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Find all $Z\in C$ such that $|z|=2$ and $Im(z^6) = 8Im(z^3).$ Practice problem from a college textbook.
Find all $Z\in\mathbb C$ such that $|z|=2$ and $Im(z^6) = 8Im(z^3).$
My method is:
$2^6\sin6\phi = 8\times2^3\sin3\phi$
$\sin6\phi = \sin3\phi$
$\sin3\phi = 2\sin3\phi\cos3\phi$
$\cos3\phi = \frac{1}{2}$
$3\phi = \frac{\pi}{3} , \frac{5\pi}{3}$
Yielding:
$\phi_{1,2,3} = \frac{\frac{\pi}{3}+2k\pi}{3} , k=0,1,2$
$\phi_{4,5,6} = \frac{\frac{5\pi}{3}+2k\pi}{3} , k=0,1,2$
This turned out wrong, because the solution provided by the textbook is
$z_1 = 2cis\frac{7}{9}\pi , z_2 = 2cis\frac{5}{9}\pi ,z_3 = 2cis\frac{2}{3}\pi .$
There are only three solutions according to the textbook.
Where did I go wrong?
|
The key equation here, as you write, is $$\sin 6\phi-\sin3\phi=0.$$
We can apply the formula for the difference of $\sin$ functions:
$$\sin 6\phi-\sin3\phi=2\sin(3\phi/2)\cos(9\phi/2)=0,$$
which immediately gives us all possible solutions:
$$\sin(3\phi/2)=0\Rightarrow \phi = \frac 23 \pi k,\quad k\in \Bbb Z, $$
$$\cos(9\phi/2)=0\Rightarrow \phi = \frac 19 \pi +\frac 29 \pi k,\quad k\in \Bbb Z, $$
so, by limiting $\phi$ to the interval $[0,2\pi)$ we obtain possible values
$$0,\quad\frac 23 \pi ,\quad\frac 43 \pi,$$
and
$$\frac 19 \pi,\quad \frac 13 \pi,\quad\frac 59 \pi,\quad\frac 79 \pi,\quad \pi,\quad\frac {11}{9} \pi,\quad\frac {13}{9} \pi,\quad\frac {5}{3} \pi,\quad\frac {17}{9} \pi.$$
You can also write these solutions as
$$0,\quad\pm\frac 23 \pi,\quad\pm\frac 19 \pi,\quad \pm\frac 13 \pi,\quad\pm\frac 59 \pi,\quad\pm\frac 79 \pi,\quad \pi.$$
|
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|
show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$ Let $a,c>0$ show that
$$\dfrac{c^2}{a}+\dfrac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$$
It seem AM-GM inequality.How?
|
Certainly an Overkill
$$\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac} \geq 9(a+c)$$
$$\Longleftrightarrow \left(\frac{c^2}{a}-a\right)+\left(\frac{a^2}{c}-c\right)- 8(\sqrt{c}-\sqrt{a})^2 \geq 0$$
$$\Longleftrightarrow (\sqrt{c}-\sqrt{a})^2 \left[(\sqrt{c}+\sqrt{a})^2 (c+a) -8ac \right] \geq 0$$
$$\Longleftrightarrow (\sqrt{c}+\sqrt{a})^2 (c+a) -8ac \geq 0$$
$$\Longleftrightarrow (a+c)^2+2(a+c)\sqrt{ac}-8ac \geq 0 \tag{*}$$
By AM-GM Inequality
$$(a+c)^2 \geq 4ac \tag{1}$$
$$2(a+c)\sqrt{ac} \geq 4ac \tag{2}$$
Thus, we see that $(*)$ follows from $(1)$ and $(2)$
|
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|
Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$ How many integer pairs $ (x, y) $ satisfy $$x^2 + 4y^2- 2xy -2x -4y -8 = 0 ?$$
As it seems at first glance, and hence, my obvious attempt was to make perfect squares but the $2xy$ term is causing much problem.
|
As a comment has said, $$(x-y-1)^2 + 3(y-1)^2 = 12$$
We have $a^2\ge 0$ for all $a\in\mathbb R$, therefore $(x-y-1)^2\ge 0$ and $(y-1)^2\ge 0$.
Notice that if $(y-1)^2\ge 9$, then $$(x-y-1)^2 + 3(y-1)^2 \ge 0+3\cdot 9>12$$
You'll get $(y-1)^2\in\{0,1,4\}$, i.e. $y-1\in\{0,\pm 1,\pm 2\}$,
i.e. $y\in\{-1,0,1,2,3\}$. Checking these $5$ cases leaves $5$ quadratic equations.
|
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|
Finding roots of $3^x+4^x+5^x-6^x=0$.
How many real roots of $3^x+4^x+5^x-6^x=0$ exist?
Could anyone please tell both the graphical and a non graphical way?
For graphical way
I am not even able to find critical points.
Could just find one root by hit and trial that is $3$.
Answer is given to be one real root.
thanks in advance
|
As indicated in Joey's comment, the function is monotonically decreasing from $\infty$ to $-1$, and so there is only
one solution, the $x=3$ you already found.
However, if the input data were different it might have been difficult to estimate a solution, around which to graph or start an iteration.
Therefore let me take this as an example to expose a possible approach that
leads to a polynomial approximation that might be useful to find a first "location" of the zero.
$$
\begin{gathered}
1 = \left( {\frac{{4 - 1}}
{6}} \right)^{\,x} + \left( {\frac{4}
{6}} \right)^{\,x} + \left( {\frac{{4 + 1}}
{6}} \right)^{\,x} \hfill \\
\left( {1 + \frac{1}
{2}} \right)^{\,x} = 1 + \left( {1 - \frac{1}
{4}} \right)^{\,x} + \left( {1 + \frac{1}
{4}} \right)^{\,x} = 1 + \left( {1 + \frac{1}
{4}} \right)^{\,x} \left( {1 + \left( {\frac{3}
{5}} \right)^{\,x} } \right) \hfill \\
\sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)\left( {\frac{1}
{2}} \right)^{\,k} } = 1 + \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
x \\
k \\
\end{gathered} \right)\left( {1 + \left( { - 1} \right)^{\,k} } \right)\left( {\frac{1}
{4}} \right)^{\,k} } = 1 + 2\sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
x \\
2k \\
\end{gathered} \right)\left( {\frac{1}
{4}} \right)^{\,2k} } \hfill \\
\end{gathered}
$$
Finally note that the second expression can be easily rewritten around the
approximation found ($x_0$)
$$
\left( {\frac{3}
{2}} \right)^{\,x_{\,0} } \left( {1 + \frac{1}
{2}} \right)^{\,x - x_{\,0} } = 1 + \left( {\frac{3}
{4}} \right)^{\,x_{\,0} } \left( {1 - \frac{1}
{4}} \right)^{\,x - x_{\,0} } + \left( {\frac{5}
{4}} \right)^{\,x_{\,0} } \left( {1 + \frac{1}
{4}} \right)^{\,x - x_{\,0} }
$$
and substituting the binomials with their first order development we
obtain a linear recursion in ${\delta x}$, which corresponds to
the tangent method
$$
0 = 1 + \left( {\frac{3}
{4}} \right)^{\,x_{\,0} } \left( {1 - \frac{{\delta x}}
{4}} \right) + \left( {\frac{5}
{4}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}}
{4}} \right) - \left( {\frac{3}
{2}} \right)^{\,x_{\,0} } \left( {1 + \frac{{\delta x}}
{2}} \right)
$$
|
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Shank's Baby-Step Giant-Step for $3^x \equiv 2 \pmod{29}$ Problem: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.
I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$
$$3^5 \equiv 11 \pmod {29}$$
$$3^6 \equiv 4 \pmod {29}$$
Then I have calculated $3^{-1} \equiv 10 \pmod {29}$ and started calculating second list:
$$2 \cdot 3^{-6} \equiv 2 \cdot 10^6 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-12} \equiv 2 \cdot {3^{-6}}^2 \equiv 2 \cdot 15^2 \equiv 2 \cdot 22 = 44 \equiv 15 \pmod {29}$$
$$2 \cdot 3^{-18} \equiv 2 \cdot {3^{-12}}^2 \equiv 2 \cdot 22^2 \equiv 2 \cdot 20 = 40 \equiv 11\pmod {29}$$
And now I can stop. I can see that:
$$ 3^5 \equiv 11 \pmod {29}\ and\ 2 \cdot 3^{-18} \equiv 11 \pmod {29}$$
Therefore $x = 5 + 18 = 23$
But when I plugin $x=23$ above I get that $3^{23} \equiv 8 \pmod {29}$.
So where am I wrong?
|
$$3^x \equiv 2 (\text{ mod } 29)$$
$\varphi(29) = 28$
$m = \lceil \sqrt{28} \rceil = 6$
$\begin{array}{c|ccccc}
j& 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
3^j & 1 & 3& 9 & 27 & 23 & \color{#f00}{11}
\end{array}$
$3^{-6}= 3^{28 -6 } = 3^{22} = 22$
$\begin{array}{c|ccccc}
i & 0 & 1 & 2\\
\hline
2\cdot 22^i & 2 & 15 & \color{#f00}{11}
\end{array}$
Also... $$x = i\cdot m + j = 2\cdot 6 + 5 = 17 $$
$$3^{17} \equiv 2 (\text{ mod } 29)$$
See Baby-step Giant-step
|
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|
Combinations with a condition of limited length If there was a mold of size $N$ and $N$ uniquely sized bricks $(N, N-1,\cdots, 1)$, in how many ways the bricks could be places into the mold? The order doesn't matter.
For examples, for $N = 5$ there are $9$ possible combinations:
$[5], [4+1], [4], [3+2], [3+1], [3], [2+1], [2], [1].$
Note: duplicates (e.g. [2+2], [1+1]) are not allowed because all the bricks are unique. Here are some more combinations with small N: $1 \to 1, 2 \to 2, 3 \to 4, 4 \to 6, 5 \to 9, 6 \to 13, 7 \to 18, 8 \to 24, 9 \to 32$...
|
To answer this question, start by considering how many combinations there are of non-negative integers $n$ and $m$ such that $n+m=N$:
If $N$ is even we have $\frac{N+2}{2}$ combinations. If $N$ is odd we have $\frac{N+1}{2}$ combinations.
To get all possible combinations sum together the result above for all numbers going from $N$ to $1$:
If $N$ is even we have
\begin{align}C_{even}&=\frac{N+2}{2}+\frac{N-1+1}{2}+\frac{N-2+2}{2}+\frac{N-3+1}{2}+\cdots+\frac{2+2}{2}+\frac{1+1}{2}\\
&= \frac{1}{2}\left(\frac{(N+1)}{2}\times N+\frac{N}{2}\times 2+\frac{N}{2}\times 1\right)\\
&= \frac{N(N+1)+3N}{4}.
\end{align}
If $N$ is odd we find
$$C_{odd}= \frac{N(N+1)+3N-1}{4}.$$
|
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|
Two factoring problems that I can't figure out I have two problems I can't seem to factor.
*
*$8a^3+27b^3+2a+3b$
I first tried it by grouping. I grouped the first two terms together and used sum of two cubes to get
$$(2a+3b)(4a^2-6ab+9b^2)$$ and then I added the leftover $2a+3b$ to get:
$$(2a+3b)(4a^2-6ab+9b^2) + 2a+3b$$
But the answer is $(2a+3b)(4a^2-6ab+9b^2 + 1)$ and I'm not quite sure how you got the $+1$ from the $2a+3b$.
*$x^3+3x^2+3x+1$
The answer is $(x+1)^3$ but I don't know how to get there. I tried factoring by grouping, but that wasn't right, and I'm stuck as to how to solve it.
Thanks!
|
.For the first question, you were very close to the answer. In fact, when you write $(2a+3b)(4a^2+9b^2-6ab) + 2a+3b$, all you need to notice is that the $2a+3b$ at the end can also be taken into the bracket, using the distributive law:
$$
(2a+3b)(4a^2+9b^2-6ab) + 2a+3b = (2a+3b)(4a^2+9b^2-6ab) + 1(2a+3b) \\ = (2a+3b)(4a^2+9b^2-6ab + 1)
$$
Using $xy+ xz = x(y+z)$, where $x=2a+3b, y = 4a^2+9b^2-6ab,z=1$.
Hence the answer follows.
For the second one, yes it is tricky, but here is the catch, or rather a heuristic in these kind of problems:
Suppose you are asked to factorize some complicated cubic, say $x^3+ax^2+bx+c$, or any other complicated expression. Then, most likely, the roots are integers, which are factors of the constant in the expression.
For example, here $x^3+3x^2+3x+1$ is given, and the constant is $1$. Hence, the roots, if they are integers, are all factors of $1$. Then, they can only be $-1$ or $1$. You can use the remainder theorem to check which of these is a factor, by substituting $-1$ and $1$ in the expression and checking when the value is zero. Turns out for $-1$ that the expression is $0$, so $(x-(-1)) = (x+1)$ is a factor of your expression. Then, go on and divide by $1$, you will get $x^2+2x+1$, which is $(x+1)^2$. So collecting everything, we have $x^3+3x^2+3x+1=(x+1)^3$.
It's good to do one more example. Suppose someone asks you to factor $x^3-x^2-4x+4$. Using the heuristic, you know that the roots must divide $4$. Hence, the roots can only be $\pm 1,\pm 2,\pm4$. Use remainder theorem, and check that substituting $1$ gives $0$. Hence $x-1$ is a factor, and on dividing you get $x^2-4$, which is $(x-2)(x+2)$. Hence the complete answer is $(x-1)(x+2)(x-2)$.
|
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|
For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roots are weighing me down! And $k$? My mind's not where it's supposed to be today. Thanks in advance for posting a solution!
|
Hint :
$$\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}=\frac1{\sqrt{2x+5}+\sqrt{x+7}}\xrightarrow[x\to2]{} ...$$
|
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
|
hint: $a^4+b^4 = a^4+2a^2b^2+b^4-2a^2b^2=(a^2+b^2)^2 - 2a^2b^2$
|
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|
How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd? How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd?
I know that it will always be whole, and that it will never be $\equiv3$ if $n$ is even. But I want to know how to prove that it will always be $\equiv3$ if $n$ is odd...
|
$(a \pm b)^n = a^n \pm n*a^{n-1}b + ...... + (-1)^i*{n \choose i}a^ib^{n-1}+ ....$.
So $(a -b)^n + (a+ b)^n = 2a^n + (n - n)a^{n-1}b + ...... (1 + (-1)^i){n \choose i}a^ib^{n-1}+....$.
$1 + (-1)^i = 0$ if $i$ is odd; $2$ if $i$ is even.
so $(a -b)^n + (a+ b)^n = 2(a^n + {n \choose 2}a^{n-2}b^2 + ...... {n \choose 2k}a^{n-2k}b^{2k}+....)$
So $N= \frac{(7 + \sqrt{48})^n + (7-\sqrt{48})}{2}= \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}\sqrt{48}^{2k} = \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}48^k$.
${n \choose 2k}7^{n-2k}48^k \equiv 0 \mod 4$ if $k \ne 0$. (Because $48\equiv 0 \mod 4$).
So $N= \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}48^k \equiv {n \choose 0}7^n*48^0 = 7^n \equiv (-1)^n \mod 4$.
If $n$ is even $N \equiv (-1)^n \equiv 1 \not \equiv 3 \mod 4$.
If $n$ is odd $N \equiv (-1)^n \equiv -1 \equiv 3 \mod 4$
|
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How to find $\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$ without using L'Hospital's Rule? I have to find $$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$
We are not allowed to use L'Hospital's rule, any suggestions would be beneficial!
I have tried multiplying by the conjugate (both the numerator and the denominator).
I have tried using trig substitution.
|
Using the definition of the derivative of $\sin(x)$ at $x = \dfrac{\pi}{6}$, we get
\begin{align}
\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}
&=\lim_{x \to \frac{\pi}{6}}
\frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}} \\
&= \sin'(\frac{\pi}{6}) \\
&= \cos(\frac{\pi}{6}) \\
&= \dfrac{\sqrt 3}{2}
\end{align}
|
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|
Rational equation my question
İf $ x+\frac{1}{x^2}=3$
then find $( x^2 -\frac{1}{x})^2 $ .
I tried factoring, taking squares of both sides and some other things that did not work. what should i do?
|
Contrary to $x+\frac{1}{x^2}=3$, the expresion $x-\frac{1}{x^2}$ cannot be constant because if $(x-\frac{1}{x^2})^2=a^2$ then one would have
$$x^3-3x^2+1=0\\x^3\mp ax^2-1=0$$ from which $$x=\pm\sqrt{\frac{2}{3\mp a}}$$ and this is not compatible with the three distinct possible (real) values of $x$.
An easy calculation gives $$\left(x-\frac{1}{x^2}\right)^2=\left(3-\frac{2}{x^2}\right)^2$$
|
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|
Pythagorean Identities Simplify.
$\tan^2\frac{\pi}{3}-\cot(\frac{\pi}{4})+2\csc(\frac{\pi}{6})$
I am trying to help my little sister with homework and am rusty on my Pythagorean Identities.
|
$$
\tan^2\frac{\pi}{3}-\cot(\frac{\pi}{4})+2\csc(\frac{\pi}{6})=\tan^2\frac{\pi}{3}-\frac{1}{\tan\frac{\pi}{4}}+2\frac{1}{\sin\frac{\pi}{6}}=\\ =\sqrt{3}^2-1+2\cdot 2=3-1+4=6
$$
|
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|
Orthodiagonal quadrilateral Is it always possible to construct an orthodiagonal quadrilateral such that the diagonals and perimeter are fixed? More specifically, given 2 fixed diagonals, how is the perimeter of the quadrilateral bounded?
|
In the above configuration, let $OA=a,OB=b,OC=c,OD=d$ and assume $b>d$.
We may show that if both $b$ and $d$ are replaced by $\frac{b+d}{2}$ (that brings $B$ to $B'$ and $D$ to $D'$), the perimeter of $ABCD$ decreases. This is a consequence of the inequalities
$$ \sqrt{a^2+b^2}+\sqrt{a^2+d^2}\geq 2\sqrt{a^2+\left(\frac{b+d}{2}\right)^2} $$
$$ \sqrt{c^2+b^2}+\sqrt{c^2+d^2}\geq 2\sqrt{c^2+\left(\frac{b+d}{2}\right)^2} $$
that follow from the convexity of the functions $g_a(x)=\sqrt{a^2+x^2}$ and $g_c(x)=\sqrt{c^2+x^2}$.
