$ has four distinct real roots,
for example, observing that
\begin{align}
f_2(-2)&=3>0
,\\
f_2(-1)&=-1<0
,\\
f_2(0)&=1>0
,\\
f_2(1)&=-3<0
,\\
f_2(3)&=43>0
,
\end{align}
Now consider
\begin{align}
f_2(x)f_2(-x)&=
(x^2)^4-8(x^2)^3+18(x^2)^2-9(x^2)+1
\tag{4}\label{4}
.
\end{align}
Expression \eqref{3}
is a polynomial in $x^2$,
its roots are squares of the roots of\eqref{1},
hence the sought sum
of the squares of distinct real roots of \eqref{1}
is $8$ (negated coefficient at $(x^2)^3$ in \eqref{4}).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2920549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
}
|
Find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$ as $n \rightarrow \infty$ The question tells me find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$.
I thought that I would use an idea similar to the one given below:
Exercise 1.8.4. $a_n=\sqrt[n]n.$
Solution. If we take $b=\sqrt[n]n$ and apply $(1.14)$, we obtain $$n>\frac{n(n-1)}2(\sqrt[n]n-1)^2.$$ It follows that $$0<\sqrt[n]n-1<\sqrt{\frac2{n-1}}\to0,$$ so $\lim{\sqrt[n]n}=1$.
(Image that replaced the text)
but with $b = n^{14/n}$ and using the $14^{\text{th}}$ term of the binomial theorem which is $$\frac{n(n-1)(n-2)\cdots(n-13)}{14!} (n^{14/n} - 1)^{14}$$ but I got stucked.
Could anyone help me please?
|
Assuming that you could be interested by more than the limit itself.
Considering Faulhaber polynomials
$$S_n=\sum_{i=1}^n i^7=\frac {a^2(6a^2-4a+1)} 3 \qquad \text{where} \qquad a=\frac 12 n (n+1)$$ you can write
$$S_n=\frac{n^8}8 \left(1+\frac{4}{n}+\frac{14}{3 n^2}-\frac{7}{3 n^4}+\frac{2}{3
n^6} \right)$$
$$\log(S_n)=\log\left(\frac{n^8}8\right)+\log \left(1+\frac{4}{n}+\frac{14}{3 n^2}-\frac{7}{3 n^4}+\frac{2}{3
n^6} \right)$$ Now, using Taylor expansion
$$\log(S_n)=\left(8 \log \left({n}\right)-\log (8)\right)+\frac{4}{n}-\frac{10}{3
n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac 1 n \log(S_n)=\frac{8 \log \left({n}\right)-\log (8) }n+\frac{4}{n^2}+O\left(\frac{1}{n^3}\right)$$
Using Taylor again
$$S_n^{\frac 1n}=e^{\frac 1 n \log(S_n)}=1+\frac{8 \log \left({n}\right)-\log (8) }n+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2924638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the maximum value of $f(z)=|z^3-z+2|$ on the unit circle $|z|=1$
Let $f:\mathbb{C} \rightarrow \mathbb{R}$ be defined by $$f(z)=|z^3-z+2|\,.$$ What is the maximum value on the unit circle $|z|=1$ ?
My approach is as follows:
$z=e^{i\theta}$ as it is mentioned that the point lie on the unit circle.
$$f(z)=|1+\cos3\theta+i\sin3\theta +1-\cos\theta-i\sin\theta|\,.$$
Therefore, $f(z)=|t|$, where
$$t=2\cos^2\frac{3\theta}{2}+2i\sin\frac{3\theta}{2}\cos\frac{3\theta}{2}+
2\sin^2 \frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\,.$$
$$t=2\cos\frac{3\theta}{2}\left(\cos\frac{3\theta}{2}+i\sin\frac{3\theta}{2}\right)-2i\sin \frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)\,.$$
$$t=2\cos\frac{3\theta}{2}\,e^\frac{i3\theta}{2}-
2i\sin \frac{\theta}{2}\,e^\frac{i\theta}{2}\,.$$
I am not able to proceed from here.
Note: Please don't mark it as duplicate because the question in the website is same but my approach is different.
|
Let $z:=\exp(\text{i}\theta)$ and $t:=\cos(\theta)$. Then,
$$\begin{align}|z^3-z+2|^2&=\Big|\big(\cos(3\theta)-\cos(\theta)+2\big)^2+\text{i}\big(\sin(3\theta)-\sin(\theta)\big)\Big|^2
\\&=4\cos(3\theta)-2\cos(2\theta)-4\cos(\theta)+6
\\&=4(4t^3-3t)-2(2t^2-1)-4t+6
\\&=16t^3-4t^2-16t+8=:g(t)\,.
\end{align}$$
Thus, we want to maximize $g(t)$ subject to $t\in[-1,+1]$. Note that
$$g'(t)=48t^2-8t-16=8(6t^2-t-2)=8(2t+1)(3t-2)\,.$$
That is, optimizing points of $g(t)$ in $[-1,+1]$ are $t=-\dfrac12$, $t=\dfrac23$, and the boundary points $t\in\{-1,+1\}$.
Note that $g(-1)=4$, $g(+1)=4$, $g\left(-\dfrac12\right)=13$, and $g\left(\dfrac23\right)=\dfrac{8}{27}$. We conclude that the minimum of $g(t)$ on $[-1,+1]$ is $\dfrac{8}{27}$, which is attained when $t=\dfrac23$, whereas the maximum of $g(t)$ on $[-1,+1]$ is $13$, which is attained when $t=-\dfrac12$. Translating this back to $\theta$, we see that $\theta=\dfrac{2\pi}{3}$ and $\theta=\dfrac{4\pi}{3}$ satisfy $\cos(\theta)=-\dfrac12$. Thus,
$$z=\exp\left(\frac{2\pi\text{i}}{3}\right)=\dfrac{-1+\sqrt{3}\text{i}}{2}\text{ and }z=\exp\left(\frac{4\pi\text{i}}{3}\right)=\dfrac{-1-\sqrt{3}\text{i}}{2}$$
maximize $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$. The maximum value of $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$ is $\sqrt{13}$.
Similarly, the minimum value of $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$ is $\sqrt{\dfrac{8}{27}}=\dfrac{2\sqrt{2}}{3\sqrt{3}}$. The minimum is attained when
$$z=\frac{2+\sqrt{5}\text{i}}{3}\text{ and }z=\frac{2-\sqrt{5}\text{i}}{3}\,.$$
(This happens when $\theta=\text{arccos}\left(\dfrac23\right)$ and when $\theta=2\pi-\text{arccos}\left(\dfrac{2}{3}\right)$.)
|
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|
Prove by induction that $\sqrt{1}+\sqrt{2}+...+\sqrt{n} \geq \frac{2}{3}n\sqrt{n}$ The base step is pretty obvious: $1 \geq \frac{2}{3}$.
Then we assume that $P(k)$ is true for some $k \in \mathbb{Z}^{+}$ and try to prove $P(k+1)$. So I have
$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$
by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than $\frac{2}{3}(k+1)\sqrt{k+1}$.
Would appreciate any help!
|
For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 = 4k^3 - 3k + 1
$$
The rest should follow since $k \geq \frac1{3}$, since the induction is over the positive integers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given $(m-2)x^2-4x+3=3-x^2+2nx$, compute the value of $m$
Given $$(m-2)x^2-4x+3=3-x^2+2nx$$ Compute the value of $m$.
$$(m-2)x^2-4x+3=3-x^2+2nx
\\mx^2+(-4m-4)x+7=-x^2+2nx+3
\\m=-1, n=0$$
The answer is $m=1,n=-2$, please tell me where did I go wrong
|
Hint:
The coefficient of $x^2$ on the left of the equal sign is $(m-2)$ and on the right it is $-1$
|
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|
Partial decomposition of $\frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$ Recently I have come arcross the following fraction
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}$$
Hence I have encountered this fraction within a task of integration I want to do a partial decomposition. First of all I rewrote it as following
$$\dfrac{z^2}{(2z^2+3)((s^2+1)z^2+1)}=\dfrac{Az+B}{2z^2+3}+\dfrac{Cz+D}{(s^2+1)z^2+1} $$
which yields to a system of equations for $A,B,C$ and $D$
$$\begin{align}
0&=A(1+s^2)+3C\\
0&=B(1+s^2)+3D\\
0&=s^2A+2C\\
1&=s^2B+2D
\end{align}$$
Solving this system gives $A=C=0$ and $B=\dfrac{3}{s^2-2}$ and $D=-\dfrac{s^2-1}{s^2-2}$. By plugging in these values we arrive at
$$\dfrac{3}{(s^2-2)(2z^2+3)}-\dfrac{s^2-1}{(s^2-2)((s^2+1)z^2+1)}=\dfrac{s^2(z^2-3)+5z^2+6}{(s^2-2)(2z^2+3)((s^2+1)z^2+1)}$$
which is in fact not the wanted fraction. From the original task(solution to Problem 2 by ysharifi) I am going through I know, that the right partial decomposition should look like
$$\dfrac1{3s^2+1}\left(\frac3{2z^2+3}-\frac1{(s^2+1)z^2+1}\right)$$
but honestly speaking I have no clue how to get to this. Could someone please go through the whole process with me. I am quite confused right now.
Thanks in advance.
|
Based on your demand, here's another method using residues
Let
\begin{equation}
F(z)= \frac{z^2}{(2z^2+3)((s^2+1)z^2+1)}
=
\frac{\frac{z^2}{2(s^2+1)}}{(z^2 + \frac{3}{2})(z^2 + \frac{1}{s^2+1})}
\end{equation}
\begin{equation}
F(z)
= \frac{\frac{z^2}{2(s^2+1)}}{\Pi_{i=0}^3(z-z_i)}
=
\frac{r_0}{z-z_0}
+
\frac{r_1}{z-z_1}
+
\frac{r_2}{z-z_2}
+
\frac{r_3}{z-z_3}
\end{equation}
where
\begin{equation}
r_k = \lim_{z \rightarrow z_k}
(z-z_k)F(z)
\end{equation}
The poles $z_k$ are the roots of the denominator, i.e.
$
z^2 = -\frac{3}{2}
$
that gives the first two poles
$
z_{0,1} = \pm \frac{\sqrt{3}}{\sqrt{2}}i
$
and similarly, we compute the second two poles as
$
z_{3,4} = \pm \sqrt{\frac{1}{s^2+1}}i
$
Now, let's compute $r_i$'s, as
\begin{align}
r_0 = \lim_{z \rightarrow z_0}
(z-z_0) F(z)
&=
\frac{\frac{z_0^2}{2(s^2+1)}}{(z_0-z_1)(z_0-z_2)(z_0-z_3)} \\
r_1 = \lim_{z \rightarrow z_1}
(z-z_1) F(z)
&=
\frac{\frac{z_1^2}{2(s^2+1)}}{(z_1-z_0)(z_1-z_2)(z_1-z_3)} \\
r_2 = \lim_{z \rightarrow z_2}
(z-z_2) F(z)
&=
\frac{\frac{z_2^2}{2(s^2+1)}}{(z_2-z_0)(z_2-z_1)(z_2-z_3)} \\
r_3 = \lim_{z \rightarrow z_3}
(z-z_3) F(z)
&=
\frac{\frac{z_3^2}{2(s^2+1)}}{(z_3-z_0)(z_3-z_1)(z_3-z_2)}
\end{align}
After you replace and compute your $r_i$'s, you get $F(z)$ in the most decomposed form. Then, you could combine $(z-z_0)(z-z_1) = 2z^2 + 3$ and $(z-z_1)(z-z_2) = (s^2+1)z^2 + 1$
|
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|
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that
$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$
My try:
First I wrote the inequality as
$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125. $$
Then I noted that
$$a^2+\frac{b}{a}\geq a^2+\frac{2}{a}, \\ b^2+\frac{c}{b}\geq b^2+\frac{2}{b}, \\ c^2+\frac{a}{c}\geq c^2+\frac{2}{c}. $$
But I don't know how this can help me.
|
Another way.