The minimum perimeter is so achieved by the rhombus and the maximum perimeter is achieved by a degenerate orthodiagonal quadrilateral that is a right triangle. If we call $d_1$ and $d_2$ the lengths of the diagonals,
$$\boxed{ 2\sqrt{d_1^2+d_2^2}\leq p(ABCD)\leq d_1+d_2+\sqrt{d_1^2+d_2^2} } $$
follows from the Pythagorean theorem. By continuity, any perimeter belonging to such interval is achievable.
|
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An inequality where we have to show that is works for all positive x, and describe equality and what Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I was thinking about multiplying both sides by x and then squaring both sides, but I don't think I have the right idea...
|
Multiplying by $x$ and squaring both sides gives
$$
4x^3 + 4x^{\frac{3}{2}} + 1 \geq 9x^2 \Leftrightarrow
$$
$$
4x^3 + 4x^{\frac{3}{2}} + 1 - 9x^2 \geq 0.
$$
I do not see an easy way to factor this, but wolfram tells us that the factors are
$$
(\sqrt{x} - 1)^2(2\sqrt{x} + 1)(2x^{\frac{3}{2}} + 3x + 1) > 0.
$$
I'd suggest the following approach instead.
We have that
$$
\frac{1}{x} \geq 3 - 2\sqrt {x} \Leftrightarrow
$$
$$
\frac{1}{x} + 2\sqrt{x} \geq 3.
$$
Let
$$
f(x) = \frac{1}{x} + 2\sqrt{x}, \ x \in (0,\infty)
$$
then
$$
f' (x) = \frac{-1}{x^2} + \frac{1}{\sqrt{x}}.
$$
Set equal to zero, which gives
$$
x^{\frac{3}{2}} = 1
$$
$$
f'(1/2) < 0 , f(2) > 0
$$
Hence we have a min, at $x=1$ for $x \in (0, \infty)$
$$
f(1) = 1 + 2\sqrt{1} = 3.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that $\tan{(\pi/7)} \tan{(2\pi/7)}\tan{(3\pi/7)}=\sqrt{7}$ I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do
|
If $\theta = \frac{k\pi}{7}$ where $k = 1, 2, 3$, then $7\theta = k\pi$ and hence $4\theta = k\pi - 3\theta$. Thus $\tan(4\theta) = -\tan(3\theta)$. Expanding, and writing $t = \tan\theta$, we get
$$ \frac{4t-4t^3}{1-4t^2 + t^4} = -\frac{3t-t^3}{1-3t^2} $$
Simplifying we get
$$t^6 - 7t^4 + \cdots -7 = 0$$
The roots are $\tan \pi/7, \tan 2\pi/7, \ldots, \tan 6\pi/7$. Noting that $\tan 2\pi/7 = - \tan 5\pi/7, \tan 4\pi/7 = -\tan 3\pi/7$ etc, the roots are $\pm \tan \pi/7, \pm \tan 2\pi/7, \pm \tan 3\pi/7$. Thus the product of the roots is
$$-\tan^2\pi/7 \tan^2 2\pi/7 \tan^2 3\pi/7 = -7$$ and since $\tan k\pi/7$ for $k = 1, 2, 3$ are all positive, we get
$$ \tan \pi/7 \tan 2\pi/7 \tan 3\pi/7 = \sqrt{7}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
Suppose that $$ax^3+8x^2+bx+6=(x^2-2x-3)(ax-2).$$ And
$$(x^2-2x-3)(ax-2)=ax^3-(2a+2)x^2+(4-3a)x+6.$$
So clearly $a=-5, b=19$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving a determinant equality
Prove
$$\begin{vmatrix}
2bc-a^2 & c^2 & b^2 \\
c^2 & 2ac-b^2 & b^2 \\
b^2 & a^2 & 2ab-c^2\\
\end{vmatrix}
=(a^3+b^3+c^3-3abc)^2$$
My attempt:
I tried using the well-known result that
$$\begin{vmatrix}
bc-a^2 & ca-b^2 & ba-c^2 \\
ac-b^2& ab-c^2 & bc-a^2 \\
ba-c^2& bc-a^2 & ca-b^2\\
\end{vmatrix}
=\begin{vmatrix}
a & b & c \\
b& c & a \\
c& a & b\\
\end{vmatrix}^2
=(a^3+b^3+c^3-3abc)^2$$
But I tried using many properties of determinant but I was unable to bring the l.h.s. to the form $$\begin{vmatrix}
bc-a^2 & ca-b^2 & ba-c^2 \\
ac-b^2& ab-c^2 & bc-a^2 \\
ba-c^2& bc-a^2 & ca-b^2\\
\end{vmatrix}$$
Please help this one out. Thanks.
|
Hint: Observe
\begin{align}
\begin{bmatrix}
2bc-a^2 & c^2 & b^2\\
c^2 & 2ac-b^2 & a^2\\
b^2 & a^2 & 2ab-c^2
\end{bmatrix}
=
\begin{bmatrix}
a & b & c\\
b & c & a\\
c & a & b
\end{bmatrix}
\begin{bmatrix}
-a & -b & -c\\
c & a & b\\
b & c & a\\
\end{bmatrix}
\end{align}
Edit: Just take the determinant.
|
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|
Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatorics identitie.
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$
Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$
I think it's beautiful. I just wanted to ask, do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.
|
Nice proof! To address your question,
Do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.
You are right to be worried about circularity! However, which concepts or theorems depend on which others is a matter of certain flexibility. Often, we take one thing to be the definition of a concept, and then have to prove the other properties as theorems -- but we could alternately have used some other property as a definition, and then the original definition would have to be a theorem.
Specific to your case, I have seen definitions of $\sin$ and $\cos$ where we start by defining $\sin$ using arclength, then we define $\cos x$ to satisfy $\cos^2 x + \sin^2 x = 1$. If we take this approach, certainly, there is nothing to prove. However, this is not the only possible approach! It is also common to define $\sin$ and $\cos$ using their Taylor series. Under this approach, you have given a very nice proof that $\sin^2 x + \cos^2 x = 1$.
Your proof could also be valid if we define $\sin$ and $\cos$ to be a basis of functions satisfying $f''(x) = -f(x)$.
In summary, it depends on what you define $\sin$ and $\cos$ to be; however, your proof is not necessarily circular. And it is a nice example of deriving a result about some functions from their Taylor series.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.
|
If no two of $x,y,z$ are equal then all of them are unequal. Assume $x>y>z.$ From $(1) and (2)$ $y - z + y(x - z) = 0 $. This implies $ y < 0 $(since $y - z > 0, x - z > 0)$.....$(4)$. From $(2) and (3)$:$ z -x + z(y - x) = 0 $
This implies $z < 0$ (since $z - x < 0, y - x < 0)$ $......(5)$.
From $(3) and (1)$: $x - y + x(z - y) = 0 $
This implies $x > 0$(since $x - y > 0, z - y < 0$)$.........(6)$
(4) and (6) contradicts (1), which has (1 + x)y = 1
→ our assumption that $x, y, z$ are all unequal is incorrect and
some two of them are equal. If $x = y$, $(2)$ becomes
$(1 + x)z = 1 = (1 + z)x$ (by $(3)$)
→$1 - z = xz = 1 - x → z = x→ x = y = z$.
If $y = z$, it similarly follows that $x = y = z$.
So we have: $tan²A = tan²B = tan²C$
→$A = B = C →sinA = sinB = sinC$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Combinatorial sum identity for a choose function $\sum\limits_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ I want to show that the following holds:
$$\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}.$$
I have an idea of what is going on here. On the RHS we are selecting $r+s+1$ elements from the set $[m+n+1]$. Another way of doing this is to partition the set $[m+n+1]$ into two sets in several ways by shifting $k$ elements from one set to the other. For example, we can partition $[m+n+1]$ into a set with $m+k$ elements and a set with $n+1-k$ elements. From the set with $m+k$ elements we can select $r$ elements and from the set with $n+1-k$ elements we can select $s+1$. However, this is not matching the LHS. I believe the problem is that we don't know which $k$ elements are being shifted around.
I thought rewriting the sum as follows would help:
$$\sum_{k=0}^{m+n} \binom{k}{r} \binom{n+m-k}{s}.$$
I just don't understand why the "$+1$" on the top and bottom of the RHS disappear. I have spent several hours on this problem, and I have run out of ideas. Please help, I truly appreciate it.
|
Suppose we seek to evaluate
$$\sum_{k=-m}^n {m+k\choose r} {n-k\choose s}
= \sum_{k=0}^{m+n} {k\choose r} {m+n-k\choose s}.$$
Here we may assume that $r$ and $s$ are non-negative. We introduce
$${m+n-k\choose s} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-k-s+1}} \frac{1}{(1-z)^{s+1}} \; dz.$$
This integral controls the range, being zero when $k\gt m+n$ and we
may extend the range of $k$ to infinity. We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-s+1}} \frac{1}{(1-z)^{s+1}}
\sum_{k\ge 0} {k\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-s+1}} \frac{1}{(1-z)^{s+1}}
\sum_{k\ge r} {k\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-s+1}} \frac{z^r}{(1-z)^{s+1}}
\sum_{k\ge 0} {k+r\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-s+1}} \frac{z^r}{(1-z)^{s+1}}
\frac{1}{(1-z)^{r+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-r-s+1}} \frac{1}{(1-z)^{r+s+2}}
\; dz.$$
We get for the answer
$${m+n-r-s+r+s+1\choose r+s+1}
= {m+n+1\choose r+s+1}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$
If $A=\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}$ and $B = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and $a,b,c>0.$ Then prove that $A\geq 1+B$
$\bf{My\; Try::}$We can write $A$ and $B$ as $$=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} = \frac{3}{A}$$ and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{3}{B}$$
Using $\bf{cauchy \; schwarz }$ Inequality
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geq \frac{3^2}{1+a+1+b+1+c} = \frac{9}{3+a+b+c}$$
Now How can i solve after that , Help Required, Thanks
|
More way.
We'll rewrite our inequality in the following form
$$3\sum\limits_{cyc}\frac{1}{a(1+a)}\geq\sum\limits_{cyc}\frac{1}{1+a}\sum\limits_{cyc}\frac{1}{a}$$
which is Rearrangement.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit
I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think this is an incorrect use of the comparison test as I get divergence. I know that the limit is $\frac{1}{4}$ but I am not sure how to use the comparison test to then apply the limit.
|
Given that the rising and falling factorials are defined as
$$
\begin{array}{l}
n^{\,\overline {\,m\,} } = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\
n^{\,\underline {\,m\,} } = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\
n^{\,\underline {\, - \,m\,} } = \frac{1}{{\left( {n + m} \right)^{\,\underline {\,m\,} } }} = \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m\,} } }} \\
\end{array}
$$
where here we consider $n$ and $m$ integers,
and that it is not difficult to demonstrate that
$$
\begin{gathered}
\Delta _{\,n} \,n^{\,\underline {\,m\,} } = \left( {n + 1} \right)^{\,\underline {\,m\,} } - n^{\,\underline {\,m\,} } = m\,n^{\,\underline {\,m - 1\,} } \hfill \\
\sum\limits_{k = a}^{b - 1} {k^{\,\underline {\,m\,} } } \quad \left| {\;a < b} \right.\quad = \frac{1}
{{m + 1}}\left( {b^{\,\underline {\,m + 1\,} } - a^{\,\underline {\,m + 1\,} } } \right) \hfill \\
\end{gathered}
$$
Then
$$
\begin{gathered}
\sum\limits_{k = 1}^\infty {\frac{1}
{{k\left( {k + 1} \right)\left( {k + 2} \right)}}} = \sum\limits_{k = 0}^\infty {\frac{1}
{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}} = \sum\limits_{k = 0}^\infty {k^{\,\underline {\, - 3\,} } } = \mathop {\lim }\limits_{n\; \to \;\infty } \sum\limits_{k = 0}^{n - 1} {k^{\,\underline {\, - 3\,} } } = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {n^{\,\underline {\, - 2\,} } - 0^{\,\underline {\, - 2\,} } } \right)} \right) = \hfill \\
= \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1}
{2}\left( {\frac{1}
{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \frac{1}
{{1 \cdot 2}}} \right)} \right) = \frac{1}
{4} \hfill \\
\end{gathered}
$$
|
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"url": "https://math.stackexchange.com/questions/1941502",
"timestamp": "2023-03-29T00:00:00",
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|
$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$
Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that:
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc.
$$
Let us consider the following proofs.
$$
a^{2}+b^{2}+c^{2} \geq ab+bc+ca
$$
By the Arithmetic Mean-Geometric Mean Inequality we have
$$
a^{2}+b^{2} \geq 2ab,\ \ b^{2}+c^{2} \geq 2bc,\ \ c^{2}+a^{2} \geq 2ca \tag{1}
$$
If we add together all the inequalities $(1)$, we obtain
$$
2a^{2}+2b^{2}+2c^{2} \geq 2ab+2bc+2ca
$$
By dividing both side by $2$, the result follows.
Now let us consider,
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq 6abc \tag{2}
$$
I already have proved $(2)$
Then,
We are given,
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3 \implies bc+ac+ab=3abc \tag{3}
$$
Notice that we have
$$
a^{2}+b^{2}+c^{2} \geq bc+ac+ab=3abc
$$
So,
$$
a^{2}+b^{2}+c^{2} \geq 3abc \tag{4}
$$
Let us multiply both side of $(4)$ by $\displaystyle\frac{2}{3}$, yield
$$
\frac{2}{3}(a^{2}+b^{2}+c^{2}) \geq 2abc
$$
Here where I stopped. Would someone help me out ! Thank you so much
|
We note that
\begin{align*}
ab(a+b)+bc(b+c)+ac(a+c) &= a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\\
&=\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+
\frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\\
&\ge \frac{4(a+b+c)^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})} \qquad \mbox{(by the Schwarz inequality)}\\
&=\frac{2}{3}(a+b+c)^2\\
&=\frac{2}{3}(a^2+b^2+c^2+2ab+2bc+2ac)\\
&=\frac{2}{3}(a^2+b^2+c^2) + \frac{4}{3}(ab+bc+ca)\\
&=\frac{2}{3}(a^2+b^2+c^2) + 4abc.
\end{align*}
Here, we employed the Schwarz inequality of the from
\begin{align*}
&\ (a+b+b+c+a+c)^2 \\
=&\ \left(\frac{a}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{b}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}+\frac{b}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{b}}}\sqrt{\frac{1}{b}}+\frac{a}{\sqrt{\frac{1}{c}}}\sqrt{\frac{1}{c}}+\frac{c}{\sqrt{\frac{1}{a}}}\sqrt{\frac{1}{a}}\right)^2\\
\le&\ \left(\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+
\frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\right)\left(2\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \big)\right).
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to find analytic function?
If $v=\Large \frac{x-y}{x^2+y^2}$ then the analytic function $f(z)=u(x, y)+iv(x, y)$ is
a) $z+c\qquad $ b) $z^{-1}+c\qquad $ c) $\frac{1-i}{z}+c\qquad $ d) $\frac{1+i}{z}+c\qquad $
my try: i computed
$$\frac{\partial v}{\partial x}=\frac{y^2-x^2+2xy}{(x^2+y^2)^2}$$
$$\frac{\partial v}{\partial y}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}$$
for analytic function $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ so i get
$$\frac{\partial u}{\partial x}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}$$
$$\int \partial u=\int \frac{y^2-x^2-2xy}{(x^2+y^2)^2}\partial x$$
$$u=\int\left(\frac{1}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2} -\frac{2xy}{(x^2+y^2)^2}\right)\partial x$$
i got stuck here i don't know how to find $u$ & then $f(z)$. can somebody please give some easy method to find analytic function $f(z)$, thank you very much
|
Let us suppose $u$ is of a similar form to $v$.
$$u = \frac{ax + by}{x^2 + y^2} + c$$
$$\frac{\partial u}{\partial x} = \frac{a(x^2 + y^2) - 2x(ax+by)}{(x^2 + y^2)^2} = \frac{ay^2 -ax^2 - 2bxy}{(x^2 + y^2)^2}$$
Equating coefficients with $\frac{\partial v}{\partial y}$ gives $a = 1,b = 1$. At this point it is worth checking the other Cauchy-Riemann equation to confirm the function is indeed analytic. It remains to put the function into the correct form:
$$f(z) = \frac{x+y}{x^2 +y^2} + i\frac{x-y}{x^2+y^2} + c= \frac{(1+i)\bar{z}}{z\bar{z}} + c = \frac{1+i}{z} + c$$
|
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If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quits very ugly.
Maybe there is something nice? Thank you!
|
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}\iff\sqrt{a+b+c+15}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\right)\ge9$$
Making $c=\frac{1 } {ab}$ the expression becomes
$$f(a,b)= \left(\sqrt{\frac{a^2b}{ab^2+1}}+\sqrt{\frac{ab^2}{a^2b+1}}+\sqrt{\frac{1}{ab(a+b)}}\right)\sqrt{\frac{ab(a+b+15)+1}{ab}}\ge9$$ for all positive $a,b$.
It follows $$\sqrt{\frac{a^3b+a^2b^2+15a^2b+a}{ab^2+1}}+\sqrt{\frac{ab^3+a^2b^2+15ab^2+b}{a^2b+1}}+\frac{1}{ab}\sqrt{\frac{ab(a+b+15)+1}{a+b}} \ge9$$
It is clear $f(x,y)$ has no maximum and, in order to prove the inequality, we want to get the minimum of $f(x,y)$.
This minimum can be calculated as usually for two variables ($f_x(x,y)=0$ and $f_y(x,y)=0$, etc).
We calculate as follows: since $f(a,b)=f(b,a)$ the minimum of $f(a,b)$ is equal to the minimum of $f(a,a)$ where $a\gt 0$. Hence we calculate the minimum of the function of one variable
$$f(x,x)=2\sqrt{\frac{2x^4+15x^3+x}{x^3+1}}+\frac{1}{x^2}\sqrt{\frac{2x^3+15x^2+1}{2x}}$$
The calculation is straightforward although somewhat tedious giving the minimum $9$ for $x=1$. For further explanation, see figure below wherein the calculation (Wolfram) and the graph of the function (Desmos) confirm the result. Thus this minimum is attained with $a = b = c = 1$ and the proposed inequality is valid for all positive with $abc=1$.
|
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Fourier series of $f(x)=\pi-x$ $$f(x)=\pi-x \qquad x \in [0,2 \pi[$$
$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$
$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$
$$\frac{1}{\pi} \ \Big( \ \Big[\frac{\pi}{n} \ \sin(nx) \Big]_0^{2 \pi}-\Big[\frac{x}{n} \ \sin(nx)+\frac{1}{n^2} \ \cos(nx) \Big]_0^{2 \pi} \Big)=0 $$
$$b_n=\frac{1}{\pi} \ \Big( \ \Big[\ -\frac{\pi}{n} \ \cos(nx) \Big]_0^{2 \pi}-\Big[-\frac{x}{n} \ \cos(nx)+\frac{1}{n^2} \ \sin(nx) \Big]_0^{2 \pi} \Big)=\frac{2}{n}$$
$$f(x)=2 \sum_{n=1}^{+\infty} \frac{1}{n} \ \sin(nx)$$
Is it correct?
|
Any function $F(x)$ defined on $[0,2\pi]$ expands as the Fourier series
$$ F(x) = \dfrac{A_0}2 + \sum^\infty_{n=1}A_n\cos(nx) + \sum^\infty_{n=1} B_n \sin(nx),$$
$$\text{where} \qquad A_0 = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)dx, \quad
A_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\cos(nx)dx, \quad
B_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\sin(nx)dx. $$
Thus, for $f(x) = \pi-x$, these coefficients are computed as follows:
$$\begin{split}
a_0 &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)dx = 0, \\
a_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\cos(nx)dx
= \frac{1-\cos(2n\pi)-n\pi\sin(2n\pi)}{n^2\pi}, \\
b_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\sin(nx)dx
= \frac{2\cos(n\pi)\big[n\pi\cos(n\pi)-\sin(n\pi)\big]}{n^2\pi}.
\end{split}$$
Therefore, its Fourier series is given by
$$\sum_{n=1}^\infty \frac1{n^2\pi} \Big\{ \big[ 1 -\cos(2n\pi) -n\pi\sin(2n\pi) \big] \cos(nx) + 2\cos(n\pi) \big[ n\pi\cos(n\pi) -\sin(n\pi) \big] \sin(nx) \Big\}.$$
|
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|
Algebraic inequalities with equality condition Let $a,b,c$ be positive real numbers such that $abc(a+b+c) =3$. Prove that $$(a+b)(b+c)(c+a)\ge 8$$ and determine when equality holds.
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9uv^2-w^3\geq8$, which is a linear inequality of $v^2$
and the condition does not depend on $v^2$.
Thus it remains to prove our inequality for an extremal value of $v^2$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x-w^3=0$ or $3v^2x=-x^3+3ux^2+w^3$.
Thus, the graph of $y=3v^2x$ and the graph of $y=-x^3+3ux^2+w^3$ have three common points.
Thus, an extremal value of $v^2$ holds for the case, when a line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$, which happens for equality case of two variables.
Id est, it remains to prove our inequality for $b=a$ and the condition gives $c=\frac{\sqrt{a^4+3}-a^2}{a}$.
Thus, we need to prove that $2a\left(a+\frac{\sqrt{a^4+3}-a^2}{a}\right)^2\geq8$, which is
$a^4-4a+3\geq0$, which is AM-GM.
Done!
|
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|
Minimum value of $2^{\sin^2x}+2^{\cos^2x}$ The question is what is the minimum value of
$$2^{\sin^2x}+2^{\cos^2x}$$
I think if I put $x=\frac\pi4$ then I get a minimum of $2\sqrt2$. But how do I prove this?
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To minimise
\begin{align}
y & = 2^{\cos^2 x} + 2^{\sin^2 x} \\
\frac{dy}{dx} & = \frac{d(2^{\cos^2 x})}{dx} + \frac{d(2^{\sin^2 x})}{dx}
\end{align}
we want $\frac{dy}{dx} = 0$. For the first summand we have
\begin{align}
f(x) & = 2^{\cos^2 x} \\
\ln f(x) & = \cos^2 x\ln 2 \\
\frac{d(\ln f(x))}{dx} & = \frac{d(\cos^2 x\ln 2)}{dx} \\
\frac{f'(x)}{f(x)} & = -2\ln 2\cos x\sin x \\
\frac{d(f(x))}{dx} & = -2^{\cos^2 x}2\ln 2\cos x\sin x
\end{align}
Analogous for the second summand we have $\frac{d(2^{\sin^2 x})}{dx} = 2^{\sin^2 x}2\ln 2\sin x\cos x$ (check). Now, we simply put these back in our original equation
\begin{align}
\frac{dy}{dx} & = -2^{\cos^2 x}2\ln 2\cos x\sin x + 2^{\sin^2 x}2\ln 2\sin x\cos x \\
2^{\cos^2 x}2\ln 2\cos x\sin x & = 2^{\sin^2 x}2\ln 2\sin x\cos x
\end{align}
Here, verify that $x = k\frac{\pi}{2}, k = 1, 2, 3, \dots$ are maxima. Accounting for those, we can now safely cross out
\begin{align}
2^{\cos^2 x} & = 2^{\sin^2 x} \\
\cos^2 x & = \sin^2 x \\
\cos^2 x & = 1 - \cos^2 x \\
\cos^2 x & = \frac{1}{2} \\
\cos x & = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2} \\
x & = \arccos{\pm\frac{\sqrt{2}}{2}} \\
x & = \pm\frac{\pi}{4}
\end{align}
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|
How to approach Inequalities that can be solved using Rearrangement Inequality? I am learning Inequalities and today I've encountered the famous Rearrangement Inequality. I'll state the theorem:
Consider any two collections of real numbers in increasing order, $$a_1\leq a_2\leq....a_n$$ and $$b_1\leq b_2\leq....b_n$$ Then for any permutation ($a_1',a_2',....a_n'$) of ($a_1,a_2....,a_n$) it happens that: $$\sum_{i=1}^{n}a_ib_i\geq\sum_{i=1}^{n}a_i'b_i\geq\sum_{i=1}^{n}a_{n-i+1}b_i$$
Now, I understand what the theorem is saying, but I am finding it difficult to apply this theorem while solving problems. So I would like to know the thought process involved in solving these problems. In particular,
1) How does one identify inequalities that can be easily solved by the Rearrangement Inequality?
2)What are some standard ways in which one applies the Rearrangement Inequality? For example, we may prove the given inequality using a direct application of RI or we may use two inequalities (derived using RI) and then prove the result.
I've listed a few problems, for starters:
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1.61
$(a,b,c)$ and $(a^2,b^2,c^2)$ are the same ordered.
Thus, by Rearrangement $\sum\limits_{cyc}(a^2\cdot a)\geq\sum\limits_{cyc}a^2\cdot b$, which gives
$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$
In another hand, $(ab,ac,bc)$ and $\left(\frac{1}{c^2},\frac{1}{b^2},\frac{1}{a^2}\right)$ are the same ordered.
Thus, by Rearrangement $\sum\limits_{cyc}\left(ab\cdot\frac{1}{c^2}\right)\geq\sum\limits_{cyc}\left(ab\cdot\frac{1}{a^2}\right)=\sum\limits_{cyc}\frac{a}{c}$ or
$$a^3b^3+a^3c^3+b^3c^3\geq a^2b+b^2c+c^2a$$
and we are done!
1.62.
Let $a^2c=x$, $b^2a=y$ and $c^2b=z$.
Hence, we need to prove that $x^2+y^2+z^2\geq xy+xz+yz$,
which is Rearrangement because $(x,y,z)$ and $(x,y,z)$ are the same ordered.
1.63.
$\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$ are the same ordered.
1.64.
$(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered.
Thus, $2\sum\limits_{cyc}\frac{a}{b+c-a}\geq\sum\limits_{cyc}\left(\frac{b}{b+c-a}+\frac{c}{b+c-a}\right)=\sum\limits_{cyc}\frac{b+c}{b+c-a}$, which gives
$\sum\limits_{cyc}\frac{a}{b+c-a}\geq\sum\limits_{cyc}\frac{b+c-a}{b+c-a}=3$.
Done!
|
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|
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
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For any polynomial $f(x) = \sum\limits_{k=0}^{\deg f} a_k x^k$, we will use $[x^k] f(x)$ to denote $a_k$, the coefficient of $x^k$ in $f(x)$. Notice
$$\binom{n-k}{k} = [x^k](1+x)^{n-k} = [x^n] (x+x^2)^{n-k}$$
We have
$$\require{cancel}
\begin{align}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
&= \sum_{k=0}^n (-1)^k \binom{n-k}{k}\\
&= \sum_{k=0}^n [x^n] (x+x^2)^{n-k}(-1)^k
= [x^n]\left(\sum_{k=0}^n (x+x^2)^{n-k}(-1)^k\right)\\
&= [x^n] \frac{(x+x^2)^{n+1}-(-1)^{n+1}}{1+x+x^2}
= [x^n] \frac{[\color{red}{\cancel{\color{gray}{(x+x^2)^{n+1}}}} + (-1)^n](1-x)}{1-x^3}\\
&= (-1)^n [x^n] \frac{1-x}{1-x^3}\\
&= (-1)^n [x^n]\left((1 - x) + x^3(1-x) + x^6(1-x) + \cdots\right)
\end{align}
$$
Based on last expression, it is clear the sum depends only on $n \pmod 6$.
Its value is given by following formula:
$$\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
= \begin{cases}
+1, & n \equiv 0, 1 \pmod 6\\
-1, & n \equiv 3, 4 \pmod 6\\
0, & n \equiv 2, 5 \pmod 6
\end{cases}$$
|
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|
Solving an ordinary differential equation nonlinear I have the following
$$ y' = \frac{ xy }{x^2 + y^2} $$
My approach would be to rewrite this in terms of $x$
$$ x' = \frac{x}{y} + \frac{y}{x} $$
And then let $u = \frac{y}{x} \implies u' = \frac{x + y x'}{x^2} = \frac{1}{x} + \frac{u}{x} x' = \frac{1}{x} + \frac{u}{x} \left( \frac{1}{u} + u\right)$. Therfore,
$$ u' = \frac{1}{x} + \frac{1}{x} + \frac{u^2}{x} = \frac{2 + u^2}{x}$$
Thus,
$$ \int \frac{ d u}{2 + u^2} = \int \frac{ dx}{x} \implies \frac{1}{\sqrt{2}} \arctan (u/ \sqrt{2}) = \ln |x| + C$$
So general solution ( with $u = y/x$) is
$$ \boxed{ \arctan \left( \frac{y}{x \sqrt{2}} \right) = \sqrt{2} \ln |x| + \sqrt{2} +C }$$
is this sufficient? is there an easier way to solve this ODE?
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$$y' = \frac{ xy }{x^2 + y^2}=\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}$$
let $y=ux$
$$y'=u+xu'$$
so
$$u+xu'=\frac{u}{1+u^2}$$
$$xu'=\frac{-u^3}{1+u^2}$$
$$-\frac{dx}{x}=\frac{u^2+1}{u^3}du$$
$$-\frac{dx}{x}=\frac{1}{u}du+\frac{1}{u^3}du$$
|
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Find the sum $\sum_{n=1}^{\infty} \frac{4n}{n^4+2n^2+9}$ Find the sum
$$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$
By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that
$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2+n)}$ for all $n\ge n_0$, $n_0$ is large enough.
And $\sum_{n=1}^{\infty} \dfrac{5}{6(n^4+n^2)}= \sum_{n=1}^{\infty}\dfrac{5}{6(n^2+n)}=\dfrac{5}{6}$.
But it is not enough to confirm that the given series converges to $\dfrac{5}{6}$. Can someone help me, please? Thanks in advanced.
|
HINT:
$$(n^2)^2+3^2+2n^2=(n^2+3)^2-(2n)^2=(n^2+2n+3)(n^2-2n+3)$$
$$(n^2+2n+3)-(n^2-2n+3)=?$$
Now if $f(m)=m^2-2m+3,$
$f(m+2)=(m+2)^2-2(m+2)+3=m^2+2m+3$
|
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If $ X$ is the incentre of $\Delta ABY $ show that $∠CAD=90°$. In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$. If $X$ is the incentre of triangle $ABY$ , show that $∠CAD = 90°.$
|
Look at quadrilateral $XCYD$.
*
*$\angle\, XDY = \angle \, XAY = \alpha$ as inscribed in a circle.
*$\angle \, BAC = \angle \, BAX = \angle \, XAY = \alpha$ since $AC$ passes through the incenter $X$ of triangle $ABY$ and therefore $AC$ is the interior angle bisector of angle $\angle \, BAY$.
*$\angle \, XDC = \angle \, BDC = \angle \, BAC = \alpha$ as inscribed in a circle.
*Hence $\angle \, XDC = \angle \, XDY = \alpha$.
*Analogously, $\angle \, XCD = \angle \, XCY = \beta$.
*In triangle $CDX$ angles sum up to $180^{\circ}$ so $$180^{\circ} = \angle \, CXD + \angle \, XCD + \angle \, XDC = \angle \, CXD + \alpha + \beta$$ and thus $\angle \, CXD = 180^{\circ} - \alpha - \beta$.
*In quadrilateral $XCDY$ angles sum up to $360^{\circ}$ so $$360^{\circ} = \angle \, CXD + \angle \, XCY + \angle \, XDY + \angle \, CYD = \angle \, CXD + \alpha + \beta + \angle \, CYD = $$ $$= 180^{\circ} + \angle \, CYD $$ therefore $\angle \, CYD = 180^{\circ}$ and thus $Y$ lies on $CD$.
*$\angle \, AXD = \angle \, BXC$ since $X$ is the intersection point of $AC$ and $BD$.
*$\angle \, AYD = \angle \, AXD$ and $\angle \, BYC = \angle \, BXC$ as inscribed in corresponding circles. Hence $\angle \, AYD = \angle \, BYC$
*$\angle \, XYA = \angle \, XYB$ since $YX$ is angle bisector of $\angle \, AYC$. Hence
$$\angle \, XYD = \angle \, XYA + \angle \, AYD = \angle \, XYB + \angle \, BYC = \angle \, XYC$$
*But $180^{\circ} = \angle \, XYD + \angle \, XYC = 2 \, \angle \, XYD $ so $\angle \, XYD = 90^{\circ}$.
*As quadrilateral $AXYD$ is inscribed in a circle, $180^{\circ} = \angle \, XAD + \angle \, XYD = \angle \, XAD + 90^{\circ}$ so $\angle \, XAD = 90^{\circ}$.
*$X \in AC$ so $\angle \, CAD = \angle \, XAD = 90^{\circ}$.
|
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Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving this integral? Maybe there any way to simplify it?
|
One step that can simplify the partial fraction decomposition is to first perform an integration by parts with the choice $$u = -(2x)^{-3}, \quad du = \frac{3}{8}x^{-4} \, dx, \quad dv = -\frac{(2x)^3}{(x^4-1)^{3}} \, dx, \quad v = (x^4-1)^{-2},$$ giving $$\int \frac{dx}{(x^4-1)^3} = -\frac{1}{8x^3(x^4-1)^2} - \frac{3}{8} \int \frac{dx}{x^4 (x^4-1)^2}.$$ Partial fraction decomposition on $(x^2 (x^4-1))^{-2}$ results in a simpler integrand with lower-degree polynomial denominator terms:
$$\frac{1}{x^4(x^4-1)^2} = \frac{1}{x^4}+\frac{3}{4 \left(x^2+1\right)}+\frac{1}{4(x^2+1)^2}+\frac{7}{16 (x+1)}+\frac{1}{16 (x+1)^2}-\frac{7}{16(x-1)}+\frac{1}{16 (x-1)^2},$$ all of which are trivially integrable with the exception of the third term, which may be handled with the usual trigonometric substitution $x = \tan \theta$, $dx = \sec^2 \theta \, d\theta$, resulting in the transformed integrand $\frac{1}{4} \cos^2 \theta$, which again is easily handled.
|
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How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contained in the answer book, is as follows:
\begin{align*}
x^n-y^n &= (x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&= x(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})\\
&\qquad -[y(x^{n-1}+x^{n-2}y+{...}+xy^{n-2}+y^{n-1})]&\Rightarrow \mathbf{Equation1}\\
&=x^n+x^{n-1}y+\dotsb+x^2y^{n-2}+xy^{n-1}\\
&\qquad -[x^{n-1}y+x^{n-2}y^2+xy^{n-1}+y^n]&\Rightarrow \mathbf{Equation2}\\
&=x^n-y^n
\end{align*}
While I believe that the distributive law was used to arrive at Equation 1, I do not understand how Equation 2 was arrived at.
I have tried to solve this independently to no avail. I cannot seem to understand how $x^n$ and $y^n$ came about in Equation 2 for example.
To summarise the question, what principle was Equation 2 based upon? And how was this applied in the above problem.
|
The simplest way is to prove first
$$1-t^n=(1-t)(1+t+\dots+t^{n-1})$$
by induction.
The formula is trivial for $n=1$. So suppose the formula is valid for some $n\ge 1$, and consider the exponent $n+1$.
\begin{align}
1-t^{n+1}&=(1-t^n)+(t^n-t^{n+1})\\
&=(1-t)(1+t+\dots+t^{n-1})+t^n(1-t)\\
&=(1-t)(1+t+\dots+t^{n-1+t^n}),
\end{align}
which proves the inductive step.
Now for the general formula: set $y=tx$, and substitute in the expression:
\begin{align}
x^n-y^n&=x^n(1-t^n)=x^n(1-t)(1+t+\dots+t^{n-1})\\
&=x(1-t)\cdot x^{n-1}(1+t+\dots+t^{n-1})\\
&=(x-tx)(x^{n-1}+x^{n-2}tx+\dots +(tx)^{n-1})\\
&=(x-y)(x^{n-1}+x^{n-2}y+\dots +y^{n-1}).
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate limit involving $\sin$ function Calculate the following limit:
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$
I tried applying L'Hospital's rule, but it got too messy.
Thank you in advance!
|
Here is a more or less elementary calculation of the limit.
A quite commonly known limit is
$$
\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0 }{x-0}=\sin'(0)=\cos(0)=1.
\tag{1}\label{1}
$$
Also, you may prove with L'Hopital that
$$
\lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{1-\cos x}{3x^2}=\lim_{x\to 0}\frac{\sin x}{6x}=\frac{1}{6}\lim_{x\to 0}\frac{\sin x}{x}\stackrel{\eqref{1}}=\frac{1}{6}\cdot 1=\frac{1}{6}.
\tag{2}\label{2}
$$
Using this, we see
$$
\lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{x^3}\\
=\lim_{x\to 0}\left[\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\cdot\left(\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\right]\\
=\left(\lim_{x\to 0}\frac{\sin^{(n)}x-\sin^{(n+1)} x}{\left(\sin^{(n)}x\right)^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin^{(n)}x}{\sin^{(n-1)}x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin^{(n-1)}x}{\sin^{(n-2)}x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin^{(1)}x}{\sin^{(0)}x}\right)^3\\
=\left(\lim_{x\to 0}\frac{x-\sin x}{x^3}\right)\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdot\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\cdots\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^3\stackrel{\eqref{1}\&\eqref{2}}=\frac{1}{6}.