By AM-GM for all $x\geq2$ we obtain:
$$x^3+16=x^3+8+8\geq3\sqrt[3]{x^3\cdot8^2}=12x.$$
Thus, $$\prod_{cyc}(a^3+b)\geq\prod_{cyc}(12a-16+b)=\prod_{cyc}(5a+7a+b-16)\geq\prod_{cyc}5a=125abc.$$
|
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|
The probability of getting a 7 This problem is probably too easy for all you math geniuses out there, but I am having trouble understanding it, so here it is:
An athlete has 10 cards, numbered 1 to 10. Every day, he picks a card randomly and does that number of pushups. He doesn't replace the card, and picks until all the cards are gone. If after a certain number of tries, the athlete has done a total of 12 pushups, then the probability of him/her picking a 7 as the very next card is $\frac{a}{b}$. What is a + b?
I know that I have to count the possibilities of getting a 12:
*
*2 cards
The possibilities would be 2 and 10, 3 and 9, 4 and 8, 5 and 7, which would make 4 possibilities.
*
*3 cards
1, 2, 9 | 1, 3, 8 | 1, 4, 7 | 1, 5, 6 | 2, 3, 7 | 2, 4, 6 | 3, 4, 5 | which would be 7 possibilities.
*
*4 cards
1, 2, 3, 6 | 1, 2, 4, 5 | which is 2 possibilities.
There cannot be more than 4 cards: 1, 2, 3, 4, 5 would make 15.
The total number of possibilities for getting a 12 is 17, and out of those, there are 3 of them that will pick the card 7, which means 14 out of the 17 possibilities will be valid for the problem.
I don't know how to continue from here. Should I solve for the probability each of the three cases listed above separately, then add them, or is there a better method?
|
$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
*
*The number of decks for which the first $2$ cards sum to $12$ is $(4)(2!)(8!)$, where
*
*The factor $4$ counts the ordered pairs $(2,10),\;(3,9),\;(4,8),\;(5,7)$.
*The factor $2!$ allows each pair to be permuted.
*The factor $8!$ counts the permutations of the remaining cards.
*The number of decks for which the first $2$ cards sum to $12$ and the next card is $7$ is $(3)(2!)(7!)$, where
*
*The factor $3$ counts the ordered pairs $(2,10),\;(3,9),\;(4,8)$.
*The factor $2!$ allows each pair to be permuted.
*The factor $7!$ counts the permutations of the cards remaining after the $7$.
$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
*
*The number of decks for which the first $3$ cards sum to $12$ is
is $(7)(3!)(7!)$, where
*
*The factor $7$ counts the ordered triples $(1,2,9),\;(1,3,8),\;(1,4,7)\;(1,5,6),\;(2,3,7),\;(2,4,6),\;(3,4,5)$.
*The factor $3!$ allows each triple to be permuted.
*The factor $7!$ counts the permutations of the remaining cards.
*The number of decks for which the first $3$ cards sum to $12$ and the next card is $7$ is $(5)(3!)(6!)$, where
*
*The factor $5$ counts the ordered triples $(1,2,9),\;(1,3,8),\;(1,5,6),\;(2,4,6),\;(3,4,5)$.
*The factor $3!$ allows each triple to be permuted.
*The factor $6!$ counts the permutations of the cards remaining after the $7$.
$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
*
*The number of decks for which the first $4$ cards sum to $12$ is
is $(2)(4!)(6!)$, where
*
*The factor $2$ counts the ordered quadruples $(1,2,3,6),\;(1,2,4,5)$.
*The factor $4!$ allows each quadruple to be permuted.
*The factor $6!$ counts the permutations of the remaining cards.
*The number of decks for which the first $4$ cards sum to $12$ and the next card is $7$ is $(2)(4!)(5!)$, where
*
*The factor $2$ counts the ordered quadruples $(1,2,3,6),\;(1,2,4,5)$.
*The factor $4!$ allows each quadruple to be permuted.
*The factor $5!$ counts the permutations of the cards remaining after the $7$.
$$
\overline{
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}
$$
Hence, the probability that the next card is $7$, given that the previous cards sum to $12$ is
$$
\frac
{\;\,(3)(2!)(7!)+(5)(3!)(6!)+(2)(4!)(5!)}
{\;\,(4)(2!)(8!)+(7)(3!)(7!)+(2)(4!)(6!)}
=
\frac{8}{79}
$$
|
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|
A special case of the Fermat-Torricelli point in a triangle $\triangle{ABC}$ is an isosceles right triangle; its legs are of length $s = 30\sqrt{5}$ and hypotenuse is of length $30\sqrt{10}$. The Fermat-Torricelli point $P$ must be along the median from the vertex $C$ of the right angle of the triangle.
$\left\vert\overline{CP}\right\vert = x$, and $\left\vert\overline{AP}\right\vert = y = \left\vert\overline{BP}\right\vert$, and
\begin{equation*}
\mathrm{m}\angle{APC} = 180 - \theta = \mathrm{m}\angle{BPC} .
\end{equation*}
According to the Law of Cosines,
\begin{equation*}
s^{2} = x^2 + y^2 + 2xy\cos(\theta) .
\end{equation*}
According to the Law of Sines,
\begin{equation*}
y = \frac{s/\sqrt{2}}{\sin(\theta)} .
\end{equation*}
According to the Pythagorean Theorem,
\begin{equation*}
\left(s/\sqrt{2} - x\right)^2 + \left(s/\sqrt{2}\right)^2 = y^2 .
\end{equation*}
So,
\begin{equation*}
y = \sqrt{x^{2} - 30\sqrt{10} \, x + 4500} .
\end{equation*}
\begin{equation*}
\left\vert\overline{AP}\right\vert + \left\vert\overline{BP}\right\vert + \left\vert\overline{CP}\right\vert
= x + 2y
= x + 2\sqrt{x^{2} - 30\sqrt{10} \, x + 4500} .
\end{equation*}
The minimum value of this function is
\begin{equation*}
30\sqrt{5\left(2 + \sqrt{3}\right)}
\end{equation*}
at
\begin{equation*}
15\sqrt{10} - 5\sqrt{30} .
\end{equation*}
\begin{equation*}
\left\vert\overline{CP}\right\vert = x
= 15\sqrt{10} - 5\sqrt{30} \approx 20.05
\end{equation*}
and
\begin{equation*}
\left\vert\overline{AP}\right\vert = \left\vert\overline{AP}\right\vert
= y
= \frac{1}{2} \left(30\sqrt{5\left(2 + \sqrt{3}\right)}
- \left(15\sqrt{10} - 5\sqrt{30}\right)\right) \approx 54.77 .
\end{equation*}
My Concern
I am told that the distances between the vertices and the Fermat-Torricelli Point are natural numbers.
|
I've found no errors in your calculations.
If we can use the fact that the point $P$ satisfies
$$\angle{APB}=\angle{BPC}=\angle{CPA}=120^\circ$$
then, we can easily see that
$$|\overline{CP}|=x=15\sqrt{10}-5\sqrt{30},\qquad|\overline{AP}|=|\overline{BP}|=y=10\sqrt{30}$$
which are the same as what you've got.
|
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|
how to prove that this sequence converges to $0$? $0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$
When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to find its limit.
|
Let's study
$$x_{n+1}^{-2} - x_n^{-2} = \left( x_n - \frac{4x_n^3}{n+1} \right)^{-2} - x_n^{-2} = x_n^{-2} \left( \left( 1- \frac{4x_n^2}{n+1}\right)^{-2} -1\right)$$
So
$$x_{n+1}^{-2} - x_n^{-2} = x_n^{-2} \left( \frac{8x_n^2}{n+1} + o\left(\frac{x_n^2}{n+1} \right) \right) = \frac{8}{n+1} + o\left( \frac{1}{n+1} \right)$$
So you get that
$$x_{n+1}^{-2} - x_n^{-2} \sim \frac{8}{n+1}$$
The series $\sum \frac{8}{n+1}$ diverges, so we can sum to obtain
$$x_{n+1}^{-2} \sim \sum_{k=1}^n \frac{8}{k+1}$$
You obtain finally
$$x_{n+1} \sim \frac{1}{\sqrt{\sum_{k=1}^n \frac{8}{k+1}}}$$
In particular, the limit is indeed $0$.
|
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|
Roots of $2x^3-4x+1$ I'm having difficulty getting the solution to the cubic equation $2x^3-4x+1=0$ and from http://www2.trinity.unimelb.edu.au/~rbroekst/MathX/Cubic%20Formula.pdf it claims that the general solution to $Ax^3+Bx^2+Cx+D=0$ is $$p+q+r=-B/A$$
$$pq+qr+rp=C/A$$
$$pqr=-D/A$$
where $p,q,r$ are the roots.
I also tried http://www2.trinity.unimelb.edu.au/~rbroekst/MathX/Cubic%20Formula.pdf and very carefully followed their technique (which looks at first glance different but must obviously be equivalent) however it didn't line up with what I got from Wolframalpha. So I'm just looking to see where I messed up or how others would solve this.
Here's my attempt:
$$p+q+r=0$$
$$pq+qr+rp=-2$$
$$pqr=-1/2$$
From the first we get that $$p=-q-r$$
then from the second
$$p(q+r) = -2-qr$$
then from the two of these we get
$$-p^2 = -2-qr$$
equivalently
$$0=p^3 -2p+1/2$$
Wait! What?! A cubic to solve a cubic..... cubics all the way down!? Obviously not.... so if it's possible to solve this apart from numerical techniques, I would be really interested in such an answer. Thank you
|
We have a cubic equation $2x^3 - 4x + 1 = 0$, what $x$ is its root ?
May you can solve this by formula $\cos 3t = 4\cos^3 t - 3\cos t$. But I hope that you understood the means and enjoys from apply them on many cubic equations. Indeed, you will solve this cubic equation.
The cubic equation is $x^3 - 2x + \frac{1}{2} = 0$ which have form $t^3 + pt + q = 0$. A cubic equation of this class can be got from $x^3 + ax^2 + bx + c = 0$ if replace the $x$ by $t - \frac{a}{3}$ and then the coefficients $$p = \frac{3b - a^2}{3}\quad \mathrm{and}\quad q = \frac{2a^2 - 9ab + 27c}{27}.$$
Just as $(x + a)^3 = x^3 + 3ax^2 + 3a^2x + a^3$, if $z = e^{it} = x + iy$, then $$z^3 = e^{i3t}= x^3 + i3yx^2 - 3y^2x - iy^3 = (4\cos^3 t - 3\cos t) + i(-4\sin^3 t + 3\sin t).$$
If $z = e^{it} = x + iy$ and $\zeta = x^3 - iy^3$, then it is $4\zeta - 3\overline{z} - z^3 = 0$. Just as this, if $z = re^{it} = rx + iry$ and $\zeta = r^3x^3 - ir^3y^3$, then $4\zeta^3 - 3r^2\overline{z} - z^3 = 0$.
For your cubic equation $4x^3 - 8x + 2 = 0$, you know
$$r = \sqrt{\frac{8}{3}}\quad \mathrm{and} \quad t = \frac{1}{3}\arg(-2\sqrt{\frac{3}{8}}^3) + \frac{2n\pi}{3} = \frac{1}{3}\arg(-\frac{3}{8}\sqrt{\frac{3}{2}})+ \frac{2n\pi}{3}$$ where $n \in \mathbb{Z}$.