\tag{3}\label{3}
$$
Where $\sin^{(n)}x:=\overbrace{\sin(\sin(...\sin(x)...))}^{n\text{ times}}$ and $\sin^{(0)}x:=x$. We used the fact that if for two functions $f,g$ we have
$$
\lim_{x\substack{\to\\ \neq}a}g(x)=b\qquad\text{and}\qquad \lim_{x\substack{\to\\ \neq}b}f(x)=l
$$
and $g(x)\neq b$ in a neighborhood of $a$ then
$$
\lim_{x\substack{\to\\ \neq}a}f(g(x))=l.
$$
Finally we conclude
$$
\lim_{x\to 0}\frac{x-\sin^{(n)} x}{x^3}=\lim_{x\to 0}\left[\frac{x-\sin^{(1)}x+\sin^{(1)}x-\sin^{(2)}x+...+\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right]\\
=\left(\lim_{x\to 0}\frac{x-\sin^{(1)}x}{x^3}\right)+\left(\lim_{x\to 0}\frac{\sin^{(1)}x-\sin^{(2)}x}{x^3}\right)+...+\left(\lim_{x\to 0}\frac{\sin^{(n-1)}-\sin^{(n)} x}{x^3}\right)\\
=\overbrace{\frac{1}{6}+...+\frac{1}{6}}^{n\text{ times}}\stackrel{\eqref{3}}=\frac{n}{6}
\tag{4}\label{4}
$$
and therefore
$$
\lim_{x\to 0}\frac{x-\sin^{(150)} x}{x^3}\stackrel{\eqref{4}}=\frac{150}{6}=25.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which of following is/are true? In expansion $(x^2+1+\frac{1}{x^2})^n$ , n $\in \mathbb{N}$
1.number of terms is $2n+1$
2.coefficient of constant term is $2^{n-1}$
3.coefficient of $x^{2n-2}$ is $n$
4.coefficient of $x^2$ is $n$
I tried by taking lcm and writing as $\frac{(1+x^2+x^4)^n}{x^{2n}}$. How do i proceed? thanks
|
Usign the Newton formula we have:
$\left(x^2+1+\frac{1}{x^2}\right)^n=(x^2+1)^n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x^2)^{n-1}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x)^{2n-2}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
i.e., concule that 3. is correct.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fourier series for $\cosh(x)$ Find the odd Fourier series for the periodic function whose period is $2\pi$, and which is equal to $\cosh{x}$ in the range $0<x<\pi$. Hence show that
$\frac{\pi}{4}sech{(\frac{\pi}{2}})=\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)+(2n+1)^{-1}}$.
$a_n=\frac{1}{2\pi}\int^{\pi}_0(e^x+e^{-x})\cos{nx}$ $dx$
$=\frac{sinh (\pi)}{\pi(1+n^2)}(-1)^n$
$a_0=\frac{1}{\pi} sinh(\pi)$
$b_n=\frac{1}{2\pi}\int^{\pi}_0cosh(x). \sin{nx}$ $dx$
$b_n=\frac{\cosh(\pi)}{\pi}[\frac{2n}{1+n^2}[1-(-1)^n]]$
$1-(-1)^n =$
\begin{cases}
0, & \text{if $n$ is even} \\[2ex]
2, & \text{if $n$ is odd}
\end{cases}
$\therefore b_n=\frac{4(2k+1) cosh(\pi)}{\pi[1+(2k+1)^2]}$
For Odd series,
$cosh(x)=\frac{4}{\pi}cosh(\pi) \sum^{\infty}_{n=1}\frac{(2k+1)}{1+(2k+1)^2}\sin{nx}$
Putting $x=\frac{\pi}{2}$
$\frac{\pi}{4}[\frac{cosh(\frac{\pi}{2})}{\cosh(\pi)}=\sum^{\infty}_{n=1}\frac{1}{(2k+1)+(2k+1)^{-1}}$
I am a bit far from what is required.
P.S I need to know if all my Fourier constants are correct, because I have got other similar problems to solve, this time with $a_n$ and $a_0$
|
$$\int_{0}^{\pi}\cosh(x)\sin(nx)\,dx = \frac{n}{n^2+1}\left(1-(-1)^n \cosh\pi\right)\tag{1}$$
leads to
$$ \cosh(x) = \frac{2}{\pi}\sum_{n\geq 1}\frac{n\sin(nx)}{n^2+1}(1-(-1)^n\cosh \pi) \tag{2} $$
for any $x\in(0,\pi)$. By evaluating both sides of $(2)$ at $x=\frac{\pi}{2}$,
$$ \cosh\frac{\pi}{2}=\frac{2}{\pi}\sum_{m\geq 0}\frac{(2m+1)(-1)^m}{(2m+1)^2+1}(1+\cosh \pi)\tag{3}$$
follows, and by rearranging both sides of $(3)$, we get
$$\sum_{m\geq 0}\frac{(-1)^m}{(2m+1)+(2m+1)^{-1}} = \frac{\pi\cosh\frac{\pi}{2}}{2(1+\cosh\pi)}=\color{red}{\frac{\pi}{4\cosh\frac{\pi}{2}}}\tag{4} $$
as wanted.
|
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|
Calculation of product of cycles Let $n \in \mathbb{N}$, $n \geq 3$ and $n = 2m$ for some $m \in \mathbb{N}$. How do I efficently get the result of the product of cycles $$\begin{pmatrix} 2 & n\end{pmatrix}\begin{pmatrix} 3 & n-1\end{pmatrix}\cdots\begin{pmatrix} m & m+2\end{pmatrix}\begin{pmatrix} 1 & 2 & \dots & n\end{pmatrix}\begin{pmatrix} 2 & n\end{pmatrix}\begin{pmatrix} 3 & n-1\end{pmatrix}\cdots\begin{pmatrix} m & m+2\end{pmatrix}$$
I could write it somehow in the typical form $$\begin{pmatrix} 1 & 2 & \dots & m & m+1 & m+2 & \dots & n-1 & n\\
1 & 2 & \dots & m & m+1 & m+2 & \dots & n-1 & n\end{pmatrix}$$ but this gets really messy. Thanks.
Edit. An idea would be induction on $m$.
|
We have $(k \; n-k+2)\; (n-k+2 \; n-k+3)\; (n-k+3 \;k-1) = (k \; k-1) $ (a middle transposition comes from a $(1 \; 2 \; \ldots \; n)$ cycle ) and it could be applied to compute this product:
$k$ goes to $k-1$ hence we have a cycle:
$ (n\; n-1 \; n-2\; \ldots \; 3 \; 2 \; 1)$.
|
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|
find partial fraction $(2x^2-1)/((x^2-1)(2 x^2+3))$
Find partial fraction of $\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}$?
My attempt:
I did google and I tried to solved it as :
Let’s first get the general form of the partial fraction decomposition.
$\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}=\cfrac{A}{(x+1)}+\cfrac{B}{(x-1)}+\cfrac{C+Dx}{(2x^2+3)}$
Setting numerators gives,
$(2x^2-1)=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+C(x^2-1)$
I stuck here, how to proceed next?
Can you explain it, please?
|
you are mistaken in the second step, the expression must be simplified as
$$2x^2-1=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+(C+Dx)(x^2-1)$$
now, compare powers of $x^3, x^2, x$ & constant terms, you will get
$$2A+2B+D=0\tag 1$$
$$-2A+2B+C=2\tag 2$$
$$3A+3B-D=0\tag 3$$
$$-3A+3B-C=-1\tag 4$$
solve all four linear equations to get the values of A, B,C & D you will get
$$A=-\frac{1}{10}, B=\frac{1}{10}, C=\frac{8}{5}, D=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1) \right ) + 1(1007)
\\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\
=& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007
\\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\end{align}
This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?
|
\begin{align}&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\\&=1007\left(\frac{12(1008)(2015)}{6}-\frac{2(1008)(3)}{2}+ 1\right)
\\&=1007\left(2(1008)(2015)-(1008)(3)+ 1\right)
\\&=1007\left(2(1007+1)(2015)-(1007+1)(3)+ 1\right)
\\&=1007\left(1007(2(2015)-3)+2(2015)-3+ 1\right)
\\&=1007\left(1007(2(2015)-3)+2(2015)-2\right)
\\&=1007\left(1007(2(2015)-3)+2(2014)\right)
\\&=1007\left(1007(2(2015)-3)+4(1007)\right)
\\&=1007^2(2(2015)-3+4)
\\&=1007^2(4031)\end{align}
|
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|
Prove that $\frac{\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{200}}{\frac{1}{10*11}+\frac{1}{11*12}+\dots+\frac{1}{19*20}}>19$ Prove that $$\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dots+\dfrac{1}{19\cdot20}}>19$$
My attempt:The denominator is $\frac{1}{20}$ using telescopic series.
$\frac{1}{10*11}+\frac{1}{11*12}+\dots+\frac{1}{19*20}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\dots-\frac{1}{20}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
I think we could Prove that the top part of fraction is bigger than $1$ but I don't know how.
|
Your calculation for the denominator (telescopic sum and not series) was the hard part, since
\begin{align}\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}&>\dfrac{\dfrac{1}{200}+\dfrac{1}{200}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}=\dfrac{\dfrac{200-10}{200}}{\dfrac1{20}}=\frac{\dfrac{19}{20}}{\dfrac1{20}}=19\end{align}
|
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|
Find the set of points on the complex plane for which $z^2 + z + 1$ is real and positive
Find the set of points on the complex plane for which $z^2 + z + 1$ is real and positive.
For being real I applied the condition: $z^2 + z + 1=\bar{z}^2 +
\bar{z} + 1$
For the expression to be positive I applied:
$z^2+z+1>0 \implies (z+\dfrac{1+\sqrt{3}i}{2})(z+\dfrac{1-\sqrt{3}i}{2})>0 \implies z \in (-\infty,\dfrac{-1-\sqrt{3}i}{2}] \cup [\dfrac{-1+\sqrt{3}i}{2},\infty)$
But my doubt is it won't be correct to say $z$ is larger or smaller than a particular complex number as that makes no sense. What will be the correct method of solving this problem?
|
$$z^2+z+1=k,\ k>0$$
$$z^2+z+(1-k)=0$$
$$z=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(1-k)}}{2\cdot1}$$
$$z=\frac{-1\pm\sqrt{4k-3}}2\tag1$$
When $0<k<\frac34$, $\sqrt{4k-3}$ is imaginary and $z=-\frac12\pm\frac{\sqrt{3-4k}}2i$; the range of the imaginary part is $(-\frac{\sqrt3}2,\frac{\sqrt3}2)$. When $k\ge\frac34$, $\sqrt{4k-3}$ is real and $z$ is given by $(1)$; its range is the whole real line.
In conclusion, the range of $z$ for which $z^2+z+1$ is real and positive is
$$\Bbb R\cup\{-\tfrac12+ai\mid a\in(-\tfrac{\sqrt3}2,\tfrac{\sqrt3}2)\}$$
|
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|
Showing equivalence $(A \Rightarrow (B\Rightarrow C))\Leftrightarrow (A\wedge B) \Rightarrow C)$ My original excercise is, to use the truthtable, but do i have to say anything here?
$\begin{array}{ccc|c@{}ccc@{}ccc@{}c@{}ccc@{}c@{}ccc@{}ccc@{}c}
A&B&C&(&A&\Rightarrow&(&B&\Rightarrow&C&)&)&\Leftrightarrow&(&(&A&\land&B&)&\Rightarrow&C&)\\\hline
1&1&1&&1&1&&1&1&1&&&\mathbf{1}&&&1&1&1&&1&1&\\
1&1&0&&1&0&&1&0&0&&&\mathbf{1}&&&1&1&1&&0&0&\\
1&0&1&&1&1&&0&1&1&&&\mathbf{1}&&&1&0&0&&1&1&\\
1&0&0&&1&1&&0&1&0&&&\mathbf{1}&&&1&0&0&&1&0&\\
0&1&1&&0&1&&1&1&1&&&\mathbf{1}&&&0&0&1&&1&1&\\
0&1&0&&0&1&&1&0&0&&&\mathbf{1}&&&0&0&1&&1&0&\\
0&0&1&&0&1&&0&1&1&&&\mathbf{1}&&&0&0&0&&1&1&\\
0&0&0&&0&1&&0&1&0&&&\mathbf{1}&&&0&0&0&&1&0&
\end{array}$
To me that's self explanatory. Do i have to write anything here?
And is it possible to do it without a truthtable?
Note: This is a homework, so please no full solutions.
Anyways I'm more interested in how to do it rather then how it is as a result.
|
Yes, it is possible to do this without truth table.
$A \Rightarrow B$ means either $A$ is false (or) $B$ is true.
So, $(A \Rightarrow B)$ $\Leftrightarrow \overline{A} \lor B$
$A \Rightarrow (B \Rightarrow C) \Leftrightarrow (A \Rightarrow (\overline{B} \lor C)) \Leftrightarrow \overline{A} \lor \overline{B} \lor C$
$\color{olive}{(A \wedge B) \Rightarrow C \Leftrightarrow (A \land B \Rightarrow C) \Leftrightarrow \overline{AB} \lor C \Leftrightarrow \overline{A} \lor \overline{B} \lor C}$
Thus, both LHS and RHS are equivalent
|
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|
Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
|
.In this question, you have to attempt to complete two squares, namely:
$$
x^4-4x^3 + 12x^2-24x+24 = x^4 -4x^3+6x^2-4x+1 + 6x^2-20x+23
\\ = (x-1)^4 + 6x^2 -20x+23
$$
Now, as it turns out, we can complete the second square, and:
$$
6x^2-20x+23 = \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$
Hence, we can rewrite the whole expression as:
$$
x^4-4x^3 + 12x^2-24x+24 = (x-1)^4 + \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$
It is a sum of positive expressions, hence is always positive, hence $P(x) = |P(x)|$.
|
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|
What number must be subtracted from the denominator of $\frac{10}{23}$ to make the result $\frac13$? Is it $\frac{10}{23} - \frac{1}x = \frac13$ or something similar?
|
Given that $\frac{10}{23} - \frac{1}{x} = \frac{1}{3}$ then multiply both sides by $3 \cdot 23 \cdot x$ to obtain $3\cdot 10 \cdot x - 3 \cdot 23 = 23 \cdot x$ which becomes $x = \frac{69}{7}$. This can be seen in the following:
\begin{align}
\frac{10}{23} - \frac{1}{x} &= \frac{10}{23} - \frac{7}{69} = \frac{30}{69} - \frac{7}{69} = \frac{23}{3 \cdot 23} = \frac{1}{3}.
\end{align}
|
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|
Use a power series to approximate the definite integral to six decimal places. Use a power series to approximate the definite integral to six decimal places.
$\int_{0}^{0.3} \frac{x^2}{1+x^4}$
I'm not sure how to find the sum of this for solving when it has x's in the numerator, this is what I assumed however.
$\frac{1}{1-x} $ $\sum_{n=0}^\infty $ x^n
$\sum_{n=0}^\infty \frac{1}{1-x^4}$
$\sum_{n=0}^\infty {(-x^4)}^n$ ==> $\frac{x}{1+x^4}$
I'm not sure if this is right nor what the next step is to get the approximate definite integral.
|
$$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$
and the last series is a series with positive terms. Since
$$ \sum_{n\geq 2}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)<\frac{1}{19}\sum_{n\geq 2}\left(\frac{3}{10}\right)^{8n+3}<10^{-11} \tag{2}$$
the first eight figures of $I$ are given by the sum appearing in the RHS of $(1)$ restricted to $n=0$ and $n=1$:
$$ \sum_{n=0}^{1}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right)=\frac{3453033133161387}{385000000000000000}=\color{green}{0.00896891}72289906\ldots\tag{3}$$
|
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|
Find the minimum value of this Expression (Without Calculus!) Problem:
Suppose that $x,y$ and $z$ are positive real numbers verifying $xy+yz+zx=1$ and $k,l$ are two positive real constants. The minimum value of the expression:
$$kx^2+ly^2+z^2$$ is $2t_0$, where $t_0$ is the unique root of the equation $2t^3+(k+l+1)t-kl=0.$
My Attempt: Let $k=a+(k-a)$ and $l=b+(l-b)$, where $a,b\in \mathbb{R^+}$ and $a\leq k$ and $b\leq l.$ Then from the following Inequalities: $$ax^2+by^2\geq2\sqrt{ab}xy$$ $$(l-b)y^2+z^2/2\geq\sqrt{2(l-b)}yz$$ $$z^2/2+(k-a)x^2\geq\sqrt{2(k-a)}zx$$ We can deduce that $$kx^2+ly^2+z^2\geq2\sqrt{ab}xy+\sqrt{2(l-b)}yz+\sqrt{2(k-a)}zx.$$ Since $xy+yz+zx=1\implies 2\sqrt{ab}=\sqrt{2(l-b)}=\sqrt{2(k-a)}.$ After this stage I get two quadratic expression for $a$ and $b$, which seems far of from the main result. Where am I going wrong?
PS.Please do not use Calculus to solve this Problem.
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Let $f = kx^2+ly^2+z^2,$ suppose $f_{\min} = 2t > 0.$ We need to prove
$$kx^2+ly^2+z^2 \geqslant 2t(xy+xz+yz),$$
equivalent to
$$kx^2+ly^2+z^2+t(x^2+y^2+z^2)\geqslant 2t(xy+xz+yz)+t(x^2+y^2+z^2),$$
or
$$(k+t)x^2+(l+t)y^2+(1+t)z^2 \geqslant t(x+y+z)^2.$$
Using the Cauchy-Schwarz inequality, we have
$$(k+t)x^2+(l+t)y^2+(1+t)z^2 \geqslant \frac{(x+y+z)^2}{\frac{1}{k+t}+\frac{1}{l+t}+\frac{1}{1+t}}.$$
Therefore, the constant $t$ must be satisfy
$$\frac{1}{k+t}+\frac{1}{l+t}+\frac{1}{1+t} = \frac{1}{t},$$
or
$$2t^3+(k+l+1)t-kl=0.$$
The proof is completed.
P/s. Many years ago I created this problem (here) but thought this was an old problem.
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|
What's the sum of all the positive integral divisors of $540$?
What's the sum of all the positive integral divisors of $540$?
My approach: I converted the number into the exponential form. And found out the integral divisors which came out to be $24$.
But couldn't find the sum...Any trick?
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Prime factorization of 540 is $2^2\cdot 3^3\cdot 5$. Also sum of divisors of 540 equals:
$$\begin{align}
\sigma\left(2^2\cdot 3^3\cdot 5\right) &=\frac {2^3-1}{2-1}\cdot\frac{3^4 - 1}{3-1}\cdot \frac{5^2 -1}{ 5-1}\\ &=7\cdot 40\cdot 6\\ &= 1680
\end{align}$$
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|
Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\
&=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right)
\end{align}$$
and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.
Once the solution
$$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$
is known, the telescoping sum can be easily derived, i.e.
$$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$
However, without knowing this a priori, how would we have approached this problem?
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You may do the following to get rid of the powers of two and transform it into a standard sum:
\begin{equation} \binom{2m}{m}=\frac{(2m)!}{m!m!}=\frac{2^m m! 1\cdot3\cdot5\cdot \cdots (2m-1)}{m!m!}=2^{2m} \frac{(1/2)(3/2)\cdots (m-1/2)}{m!}=2^{2m}\binom{m-1/2}{m}~~~~~~ (1) \end{equation}
Then your sum becomes
$$\sum_{m=0}^{n} \binom{m-1/2}{m}=\binom{n-1/2 +1}{n} =\binom{n+1-1/2}{n}=\frac{n+1}{n+1-1/2-n}\frac{n+1-1/2-n}{n+1}\binom{n+1-1/2}{n}=\frac{n+1}{1/2}\binom{n+1-1/2}{n+1}=2(n+1)\frac{1}{2^{2(n+1)}}\binom{2(n+1)}{n+1}=\frac{n+1}{2^{2{n+1}}}\binom{2n+2}{n+1},$$
where the first equality is just the usual identity $\sum_{j=0}^n \binom{r+j}{j} =\binom{r+n+1}{n}$ valid for any real $r$ and any non-negative integer $n$, and the last few steps were just trying to slightly transform the coefficient in a way that allowed application of $(1)$.