Therefore $$x = \sqrt{\frac{8}{3}}\exp(i(\frac{1}{3}\arg(-\frac{3}{8}\sqrt{\frac{3}{2}})+\frac{2n\pi}{3})),\quad n \in \mathbb{Z}.$$
But $x \in \mathbb{R}$ and the term under $\arg$ function is real, so the root $$x = \sqrt{\frac{8}{3}}\cos(\frac{1}{3}\cos^{-1}(-\frac{3}{8}\sqrt{\frac{3}{2}})+\frac{2n\pi}{3})),\quad n \in \mathbb{Z}.$$
|
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"url": "https://math.stackexchange.com/questions/2938869",
"timestamp": "2023-03-29T00:00:00",
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|
$E(X^2)$ of discrete uniform distribution I have a discrete uniform distribution (from IFoA formulae) with parameters a, b, h, where a and b are the start/end points, and h is the interval between each value. The p.d.f. is given as $\frac{h}{b-a+h}$. I manage to find out $$E[X] = \sum_{x\in X} xf(x) = \sum_{x\in X} \frac{xh}{b-a+h} = \frac{h}{b-a+h}\sum_{x\in X} x\\= \frac{h}{b-a+h}\times \frac{a+b}{2}\times \frac{b-a+h}{h} = \frac{a+b}{2}.$$ However, I cannot figure out the second moment ($E[X^2]$) and Variance by using the same method. I wonder if it is possible to find them using the same way, I know they could be found by using the M.G.F., I just want to know if it is possible.
$Var(X) = \frac{(b-a)(b-a+2h)}{12}$ is given by the formulae.
Here is what I have got so far:
$$\sum_{x\in X} x^2f(x) = \sum_{x\in X} \frac{x^2 h}{b-a+h} = \frac{h}{b-a+h}\sum_{x\in X} x^2=\frac{h}{b-a+h}\sum_{k=0}^{\frac{b-a+h}{h}}(hk+a)^2$$
edit:
\begin{align*}
E[X^2] &= \sum_{x\in X} x^2f(x)\\
&=\frac{h^3}{b-a+h}\sum_{k=0}^{\frac{b-a+h}{h}}k^2+\frac{2ah^2}{b-a+h}\sum_{k=0}^{\frac{b-a+h}{h}}k+\frac{a^2h}{b-a+h}\sum_{k=0}^{\frac{b-a+h}{h}}1\\
&=\frac{h^3}{b-a+h}\frac{(\frac{b-a}{h})(\frac{b-a}{h}+1)(2(\frac{b-a}{h})+1)}{6}+\frac{2ah^2}{b-a+h}\frac{(\frac{b-a}{h}+1)(\frac{b-a}{h})}{2}+\frac{a^2h}{b-a+h}\bigg(\frac{b-a}{h}+1\bigg)\\
&= \frac{h^3}{b-a+h}\frac{(\frac{b-a}{h})(\frac{b-a+h}{h})(\frac{2(b-a)+h}{h})}{6}+\frac{2ah^2}{b-a+h}\frac{(\frac{b-a+h}{h})(\frac{b-a}{h})}{2}+\frac{a^2h}{b-a+h}\bigg(\frac{b-a+h}{h}\bigg)\\
&= \frac{(b-a)(2(b-a)+h)}{6}+a(b-a)+a^2\\
Var(X) &= E[X^2] - E[X]^2\\
&= \frac{(b-a)(2(b-a)+h)}{6}+a(b-a)+a^2 - \bigg(\frac{a+b}{2}\bigg)^2\\
&= \frac{2(b-a)(2(b-a)+h)+12a(b-a)+12a^2}{12} - \frac{3(a+b)^2}{12}\\
&= \frac{4b^2-4ab+2bh-4ab+4a^2-2ah+12ab-12a^2+12a^2-3a^2-6ab-3b^2}{12}\\
&= \frac{(a-b)^2+2h(b-a)}{12} \\
&= \frac{(b-a)^2+2h(b-a)}{12}\\
&= \frac{(b-a)(b-a+2h)}{12}
\end{align*}
|
With your edit, you are almost there. You can simplify
$$\frac{h}{b-a+h}\sum_{k=0}^{b-a}(hk+a)^2$$
$$=\frac{h}{b-a+h}\sum_{k=0}^{b-a}\left((hk)^2+2ahk+a^2 \right)$$
$$=\frac{h}{b-a+h}\sum_{k=0}^{b-a}(hk)^2+\frac{h}{b-a+h}\sum_{k=0}^{b-a}(2ahk)+\frac{h}{b-a+h}\sum_{k=0}^{b-a}a^2$$
$$=\frac{h^3}{b-a+h}\sum_{k=0}^{b-a}k^2+\frac{2ah^2}{b-a+h}\sum_{k=0}^{b-a}k+\frac{a^2h}{b-a+h}\sum_{k=0}^{b-a}1$$
For the first term you can apply
$$\sum_{x=0}^n x^2 = \frac{n(n+1)(2n+1)}{6}$$
and apply the similar formulas that you did previously for the other 2 terms.
|
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|
Find height of a wall if at the beginning it exceeds $10$ meters and then $8$ meters When the foot of a staircase is $5$ meters from the base of a wall, it protrudes $10$ meters above the wall; and if it is $9$ meters from the base, it stands $8$ meters.
Find the height of the wall.
Using the Pythagorean theorem for the two situations I have
$$\begin{cases}(10-x)^2=5^2+y^2\\(8-x)^2=9^2+y^2,\end{cases}$$ where $x$ is the hypotenuse and $y$ is the height of the wall.
Solving that system of equations I have that $x=23$ and $y=\pm12$, but since a height is always positive, the solution is $\boxed{12~\text{meters}}$.
Is it correct?
I am not sure of the $x$ value because first it has a value but then it has another, so maybe we have to use $x_1$ and $x_2$, but then we have $3$ equations with $2$ variables, so it is not possible.
Thanks!
|
The hypotenuse is $x - 10$ and $x - 8$ where $x$ is the total length of the staircase with overhang and is the same for both.
\begin{cases}(x-10)^2=5^2+y^2\\(x-8)^2=9^2+y^2\end{cases}
$(x-8)^2 - (x-10)^2 = 56$
$(x^2 - 16x + 64)-(x^2 - 20x +100) = 56$
$4x -36 = 56$
$x = 23$
$y = \sqrt{13^2 - 5^2} = 12$
Also $y = \sqrt{15^2 - 9^2} = 12$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there any mistake in my approach for solving $ \int_0^{\pi/2} \frac{ \cos x}{3 \cos x + \sin x} \, dx $ ?? I had to evaluate this integral .
$$
\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx
$$
Here is how I proceeded
Dividing $N^r$ And $D^r$ by $\cos^3 x$
$$
\int_0^{\pi/2} \frac{ \sec^2 x}{3 \sec^2 x + \tan x \sec^2 x}\, dx \\
$$
Substituting $\tan x = t$
$$
\int_0^\infty \frac{ 1 }{(1+t^2)(t+3)} \, dt \\
$$
Then by using Partial Fractions , I got the answer as
$$
\frac{1}{10} \log (t+3) - \frac{1}{20} \log (t^2 + 1) + \frac{3}{10} \arctan (t) \biggr|_{0}^{\infty}
$$
But while substituting the limits , the answer comes out be be infinity which is wrong .
Is there any mistake in my approach ??
|
A simple method to avoid using partial fractions and improper integrals. Let
$$ A=\int_0^{\pi/2} \frac{\cos x}{3 \cos x + \sin x} \, dx, B=\int_0^{\pi/2} \frac{\sin x}{3 \cos x + \sin x} \, dx. $$
Clearly
$$ 3A+B=\frac{\pi}{2}. \tag{1}$$
Noting that
$$ A=\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d\sin x, B=-\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d\cos x $$
it is easy to see
$$ -3B+A=\int_0^{\pi/2} \frac{1}{3 \cos x + \sin x} \, d(3\cos x+\sin x)=\ln(3\cos x+\sin x)\bigg|_0^{\frac\pi2}=-\ln 3.\tag{2} $$
From (1) and (2), one has
$$A=\frac{3\pi}{20}-\frac{1}{10}\ln3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$
Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$
My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that
$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$
By substituting (1) in the LHS of the original expression,
it is straightforward to show that it is equal to the RHS.
But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of
multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to
$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$
But I was not able to show that this last expression is equal to RHS by this approach.
Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.
|
we have to show,
$$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$$
Let,$2+\sqrt{3}=p$
$$R.H.S.= -3-2\sqrt{3}$$
$$=-2(\frac{3}{2}+\sqrt{3})$$
$$=-2\bigg(\frac{3}{2}+\frac{1}{2}-\frac{1}{2}+\sqrt{3}\bigg)$$
$$=-2(2+\sqrt{3}-\frac{1}{2})$$
$$=-2(p-\frac{1}{2})$$
$$=1-2p$$
$$=\dfrac{(2-p)(1-2p)}{2-p}$$
$$=\dfrac{2-5p+2p^2}{2-p}...........(1)$$
Now,
$$L.H.S=\dfrac{\sqrt{2}+\sqrt{p}}{\sqrt{2}-\sqrt{p}}$$
$$=\dfrac{(\sqrt{2}+\sqrt{p})^2}{2-p}.............(2)$$
Now,we can conclude that if the given statement is true then,for $p=2+\sqrt{3}$,the numerators of part (1) and (2) will be equal.in reverse we can also say that the numerators will be same if $p=2+\sqrt{3}$.So,now my target is to solve for $p$ to get $2+\sqrt{3}$.
to be the statement true,
$$(\sqrt{2}+\sqrt{p})^2=2-5p+2p^2$$
$$p^2-3p-\sqrt{2}p=0$$
$$p=0,2+\sqrt{3}$$
As we got a root $p=2+\sqrt{3}$,then the claim is true.$_{[Showed]}$
NOTE: you can also prove it by substituting the value of p in (2) and getting (1).but I think the above procedure is more simple.
|
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|
Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that of a hyperbola.
|
One of the approaches may be as follows:
Suppose there exists some $a$ which is the minimum. Then:
$$
\begin{align}
x + 2\sqrt{x^2 - \sqrt2 x+1} &= a \\
2\sqrt{x^2 - \sqrt2 x+1} &= a-x
\end{align}
$$
Square both sides:
$$
(a-x)^2 = 4x^2 - 4\sqrt2x + 4
$$
Applying some transformations you can get:
$$
3x^2 + x(2a-4\sqrt2) + 4 -a^2 =0
$$
Now you want the discriminant to be equal to zero which means only one root will exist, so:
$$
D = 16a^2 - 16\sqrt2a-16 = a^2 - \sqrt2a - 1 =0
$$
The equation in terms of $a$ has two solutions:
$$
a_1 = {1-\sqrt3 \over \sqrt2} \\
a_2 = {1+\sqrt{3} \over \sqrt2}
$$
If you plug them into the initial equation only one of them is going to produce a valid statement. Therefore that will be you minimum, which appears to be ${1+\sqrt3 \over \sqrt2}$
|
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|
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
|
One way is to write $z=r^2 e^{2\theta}$ and roots will be $re^\theta$ and $re^{\pi -\theta}$.
From $z =-1+4\sqrt{3}i$, we obtain $r=7$ and $\tan{2\theta} = \frac{2\tan\theta}{1-\tan^2\theta} = -4\sqrt{3}$. Second expression gives you a quadratic equation, $2\sqrt{3}\tan^2\theta -2\sqrt{3} + \tan\theta =0$.
Roots of the above quadratic equation are $\tan\theta= \sqrt{3}/2,-2/\sqrt{3}$ which form $\tan\theta$ and $\tan(\pi-\theta)$.
Hence, square roots of $z$ are $(1+\sqrt{3}/2i)\frac{7}{\sqrt{1+3/4}} = \sqrt{7}(2+\sqrt{3}i)$ and $(1-2/\sqrt{3}i)\frac{7}{\sqrt{1+4/3}} = \sqrt{7}(\sqrt{3}-2i)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine linear map so that equation is correct Let $f:\mathbb R^3→\mathbb R^2$, $(x,y,z)^T\mapsto(x^2y-5z,2x+4yz-3z^3)^T$,
and $(x_0, y_0, z_0)^T=(-1,0,1)^T$.