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Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left side, $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = \sqrt{9\sin^250^\circ-8\sin^350^{\circ}}$$
Now How can i solve it after that , Help required, Thanks
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$$\sqrt{9-8\sin 50^\circ}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin^350^\circ}$$
$$(\text{using }\sin^2x=1-\cos^x)$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ(1-\cos^250^\circ)}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+8\sin50^\circ\cos^250^\circ}$$
$$(\text{using }2\sin x\cos x=\sin2x)$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+4\sin100^\circ\cos50^\circ}$$
$$(\text{using }2\sin x\cos y=\sin(x+y)-\sin(x-y))$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+2(\sin150^\circ+\sin50^\circ)}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-6\sin50^\circ+2\sin150^\circ}$$
$$(\text{using }\sin150^\circ=\sin30^\circ=\frac{1}{2})$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-6\sin50^\circ+1}$$
$$=\csc50^\circ\sqrt{(3\sin50^\circ-1)^2}$$
$$\text{(Taking the positive root.)}$$
$$=\csc50^\circ(3\sin50^\circ-1)$$
$$=3-\csc50^\circ$$
$$\text{So }a=3\text{ and }b=-1$$
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|
Surface area of sphere within a cylinder I have to
Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$
My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$.
$${\delta z\over \delta x}={-x\over \sqrt{a^2-x^2-y^2}}\ ;\ {\delta z\over \delta y}={-y\over \sqrt{a^2-x^2-y^2}}$$
Using the formula $$a(\Bbb{S})=\iint\limits_\Bbb{T}\sqrt{1+\left({\delta z\over \delta x}\right)^2+\left({\delta z\over \delta y}\right)^2}dx \ dy$$
I get $$a(\Bbb{S})=\iint\limits_{x^2+y^2=by}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\sqrt{1+{by\over a^2-by}}\ dy\ dx\\=a\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}{1\over\sqrt{a^2-by}}dy \ dx\\=a\int_0^{b/2}\left\{\left[{-2\sqrt{a^2-by}\over b}\right]_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\right\}\ dx$$
How to proceed now?The integral seems too bad.
OR
Is there a simpler parametrization ?
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Another approach in spherical coordinates: parametrize the surface with
\begin{cases}
x=a \sin\phi \cos\theta \\
y=a \sin\phi \sin\theta \\
z=a \cos\phi
\end{cases}
with $(\theta,\phi)\in [0,\pi]\times[0,\Phi]$, where $\Phi \in ]0,\pi]$ is the solution to
\begin{cases}
x^2+y^2+z^2=a^2 \\
x^2+y^2=by
\end{cases}
Substituting $x,y,z$ by their expressions in spherical coordinates yields
$$
\Phi = \sin^{-1}\left(\frac{b}{a}\sin\theta\right)
$$
It follows that
$$
S=\int_{0}^{\pi}\int_0^{\Phi} a^2\sin\phi\; d\phi d\theta = a^2 \int_0^{\pi}1-\cos\Phi\; d\theta
$$
With a little trigonometric work ($\cos x =\pm \sqrt{1-\sin^2x}$), we can show that
$$
\cos\Phi = \sqrt{1-\left(\frac{b}{a}\sin\theta\right)^2}
$$
Therefore
$$
S = \pi a^2 - a \int_0^{\pi} \sqrt{a^2-b^2\sin^2\theta}\; d\theta
$$
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|
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothing.
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From what you did, we want to prove that
$$\frac{1}{1+\log_32}\lt \log_32,$$
i.e.
$$\log_32\gt \frac{\sqrt 5-1}{2},$$
i.e.
$$2\gt 3^{(\sqrt 5-1)/2},$$
i.e.
$$2^2\gt 3^{\sqrt 5-1},$$
i.e.
$$3\cdot 2^2\gt 3^{\sqrt 5}$$
It is sufficient to prove that
$$12\gt 3^{2.25},$$
i.e.
$$12^2\gt 3^4\sqrt 3,$$
i.e.
$$16\gt 9\sqrt 3$$
i.e.
$$16^2\gt 81\cdot 3$$
which holds.
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|
Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T(k) + k+1 + 2$
$= \frac{1}{2}k^2 + \frac{5}{2}k -2 + k+1 +2$
$= \frac{1}{2}k^2 + \frac{7}{2}k +1$
What can I do next?
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What can I do next?
Hint. Prove that, for $k=1,2,\cdots,$
$$
\frac{1}{2}k^2 + \frac{7}{2}k +1=\frac{1}{2}(k+1)^2+\frac{5}{2}(k+1)-2.
$$
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|
Prove that $\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\geq\sqrt[3]{45(ab+ac+bc)}$
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sqrt[3]{a^2+4bc}+\sqrt[3]{b^2+4ac}+\sqrt[3]{c^2+4ab}\geq\sqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$\left(\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\right)^3\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4\geq\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4$$
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a^2+4bc)(ka+b+c)\right)^4\geq45(ab+ac+bc)\sum_{cyc}(a^2+4bc)^3(ka+b+c)^4,$$
which is nothing for all $k\geq0$.
Of course, we can use Holder with $(ka^2+b^2+c^2+mab+mac+nbc)^4$, but I think in this way even uvw will not help.
I have a proof of the following inequality.
Let $a$, $b$ and $c$ be non-negative numbers and $k=8\cos^340^{\circ}.$ Prove that:
$$\sqrt[3]{a^2+kbc}+\sqrt[3]{b^2+kac}+\sqrt[3]{c^2+kab}\geq\sqrt[3]{9(1+k)(ab+ac+bc)},$$
but it not so comforts.
Thank you!
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Assume that $ab+bc+ca > 0$. Rewrite the inequality as $\sqrt[3]{u} + \sqrt[3]{v} + \sqrt[3]{w} \ge 3$ where
$$u = \frac{27(a^2+4bc)}{45(ab+bc+ca)}, \ v = \frac{27(b^2+4ca)}{45(ab+bc+ca)}, \ w = \frac{27(c^2+4ab)}{45(ab+bc+ca)}.$$
We will use the fact that
$$\sqrt[3]{x} \ge \frac{3x(5+4x)}{5x^2+20x+2}, \ x \ge 0$$
which follows from
$$x - \Big( \frac{3x(5+4x)}{5x^2+20x+2}\Big)^3 = \frac{x(125x^2+272x+8)(x-1)^4}{(5x^2+20x+2)^3} \ge 0.$$
Using the fact above, it suffices to prove that
$$\frac{3u(5+4u)}{5u^2+20u+2} + \frac{3v(5+4v)}{5v^2+20v+2} + \frac{3w(5+4w)}{5w^2+20w+2} \ge 3$$
or $f(a,b,c) \ge 0$ where $f(a,b,c)$ is a homogeneous polynomial.
We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$. There are two possible cases:
1) $c \le b\le a$: Let $b = c + s, \ a = c+s+t; \ s, t\ge 0$.
Note that $f(c+s+t, c+s, c)$ is a polynomial with non-negative coefficients. True.
2) $c \le a \le b$: Let $a = c + s, \ b = c + s+t; \ s, t \ge 0$.
Note that $f(c+s, c+s+t, c)$ is a polynomial with non-negative coefficients. True.
We are done.
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How to calculate eigenvalues? Given is differential equation: $2y''+5y'-3y=0$
Write this equations as 1. order system and calculate eigenvalues of the matrix of this differential equation.
My idea:
$z_{1}=y$
$z_{2}=y'$
$\frac{d}{dx}\begin{pmatrix} z_{1} \\ z_{2} \end{pmatrix} =\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix} $
But how do I get matrix to calculate eigenvalues?
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$$\frac{d}{dx}\begin{pmatrix} z_{1} \\ z_{2} \end{pmatrix} =\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix}$$
You're close, simply rewrite:
$$\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix}
= \begin{pmatrix} 0z_1+ 1z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix}
=\color{blue}{\begin{pmatrix} 0 & 1 \\ \frac{3}{2} & -\frac{5}{2} \end{pmatrix}}
\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}$$
Now you want to diagonalize the blue matrix.
Can you calculate its eigenvalues? You should find $-3$ and $\tfrac{1}{2}$.
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|
What is the total sum of the cardinalities of all subsets of a set? I'm having a hard time finding the pattern. Let's say we have a set
$$S = \{1, 2, 3\}$$
The subsets are:
$$P = \{ \{\}, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\} \}$$
And the value I'm looking for, is the sum of the cardinalities of all of these subsets. That is, for this example, $$0+1+1+1+2+2+2+3=12$$
What's the formula for this value?
I can sort of see a pattern, but I can't generalize it.
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For the subsets of size $k$ one needs the number of ways how to draw $k$ distinguishable elements from $\lvert S \rvert$ available, which is given by the binomial coefficient. This is multiplied by the number of elements, thus $k$, and summed up over all possibilites:
$$
\sum_{A \in 2^S} \lvert A \rvert = \sum_{k=0}^{\lvert S \rvert} \binom{\lvert S \rvert}{k} k
$$
For your example:
$$
\sum_{A \in 2^S} \lvert A \rvert = \sum_{k=0}^3 \binom{3}{k} k = 1 \cdot 0 + 3 \cdot 1 + 3 \cdot 2 + 1 \cdot 3 = 0 + 3 + 6 + 3 = 12
$$
Setting
$$
s(n) = \sum_{k=0}^n \binom{n}{k} k
$$
we have
\begin{align}
s(n+1)
&= \sum_{k=0}^{n+1}\binom{n+1}{k} k \\
&= \sum_{k=1}^{n+1}\binom{n+1}{k} k \\
&= \sum_{k=1}^{n+1}\frac{(n+1)!}{k!(n+1-k)!} k \\
&= \sum_{k=1}^{n+1}\frac{(n+1)!}{(k-1)!(n-(k-1))!} \\
&= (n+1) \sum_{k=1}^{n+1} \binom{n}{k-1} \\
&= (n+1) \sum_{k=0}^n \binom{n}{k} \\
&= (n+1) \sum_{k=0}^n \binom{n}{k} 1^k \cdot 1^{n-k} \\
&= (n+1) (1+1)^n \\
&= (n+1) 2^n
\end{align}
This gives $s(0) = 0$ and $s(n) = n 2^{n-1}$ for $n \ge 1$ and
$$
\sum_{A \in 2^S} \lvert A \rvert = \lvert S \rvert \, 2^{\lvert S \rvert - 1}
$$
For your example this would mean $s(3) = 3 \cdot 2^2 = 3 \cdot 4 = 12$.
|
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|
Does there exist a basis for the set of $2\times 2$ matrices such that all basis elements are invertible? As the title says, I'm wondering whether there exists a basis for the set of $2\times 2$ matrices (with entries from the real numbers) such that all basis elements are invertible.
I have a gut feeling that it is false, but don't know how to prove it. I know that for a matrix to be invertible, it must be row equivalent to the identity matrix and I think I may be able to use this in the proof, but I don't know how.
Thanks in advance for any help,
Jack
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Consider the counter-example basis:
$$\beta:=\left\{\begin{pmatrix} 1&0\\0&1\end{pmatrix},\begin{pmatrix} 0&1\\1&0\end{pmatrix},\begin{pmatrix} 1&0\\1&1\end{pmatrix},\begin{pmatrix} 0&1\\1&1\end{pmatrix}\right\}$$
Let $A=\begin{pmatrix} x_1&x_2\\x_3&x_4\end{pmatrix}$, let the following equation:
$$A=a\begin{pmatrix} 1&0\\0&1\end{pmatrix}+b\begin{pmatrix} 0&1\\1&0\end{pmatrix}+c\begin{pmatrix} 1&0\\1&1\end{pmatrix}+d\begin{pmatrix}
0&1\\1&1\end{pmatrix}$$
Then
$$\begin{pmatrix}
1&0&1&0\\0&1&0&1\\0&1&1&1\\1&0&1&1\end{pmatrix}\begin{pmatrix}
a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}
x_1\\x_2\\x_3\\x_4\end{pmatrix}$$
Check $\det\begin{pmatrix}
1&0&1&0\\0&1&0&1\\0&1&1&1\\1&0&1&1\end{pmatrix}=1$ . So there is an unique solution for $A$, hence $\beta$ is a basis of $2$ by $2$ matrix, where all basis elements are invertible.
|
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Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$ Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$
Hint: Use the (weighted) Mean Value Theorem for Integrals.
The MVT for Integrals:
Suppose that $u$ is continuous and $v$ is integrable and nonnegative on $[a,b]$
Then $\int_a^b u(x)v(x)dx=u(c)\int_b^a v(x)dx$
for some $c$ in $[a,b]$.
I plan on using $u(x)$ as $x^{10}$ as it is continuous and $v(x)$ as $\frac{1}{\sqrt{1+x}}$ as it is integrable on [0,1]. I'm not sure how to go from there and find $c$.
|
Choosing $v(x) = \frac{1}{\sqrt{x+1}}$ doesn't work, I think:
I didn't use MVT directly. I used something from the proof of MVT:
$$m \int_0^1 \frac{1}{\sqrt{x+1}} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le M \int_0^1 \frac{1}{\sqrt{x+1}} dx$$
The best $m$ and $M$ I got are $0$ and $1$ resp as $0 \le x^{10} \le 1$ for $0 \le x \le 1$
So let's try
$u(x) = \frac{1}{\sqrt{x+1}}$
Now $\frac{1}{\sqrt{2}} \le \frac{1}{\sqrt{x+1}} \le 1$ for $0 \le x \le 1$
So we have
$$\frac{1}{\sqrt{2}} \int_0^1 x^{10} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le 1 \int_0^1 x^{10} dx$$
Remark: I think the 11's in the denominator may be a hint to use $u(x) = \frac{1}{\sqrt{x+1}}$
|
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"url": "https://math.stackexchange.com/questions/2001806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$ Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$
For this I think I should use De l'Hopital's rule but it takes a lot time and I can't get to answer.
Can we use the De l'Hopital's rule twice?or three times?If yes what is the limit of that?and how can we find the limit of polynomial multiplys?
|
Since we have a limit at $0$ of polynomials, only the term with the least degree matters. We have
$$
(x+1)(x+2)(x+3)(x+4)-24=24+(2\times3\times4+1\times3\times4+1\times2\times3+1\times2\times4)x+c_2x^2+c_3x^3+c_4x^4-24=50x+c_2x^2+c_3x^3+c_4x^4.
$$
Then
\begin{align}
\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}
&=\frac{50x+c_2x^2+c_3x^3+c_4x^4}{x(x+5)}\\ \ \\
&=\frac{50}{x+5}+\frac{c_2x}{x+5}+\frac{c_3x^2}{x+5}+\frac{c_4x^3}{x+5}\\ \ \\
&\to\frac{50}5=10.
\end{align}
|
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|
Boolean Simplification using Algebra Having trouble showing the following relation:
$$A\cdot B + A'\cdot B' + B\cdot C = A\cdot B + A'\cdot B' + A'\cdot C
$$
using Boolean Algebra. Any help is appreciated.
|
$$\begin{array}{rcl}(A+A')=1 &\mbox{ and } &(B+B')=1 \\
(A+A')\cdot(B+B')=1& \implies &(A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') =1\\
A\cdot B + A'\cdot B' +B\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +B\cdot C)\\
&=&(A\cdot B +0+A\cdot B\cdot C)+(0+0+0) \\
&&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + 0)\\
&=& A\cdot B + A'\cdot B\cdot C + A'\cdot B' \\
&=& A\cdot B + A'\cdot( B\cdot C + B') \\
&=& A\cdot B + A'\cdot( C+B') \\
&=& A\cdot B + A'\cdot C + A'\cdot B'\\
A\cdot B + A'\cdot B' +A'\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +A'\cdot C)\\
&=&(A\cdot B +0+0)+(0+0+0) \\
&&+ (0+0+A'\cdot B\cdot C) + (0+A'\cdot B' + A'\cdot B'\cdot C)\\
&=& A\cdot B + A'\cdot (B'+B)\cdot C + A'\cdot B'\\
&=& A\cdot B + A'\cdot C + A'\cdot B'\\
\end{array}
$$
And since they are both equal to $A\cdot B + A'\cdot C + A'\cdot B'$ they are equal to each other.
I have used, in the first half of this proof,
$$
B\cdot C + B' = C+B'
$$
which is much easier to prove if you don't already have it as a lemma.
|
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|
A square of a rational between two positive real numbers ?! Let $ a,b \in \mathbb{R}^+ ,(a<b) $. Prove that there is a rational number $ q $ such that $ a<q^2<b $, without using square root function.
Can anyone help me ?
|
Write $q=r/s$ with positive integers $r$ and $s$. We need to show that
$$as^2\lt r^2\lt bs^2$$
for some choice of $r$ and $s$. If the interval $(as^2,bs^2)$ does not contain a square, then there is some integer $n$ such that $n^2\le as^2$ while $(n+1)^2\ge bs^2$. This implies $2n+1\ge(b-a)s^2$, which implies
$$as^2\ge n^2=\left((b-a)s^2-1\over2\right)^2$$
which we can rewrite as
$$(b-a)^2s^4-2(b+a)s^2+1\le0$$
Since $b-a\not=0$, the left hand side is positive for sufficiently large integers $s$, at which point the inteval $(as^2,bs^2)$ must contain the square of some integer $r$. If you want something explicit, let $s$ be an integer greater than $\sqrt{2(b+a)}/(b-a)$:
$$\begin{align}
(b-a)^2s^4-2(b+a)s^2+1&= s^2((b-a)^2s^2-2(b+a))+1\\
&\gt s^2(2(b+a)-2(b+a))+1\\
&\ge1
\end{align}$$
|
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|
Evaluating an infinite series using partial fractions... I am having trouble evaluating an infinite series that uses partial fractions. The problem is as follows:
$$
\sum_{n = 1}^{\infty}{1 \over n\left(n + 1\right)\left(n + 2\right)}
$$
I realize that this is a telescoping series, but I am unable to find a general formula for the Sn.
After the partial fraction decomposition the problem looks like
$1/\left(2n\right) - 1/\left(n + 1\right) + 1/\left(2n + 4\right)$. I input values for $n = 1, n = 2,$ etc, and some cancel out but I am unable to determine a pattern to write to find Sn... Thanks.
I found this answer to the problem online which makes sense, up to the step where it makes Sn equal to
$1/2(1/2 - 1/\left(n+1\right) + 1/\left(n+2\right)$. I get the prior steps but I am not sure how they make that leap.
|
$$\frac{1}{n(n+1)(n+2)} = \frac{1}{(n+1)} \cdot \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right) = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$$
So
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{1(1+1)}-\frac{1}{(1+1)(1+2)}+\frac{1}{2(2+1)}-\frac{1}{(2+1)(2+2)}+\cdots\right)$$
$$=\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$$
|
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|
Use euclidean algorithm to calculate the multiplicative inverse of $5$ in $\mathbb{Z}_{12}$ I really like to know the exact way how its done. Here is what I wrote:
$5$ must have a multiplicative inverse because $\text{ gcd }(12,5)=1$
So $5x \equiv 1 \text{ mod } 12 \Leftrightarrow x \equiv 5^{-1}(\text{mod } 12)$
$$12=5 \cdot 2+2$$
$$5=2 \cdot 2+1$$
$$2=1 \cdot 2+0$$
$\Rightarrow$
$$1=5-2 \cdot 2$$
$$1=5-2 \cdot (12-5 \cdot 2)$$
$$1=5-2 \cdot 12+4 \cdot 5$$
$$1=-2 \cdot 12 + 5 \cdot 5$$
From an online calculator, I know that $5$ is its own inverse. But how do you know that from the last notation?
Please don't explain it too complicated, I have very big troubles in understanding it and I'm already very happy I was able to calculate it till here myself.
|
I will show you how to do it through continued fractions, that is essentially the same. We have:
$$\frac{12}{5}=2+\frac{2}{5}= 2 +\frac{1}{2+\frac{1}{2}} = [2;2,2]\tag{1}$$
and if $\frac{p_n}{q_n},\frac{p_{n+1}}{q_{n+1}}$ are two consecutive convergents of the same continued fraction, the difference between them is $\pm\frac{1}{q_n q_{n+1}}$. In our case $[2;2] = \frac{5}{2}$, and
$$ \frac{12}{5}-\frac{5}{2} = -\frac{1}{10} \tag{2} $$
leads to:
$$ 12\cdot 2 - 5\cdot 5 = -1,\quad 5\cdot 5 = 2\cdot 12+1 \tag{3} $$
hence $5$ is the inverse of itself $\!\!\pmod{12}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Card probability problem: Five cards will be dealt from a well-shuffled deck. Find the chance of getting an ace or a king among the 5 cards. Five cards will be dealt from a well-shuffled deck.
Find the chance of getting an ace or a king among the 5 cards.
I think question means getting at least a king or an ace, and here is my procedure:
$p$(an ace or a king among 5 cards)=$p$(at least an ace)+$p$(at least an king)-$p$(ace and king at the same time)
$p$(at least an ace)=$1-\frac{48}{52}\times\frac{47}{52}\times\frac{46}{52}\times\frac{45}{52}\times\frac{44}{52}$
$p$(at least an king)=$1-\frac{48}{52}\times\frac{47}{52}\times\frac{46}{52}\times\frac{45}{52}\times\frac{44}{52}$
$p$(ace and king at the same time)=...
Here I am stuck because there are many combination that ace and king are together. I feel like I misintepreted the question. Could someone give an insight?
|
Using 1 -
$$1 - \frac {\binom{44}{5} } { \binom{52}{5} } = 0.5821374704$$
Using 1 + 2 + 3 + 4 + 5
$$\frac {\binom{8}{1} \binom{44}{4} + \binom{8}{2} \binom{44}{3} + \binom{8}{3} \binom{44}{2} + \binom{8}{4} \binom{44}{1} + \binom{8}{5}} { \binom{52}{5} } = 0.5821374704$$
|
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|
Solving a set of linked recurrent relations I'm trying to find a method how to solve this set of linked recurrent relations.
$$
\left\{\begin{matrix}
a_{n} = 3*a_{n-1} + b_{n-1}\\
b_{n} = 2*a_{n-1} + 2*b_{n-1}\\
c_{n} = b_{n-1} + 3*c_{n-1}\\
d_{n} = a_{n-1} + 2*b_{n-1} + 3*c_{n-1}\\
e_{n} = 6*d_{n-1} + 3*e_{n-1} + f_{n-1}\\
f_{n} = 2*e_{n-1} + 2*f_{n-1}\\
g_{n} = f_{n-1} + 3*g_{n-1}\\
\end{matrix}\right.
$$
From what I've seen one method is to end up with recurrent relation only with one term: $a_{n}=c_{1}*a_{n-1}+c_{2}*a_{n-2}+...+c_{k}*a_{n-k}$ and then I can reduce the relation and find out the characteristic equation $r^{n}=c_1*r^{n-1}+c_2*r^{n-2}+...+c_k*r^{n-k}$.
I need to calculate $X_{n}=a_{n}+b_{n}+c_{n}$ and $Y_{n}=d_{n}+e_{n}+f_{n}+g_{n}$ so basically only if I can have a recurrent relation in one term separately I probably could apply that method.
Is there a simpler method to find this ?
|
It's not quite clear to me if this is equivalent to the method you suggested but in the case of nice, linear recurrences like you have, one can write:
$$ \vec{x}_n
= \begin{pmatrix}a_n\\b_n\\c_n\\d_n\\e_n\\f_n\\g_n\end{pmatrix}
= \begin{pmatrix}3 & 1 & 0 & 0 & 0 & 0 & 0
\\ 2 & 2 & 0 & 0 & 0 & 0 & 0
\\ 0 & 1 & 3 & 0 & 0 & 0 & 0
\\ 1 & 2 & 3 & 0 & 0 & 0 & 0
\\ 0 & 0 & 0 & 6 & 3 & 1 & 0
\\ 0 & 0 & 0 & 0 & 2 & 2 & 0
\\ 0 & 0 & 0 & 0 & 0 & 1 & 3
\end{pmatrix}
\begin{pmatrix}a_{n-1}\\b_{n-1}\\c_{n-1}\\d_{n-1}\\e_{n-1}\\f_{n-1}\\g_{n-1}\end{pmatrix}
= A \vec{x}_{n-1}$$
and thus compute
$$ \vec{x}_n = A^n \vec{x}_0.$$
If you let $M = S J S^{-1}$ as the Jordan normal form of $M$, you can compute
$$ \vec{x}_n = S J^n S^{-1} \vec{x}_0$$
rather quickly as $J$ will only have nonzero elements on the diagonal and superdiagonal.
Performing the computations myself gives
$$ S^{-1} \vec{x}_0 =
\begin{pmatrix}0\\-2\\-2\\54\\36\\-56\\24\end{pmatrix}$$
and so
$$ J^n S^{-1} \vec{x}_0 =
\begin{pmatrix}0\\-2\\-2\\54 \cdot 3^n + 36 \cdot n \cdot 3^{n-1}\\36 \cdot 3^n\\-56 \cdot 4^n + 24 \cdot n \cdot 4^{n-1}\\24\cdot 4^n\end{pmatrix}$$
so finally we have
$$ \vec{x}_n = S J^n S^{-1} \vec{x}_0 =
\begin{pmatrix}
2 + 4^{n+1}\\
4^{n+1} - 4\\
2 + 4^{n+1}-6\cdot 3^n\\
6(4^n - 3^n) \\
2 + 18 \cdot 3^n -56 \cdot 4^n + 24 \cdot n \cdot 4^{n-1} + 36 \cdot 4^n\\
-4 + 36 \cdot 3^n -56 \cdot 4^n + 24 \cdot n \cdot 4^{n-1} + 24 \cdot 4^n\\
2 + 54 \cdot 3^n + 36 \cdot n \cdot 3^{n-1}
-56 \cdot 4^n + 24 \cdot n \cdot 4^{n-1} \end{pmatrix}$$
Putting it all together gives
$$ X_n = a_n + b_n + c_n= 6[2\cdot 4^n - 3^n]$$
as you noted, and
$$ Y_n = d_n + e_n + f_n + g_n = 6[4^n(3n-17)+3^n(2n+17)].$$
So $Y_1 = 6$ according to the formula, which matches hand computation. If you have some further values computed, it would be nice to verify, but this at least passes an initial "sniff test".
Now that I think about it, this could have been done a lot more simply by solving for $a_n$, $b_n$, and $c_n$ explicitly using the matrix method and then using the resulting solution for $d_n$ to solve the rest.
|
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|
find $a,b,c$ such that $ \big| a + b\sqrt{2}+ c\sqrt{3} \big| <10^{-3} $ i know that linear combinations of $1$, $\sqrt{2}$, and $\sqrt{3}$ are dense in the real number line.
$$ \overline{ \mathbb{Z}[\sqrt{2},\sqrt{3}] } = \mathbb{R} $$
how quickly to these numbers converge to the entire number line. what are the smallest $a,b,c$ we can find so that
$$ \big| a + b\sqrt{2}+ c\sqrt{3} \big| <10^{-3} \text{ or }10^{-6} $$
by the pigeonhole principles we can always find $-N<a,b,c<N $ such that
$$ \big| a + b\sqrt{2}+ c\sqrt{3} \big| <N^{-2} $$
|
Here are some lower bounds for the best you can do.
If $\alpha=a+b\sqrt{2}+c\sqrt{3}$ and $|\alpha|<1$ with $a,b,c$ integers, and $abc\neq 0$, then $$|\alpha|>\frac{1}{(2|b|\sqrt{2}+1)(2|c|\sqrt{3}+1)(2|a|+1)}\tag{1}$$
I get this lower bound since: $\alpha(\alpha-2b\sqrt{2})(\alpha-2c\sqrt{3})(\alpha-2a)$ must be an integer.
If $a=0$, the same trick has that if $|\alpha|<1$ then $|\alpha|>\frac{1}{2|c|\sqrt{3}+1}$.
If $b=0$ or $c=0$, the best you can do is with the continued fractions of $\sqrt{3}$ and $\sqrt{2}$, respectively.
You get (1) in general, for all $a,b,c$.
You tend to get some "good" values when:
$$N(a,b,c)=|\alpha(\alpha-2b\sqrt{2})(\alpha-2c\sqrt{3})(\alpha-2a)|=|a^4+4b^4+9c^4-(4a^2b^2+6a^2c^2+12c^2b^2)|$$
is a small integer.
For example, $N(14,5,4)=4$ and $14-5\sqrt{2}-4\sqrt{3}$ is a good approximation.
This is because if $|\alpha|<1$ then:
$$\frac{N(a,b,c)}{(2|a|+1)(2|b|\sqrt{2}+1)(2|c|\sqrt{3}+1)}<|\alpha|<\frac{N(a,b,c)}{(2|a|-1)(2|b|\sqrt{2}-1)(2|c|\sqrt{3}-1)}$$
|
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|
Quick solution of an equation in factorials Now I realize it is not that hard but it is a test question and should have a fast way to solve as indicated by the teacher. However, I fail to see it.
If both $m$ and $n$ are two-digit natural numbers and $m! = 156 \cdot n!$ find the value of $m-n$.
|
$$
\binom{m}{n}=\frac{m!}{n!(m-n)!}=\frac{156}{(m-n)!}
$$
Thus $(m-n)!$ is a divisor of $156=2^2\cdot 3\cdot 13$. This forces $(m-n)!$ to be a divisor of $2^2\cdot3=12$ (the prime $13$ cannot appear for obvious reasons), leaving $m-n=1$, $m-n=2$ or $m-n=3$.
*
*If $m=n+1$, we have
$$
\binom{n+1}{n}=156=\binom{n+1}{1}=n+1
$$
*If $m=n+2$, we have
$$
\binom{n+2}{n}=78=\binom{n+2}{2}=\frac{(n+2)(n+1)}{2}
$$
*If $m=n+3$, we have
$$
\binom{n+3}{n}=26=\binom{n+3}{3}=\frac{(n+3)(n+2)(n+1)}{6}
$$
The first case is dismissed, because it corresponds to $n=155$.
In the third case, as $n>9$, we have $(n+3)(n+2)(n+1)>12\cdot11\cdot 10>156$.
Thus only the second case can hold and $m-n=2$; precisely, $n^2+3n-154=0$ so $n=11$.
By the way, the third equation has no integer solutions so the only cases are $n=11$, $m=13$ and $n=155$, $m=156$.
|
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|
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{My\; Try::}$ Given $$ \alpha = \arctan \bigg(\frac{2(2\sqrt{2}-1)}{1-(2\sqrt{2}-1)^2}\bigg)=\arctan \bigg(\frac{(2\sqrt{2}-1)}{2\sqrt{2}-4}\bigg)$$
and $$\beta = \arcsin\bigg(1-\frac{4}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg) = \arcsin\bigg(\frac{23}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg)$$
So $$\arcsin\bigg[\frac{23}{27}\cdot \frac{4}{5}+\frac{3}{5}\cdot \frac{10\sqrt{2}}{23}\bigg]$$
Now how can i solve it, Help required, Thanks
|
On the one hand we have
$$\tan{\alpha\over2}=2\sqrt{2}-1>\sqrt{3}=\tan{\pi\over3}\ ,$$
hence ${\alpha\over2}>{\pi\over3}$, or $$\alpha>{2\pi\over3}\ .$$
On the other hand the convexity of $$t\mapsto\arcsin t\qquad(0\leq t\leq{\pi\over2})$$ implies
$$\arcsin{1\over3}<{2\over3}\arcsin{1\over2}={2\over3}\>{\pi\over6}={\pi\over9}\ .$$
Furthermore $\arcsin{3\over5}<\arcsin{1\over\sqrt{2}}={\pi\over4}$. It follows that
$$\beta<3{\pi\over9}+{\pi\over4}={7\pi\over12}<\alpha\ .$$
|
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|
factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:
How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?
The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)
|
There are two errors in the factorization
$$x^6-64=(x^2-4)(x^4+4x-16)$$
The constant term in the quartic should be $+16$ instead of $-16$, and the linear term, $4x$, should be a quadratic term, $4x^2$. That is, the correct factorization is
$$x^6-64=(x^2-4)(x^4+4x^2+16)$$
This now agrees with the factorization $(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$ since $(x-2)(x+2)=x^2-4$ and
$$\begin{align}
(x^2-2x+4)(x^2+2x+4)&=(x^2+4-2x)(x^2+4+2x)\\
&=(x^2+4)^2-(2x)^2\\
&=x^4+8x^2+16-4x^2\\
&=x^4+4x^2+16
\end{align}$$
|
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|
$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
|
The inverse $z'$ of a point $z$ with respect to the circle centred at $a$ with radius $r$ is given by $$z' = \frac{r^2}{\overline{z}-\overline{a}}+a$$
We have $$\frac{iz+2}{4z+i} = \frac{9}{4(4z+i)}+ \frac{i}{4} = \frac{9}{16 \left(z - \frac{-i}{4} \right)} + \frac{i}{4} =\frac{9}{16 \left(\overline{z} - \frac{-i}{4} \right)} + \frac{i}{4} $$
($\because z \in \mathbb{R} \Rightarrow \overline{z} = z$)
With $a = \frac{i}{4}$ and $r = \frac{3}{4}$ its now evident that the real line is being inverted with respect to the circle $ \left|z - \frac{i}{4} \right| = \frac{3}{4}$ Since the real axis does not pass through the centre of the circle it is inverted into a circle under the given transformation.
The projection of $\frac{i}{4}$ on the real axis i.e. $z=0$ will transform into one end of a diameter of this circle ($z=0$ transforms to $-2i$), while $\frac{i}{4}$ itself will be the other end of the diameter. Hence the radius is $\frac{9}{8}$ and the centre is $-\frac{7}{8}i$
|
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|
$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up zeros?
|
By using this,
we get $\lim_{n\to \infty}(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})=\ln{2}$.
|
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|
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$ subject to $x,y,z>0$ and $x+y+z=1$
This inequality is equivalent to;
$\left(\frac{x+1}{4}\cdot\frac1x\right)\left(\frac{y+1}{4}\cdot\frac1y\right)\left(\frac{z+1}{4}\cdot\frac1z\right)\ge1$ taking logarithm of both sides we have to show that LHS is greater or equal to $0$, but by convexity of $\ln\left(\frac{x+1}{4}\right)-\ln(x)$, i.e. $\frac{d^2}{dx^2}\left[\ln\left(\frac{x+1}{4}\right)-\ln(x)\right]= \frac{1}{x^2}-\frac{1}{(x+1)^2}>0$ we have
$\sum\limits_{cyc}\ln\left(\frac{x+1}{4}\right)-\ln(x)\ge3\ln\left(\sum\limits_{cyc}\frac{x+1}{4}\right)-3\ln\left(\sum\limits_{cyc}x\right)=3\ln(1)-3\ln(1)=0$
Is this OK, or is there a much simpler way ? (I did the same as in example $4.1.2.$ here)
|
By AM-GM we have:
$$
\frac{1}{3}=\frac{1}{3}(x+y+z)\geq\sqrt[3]{xyz}\implies\frac{1}{27}\geq xyz\implies\frac{1}{27xyz}\geq 1.
$$
Using AM-GM again, with 4 terms this time, we obtain
$$
1+\frac{1}{x}=1+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}\geq 4\sqrt[4]{\frac{1}{27x^3}}.
$$
Do that for the other 2 terms and multiply. We get
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\geq 64\sqrt[4]{\left(\frac{1}{27xyz}\right)^3}\geq 64.
$$
Equalities are iff $x=y=z=\frac{1}{3}$.
|
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|
Determine the remainder when $f(x) = 3x^5 - 5x^2 + 4x + 1$ is divided by $(x-1)(x+2)$ This question arose while I was tutoring a student on the topic of the Remainder Theorem. Now, the Remainder Theorem tells us that when a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the remainder is simply $p(a)$. However, in this case we have a product of linear factors.
Using the Remainder Theorem we can see that neither $(x-1)$ nor $(x+2)$ is a factor of $f(x)$. Also, if we try to find the remainder using long division, we get a relatively ugly remainder of
$$
3(14x - 13)
$$
I assume this is not the correct approach as all other questions in this topic used the Remainder Theorem. So perhaps there is a more elegant approach?
|
Hint: the remainder will be a polynomial of degree (at most) $1$ so:
$$f(x) = (x-1)(x+2)q(x) + ax + b$$
Substitute $x=1,-2$ in the above and you get two equations in $a,b$.
[ EDIT ] For a less conventional approach (justified in the answer here) note that $(x-1)(x+2)=0 \iff x^2=-x+2$. Repeatedly using the latter substitution:
$$
\begin{align}
3x^5 - 5x^2 + 4x + 1 &= 3 (x^2)^2 \cdot x - 5(x^2) + 4x + 1 \\
&= 3(x^2-4x+4)x - 5(-x+2) + 4x +1 \\
&= 3(-x+2-4x+4)x + 9x -9 \\
&= -15(x^2)+ 18x + 9x - 9 \\
&= -15(-x+2) + 27 x - 9 \\
&= 42 x -39
\end{align}
$$
|
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|
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e^{2x} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}
\end{align*}
I don't know how to proceed with the rightmost term. So far I've been applying the Leibniz Rule whenever I've had to find the $n$ derivative of a function of the form $f(x)g(x)$, because is clear that either $f(x)$ or $g(x)$ has a derivative a $k$ derivative ($1<k<n$) equal to zero, which simplifies the expression nicely.
But here
$$\frac{e^{2x}\cos 2x}{2}$$
both functions are infinitely differentiable on $\mathbb{R}$ which makes things a bit different.