I need to determine the linear map $A∈L(\mathbb R^3,\mathbb R^2)\cong\mathbb R^{2×3}$ so that$$
f \begin{pmatrix} x \\ y \\ z \end{pmatrix}=f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}+\left\|\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}\right\| ε\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}
$$
with $\lim\limits_{\vec h \to 0} ε(\vec h)=0$.
How can this be done? I don't know how to start (except for calculating $f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}$).
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}=\begin{pmatrix} -5 \\ -5 \end{pmatrix}$
$A=\begin{pmatrix} 0 && 1 && -5\\ 2 && 4 && -9 \end{pmatrix}$
$\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix} x+1 \\ y \\ z-1 \end{pmatrix}$
Therefore:
$f\begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix}+A\begin{pmatrix} x-x_0 \\ y-y_0 \\ z-z_0 \end{pmatrix}=\begin{pmatrix}y-5z \\ 6+2x+4y-9z\end{pmatrix}$
|
Hint.
$$
f(p) = f(p_0) +\frac{\partial f}{\partial p}|_{p_0}(p-p_0) + O(|p-p_p|^2)
$$
|
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|
How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
|
Method 1:
$y = -4x^2 -2x -4$
$\frac {y}{-4} = x^2 + \frac 12x + 1$
$-\frac {y}{4}- 1 = x^2 + 2*\frac 14 x$
$-1-\frac y4 +(\frac 14)^2= x^2 + 2*\frac 14 x+ (\frac 14)^2$
$-1+\frac 1{16}-\frac y4 = (x + \frac 14)^2$
$(x+ \frac 14)^2 = -\frac {15}{16} - \frac y4 = \frac {-15 - 4y}{16}$
$x + \frac 14 = \pm \sqrt {\frac {-15-4y}{16} }= \frac {\sqrt {-15 - 4y}}4$
$x = \frac {\sqrt {-15 - 4y}}4 -\frac 14 = \frac {\sqrt{-15 - 4y} - 1}4$
Method 2:
$y = -4x^2 - 2x -4 \iff y+4 = -4x^2 -2x$
We want $-4x^2 - 2x$ to be the first to summands of $(mx + n)^2 = m^2x + 2mnx + n^2$. But $m^2$ is positive while $-4$ is negative.
So we multiple everything by $-1$.
$y+4 = -4x^2 -2x \iff -y - 4 = 4x^2 + 2x$. Now we want $4 = m^2$ and $2 = 2mn$.
Or in other words $m = \pm \sqrt 4$ and $n = \frac 1m$. Or in other words $m = \pm 2$ and $n = \frac 12$.
$-y -4 = 4x^2 + 2x = m^2x^2 + 2mnx \iff$
$-y-4 + n^2 = -y -4 + (\frac 12)^2= 4x^2 + 2x + (\frac 12)^2 = m^2x^2 + 2mnx +n^2 = (mx + n)^2 = (2x + \frac 12)^2$
$\iff -y -4 +\frac 14 = -y -\frac {15}4 = (2x + \frac 12)^2$
$\iff 2x +\frac 12 =\pm \sqrt {-y -\frac {15}{4}}$
$\iff 2x =-\frac 12 \pm \sqrt {-y-\frac {15}4}=-\frac 12 \pm \frac {\sqrt{-4y-15}}2 = \frac {-1 \pm\sqrt {-4y - 15}}2$
$\iff x = \frac {-1 \pm \sqrt{-4y -15}}4$
|
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|
First year college question on divisibility of integers I'm having a hard time with a practice question.
Given $n$ is an integer, prove $2$ divides $(n^4 -3)$ iff $4$ divides $(n^2 +3)$.
So I know since it's an iff statement, I have to show the implication going both ways. Let's start with the left side first.
There exists an integer $r$ such that $2 r = n ^ 4 - 3 $.
Here, I'm thinking in my head how can I get the equation to look like the conclusion, that is $4a = (n^2 + 3)$ for an integer $a$. I see that we can play with the $3$ on both sides.
$$3 = n^4 - 2r$$
$$3 + n^2 = n^4 + n^2 - 2r$$
So here's where I'm stuck. How can I show I can factor out a $4$ out if this right side? Thanks.
|
Trick: If $2|n^4 -3$ then $n$ is odd. So let $n = 2r + 1$ then $n^2 +3 = (2r+1)^2 + 3 = 4r^2 + 4r + 1 + 3 = 4r^2 + 4r + 4 = 4(r^2 + r + 1)$ and $4|n^2 + 3$.
Trick: If $4|n^2 + 3$ then $n$ is odd. So if $n$ is odd then $n^4$ is odd and $n^4 -3$ is even. So $2|n^4 -3$.
.....
In general to prove $4|n^2 +3$ we want to assume $n \equiv k \pmod 4$ and prove that $k^2 + 3\equiv 0\pmod 4$.
If $2|n^4 - 3$ then $k^4 \equiv 3\equiv 1 \pmod 2$ so $k \equiv 1 \pmod 2$ and $k \equiv 1,3\pmod 4$. And $k^2 + 3 \equiv 1,9 + 3 \equiv 4,12 \equiv 0 \mod 4$. So that is $\implies$.
To prove $2|n^4 -3$ we want to assume $n \equiv k \pmod 2$ and prove that $k^4 -3 \equiv 0 \pmod 2$.
If $4|n^2 +3$ then $n^2\equiv -3 \equiv 1 \pmod 4$ so $n\equiv \pm 1 \pmod 4$ so $n\equiv 1\pmod 2$. So $n^4 -3 \equiv 1^4 -3 \equiv 0 \mod 2$. So that is $\Leftarrow$.
|
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|
Stuck solving a logarithmic calculation I'm preparing for my further studies (last year of high school, preparing so I can try and join the academy that I want), and just solving problems. Got stuck on this one:
What is the value of:
$$log_4log_3log_28 + log_{\sqrt{7}+1}(8+2\sqrt{7})+log_{\sqrt[3]{7}}7\sqrt{7}$$
This is what I got so far:
$log_4log_3log_28 = log_4log_33=log_41=0$
$log_{\sqrt[3]{7}}7\sqrt{7}=log_{7^{1\over3}}(7*7^{1\over2})=3log_77^{3\over2}=3*{3\over2}log_77={9\over2}$
So
$log_4log_3log_28 + log_{\sqrt{7}+1}(8+2\sqrt{7})+log_{\sqrt[3]{7}}7\sqrt{7} \\= 0 + log_{\sqrt{7}+1}(8+2\sqrt{7}) + {9\over2}\\={9\over2}+log_{\sqrt{7}+1}(8+2\sqrt{7})$
I'm lost at what to do with $log_{\sqrt{7}+1}(8+2\sqrt{7})$
|
Hint: $$8+2\sqrt{7}=(\sqrt{7}+1)^2$$
|
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|
Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sqrt[3]{x^2})}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=\lim_{x \to -8}\left[\frac{-(x+8)}{x+8}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=-\lim_{x \to -8} \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\\&=-2.\end{align*}
But when I input this
lim\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} as x to -8
into Wolfram|Alpha, it gives the limit $0$.
Why is Wolfram|Alpha making a mistake here?
|
If you take the complex roots of $\sqrt[3]{x}$ you get $0$ as the limit, because the denominator is different from zero in this case.
So, Wolfram|Alpha did not make a mistake but just uses a different root of $\sqrt[3]{x}$.
For the real root you get $-2$:
*
*$t^3 = -x \Rightarrow \lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = \lim_{t \to 2}\frac{\sqrt{1+t^3}-3}{2-t} = -f'(2) \mbox{ for } f(t) = \sqrt{1+t^3}$
$$f'(t) = \frac{3t^2}{2\sqrt{1+t^3}}\Rightarrow \lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = - f'(2) = -2$$
|
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|
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
|
$ z - \frac{1}{z} = \frac{z^2-1}{z} $
So that, we get $\left| z - \frac{1}{z} \right| $ = $\frac{|z^2-1|}{2} $
If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:
$ \frac{|x^2-y^2-1 + i(2xy)|}{2} = \frac{\sqrt{(x^2-y^2-1)^2 + (2xy)^2}}{2} $
Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $
We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x \in [-2,2] $
$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \\ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $
So, all we have to do is looking at $25-4x^2, for \ x^2 \in[0,4] $ and say what's the min and max.
Clearly if $x=0$ then we get $25$ and the max value is $\frac{\sqrt{25}}{2} = \frac{5}{2}$
if $x^2 = 4$ we get $9$ and the min value is $\frac{\sqrt{9}}{2} = \frac{3}{2}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Let $f$ be a function $f : \mathbb{N} \to \mathbb{N}$ such that $f(2x+3y)=f(x)f(y)$, determine $f$ here what I did .
$$f(0)=f(0)^2$$
so $f(0)=1$ or $f(0)=0$
IF $f(0)=1$
we have $f(2y)=f(y)$
$$f(1)=f(2)=\ldots=f(2^n)=a$$
the equation $f(x)-a=0$ has infinitly many solutions , so $f(x)=a$
since f(0)=1 , $f(x)=1$.
i don't know how to handle the other case , i just found that $f(2x)=f(3x)=0$
|
This is related to the Frobenius problem (or coins or Mcnuggets problem) of two numbers:
If $a,b\in\mathbb{N}$ are two numbers such that $\gcd(a,b)=1$, then all numbers $n\geq(a-1)(b-1)$ are representable as a nonnegative combination of $a,b$, i.e., $n=ax+by$ with $x,y\in\mathbb{N}$.
1) If $f(0)=0$:
Since $\gcd(4,3)=1$ and $(4-1)(3-1)=2$, if $n\geq6$, then there are $x_n,y_n\in\mathbb{N}$ such that $n=4x_n+3y_n$.
Since $f(0)=0$ we get $f(2x)=f(2x+3\cdot 0)=f(x)f(0)=0$ for every $x\in\mathbb{N}$, so $f(n)=f(4x_n+3y_n)=f(2x_n)f(y_n)=0$ for all $n\geq6$.
Since $f(0)=f(2)=f(4)=0$, it just remains to show what happens with $f(1),f(3),f(5)$.
Note that $f(3)^2=f(3)f(3)=f(2\cdot 3+3\cdot 3)=f(15)=0$ (because $15\geq6$), so $f(3)=0$. Similarly, $f(5)^2=f(25)=0$, and finally $f(1)^2=f(5)=0$, so $f(1)=f(3)=f(5)=0$ and $f(x)=0$ for every $x\in\mathbb{N}$.
b) If $f(0)=1$:
Since $\gcd(2,3)=1$, for every $n\geq2$ there are $x_n,y_n\in\mathbb{N}$ such that $n=2x_n+3y_n$. Denote $a:=f(1)$. By strong induction, suppose that for every number $m<n$, $f(m)$ is some power of $a$ (we have as base cases $f(0)=a^0$, $f(1)=a^1$). Then $f(n)=f(2x_n+3y_n)$ with $0\leq x_n,y_n<n$, thus $f(n)=f(x_n)f(y_n)=a^ra^s=a^{r+s}$ for some exponents $r,s$.Note that we can do this in such a way that the exponent is never $0$ (except for $f(0)$).
On the other hand, the first number representable in two ways (with different numbers unregarding the order) as a nonnegative linear combination of $2$ and $3$ is $8=2\cdot 4+3\cdot 0=2\cdot 1+3\cdot 2$; thus $f(8)=f(4)f(0)=f(4)$ and $f(8)=f(1)f(2)$, so $f(4)=f(1)f(2)=af(2)$. But $f(4)=f(2\cdot 2 +3\cdot 0)=f(2)f(0)=f(2)$, hence $af(2)=f(2)$, so either $a=1$ or $f(2)=0$; and $f(2)=f(2\cdot 1+3\cdot 0)=f(1)=a$, so either $a=1$ or $a=0$. Noting that every other image is a power of $a$, we get either $f(x)=1$ for every $x$ or $f(0)=1$, $f(x)=0$ for every $x>0$.