My only attempt was to write its n derivative in this form
\begin{align*}
&=\frac{1}{2}\sum_{k=0}^n{n \choose k} \big(2^k e^{2x}\big)\bigg(2^{n-k}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\bigg)\\
&=\sum_{k=0}^n {n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]
\end{align*}
So
\begin{align*}
y^{(n)} &= 2^{n-1}e^{2x} -\sum_{k=0}^n{n \choose k} 2^{n-1}e^{2x}\cos \left[2x + \frac{\pi(n-k)}{2}\right]\\
&= 2^{n-1}e^{2x}\left(1 -\sum_{k=0}^n{n \choose k} \cos \left[2x + \frac{\pi(n-k)}{2}\right]\right)
\end{align*}
But the textbook's answer is
$$2^{n-1}e^{2x}\left(1 -2^{n/2}\cos \left[2x + \frac{\pi n}{4}\right]\right)$$
For some reason I have the feeling that a little of modular arithmetic has to be applied on $\frac{\pi(n-k)}{2}$
|
We know that $$\sin^2(x) = \dfrac{1 - \cos(2x)}{2}= \dfrac{1}{2} -\dfrac{1}{2}\Re(\exp(2ix)), $$ where $\Re$ is the real part of a complex number.
It follows that
$$y =\exp(2x)\sin^2(x) = \dfrac{\exp(2x)}{2} - \dfrac{1}{2}\Re(\exp(zx)), $$ with $z = 2+2i.$
Using complex-valued derivatives:
$$\dfrac{dy}{dx} = \exp(2x)- \dfrac{z}{4}\exp(zx)- \dfrac{z^{*}}{4}\exp(z^{*}x),$$ which leads to
$$\dfrac{d^ny}{dx^n} = 2^{n-1}\exp(2x) - \dfrac{z^n}{4}\exp(zx) - \dfrac{z^{*n}}{4}\exp(z^{*}x) = 2^{n-1}\exp(2x) - \dfrac{1}{2}\Re (z^n \exp(zx)).$$
|
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|
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
|
Note that
$$f(x)=\frac{1}{x^2 + 2x + 2}=\frac{1/2}{1+(x+x^2/2)}\\=\frac{1}{2}\left(1-(x+x^2/2)+(x+x^2/2)^2-((x+x^2/2)^3+o(x^3)\right).$$
In general, for any positive integer $n$,
$$2f(x)=1+\sum_{k=1}^n x^k(1+x/2)^k+o(x^n)=
\sum_{k=1}^n (-1)^k\sum_{j=0}^{n-k} \frac{1}{2^j}\binom{k}{j}x^{j+k}+o(x^n)\\
=1+\sum_{k=1}^n \frac{x^m}{2^m}\left(\sum_{k=1}^m(-2)^k\binom{k}{m-k}\right)+o(x^n).$$
P.S. Note that $a_m$, the coefficient of $x^m$, is related to the sequence A009116. It follows immediately from its generating function that it satisfies the linear sequence
$$a_{m+1}=-\left(a_m+\frac{a_{m-1}}{2}\right).$$
|
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|
Proof for rational roots Let $f(x) = a_0+a_1x+\cdots+a_nx^n$ be a polynomial of degree $n$ over $\Bbb Z$.
|
Part A:
Given $\frac pq$ is a rational root of a polynomial$$f(x)=a_nx^n+x_{n-1}x^{n-1}+\ldots a_0\tag1$$
Where $a_n\in\mathbb{Z}$. We wish to show that $p|a_0$ and $q|a_n$. Since $\frac pq$ is a root$$0=a_n\left(\frac pq\right)^n+\ldots+a_0\tag2$$
Multiplying by $q^n$, we have$$a_np^n+\ldots+a_0q^n=0\tag3$$
Examining this in module $p$, we have $a_0q^n\equiv 0(\text{mod} \text{ p})$. As $q$ and $p$ are relatively prime, we have $p|a_0$. And with the same logic for module $q$, we have $q|a_n$. Which completes the proof.
Note that this theorem is called the Rational Root Theorem!
Part B:
Roots of $9x^3+18x^2-4x-8=0$
By the Rational Root Theorem, we have the possible roots as$$\begin{align*} & \pm1\pm2\pm4\pm8\\ & \pm1\pm3\\ & \implies\pm\frac 13,\pm\frac 23,\pm\frac 43,\pm\frac 83,\pm1,\pm2,\pm4,\pm8\end{align*}\tag4$$
Testing out the points, we find that $-2$ is a root. Thus, we can factor out $x+2$ from the polynomial to get$$(x+2)(9x^2-4)=0\tag5$$
For which we see the roots as $x=-2,x=\pm\frac 23$.
Part B:
Roots of $6x^4-7x^3+8x^2-7x+2=0$
Using our handy Rational Root Theorem, we have the possible roots as$$\begin{align*} & \pm1,\pm2\\ & \pm1\pm2\pm3\pm6\\ & \implies\pm1,\pm\frac 16,\pm\frac 13,\pm\frac 12\pm2\end{align*}\tag6$$
After some handy guesswork, we see that $\frac 12$ is a root. Thus, we can factor out $2x-1$ to get$$(2x-1)(3x^3-2x^2+3x-2)=0\tag7$$
Fortunately, we can factor $3x^3-2x^2+3x-2$ by hand. We get$$3x^3-2x^2+3x-2\implies x^2(3x-2)+3x-2\implies (x^2+1)(3x-2)$$
Thus, our quartic can be factored as$$(2x-1)(3x-2)(x^2+1)=0$$
With roots $x=\frac 12,\frac 23,\pm i$
|
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|
Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) $ Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{n\pi t}{3}) $
I am unsure how to simplify this in the best form, anyone have any ideas?
|
Here is an alternate representation using the fact that
\begin{equation}
\cos\left(\dfrac{n\pi}{2}\right)-1=\begin{cases}
-1&\text{ for }n=4k-3,\,n=4k-1\\
-2&\text{ for }n=4k-2\\
\phantom{-}0&\text{ for }n=4k\\
\end{cases}
\end{equation}
\begin{eqnarray*}
G(t)&=&-\sum_{k=1}^{\infty}\left[\cos\left(\frac{4k-3}{3}\pi t\right)
+2\cos\left(\frac{4k-2}{3}\pi t\right)
+\left.\cos\left(\frac{4k-1}{3}\pi t\right)\right]\right.
\end{eqnarray*}
|
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|
How does 1 not congruent imply Fermat n=4? A natural number is said to be congruent if it is the area of a right triangle with rational sides. I've been told that Fermat actually proved his last theorem for $n=4$ by proving that number 1 is not congruent, but I can't seem to find the connection!
It is probably very easy, thanks in advance.
|
The key thing to realize is that $1$ being or not being a congruent number is not directly related to the Fermat equation $x^4 + y^4 = z^4$. It is related to $X^4 + Y^2 = Z^4$, where the second term on the left side is only a square, not a fourth power. The second equation is what Fermat used, and it is the one you'll see in any account of FLT for exponent $4$.
Any solution to $x^4 + y^4 = z^4$ in positive integers leads to a solution of $X^4 + Y^2 = Z^4$ in positive integers ($X = x, Y = y^2, Z = z$), but not conversely. So if $X^4 + Y^2 = Z^4$ has no solution in positive integers then $x^4 + y^4 = z^4$ also has no solution in positive integers. And it's $X^4 + Y^2 = Z^4$ that is closely related to $1$ being or not being a congruent number.
If $X^4 + Y^2 = Z^4$ for positive integers $X$, $Y$, $Z$, then $1$ is a congruent number: $a^2 + b^2 = c^2$ with $(1/2)ab = 1$ using the positive rational numbers $a = Y/(XZ)$, $b = 2XZ/Y$, and $c = (X^4 + Z^4)/(XYZ)$.
Conversely, if $1$ is a congruent number, meaning there are positive rational numbers $a$, $b$, and $c$ such that $a^2 + b^2 = c^2$ and $(1/2)ab = 1$, then give $a$, $b$, and $c$ a common denominator, say $D$: $a = A/D$, $b = B/D$, and $c = C/D$ for positive integers $A, B, C$, and $D$. We get $X^4 + Y^2 = Z^4$ for the rational numbers $X = 2D$, $Y = |A^2 - B^2|$, and $Z = C$. (The integer $Y$ is not $0$, since if it were $0$ then $C^2 = A^2 + B^2 = 2A^2$, making $\sqrt{2} = C/A$ rational, which is false.)
To prove $X^4 + Y^2 = Z^4$ has no solution in positive integers, Fermat used his method of descent to turn a solution $(X,Y,Z)$ in positive integers into another solution $(X',Y',Z')$ of the same equation in positive integers where $Y' < Y$, and repeating that process enough times produces a contradiction since there's no infinite decreasing sequence of positive integers. If you try to run through that descent argument with the equation $X^4 + Y^4 = Z^4$ instead, then the argument doesn't work anymore because the descent process doesn't produce a new solution of the Fermat equation. So it was essential to be working in the broader context of nonsolvability of $X^4 + Y^2 = Z^4$ in order for the descent to work.
You can also relate the nonsolvability of the Fermat equation $x^4 + y^4 = z^4$ with $2$ not being a congruent number by replacing the Fermat equation with the more general equation $X^4 + Y^4 = Z^2$, where the right side is a square rather than a fourth power. The solvability of $X^4 + Y^4 = Z^2$ in positive integers turns out to be equivalent to the solvability of $a^2 + b^2 = c^2$ and $(1/2)ab = 2$ in positive rational $a$, $b$, and $c$, which says $2$ is a congruent number.
Almost every textbook treatment of Fermat's Last Theorem for exponent $4$ proceeds through the nonsolvability in positive integers of $X^4 + Y^4 = Z^2$ rather than $X^4 + Y^2 = Z^4$, which is understandable since the second equation looks a bit awkward compared to the first. So in fact textbook proofs of FLT for exponent $4$ are implicitly based on $2$ not being a congruent number rather than $1$ not being a congruent number. The books almost never explain where the equation $X^4 + Y^4 = Z^2$ comes from.
|
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|
How many positive integers less than 1,000,000 have the sum of their digits equal to 19? How many positive integers less than $1,000,000$ have the sum of their digits equal to $19$ ?
I tried to answer it by using stars and bars combinatorics method.
The question says that sum of $6$ digit numbers { considering the number as $abcdef$ } is 19.
Hence, I can write it as $a + b + c + d + e + f = 19$ ,but this assumes that $a,b,c,d,e,f >= 0$ .
But, here each digit lies between 0 and 9. So, how stars and bars method will be modified according to this domain ?
|
The number is $<$ $1000000$ ,
$\Rightarrow$ it contains 6 digits.
Each of these digits can be one of $0,1,2,3....9$
$\Rightarrow$ problem reduces to no of integral solution to the following equation
$d_1+d_2+d_3+d_4+d_5+d_6 = 19$ where $0\leq d_i \leq 9$
Using generating function :
$$\begin{align*} & \ \ \ \left [ x^{19} \right ]\left ( 1+x+x^2+x^3+....+x^9 \right )^{6} \\ &=\left [ x^{19} \right ]\left [ \frac{1-x^{10}}{1-x} \right ]^{6}\\ &=\left [ x^{19} \right ]\left ( 1-x^{10} \right )^{6}. \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \\ &=\left [ x^{19} \right ]\left [ \sum_{r=0}^{6}.\binom{6}{r}.\left ( -x^{10} \right )^{r} \right ]. \left [ \sum_{r=0}^{\infty}\binom{5+r}{r}.x^{r} \right ] \\ &=\left ( -1 \right )^{0}*\binom{24}{19} + \left ( -1 \right )^{1}*6*\binom{14}{9} \\ &=30492\\ \end{align*}$$
NOTE:
1. $1+x+x^2+x^3+.....x^n = \frac{1-x^{n+1}}{1-x}$
2. $\frac{1}{(1-x)^n} = \sum_{r=0}^{\infty}\binom{n+r-1}{r}.x^r$
3. $\left [ x^{19} \right ]$ means coefficient of $x^{19}$ of the whole expression.
|
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|
Finding the zeroes of the function. I would like to find the zeroes of the following function given that $3-i$ is a zero of $f$:
$f(x) = 2x^4-7x^3-13x^2+68x-30$
Please explain to me how to do this problem. Thanks!
|
Let $x_1,x_2,x_3,x_4$ be the roots of:
$$f(x) = 2x^4-7x^3-13x^2+68x-30$$
Then by Vieta's formulas:
$$
\begin{cases}
x_1+x_2+x_3+x_4=\frac{7}{2} \\
x_1x_2x_3x_4 = -\frac{30}{2}
\end{cases}
$$
Now you know $x_1=3-i$, and its conjugate must be also a root $x_2=3+i$, so the above reduces to:
$$
\begin{cases}
x_3+x_4=\frac{7}{2} - 6 = -\frac{5}{2} \\
x_3x_4 = -\frac{30}{2} / 10 = -\frac{3}{2}
\end{cases}
$$
Then, again by Vieta's formulas, the remaining roots $x_3,x_4$ must be the roots of:
$$x^2 + \frac{5}{2}x-\frac{3}{2} = 0$$
|
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Solution of this summation: $\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}$ The summation in question:
$$\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}$$
I have been able to factorize $i^4+i^2+1$ as $(i^2+i+1)(i^2-i+1)$ but I doubt this will help.
What is the solution?
|
Writting $$\sum^{10}_{i=1}\frac{i}{i^4+i^2+1} = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{(i^2+i+1)-(i^2-i+1)}{(i^2+i+1)(i^2-i+1)}\bigg]$$
$$ = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{1}{i^2-i+1}-\frac{1}{i^2+i+1}\bigg]$$
Now Using Telescopic Sum (expanding summation), Then we get
$$ =\frac{1}{2}\bigg[\frac{1}{1^2-1+1}-\frac{1}{10^2+10+1}\bigg]= \frac{1}{2}\left(1-\frac{1}{111}\right) = \frac{55}{111}$$
|
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How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
|
The polynomial in invariant under any permutation of $a,b,c$.
One way to attack this sort of polynomial is express it in terms
of its elementary symmetric polynomials and try to locate/detect any patterns that are useful.
We have $3$ variables, so we have $3$ elementary symmetric polynomials.
Let us call them $A,B,C$. They can be defined using following relations:
$$
(x-a)(x-b)(x-c) = x^3 - Ax^2 + Bx - C
\quad\iff\quad
\begin{cases}
A = a+b+c,\\
B = ab+bc+ca,\\
C = abc
\end{cases}$$
With help of $A,B,C$, we have:
$$\begin{align}
& {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc\\
= & ab(a+b) + bc(b+c)+ ca(c+a) + 2abc\\
= & ab(A-c) + bc(A-a)+ ca(A-b) + 2abc\\
= & (ab+bc+ca)A - abc\\
= & AB-C\\
= & A^3 - A\,A^2 + B\,A - C\\
= & (A-a)(A-b)(A-c)\\
= & (b+c)(c+a)(a+b)
\end{align}
$$
|
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How to find the eigenvalues of the following matrix. So I have to find the eigenvalues of this matrix: $\begin{bmatrix}-7/9&-4/9&8/9\\-4/9&-1/9&2/9\\2/9&-4/9&8/9\end{bmatrix}$.
What I did is start by writing it like this: $\begin{bmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{bmatrix}$ and then row reducing it to this: $\begin{bmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\0&-1+\lambda&1-\lambda\end{bmatrix}$. Then I found the determinate of this by expanding the bottom row. So I did this: $(-7/9 - \lambda)(2/9) + 32/81$ = $(-1)(2/9 - 2/9\lambda)(-1+\lambda)$ = $(2/9\lambda - 2/9)(-1+\lambda)$
Then for the 3rd column: $(-7/9 - \lambda)(-1/9 - \lambda)$ = $\lambda^2 + 8/9\lambda + 7/81 - 16/81$ = $(\lambda^2 + 8/9\lambda -1/9)(1-\lambda)$
So then I combine the two equations: $-(1+\lambda)((2/9\lambda - 2/9) + (\lambda^2 + 8/9\lambda-1/9))$ = $-(1+\lambda)(\lambda^2+10/9\lambda-1/3)$ and then I factor the polynomial to find out what lambda is.
The problem is I don't think my answer is correct since I can't factor my degree 2 polynomial where lambda equals two whole numbers. So where did I mess up in my answer?
Sidenote: I did try this as well by factoring out 1/9 first and I STILL GOT THE WRONG ANSWER. So can someone please help me and tell why I keep getting the wrong answer because this is very confusing and I really need some help.
|
Note: I recommend you factor out a $\dfrac{1}{9}$ to reduce the possibility of errors, so repeat this approach.
We want to find the determinant of
$$\begin{vmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{vmatrix}=0$$
Expanding along the top row since there are no zeros to take advantage of, we have:
$$\left(-\dfrac{7}{9}-\lambda \right)\begin{vmatrix}-\dfrac{1}{9}-\lambda &\dfrac{2}{9}\\-\dfrac{4}{9}&\dfrac{8}{9}-\lambda\end{vmatrix} + \dfrac{4}{9} \begin{vmatrix}-\dfrac{4}{9}&\dfrac{2}{9}\\\dfrac{2}{9}&\dfrac{8}{9}-\lambda\end{vmatrix} + \dfrac{8}{9} \begin{vmatrix}-\dfrac{4}{9} &-\dfrac{1}{9}-\lambda\\\dfrac{2}{9}&-\dfrac{4}{9}\end{vmatrix}$$
Hopefully, you can take it from here and arrive at:
$$\lambda - \lambda^3 = 0 \implies \lambda_{1,2,3} = -1, 0, 1$$
Update Expanding the three determinants, we have
$$\left(\lambda-\frac{7}{9}\right) \left(\lambda^2-\frac{7 \lambda}{9}\right)+\frac{4}{9} \left(\frac{4 \lambda}{9}-\frac{4}{9}\right)+\frac{8}{9} \left(\frac{2 \lambda}{9}+\frac{2}{9}\right)$$
This reduces to
$$\dfrac{49 \lambda}{81}-\lambda^3 + \dfrac{16 (\lambda-1)}{81} + \dfrac{16 (\lambda+1)}{81} = \dfrac{49 \lambda}{81}-\lambda^3 + \dfrac{32 \lambda}{81} = \lambda - \lambda^3$$
|
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|
Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$ Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$
Here i provide proof found in book. But i don't understand a lot from this proof.
$\overline x$ denotes an element in the quotient group involved.
Here we have $(2 − 5j)(2 − 5j^2) = 4 − 10(j + j^2) + 25j^3 = 4 + 10 + 25 = 39.$ Hence $39\cdot \overline 1 = \overline {39} = \overline{(2 − 5j) · (2 − 5j^2)} = \overline 0.$ Then the characteristic of $R$ is finite and divides
$39.$ Therefore the characteristic of $R$ is $1, 3, 13$ or $39.$ Now
let $c = \mbox{char}(R) > 0.$ Since $c \cdot 1_R$ lies in the ideal
$(2 − 5j),$ then $c = (2 − 5j)(a + bj)$ for some $a, b, \in Z.$ Hence
$|c|^2 = |2 − 5j|^2|a + bj|^2,$ so $$c^2 = 39(a^2 + b^2 − ab)$$ and
therefore $39|c^2.$ The only value (among $1, 3, 13$ and $39$) for
which it is possible is $c = 39.$ Thus $\mbox{char}(R) = 39.$
Why $j+j^2 = -1?$
Why $|2 − 5j|^2|a + bj|^2= 39(a^2 + b^2 − ab)?$
|
$j^3 = 1$ impies $0=j^3-1=(j-1)(j^2+j+1)$.