Observe that these functions indeed satisfy the given condition.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Uniform convergence of $\sum\limits_{n=1}^∞n^{-x}(e^{\frac{x}{n^2}}-1)$
Pointwise and uniform convergence of the following series of functions:
$$\sum_{n=1}^{\infty} n^{-x}\left(e^{\frac{x}{n^2}}-1\right).$$
Now, the series of function converges pointwise as $x \in (-1, +\infty)$, because, as $n \to \infty$, $$f_n(x)=\mathcal{O}\left(\frac{x}{n^{x+2}}\right). \quad \forall x\in \mathbb{R}$$
I am having some problems with the uniform convergence because I could not find the supremum. Any suggestions?
|
$\def\e{\mathrm{e}}\def\peq{\mathrm{\phantom{=}}{}}$First, since$$
\frac{1}{n^x} \left(\exp\left( \frac{x}{n^2} \right) - 1 \right) \sim \frac{x}{n^{x + 2}}\quad (n → ∞)
$$
for any fixed $x$, then $\displaystyle \sum_{n = 1}^∞ \frac{1}{n^x} \left( \exp\left( \frac{x}{n^2} \right) - 1 \right)$ converges iff $x + 2 > 1$, i.e. $x \in (-1, +∞)$.
Next, to prove that the series does not converge uniformly for $x \in (-1, +∞)$, it suffices to prove the following lemma.
Lemma: Given that $\{u_n(x)\} \subset C([a, b])$. If $\displaystyle \sum_{n = 1}^∞ u_n(x)$ converges pointwise for $x \in (a, b]$, but $\displaystyle \sum_{n = 1}^∞ u_n(a)$ diverges, then $\displaystyle \sum_{n = 1}^∞ u_n(x)$ does not converge uniformly for $x \in (a, b]$.
Proof: Suppose that $\displaystyle \sum_{n = 1}^∞ u_n(x)$ converges uniformly for $x \in (a, b]$. For any fixed $ε > 0$, there exists $N \geqslant 1$ such that for any $m > n \geqslant N$,$$
\left| \sum_{k = n + 1}^m u_k(x) \right| = \left| \sum_{k = 1}^m u_k(x) - \sum_{k = 1}^n u_k(x) \right| < ε, \quad \forall x \in (a, b]
$$
which implies$$
\left| \sum_{k = n + 1}^m u_k(a) \right| = \lim_{x → a^+} \left| \sum_{k = n + 1}^m u_k(x) \right| \leqslant ε.
$$
Thus $\displaystyle \sum_{n = 1}^∞ u_n(a)$ converges, a contradiction.
Back to the question. Since the series diverges for $x = -1$, then it does not converge uniformly for $x \in (-1, 0]$ by the lemma, and thus not for $x \in (-1, +∞)$.
Finally, it will be proved that for any $δ > -1$, this series converges uniformly for $x \in [δ, +∞)$. It suffices to prove for $-1 < δ < 0$.
For $x \in [0, 2]$ and $n \geqslant 2$, note that$$
e^{-t} - 1 \geqslant -t \Longrightarrow \e^t - 1 \leqslant t\e^t, \quad \forall t \in \mathbb{R}
$$
thus\begin{align*}
&\peq \sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right)\\
&\leqslant \sum_{k = n}^∞ \frac{1}{k^x} · \frac{x}{k^2} \exp\left( \frac{x}{k^2} \right)= \sum_{k = n}^∞ \frac{x}{k^2} · \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^x\\
&\leqslant \sum_{k = n}^∞ \frac{x}{k^2} · \left( \frac{1}{2} \exp\left( \frac{1}{4} \right) \right)^x \leqslant \sum_{k = n}^∞ \frac{x}{k^2} \leqslant \sum_{k = n}^∞ \frac{2}{k^2}.
\end{align*}
Since $\sum\limits_{n = 2}^∞ \dfrac{2}{n^2} < +∞$, then the given series converges uniformly for $x \in [0, 2]$.
For $x \in (2, +∞)$ and $n \geqslant 2$, note that $\dfrac{1}{k} \exp\left( \dfrac{1}{k^2} \right) \leqslant \dfrac{1}{2} \exp\left( \dfrac{1}{4} \right) < 1$ for $k \geqslant 2$, thus\begin{align*}
&\peq \sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right)\\
&\leqslant \sum_{k = n}^∞ \frac{1}{k^x} \exp\left( \frac{x}{k^2} \right) = \sum_{k = n}^∞ \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^x \leqslant \sum_{k = n}^∞ \left( \frac{1}{k} \exp\left( \frac{1}{k^2} \right) \right)^2.
\end{align*}
Because$$
\sum_{n = 2}^∞ \left( \frac{1}{n} \exp\left( \frac{1}{n^2} \right) \right)^2 = \sum_{n = 2}^∞ \frac{1}{n^2} \left( \exp\left( \frac{2}{n^2} \right) - 1 \right) + \sum_{n = 2}^∞ \frac{1}{n^2} < +∞,
$$
then the given series converges uniformly for $x \in (2, +∞)$.
For $x \in [δ, 0)$ and $n \geqslant 1$, since$$
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| = \sum_{k = n}^∞ \frac{1}{k^x} \left( 1 - \exp\left( \frac{x}{k^2} \right) \right) \leqslant \sum_{k = n}^∞ \frac{1}{k^δ} \left( 1 - \exp\left( \frac{δ}{k^2} \right) \right),
$$
and $\displaystyle \sum_{n = 1}^∞ \frac{1}{n^δ} \left( 1 - \exp\left( \frac{δ}{n^2} \right) \right) < +∞$, then the given series converges uniformly for $x \in [δ, 0)$.
Therefore for any $ε > 0$, there exists $N_1, N_2, N_2 \geqslant 1$ such that\begin{gather*}
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε, \quad \forall n \geqslant N_1,\ x \in [δ, 0)\\
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε, \quad \forall n \geqslant N_2,\ x \in [0, 2]\\
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε. \quad \forall n \geqslant N_3,\ x \in (2, +∞)
\end{gather*}
Taking $N = \max(N_1, N_2, N_3)$ yields$$
\sum_{k = n}^∞ \left| \frac{1}{k^x} \left( \exp\left( \frac{x}{k^2} \right) - 1 \right) \right| < ε. \quad \forall n \geqslant N,\ x \in [δ, +∞)
$$
Thus the given series converges uniformly for $x \in [δ, +∞)$.
|
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"url": "https://math.stackexchange.com/questions/2952960",
"timestamp": "2023-03-29T00:00:00",
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|
Sum of $\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - ...$ I have to find the sum of :
$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$
So far I have :
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$
which is very close to $\ln(1+x)$... but I just can't figure out what I have to do from there.
|
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)} &= \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) \, (-1)^{n-1} \, x^{n+1} \\
&= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} + \sum_{n=1}^{\infty} \frac{(-1)^n \, x^{n+1}}{n+1} \\
&= x \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} + \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} \\
&= (x+1) \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n}}{n} - x \\
&= (x+1) \, \ln(1+x) - x.
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the remainder when $13^{13}$ is divided by $25$. Find the remainder when $13^{13}$ is divided by $25$.
Here is my attempt, which I think is too tedious:
Since $13^{2} \equiv 19 (\text{mod} \ 25),$ we have $13^{4} \equiv 19^{2} \equiv 11 (\text{mod} \ 25)$ and $13^{8} \equiv 121 \equiv 21 (\text{mod} \ 25).$
Finally, we have $13^{8+4} \equiv 13^{12} \equiv 21\times 11 \equiv 231 \equiv 6 (\text{mod} \ 25)$ and hence $13^{13} \equiv 3 (\text{mod} \ 25).$
Is there a less tedious way to find the remainder? Thank you.
|
$13$ is coprime with $25$, so Euler-Fermat tells us that
$$
13^{\varphi(25)}\equiv1\pmod{25}
$$
Since $\varphi(25)=20$, this is not much of a help.
You can try with repeated squares: $13=1+4+8$; since
$$
13^2\equiv 19,\quad
13^4\equiv 19^2\equiv11\quad
13^8\equiv 11^2\equiv21
$$
we get
$$
13^{13}\equiv 13\cdot 11\cdot 21\equiv3\pmod{25}
$$
which actually is what you did. I don't think there's much easier methods.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
An analytical solution of a tricky integral Can anyone propose a method (some methods) to determine the following indefinite integral?
$$I=\int\frac{\textrm{d}x}{\sqrt[3]{\sqrt{1+x^{2}}-x}}.$$
I think an analytical solution should be possible...
|
Using the substitution $x = \sinh t$, we have $dx = \cosh t dt$.
\begin{align}
I &= \int\frac{dx}{\sqrt[3]{\sqrt{x^2 + 1}-x}}\\
&= \int\frac{\cosh t}{\sqrt[3]{\cosh t - \sinh t}}\,dt\\
&= \int e^{t/3}\cosh t \,dt\\
&= \frac{1}{2}\int\left(e^{4t/3} + e^{-2t/3}\right)\,dt\\
&= \frac{1}{2}\left(\frac{3}{4}e^{4t/3} - \frac{3}{2}e^{-2t/3}\right) + C
\end{align}
Recalling our substitution, which implies that $x = \operatorname{arsinh}t$, and using that
$$\exp(\beta\operatorname{arsinh}x) = \left(\sqrt{x^2+1}+x\right)^\beta$$
we obtain that
$$I = \frac{3}{8}\left(\sqrt{x^2+1}+x\right)^{4/3} - \frac{3}{4}\left(\sqrt{x^2+1}+x\right)^{-2/3} + C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $A=PDP^T$, does $P$ have to be orthogonal. A matrix A is called orthogonally diagonalizable if $A=PDP^{-1}$ and $A=PDP^{T}$, where $D$ is diagonal. Therefore, $P^{-1}=P^T$ and thus $P$ is an orthogonal matrix. If you are only given the fact that $A=PDP^T$, and $D$ is diagonal, is it guaranteed that $P$ is orthogonal and $A=PDP^{-1}$ also?
|
No, your assertion is not true. Choose $$
P = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix},
$$ and let $$D = I_2 = \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}$$ be the identity matrix (which is obviously diagonal) and $A$ defined by $$A = PDP^T = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix} \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}^T = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix} \begin{pmatrix}
1 & 0 \\ 1 & 1
\end{pmatrix} = \begin{pmatrix}
2 & 1 \\ 1 & 1
\end{pmatrix}.
$$
But $$P^{-1} = \begin{pmatrix}
1 & -1 \\ 0 & 1
\end{pmatrix} \neq P^T,$$
so $P$ is not orthogonal, and $$
PDP^{-1} = PP^{-1} = I_2 \neq A.
$$
|
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|
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
|
It's not hard to see that $x^4+x^3-1=0$ has two real and two complex roots. If we let these be $a$, $b$, $c+di$, and $c-di$, then we have $a+b+2c=-1$ and $ab(c^2+d^2)=-1$ from the $x^3$ and constant coefficients, $ab+(c^2+d^2)+2c(a+b)=0$ from the (missing) $x^2$ coefficient, and
$${1\over a}+{1\over b}+{1\over c+di}+{1\over c-di}=0$$
from the (missing) $x$ coefficient, which simplifies first to
$${a+b\over ab}+{2c\over c^2+d^2}=0$$
then to
$${a+b\over ab}+(1+a+b)ab=0$$
from which we obtain
$$a+b={-(ab)^2\over1+(ab)^2}$$
Plugging this into the equation $ab+(c^2+d^2)+2c(a+b)=0$, we have
$$ab-{1\over ab}+\left(1-{(ab)^2\over1+(ab)^2} \right){(ab)^2\over1+(ab)^2}=0$$
Writing $p=ab$, this simplifies first to
$${p^2-1\over p}+{1\over1+p^2}\cdot{p^2\over1+p^2}=0$$
Clearing denominators leaves $(p^2+1)(p^4-1)+p^3=0$, which expands out to
$$p^6+p^4+p^3-p^2-1=0$$
Thus $p=ab$ is one solution to $x^6+x^4+x^3-x^2-1=0$
Remark: This answer assumes that where the OP referred to $a$ and $b$ as "the" two solutions of the quartic, they meant the two real solutions, and where they asked for "the" solution of the sextic, they meant to ask for a solution.
|
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|
Finding the minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$ Minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$.
$(2+\sqrt3)^2$= $7+4\sqrt3$
$\implies$ $((2+\sqrt3)^2-7)^2$=$48$
$\implies$ $((2+\sqrt3)^4-14(2+\sqrt3)^2+1=0$
So $2+\sqrt3$ is a root of $x^4-14x^2+1$. Did i do this correct so far? Any tips on factoring this into quadratics? Thanks!
|
This strategy can be applied in lots of situations:
\begin{align}
\alpha = 2+\sqrt 3 &\implies \alpha - 2 = \sqrt 3\\
&\implies (\alpha-2)^2 = 3\\
&\implies \alpha^2 - 4\alpha + 1 = 0
\end{align}
so $\alpha$ vanishes on $x^2-4x+1$. It is irreducible since it has no rational roots, so it is the minimal polynomial.
Your $x^4-14x^2+1$ can be factored as $(x^2-4x+1)(x^2+4x+1)$ and it has roots $\pm 2\pm\sqrt 3$.
To get that factorization, you can first observe that your polynomial has no rational roots (and therefore no linear factors), so it's of the form $(x^2+ax+b)(x^2+cx+d)$. Expanding and comparing coefficients would give you factorization.
|
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|
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.
Since $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.
$$
\begin{aligned}
\lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\
&= \frac{5}{6}
\end{aligned}
$$
And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?
|
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
&& 1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9} + \cdots\\
&=&1 + \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9} + \cdots\\
&=& \int_0^1\frac{1+x^2-2x^3}{1-x^4}dx=\int_0^1\frac{(1-x)(1+x+2x^2)}{(1-x)(1+x)(1+x^2)}dx\\
&=&
\int_0^1\left(\frac{1}{1+x}+\frac{x}{1+x^2}\right)dx\\
&=&\left(\ln(1+x)+\frac{1}{2}\ln(1+x^2)\right)\Big|_0^1=\frac{3}{2}\ln 2.
\end{eqnarray*}
|
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|
Prove that $2z^4-3z^3+3z^2-z+1=0$ has exactly one complex root in each of the four quadrants.
I am trying to show that $$p(z)=2z^4-3z^3+3z^2-z+1=0$$ only has a single root in all four quadrants.
From two previously related posts, I have shown that $p(z)$ does not have a root on neither the imaginary or real axes. We also know that as the coefficients of $p(z)$ are real, then the roots of $p(z)$ occur in complex conjugate pairs. Hence we have roots $$w_1=a\pm ib \ \ \text{and} \ \ w_2=c\pm id \ \ \text{where} \ \ a,b,c,d\in\mathbb{R} \ \ \text{and} \ \ a,b,c,d\neq 0.$$
Ideally if we could show that $\Re(w_1)=-\Re(w_2)$, then we could conclude the result. But the sum of the roots is $\frac{3}{2}$ and not $0$.
How can we show only a single root exists in each quadrant?
edit
In consultation with my professor, he suggested using a corollary of Cauchy's argument principle:
"If $f\in H(\Omega)$, $\Omega$ is a domain, $\gamma:[a,b]\rightarrow\Omega$ is a simple closed contour, $f(\gamma(t))\neq 0 \ \ \forall t\in [a,b]$, then the number of zeros of $f$ in Int($\gamma$) (counting multiplicities) is equal to the number of times $f\circ\gamma$ winds around $0$".
We can consider $\gamma(t)=Re^{it}$ where $t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $R>0$. We would expect the result of this to be $2$, meaning that there is a root $w$ in the first quadrant and $\overline{w}$ in the forth quadrant.
|
Let's do a proof by contradiction. I'm going to start from Batominovski's (1-4),
\begin{eqnarray}
a+c &=& \frac{3}{4}\tag{1}\\
x+y &=& \frac{3}{2}\tag{2} -4m\\
cx + ay &=& \frac{1}{4}\tag{3} \\
xy &=& \frac{1}{2}\tag{4}
\end{eqnarray}
and assume $m = ac \ge 0$. Since $m \ge 0$, it follows from (2) that $x + y \le 3/2$, and from (4) we have
$$
x + \frac{1}{2x} \le \frac{3}{2}\Longrightarrow2x^2+1\le3x \Longrightarrow(2x - 1)(x-1) \le 0
$$
Thus $1/2 \le x \le 1$, and from (4) we have $1/2\le y \le 1$. Since, $x,y\ge 1/2$, from (3) and (1) we have
$$
\frac{1}{4} = cx + ay \ge\frac{c}{2} + \frac{a}{2} = \frac{a+c}{2} = \frac{3}{8}
$$
Since $1/4 <3/8$, we have our contradiction. $m = ac<0$, and thus $c < 0 < a$ and the four roots $a+bi$, $a-bi$, $c+di$, and $c-di$ lie in different quadrants.
|
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|
Solving $e^{2z}+e^z +1=0$ Am I doing this right?
$e^{2z}+e^z +1=0$. Let $x=e^z$ so the original equation translates to \begin{align}x^2+x+1=0\end{align}
Using the quadratic formula for real numbers, we get $\frac{-1}{2}+\frac{i\sqrt{3}}{2}, \frac{-1}{2}+\frac{i\sqrt{3}}{2}$. Equating $e^z$ with both values, let $z=r(\cos\theta +i\sin\theta)$ and we have
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=\frac{-1}{2}+\frac{i\sqrt{3}}{2}\\
e^x=1 \quad \text{and}\quad y&=\frac{2\pi}{3}+2\pi k\quad k \in \Bbb{Z}\\
x&=0 \quad \text{and} \quad y=\frac{2\pi}{3}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{4\pi}{3})+i\sin\frac{4\pi}{3})=\frac{-1}{2}+\frac{i\sqrt{3}}{2}\\
e^x=1 \quad \text{and}\quad y&=\frac{4\pi}{3}+2\pi k\quad k \in \Bbb{Z}\\
x&=0 \quad \text{and} \quad y=\frac{4\pi}{3}\\
\end{align}
Therefore, the solutions are this values, again, $\frac{-1}{2}+\frac{i\sqrt{3}}{2}, \frac{-1}{2}+\frac{i\sqrt{3}}{2}$.
|
Beware the notations ($x$) clash.
$\color{red}{\text{Suppose there exists }} z \in \mathbb{C} \text{ such that } e^{2z}+e^z+1=0.$
Let $u=e^z$. $$u^2+u+1=0$$Then, $e^z=u$, with $u\in\{e^{i\frac{2\pi}{3}},e^{i\frac{4\pi}{3}}\}$. Then, $\color{red}{\text{necessarily, }}z=i\left(\frac{2\pi}{3}+2n\pi\right) \text{ or } i\left(\frac{4\pi}{3}+2n\pi \right)$, with $n\in \mathbb{Z}.$
Conversely, $\color{red}{\text{these}}$ solutions are suitable:$\{z\in\mathbb{C}:e^{2z}+e^z+1=0\}=\{i\left(\frac{2\pi}{3}+2n\pi\right),i\left(\frac{4\pi}{3}+2n\pi \right):n\in \mathbb{Z}\}$
|
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|
how to find a, b that satisfy $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ How can I find those $a$ and $b$ in $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ ?
[my attempt] Since the denominator is $x^2$, it will be $0$. And $e^{-2x}$ is 1 so I get $1-\frac{1+ax}{1+bx} = 0$. I get $a=b$ but Im not sure this is right. Am I taking a wrong way?
|
As already notice the case $\lim_{x \to \infty} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ is trivial for $$\lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$$
we have by $e^{-2x}=1-2x+2x^2+o(x^2)$
$$\frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}
=\frac{e^{-2x}(1+bx) -(1+ax)}{x^2(1+bx)}
=\frac{(1-2x+2x^2+o(x^2))(1+bx) -(1+ax)}{x^2+o(x^2)}=$$
$$=\frac{(b-a-2)x+(2-2b)x^2+o(x^2)}{x^2+o(x^2)}$$
and the limit is zero for
*
*$2-2b=0 \implies b=1$
*$b-a-2=0 \implies a=-1$
|
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|
Study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^25^2\cdots (4n-3)^2}{3^27^2\cdots(4n-1)^2}$ I need to study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}$.
Now, I think we can do it by using the fact that if we have a series $\sum_{n=1}^{\infty}a_n$ and we can find $b_n$ so that $a_n<b_n$ then:
if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ i convergent
or
if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent
What I did so far is:
I said that $\frac{1}{3}<\frac{3}{5},\frac{5}{7}<\frac{7}{9},...,\frac{4n-3}{4n-1}<\frac{4n-1}{4n+1}$
and we get that
$\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}<\frac{3*7*...*(4n-1)}{5*9*...*(4n+1)}$
and if we multiply both sides with $\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}$
we get that
$\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}<\frac{1}{4n-1}$ but that does not really help me.
|
We are interested in
$$ \sum_{n\geq 1}\left[\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}\right]^2 $$
and by Gautschi's inequality the main term of this series behaves like $\frac{K}{n}$ as $n\to +\infty$, hence the given series is divergent by asymptotic comparison with the harmonic series. Namely $K=\frac{1}{2\pi^2}\Gamma\left(\frac{3}{4}\right)^4$.
Elementary alternative: you may prove by induction that for any $n\in\mathbb{N}^+$
$$\left(\prod_{k=1}^{n}\frac{4k-3}{4k-1}\right)^2 \geq \frac{1}{9n} $$
holds (since $\left(\frac{4n+1}{4n+3}\right)^2-\frac{n}{n+1}=\frac{1}{(n+1)(4n+3)^2}>0$) and the conclusion is just the same.
|
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|
How to calculate $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ How do I calculate the $\frac{I_{n+2}}{I_n}$ of $I_n = \int_{\frac {-\pi}{2}}^\frac{\pi}{2} cos^n \theta d\theta$ ?
[my attempt]:
I could calculate that $nI_n = 2cos^{n-1}\theta sin\theta+2(n-1)\int_0^\frac{\pi}{2}I_{n-2}d\theta$ but how do I caluclate the $I_{n+2}$?
I am stacking there...
|
\begin{align}
I_n
&= \int_{-\pi/2}^{\pi/2}\cos^n\theta\ d\theta \\
&= 2\int_{0}^{\pi/2}\cos^n\theta\ d\theta \\
&= {\bf B}\left(\dfrac{1}{2},\dfrac{n+1}{2}\right) \\
&= \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+1}{2}\right)}{\Gamma\left(\dfrac{n}{2}+1\right)} \\
\dfrac{I_{n+2}}{I_{n}}
&= \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+3}{2}\right)}{\Gamma\left(\dfrac{n+2}{2}+1\right)}
\dfrac{\Gamma\left(\dfrac{n}{2}+1\right)}{\Gamma\left(\dfrac{1}{2}\right)\Gamma\left(\dfrac{n+1}{2}\right)} \\
&= \color{blue}{\dfrac{n+1}{n+2}}
\end{align}
|
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|
find all integers $n$ such that $2^{n-1}*n+1$ is a perfect square. Clearly $n=0$
and i found that also $n=5$ gives a perfect square
And By representing the two functions , we found that there are only two solutions that are $n=0,5$
But I don't know how to prove that using elementary number theory.
|
Suppose $2^{n-1} n + 1 = x^2$ with $n > 5$. Then $2^{n-1} n = (x-1)(x+1)$ is a product of two integers that differ by $2$. Corresponding to this we must have $x-1 = 2^k u$ and $x+1 = 2^{n-1-k} v$ where $uv = n$. Now $\min(k,n-1-k) \le 2$, otherwise $x+1$ and $x-1$ would differ by at least $4$.
If $k \le 2$, $x+1 \ge 2^{n-3}$ and $x-1 \ge 2^{n-3}-2$ so
$$2^{n-1} n = (x+1)(x-1) \ge 2^{2n-6} - 2^{n-2} $$
which simplifies to
$$ n + 1/2 \ge 2^{n-5}$$
and it is easy to see that this is false if $n \ge 9$.
Similarly if $n-1-k \le 2$, $ x-1 \ge 2^k \ge 2^{n-3}$ and
$x+1 \ge 2^{n-3}+2$ so
$$ 2^{n-1} n \ge 2^{2n-6} + 2^{n-2}$$
and this is false if $n$ is too big.
|
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|
Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$ Determine the set of all $x∈ℝ$ satifying $x^2 + |x-1| ≤ 1$
My working out so far:
$|x-1|$ =
$x-1$ when $x≥1$
$1-x$ when $x<1$
Then we compute $x^2 + |x-1|$ as follows:
$x^2 + |x-1|$ =
$x^2 +x -1$ when $x≥1$
$x^2 - x + 1$ when $x<1$
So we have the following 2 cases:
i) When $x≥1$, we get $x^2 + |x-1| = x^2 +x -1$ and substituting this into the inequality $x^2 + |x-1| ≤ 1$ gives $x^2 + x-2≤0$. Hence we get $x≤1$ when $x≥1$.
ii) When $x<1$ we get $x^2 + |x-1| = x^2 -x +1$ and substituting into the inequality we get $x^2 - x +1 ≤ 1$ therfore $x^2-x ≤ 0$, factorising we get $x(x-1)≤0$, so we have $x≤0$ and $x≤1$
Not sure if this is correct, I feel there's a contradiction in the second part. I would really appreciate the help.
|
You have almost got it right but the final conclusion is as follows: for $x \geq 1$ we get $x=1$ and for $x <1$ the answer is $0\leq x <1$. Hence the final answer is $0\leq x \leq 1$.
|
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|
Geometrically proving the half-angle formula for sine from a particular diagram
I have attached a picture of the diagram I am using to prove the trig identity $\sin(\frac{\alpha}{2})= \sqrt{(\frac{1-\cos\alpha}{2})}$. I have that $\sin \alpha = \frac{DG}{OD}$ and $\sin\frac{\alpha}{2}=\frac{DE}{OD}$ as well as $\sin\frac{\alpha}{2}=\frac{EF}{OE}=EF$. I'm unsure of what segment could even represent $\sqrt{\frac{1-\cos\alpha}{2})}$?
|
Construct $DF$
let $DF$ intersect $OE$ at $P$
$\triangle OPF \cong \triangle OPD$
$\angle OPD\cong \angle OPF$ is a right angle.
$\sin \frac a2 = \frac {\text {opposite}}{\text{hypotenuse}} = \frac {PF}{OF} = \frac {PD}{OD}$
$OF = OD = \cos \frac a2\\
DP = PF = \cos \frac a2\sin \frac a2$
$DF= DP + PF = 2\cos \frac a2\sin \frac a2$
by the Pythagorean theorem $DF^2 = FG^2 + GD^2$
$OF = \cos \frac a2\\OG = OD\sin a= \cos \frac a2 \cos a\\FG = OF - OG = \cos a2 (1-cos a)\\GD = OD\sin a = \cos \frac a2\sin a$
$(2\cos \frac a2\sin \frac a2)^2 = (\cos \frac a2)^2(1-\cos a)^2 + ((\cos \frac a2)^2(\sin a)^2$
We can divide through by $(\cos \frac a2)^2$
$(2\sin \frac a2)^2 = (1-\cos a)^2 + (\sin a)^2$
$4\sin^2 \frac a2 = 1-2\cos a + cos^2 a + \sin^2 a = 2-2\cos a\\
\sin \frac a2 = \sqrt {\frac{1-\cos a}{2}}$
|
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|
Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^5-k)$ is true. Then there is $m \in \mathbb{Z}^+$ such that $$k^5 - k = 5m.$$ We must show that this statement is true for $n = k+1$, i.e. show that there is $\ell \in \mathbb{Z}^+$ such that $$(k+1)^5 - (k+1) = 5\ell.$$ Note that $(k+1)^5 - (k+1)$ expands as $k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$.
We can try to find a polynomial $P(x)$ such that $$(k^5-5) + P(x) =(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$$ so as to try to add $P(x)$ to both sides of the assumption. We find that
$$\begin{align}P(x) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - (k^5-5) \\
&=5k^4 + 10k^3 + 10k^2 + 5k \end{align}$$ and we can also observe that since $k\in\mathbb{Z}^+$, we have that $\frac{1}{5}P(x) = k^4 + 2k^3 + 2k^2 + k$ is a positive integer. Thus we add this to both sides of our assumption $$ \begin{align}
k^5 - k &= 5m \\
(k^2 - k) + P(x) &= 5m + P(x) \\
(k+1)^5 - (k+1) &= 5\left(m + \tfrac{1}{5}P(x)\right)
\end{align}.$$ Since $\frac{1}{5}P(x),m \in \mathbb{Z}^+$, it follows that $m + \tfrac{1}{5}P(x) \in \mathbb{Z}^+$. Thus $5 \left| \big[ (k+1)^5 - (k+1) \big] \right.$
By PMI, $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$.
My questions
*
*Is this proof valid?
*What other ways can this be proved by induction? The polynomial expansions took a while to deal with, so I was wondering if there are any alternate methods. (Just FYI, I only have college first-year-level knowledge)
|
You were close to a shorter proof after you expanded $(k+5)^5-(k+1)$, since
\begin{align*}
(k+5)^5 - (k+1) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k \\
&= (k^5-k) + (5k^4 + 10k^3 + 10k^2 + 5k),
\end{align*}
and by induction hypothesis, $5$ divides $k^5-k$ and clearly $5$ divides each term in $5k^4 + 10k^3 + 10k^2 + 5k$, so $5$ divides $(k+5)^5 - (k+1)$. The moral of the story is that sometimes it's a good idea not to cancel things out, but to try to find a way to shift them around.
Edit: I see this is effectively what you ended up with anyway, but finding ways to shorten proofs and make them clearer is a good thing to strive for.
|
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|
Prove that $\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$ is a root for $x^3+\sqrt[3]{6}x^2-1$
Prove that
$$c=\sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$
is a root for
$$P(x)=x^3+\sqrt[3]{6}x^2-1$$
Source: list of problems for math contest preparation.
I have no clue on how to approach the problem. Most surely not by direct substitution, but I'm not seeing how. Some hint will be appreciated.
|
It's not out of the question to compute
$$(1-\sqrt[3]2+\sqrt[3]4)^2=1+\sqrt[3]4+2\sqrt[3]2+2(\sqrt[3]4-\sqrt[3]2-2)=3(\sqrt[3]4-1)$$
so that
$$c^2=\left(1-\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9 \right)^2={3(\sqrt[3]4-1)\over3\sqrt[3]3}={\sqrt[3]4-1\over\sqrt[3]3}$$
and, since $\sqrt[3]{54}=3\sqrt[3]2$,
$$c+\sqrt[3]6={1-\sqrt[3]2+\sqrt[3]4+3\sqrt[3]2\over\sqrt[3]9}={1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}$$
Finally,
$$(\sqrt[3]4-1)(1+2\sqrt[3]2+\sqrt[3]4)=\sqrt[3]4+4+2\sqrt[3]2-1-2\sqrt[3]2-\sqrt[3]4=3$$
tells us that
$$c^3+\sqrt[3]6c^2-1=c^2(c+\sqrt[3]6)-1={\sqrt[3]4-1\over\sqrt[3]3}\cdot{1+2\sqrt[3]2+\sqrt[3]4\over\sqrt[3]9}-1={3\over3}-1=0$$
|
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|
Show that any root $z$ of $z^4+z+3=0$ satisfies $|z|>1$ Show that any root $z$ of $z^4+z+3=0$ satisfies $|z| > 1$
My working is as such:
Let $|z| \leq 1$.
Then consider $|z^4+z| \leq |z^4| + |z| \leq 1 + 1 = 2$.
So, $|z^4 + z| \leq 2$.
Thus $-2 \leq z^4 + z \leq 2$. So $1 \leq z^4+z+3 \leq 5$.
And so $z^4+z+3 \neq 0$.
Hence if $|z| \leq 1$ there is no root $z$, and so any root $z$ must satisfy $|z| < 1$.
Is this correct?
|
It's better to use Rouché's theorem, with $f(z)=z^4+z$ and $g(z)=3$ then on $z=1$ we have
$$|f(z)|=|z^4+z^3|\leq2<3$$
then the number of zeros $z^4+z+3$ and $3$ are the same.
|
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|
The distances from a point to the corners of a rectangle are $6$, $7$, $9$, and (integer) $d$. Find $d$.
Christina is standing in a rectangular garden. Her distances from the corners of the garden are $6$ meters, $7$ meters, $9$ meters, and $d$ meters, where $d$ is an integer. How to find $d$?
Can someone lend me your hand on it?
|
Let $P$ be a point inside a rectangle $XYZW$ such that $\{PX,PY,PZ\}=\{6,7,9\}$ and $PW=d$ is an integer. Suppose that the projections of $P$ onto $XY$, $YZ$, $ZW$, and $WX$ are $A$, $B$, $C$, and $D$, respectively. Write
$$x:=XA=WC\,,\,\,y:=YA=ZC\,,\,\,z:=YB=XD\,,\text{ and }w:=ZB=WD\,.$$
Therefore,
$$PX^2=x^2+z^2\,,\,\,PY^2=y^2+z^2\,,\text{ and }PZ^2=y^2+w^2\,.$$
This gives
$$d^2=PW^2=x^2+w^2=(x^2+z^2)+(y^2+w^2)-(y^2+z^2)=PX^2+PZ^2-PY^2\,.$$
The only possible choices for $(PX,PY,PZ)$ such that $d$ is an integer are $$(PX,PY,PZ)=(6,9,7)\text{ or }(PX,PY,PZ)=(7,9,6)\,,$$
for which $d=2$. Below is an example of a possible configuration.
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|
Solve equation in prime numbers
Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$
I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1\leq p^2 q^2$ for $p,q >2 $ but how to prove it.
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My solution
Case 1. Let $p \mod 3=q \mod 3=1$. Then $p^3+q^3+1=3=0 \mod 3 \neq p^2 q^2=1 \mod 3$
Case 2 $p \mod 3=q \mod 3=2$. Then $p^3+q^3+1=2+2+1=2 \mod 3 \neq p^2 q^2=1 \mod 3$
Сase 3. $p \mod 3=1, q \mod 3=2$. Then $p^3+q^3+1=1+2+1=1 \mod 3 \neq p^2 q^2=1 \cdot 2 =2 \mod 3$
So, must be, say $p=0 \mod 3 \implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=\left( q-2 \right) \left( {q}^{2}-7\,q-14 \right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2991909",
"timestamp": "2023-03-29T00:00:00",
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|
Prove the matrix is positive Consider the matrix $A=\begin{bmatrix}
1 & 1/2 & 1/3 &\dots &1/n \\
1/2 & 1/3 & 1/4 &\dots &1/(n+1) \\
\vdots & \vdots & \vdots & \vdots & \vdots\\
1/n & 1/(n+1) & 1/(n+2) & \dots& 1/(2n-1)
\end{bmatrix}$
Prove that $A$ is positive.
My work: $A$ is diagonalisable, symmetric but I can't seem to put these facts togheter to help me. I tried to prove by induction (a naive attempt) that the determinant of its minors is always positive but knowing $det(A^{k,k})>0$ there is no information of $det(A^{k+1,k+1}).
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tried two , Sylvester Inertia
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 \\
\frac{ 1 }{ 3 } & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
60 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
60 & 30 & 20 \\
30 & 20 & 15 \\
20 & 15 & 12 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
\frac{ 1 }{ 2 } & 1 & 0 & 0 \\
\frac{ 1 }{ 3 } & 1 & 1 & 0 \\
\frac{ 1 }{ 4 } & \frac{ 9 }{ 10 } & \frac{ 3 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
420 & 0 & 0 & 0 \\
0 & 35 & 0 & 0 \\
0 & 0 & \frac{ 7 }{ 3 } & 0 \\
0 & 0 & 0 & \frac{ 3 }{ 20 } \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } \\
0 & 1 & 1 & \frac{ 9 }{ 10 } \\
0 & 0 & 1 & \frac{ 3 }{ 2 } \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
420 & 210 & 140 & 105 \\
210 & 140 & 105 & 84 \\
140 & 105 & 84 & 70 \\
105 & 84 & 70 & 60 \\
\end{array}
\right)
$$
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{
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"url": "https://math.stackexchange.com/questions/2996169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
maximal surface of a cone inscribed in a sphere with radius 1 Amongst all the cones inscribed in a sphere of radius 1, I have to find that one of maximal surface.
Every cone is characterized by a value for α, the opening angle of the cone,
with $\alpha \in [0, \frac{\pi}{2}]$ (the angle between the vertical line passing through the centre of the sphere and the apotema con the cone)
Applying some properties of the rectangular triangles:
$r$ = radius of the base = $2\sin \alpha \cos \alpha$
and $h$ = height of the cone = $2 (\cos \alpha )^2$.
Then the surface of the cone is:
$$S(\alpha)=4 \pi \sin \alpha (\cos \alpha)^2+4 \pi (\sin \alpha)^2 (\cos \alpha)^2=4 \pi (\sin \alpha (\cos \alpha)^2+(\sin \alpha)^2 (\cos \alpha)^2)=
4 \pi (-(\sin \alpha)^4-(\sin \alpha)^3+(\sin \alpha)^2+\sin \alpha)$$
$$S(\alpha)'=\cos \alpha (-4 (\sin \alpha)^3-3 (\sin \alpha)^2 +2\sin \alpha +1)=0 \iff \alpha=0 \lor \sin\alpha \approx 0,64 $$
$0$ is the minimum solution , while $0,64$ is our solution.
Then $S_{max}=4 \pi (-0,64^4+0,64^3+0,64^2+0,64) =14,37$
In the book the suggested solution is $\frac{\pi(107+51 \sqrt{17})}{128}$.
Can someone help me to understand how did it find it?
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I'll start from the equation $S'(\alpha)=0$. $$\begin{aligned}S'(\alpha)&=\cos \alpha (-4 (\sin \alpha)^3-3 (\sin \alpha)^2 +2\sin \alpha +1)\\&=\cos\alpha(\sin\alpha+1)(1+\sin\alpha-4\sin^2\alpha)\end{aligned}$$
Thus, equation $S'(\alpha)=0$ is equal to $\cos\alpha=0$ or $\sin\alpha=-1$ or $1+\sin\alpha-4\sin^2\alpha=0$.
First equation has solution in region $\left[0;\frac\pi2\right]$ and is equal to $\frac\pi2$. The second equation hasn't any solution in permissible region. The third equation can be solved by substitution $t=\sin\alpha$. So we get $$1+t-4t^2=0$$ and $$t_{1,2}=\frac{1\pm\sqrt{17}}8$$
Since $0\leq\alpha\leq\frac\pi2$, $0\leq\sin\alpha\leq1$ and $t_1=\frac{1-\sqrt{17}}8$ is out of permissible region. Thus, $\alpha=\frac\pi2$ and $\alpha=\arcsin\frac{1+\sqrt{17}}8$ are the only solutions of $S'(\alpha)=0$.
Finally we get $S(\frac\pi2)=0$ and $S\left(\arcsin\frac{1+\sqrt{17}}8\right)=\frac{\pi(107+51 \sqrt{17})}{128}
$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Problems with proof by induction $\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$?
$$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Prove for $n=1$: $$\frac1{1\times2}=\frac1{1+1}=\frac12$$
Hip: $$\frac1{1\times2} + \frac1{2\times3} + \dots + \frac1{n(n+1)} = \frac1{n+1}$$
Demonstration: $$\frac1{n+1} + \frac1{(n+1)(n+2)}=\dots=\frac1{(n+1)+1}$$
My problem is that I can't find the correct algebra steps to get from the beginning of the demonstration to the end of the demonstration.
|
Actually, this is incorrect. The correct answer should be $1-\frac1{n+1}$.
We can show that the above is true for $n=1$ easily. Now let us show $1-\frac1{n+1}+\frac1{(n+1)(n+2)}=1-\frac1{n+2}$
We can prove that $\frac1{n+1}-\frac1{n+2}=\frac {n+2}{(n+1)(n+2)}-\frac {n+1}{(n+1)(n+2)}=\frac{n+2-n-1}{(n+1)(n+2)}=\frac1{(n+1)(n+2)}$ and vice versa. The $\frac1{n+1}$ terms cancel out, giving us $1-\frac1{n+2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$ Can anyone give me a proof or a reference for a proof that there exists a prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$, for $n$ sufficiently large?
I am reading a lecture where the professor uses this fact but does not provide any reference.
Thank you!
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I think this gets you in the ball park.
Number of primes $<x\approx \frac{x}{ln(x)}$
$\frac{\sqrt{x}}{ln(\sqrt{x})}=\frac{2}{\sqrt{x}}\frac{x}{ln(x)}$
$\frac{x^{1/3}}{ln(x^{1/3})}=\frac{3}{x^{2/3}}\frac{x}{ln(x)}$
So it suffices to prove $(\frac{2}{\sqrt{x}}-\frac{3}{x^{2/3}})\frac{x}{ln(x)}>1$
for sufficient large x.
Perhaps the ratios could help mitigate relative error issues.
We want: $$\pi(\sqrt{x})-\pi(x^{\frac{1}{3}})>1$$
$$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$$
$$\pi(\sqrt{x})>2\pi(x^{\frac{1}{3}})$$
Taking ratios of the approximations above,
$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}\approx \frac{2}{3}x^{\frac{1}{6}}$
$\frac{2}{3}x^{\frac{1}{6}}\pi(x^{\frac{1}{3}})>\pi(x^{\frac{1}{3}})+1$
$\frac{2}{3}x^{\frac{1}{6}}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$
$\frac{2}{3}x^{\frac{1}{6}}>2$
$x>729$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\binom{n}{1}^2+2\binom{n}{2}^2+\cdots +n\binom{n}{n}^2=n\binom{2n-1}{n-1}$
Prove that
$$
\binom{n}{1}^2+2\binom{n}{2}^2+\cdots + n\binom{n}{n}^2
= n \binom{2n-1}{n-1}.
$$
So
$$
\sum_{k=1}^n k \binom{n}{k}^2
= \sum_{k=1}^n k \binom{n}{k}\binom{n}{k}
= \sum_{k=1}^n n \binom{n-1}{k-1} \binom{n}{k}
= n \sum_{k=0}^{n-1} \frac{(n-1)!n!}{(n-k-1)!k!(n-k-1)!(k+1)!}
= n^2 \sum_{k=0}^{n-1} \frac{(n-1)!^2}{(n-k-1)!^2k!^2(k+1)}
=n^2 \sum_{k=0}^{n-1} \binom{n-1}{k}^2\frac{1}{k+1}.
$$
I do not know what to do with $\frac{1}{k+1}$, how to get rid of that.
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$$k \binom nk^2=\binom nk\cdot k\binom nk$$
For $k\ge1,$ $$k\binom nk=k\cdot\dfrac{n!}{k!\cdot(n-k)!}=n\dfrac{(n-1)!}{(k-1)!\{n-1-(k-1)\}!}=n\binom{n-1}{k-1}$$
Now in the identity $(1+x)^{2n-1}=(x+1)^n(1+x)^{n-1},$ compare coefficients of $x^n$
$$\binom{2n-1}n=\sum_{k=0}^n\binom nk\binom{n-1}{k-1}$$
|
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|
$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot \cdot\frac{99}{100})<\frac{1}{10}$. Show that
$$\frac{1}{15}<(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot \cdot \cdot\cdot\frac{99}{100})<\frac{1}{10}$$
My attempt: This problem is from a text book where is introduced as: https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I knew the sum or a product of the given numbers. Then I will use AG inequality, but I don't know how.
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Let $S_n=\prod_{k=1}^n\frac{2k-1}{2k}$. From Wallis' product, we have
$$\prod_{k=1}^\infty\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)=\frac{\pi}{2}.$$
Since each term in the product above is greater than $1$, this shows that
$$\prod_{k=1}^n\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)<\frac{\pi}{2}$$
for all $n$. That is,
$$(2n+1)S_n^2=(2n+1)\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2=\prod_{k=1}^n\left(\frac{2k-1}{2k}\cdot\frac{2k+1}{2k}\right)>\frac{2}{\pi}.$$
Therefore,
$$S_n>\frac{1}{\sqrt{\pi\left(n+\frac12\right)}}.$$
Similarly, Wallis' product also implies that
$$\prod_{k=2}^\infty\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)=\frac{4}{\pi}.$$
Since each term in the product above is greater than $1$, this shows that
$$\prod_{k=2}^n\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)<\frac{4}{\pi}$$
for all $n$. That is,
$$2nS_n^2=2n\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^2=\frac12\prod_{k=2}^n\left(\frac{2k-1}{2k-2}\cdot\frac{2k-1}{2k}\right)<\frac12\left(\frac{4}{\pi}\right).$$
Therefore,
$$S_n<\frac{1}{\sqrt{\pi n}}.$$ Hence,
$$\frac{1}{\sqrt{\pi\left(n+\frac12\right)}}<S_n<\frac1{\sqrt{\pi n}}$$
for every $n$.
|
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|
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$ In a proof, the author states that it is clear that:
Given $x\geq 1$ and $ n-x \geq 1$ and finally also $n\geq 2$
$${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$$
This is not immediately clear to me. Of course, If I have $n$ objects and I split them up in $x$ and $n-x$ objects and then I choose to form pairs, amongst these
two subsets, I end up with fewer pairs than if I would consider the bigger set, so it is certainly smaller than $ n \choose 2$.
I just don't see how it is smaller than $ {n-1\choose 2}$.
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Writing them out,
${x\choose 2}+{n-x\choose 2} \leq {n-1\choose 2}$
becomes
$x(x-1)+(n-x)(n-x-1) \le (n-1)(n-2)
$
or
$x^2-x+n^2-n(x+x+1)+x^2-x
\le n^2-3n+2
$
or
$2x^2-2x
\le n(2x+1-3)+2
$
or
$2x^2-2x
\le n(2x-2)+2
$
or
$x^2-x
\le n(x-1)+1
$
or
$x(x-1)
\le n(x-1)+1
$
or
$(x-n)(x-1)
\le 1
$
which is true for
$x \ge 1$
and
$x \le n-1$.
This is false for $x=0$
or
$x=n$
where this algebra does not work.
|
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|
Find all the intervals in which $ -x^4 + x + 3 \ge 0 $ How do I find all the intervals in which $$-x^4 + x + 3 \ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - x\sqrt[3]{4})(1 + x \sqrt[3]{4} + (x \sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/\sqrt[3]{4}$.
For $4^\frac{2}{3}x^2 + x\sqrt[3]{4} + 1$ we have $\Delta = 4^\frac{2}{3}-4\cdot4^\frac{2}{3}\cdot1 < 0 $ which means that there are no solutions in $\mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/\sqrt[3]{4}, \infty)$ and increasing on $(-\infty, 1/\sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
|
You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0\implies x^3=\frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $\dfrac1{\sqrt[3]4}$ we find
$$-\dfrac1{4\sqrt[3]4}+\dfrac1{\sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of $\dfrac1{\sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.