$j^2 \ne 1$ implies $j \ne 1$ and so $j^2+j+1=0$.
$|\cdot|^2$ here is just the complex number absolute value, $|z|^2 = z \bar z$, which implies $|zw| = |z| \, |w|$.
Using that $\bar j = j^2 = -1-j$, it is easy to compute $|a + bj|^2 = a^2 + b^2 − ab$.
|
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|
Show $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges I'm trying to show that $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges. Using the Limit Ratio Test for Series, we want to show that $\lim_{n\to \infty} \left\lvert \frac{a_{n+1}}{a_n}\right\lvert<1$. However, I'm having trouble finding said limit (I know that it is equal to $\frac{1}{2}$, but I don't know how show it). Thanks in advance
|
If $a_n = \left(\frac{n}{2n-1}\right)^n$, then
\begin{align} \frac{a_{n+1}}{a_n} &= \left(\frac{n+1}{2(n+1)-1}\right)^{n+1}/\left(\frac{n}{2n-1}\right)^n \\
&= \left(\frac{n+1}{2n+1}\right)\frac{\left(\frac{n+1}{2n+1}\right)^n}{\left(\frac{n}{2n-1}\right)^n} \\
&= \left(\frac{n+1}{2n+1}\right)\left(\frac{(n+1)(2n-1)}{n(2n+1)}\right)^n \\
&=\left(\frac{n+1}{2n+1}\right)\left(\frac{2n^2+n-1}{2n^2+n}\right)^n.
\end{align}
Now, $\frac{2n^2+n-1}{2n^2+n}<1$, so it follows that $\lim\limits_{n\rightarrow\infty}{\left(\frac{2n^2+n-1}{2n^2+n}\right)^n}\le 1$. (It is not too hard to show, in fact, that the limit is exactly $1$.) Hence
$$\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} = \left(\lim\limits_{n\rightarrow\infty}{\frac{n+1}{2n+1}}\right)\left(\lim\limits_{n\rightarrow\infty}{\left(\frac{2n^2+n-1}{2n^2+n}\right)^n}\right)\le\frac{1}{2}\cdot 1 = \frac{1}{2} $$
i.e. $\lim\limits_{n\rightarrow\infty}{\left|\frac{a_{n+1}}{a_n}\right|} < 1$.
|
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|
52 cards, 5 picked, 3 of same suit 5 cards are drawn from a normal deck of cards (52). What is the probability that 3, and only 3, of the cards are of the same suit?
I'm wondering if my reasoning is sound:
C(13,1) * C(4,3) * C(12,2) * C(4,1)^2
-------------------------------------
C(52,2)
|
3 solutions.
Solution1: Exclude 1 suit from 52
Solution2: Consider cases for pair or not a pair for the 2 cards left
Solution3: Consider cases for cards of same suit or not of the same suit for the 2 cards left
Solution2:
4 ways for choose 3 cards from 13. (same suit)
for remaining 2 cards, we have 2 cases to consider:
1. can either have a pair. ( 3 choose 2) from 3 suits left.
2. not have a pair. (3 choose 1) 3 suits left.
$$\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot \left( {13 \choose 2} \cdot {3 \choose 1}^2 + {13 \choose 1 } \cdot {3 \choose 2}\right) }{{52 \choose 5}}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (702+39)}{52 \choose 5}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (741)}{52 \choose 5}$$
this would give the same solution as $${39 \choose 2} = 741$$
(exclude 1 suit from 52 cards)
Solution3:
$$\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot \left( {13 \choose 2} \cdot {3 \choose 1} + {13 \choose 1 }^2 \cdot {3 \choose 2}\right) }{{52 \choose 5}}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (234+507)}{52 \choose 5}$$
$$=\frac{{4 \choose 1} \cdot {13 \choose 3} \cdot (741)}{52 \choose 5}$$
this would give the same solution as $${39 \choose 2} = 741$$
(exclude 1 suit from 52 cards)
|
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|
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$.
So if we substitute $z=1-x-y$ into the equation $x^2+y^2=z$ we get $x^2+y^2=1-x-y$ which can be written in the form $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=1$.
Is this the curve of intersection? Because it seems to me (by geometrical intuition) that the curve should be an ellipse. What is the logical flaw here. Help please!
|
After the coments, a possible parametrization is
$$r(t)=\left(\,-\frac12+\sqrt\frac32\,\cos t\,,\,\,-\frac12+\sqrt\frac32\,\sin t\,,\,\,2-\sqrt\frac32(\cos t+\sin t)\,\right)\;,\;\;0\le t\le 2\pi$$
Observe that we have above a general vector of the form $\;(x,y,z)\;$ , where for the third coordinate we have either $\;z=x^2+y^2\;$ or $\;z=1-x-y\;$ .
|
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how can I prove this identity How can I prove this identity
$$\cos (3x/2)-\cos(x/2)=-2 \sin (x/2) \sin(x)$$
I try this but does not work with me $$\cos(3x/2) = \cos(x + x/2) = \cos (x) \cos (x/2) - \sin (x) \sin (x/2)$$
I do not know what to do after this step.
|
We have $$\cos(A+B) -\cos(A+B) = (\cos A\cos B -\sin A\sin B)-(\cos A\cos B +\sin A\sin B) = -2\sin A\sin B. $$ Let $A+B = C$ and $A-B = D$. Our equation then becomes $\cos C -\cos D = -2\sin (\frac{C+D}{2})\sin (\frac{C-D}{2})$. With $C=\frac{3x}{2}$ and $D=\frac{x}{2}$, we have $$ \cos(\frac{3x}{2})-\cos(\frac{x}{2}) = -2\sin (x) \sin (\frac{x}{2})$$
|
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|
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $\ln(2)=0.693$.
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
$$
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
$$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$
I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?
|
As others have pointed out, rearrangement is not allowed, so I will give you the most extreme case:
$$S=1-\frac12-\frac14-\frac16-\frac18+\frac13+\frac15+\frac17+\frac19+\frac1{11}+\frac1{13}+\dots$$
The general pattern is simple. Start at $1$. Then add up the even terms until the sum is less than $0$. Then add up the odd terms until the sum is more than $1$. Then add up the even terms until the sum is less than $0$ again...
Clearly if we repeat this process indefinitely, we have rearranged it so that the sum never converges and oscillates between $0$ and $1$.
|
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|
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$?
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$
If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1}{x+y}dxdy$
Is it possible to see it from Taylor ?
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We have $$\frac{2a+2a+3}{2a+2b+1} = 1+ \frac{2}{2a+2b+1} = 1+\frac{1}{a+b+1/2}.$$
The Maclauren series for $\log(1-x)=-x-x^2/2-x^3/3-\cdots.$ So your left side
$$=\log\left(1-\frac{-1}{a+b+1/2}\right) = \frac{1}{a+b+1/2} - \frac{1}{2}\left(\frac{1}{a+b+1/2}\right)^2+\cdots . $$
The series is alternating so the above is
$$> \frac{1}{a+b+1/2}>\frac{1}{a+b+1}.$$
|
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Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$
Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.
I would like to see a solving method without l'Hopital or Taylor expansion.
|
\begin{align*}
L &= \lim_{x\to \infty} (a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}) \\[5pt]
&= \lim_{x\to \infty} \sqrt{x}
\left(
a\sqrt{1+\frac{1}{x}}+2b\sqrt{1+\frac{1}{2x}}+3c\sqrt{1+\frac{1}{3x}}
\right) \\[5pt]
&= \lim_{x\to \infty} \sqrt{x}
\left[
a\left( 1+\frac{1}{2x} \right)+
2b\left( 1+\frac{1}{4x} \right)+
3c\left( 1+\frac{1}{6x} \right)
\right] \\[5pt]
&=
\left \{
\begin{array}{ccl}
\infty &,& a+2b+3c \ne 0 \\
0 &, & a+2b+3c=0
\end{array}
\right.
\end{align*}
For one case of $(a,b,c)=(1,1,-1) \implies a+2b+3c=0$.
See the comparison of
$\color{blue}{y=a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}}$ and
$\color{red}{y=\dfrac{a+b+c}{2\sqrt{x}}}$ below:
|
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|
Use Mathematical Induction to prove equation? Use mathematical induction to prove each of the following statements.
let $$g(n) = 1^3 + 2^3 + 3^3 + ... + n^3$$
Show that the function $$g(n)= \frac{n^2(n+1)^2}{4}$$ for all n in N
so the base case is just g(1) right? so the answer for the base case is 1, because 4/4 = 1
then for g(2) is it replace all of the n's with n + 1 and see if there is a concrete answer?
|
Assuming you are asking what the inductive step is, you simply need to assume that $g(n) =\frac{n^2(n+1)^2}{4}$ and then use it to evaluate g(n+1) like so:
\begin{equation}
g(n+1) = g(n) + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{n^2(n+1)^2}{4} + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{4(n+1)^3 + n^2(n+1)^2}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(4(n+1) + n^2)}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(n^2 + 4n + 4)}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(n+2)^2}{4}
\end{equation}
Q.E.D.
|
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Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
|
$x=0\iff y=0.$ And $x=y=0$ is a solution.
If $x\ne 0\ne y$ then $$2x^2+y^2=3x^2y\implies x^2(2-3y^2)=-y^2\ne 0.$$ This implies $x^2$ divides $y^2.$ Now $y^2/x^2$ is an integer for non-zero integers $x,y$ only if $x$ divides $y$. So we have $$y=zx$$ with integer $z.$
So $2x^2+z^2x^2=3x^3z$. Since $x\ne 0$ we have $2+z^2=3xz.$... So $$z(z-3x)=-2$$ which implies that $z$ divides $2$.
Putting $y=zx$ with $z\in \{\pm 1 ,\pm 2\}$ into the original equation gives us the non-zero solutions :$$(y=1, x=\pm 1), (y=2,x=\pm 1).$$ For example if $z=1$ then $y=x$ and $3x^2=2x^2+y^2=3x^2y=3x^3\ne 0$ implies $1=x=y.$
|
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|
Distributing $8$ different articles among $7$ boys
Problem Statement:-
Find the number of ways in which $8$ different articles can be distributed among $7$ boys, if each boy is to receive at least one article.
Attempt at a solution:-
First, start by numbering the boys from $1$ to $7$, which can be done in $7!$ ways.
After the boys have been numbered then we resume our task of distributing the articles to the boys. As, each boy is supposed to at least get a single article.
Hence,
$\text{Ways of giving $1$ article to the $1^{\text{st}}$ boy= $8\choose1$}\\
\text{Ways of giving $1$ article to the $2^{\text{nd}}$ boy= $7\choose1$}\\
\text{Ways of giving $1$ article to the $3^{\text{rd}}$ boy= $6\choose1$}\\
\vdots\\
\text{Ways of giving $1$ article to the $7^{\text{th}}$ (and the last) boy= $2\choose1$}$
Now remains the last article, which can go to any one boy of the $7$ boys in ${{7}\choose{1}}=7$ ways
So, the number of ways of distributing $8$ articles among $7$ boys$=$
$$7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}$$
But, the arrangement of the two things that the $1^{\text{st}}$ boy gets is of no relevance so we gotta remove the consideration of the two objects being permuted among themselves.
Hence, the total the number of ways of distributing $8$ articles among $7$ boys$=$
$$\frac{7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}}{2!}$$
The textbook gives the answer to the question as
$$7\times \left( \binom{8}{2} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} \times \binom{1}{1}\right)$$
Where am I going wrong
|
The main problem here is that you have ordered the boys and also regarded the distribution of the presents as extra options.
Probably the simplest way to get to the answer is to select one boy to get two presents ($7$ options), he chooses those two ($\binom 82$) then each boy chooses in turn ($6!$ options) :
$$7\binom 82 \binom 61 \binom 51 \binom 41 \binom 31 \binom 21 \binom 11 = 7\binom 82 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 7!\binom 82 $$
Another way to view the last formulation is that you pick two presents to "bundle", then distribute the 7 packages to the 7 boys.
|
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|
Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}+\sqrt{\frac{8ab+8bc+9ac}{(2a+c)(a+2c)}}+\sqrt{\frac{8ac+8bc+9ab}{(2a+b)(a+2b)}}\geq5$$
I tried Holder, but without success.
For example, Holder even with $(ka^2+b^2+c^2+nab+nac+mbc)^3$ does not help.
SOS or C-S seems very ugly here.
The equality occurs also for $(a,b,c)=(1,1,0)$, which adds a problems.
|
There is the following estimation of Nguyenhuyen_AG.
$$\sqrt{\frac{8a(b+c)+9bc}{(2b+c)(2c+b)}} \geqslant\frac{17a^2-b^2-c^2+20(ab+ac+bc)}{3(a^2+b^2+c^2+4(ab+bc+ca))}$$
|
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"url": "https://math.stackexchange.com/questions/2057006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove Fibonacci Identity using generating functions I have the following summation identity for the Fibonacci sequence.
$$\sum_{i=0}^{n}F_i=F_{n+2}-1$$
I have already proven the relation by induction, but I also need to prove it using generating functions, but I'm not entirely sure how to approach it.
I do know that the generating function for the fibonacci sequence is $$F(x) = \dfrac{1}{1-x-x^2}$$
But, I'm not entirely sure if that applies here. Any help would be appreciated!
|
Since a generating function for the Fibonacci numbers $(F_n)_{n\geq 0}=(1,1,2,3,5,8,\ldots)$ is
\begin{align*}
\frac{1}{1-x-x^2}=1+x+2x^2+3x^3+5x^4+8x^5+\cdots
\end{align*}
and multiplication of a generating function $A(x)$ with $\frac{1}{1-x}$ results in summing up the coeffcients
\begin{align*}
\frac{1}{1-x}A(x)&=\frac{1}{1-x}\sum_{n=0}^\infty a_nx^n\\
&=\sum_{n=0}^\infty\left(\sum_{i=0}^n a_i\right)x^n
\end{align*}
we can show the following:
\begin{align*}
\sum_{i=0}^n F_i=F_{n+2}-1\qquad\qquad n\geq 0\tag{1}
\end{align*}
A generating series of the LHS of (1) is
\begin{align*}
\sum_{n=0}^\infty\left(\sum_{i=0}^n F_i\right)x^n=\frac{1}{1-x}\cdot\frac{1}{1-x-x^2}
\end{align*}
A generating series of the RHS of (1) is
\begin{align*}
\sum_{n=0}^\infty \left(F_{n+2}-1\right)x^n&=\sum_{n=0}^\infty F_{n+2}x^n-\sum_{n=0}^\infty x^n\\
&=\sum_{n=2}^\infty F_n x^{n-2}-\frac{1}{1-x}\tag{2}\\
&=\frac{1}{x^2}\left(\frac{1}{1-x-x^2}-1-x\right)-\frac{1}{1-x}\tag{3}\\
&=\frac{1}{(1-x)(1-x-x^2)}\\
\end{align*}
and the claim follows.
Comment:
*
*In (2) we shift the index $n$ by two to start from $n=2$ and use the formula for the geometric series expansion.
*In (3) we use the generating function series of the Fibonacci numbers and do some simplifications in the line after.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2060740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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|
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am unable to do by hand.
|
Substituting $x\mapsto\frac1x$, we get
$$
\begin{align}
\int_1^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{x\,\mathrm{d}x}{1+x^5}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{\left(x^2+x\right)\mathrm{d}x}{1+x^5}\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\left(x^{5k+1}+x^{5k+2}\right)\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\left(\frac1{5k+2}+\frac1{5k+3}\right)\\
&=\sum_{k=1}^\infty\left(\frac1{10k-8}+\frac1{10k-7}-\frac1{10k-3}-\frac1{10k-2}\right)\\
&=\frac1{10}\sum_{k=1}^\infty\left(\frac1{k-\frac8{10}}+\frac1{k-\frac7{10}}-\frac1{k-\frac3{10}}-\frac1{k-\frac2{10}}\right)\\[3pt]
&=\frac1{10}\left(H_{-\frac3{10}}+H_{-\frac2{10}}-H_{-\frac8{10}}-H_{-\frac7{10}}\right)\\[9pt]
&=\frac\pi{10}\left(\cot\left(\frac{3\pi}{10}\right)+\cot\left(\frac{2\pi}{10}\right)\right)\\[6pt]
&=\frac\pi5\sqrt{2-\frac2{\sqrt5}}
\end{align}
$$
Using $(5)$ from this answer, which is probably out of scope.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2061215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Combinatorial coefficients squared If $C_0,C_1,C_2,...,C_n$ are the combinatorial coefficients in the expansion of $(1+x)^n$ then prove that:
$$1C_0^2+3C_1^2+5C_3^2+...+(2n+1)C_n^2=\dfrac{(n+1)(2n)!}{n!n!}=(n+1)\binom {2n}n$$
I am able to compute the linear addition but not the squares of coefficients.
Thanks!
|
Note that
$C_r=\binom nr$.
It is a well-known result that $\sum_{r=0}^n\binom nr^2=\binom {2n}n$ which can be proven easily using the Vandermonde identity.
For even $n$:
The number elements is odd, index from $0$ to $(\frac n2-1)$, $n$, and from $(\frac n2+1)$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^{\frac n2-1}\binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned}
&\begin{array}r
1\binom n0^2
&+3\binom n1 ^2
&+\cdots
&+(n-1)\binom n{\frac n2-1}^2\\
& & & & +(n+1) \binom n{\frac n2} ^2\\
+(2n+1)\binom nn^2
&+(2n-1)\binom n{n-1}^2
&+\cdots
&+(n+3)\binom n{\frac n2+1}^2
\\
=(2n+2)\binom n0^2
&+(2n+2)\binom n1^2
&+\cdots
&+(2n+2)\binom n{\frac n2-1}^2
&+(n+1)\binom n{\frac n2}^2
\\
\end{array}\\
&=(2n+2)\cdot \displaystyle\sum_{r=0}^{\frac n2-1}\binom nr^2+(n+1)\binom n{\frac n2}\\
&=(2n+2)\cdot \frac 12 \left[\displaystyle\sum_{r=0}^{n}\binom nr^2-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\
&=(n+1)\left[\binom {2n}n-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\
&=\color{red}{(n+1)\binom {2n}n}
\end{aligned}$$
__
For odd $n$:
The number elements is even, index from $0$ to $\frac {n-1}2$, and from $\frac {n+1}2$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^\frac {n-1}2 \binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned}
&\begin{array}r
1\binom n0^2
&+3\binom n1 ^2
&+\cdots
&+n\binom n{\frac {n-1}2}^2\\
+(2n+1)\binom nn^2
&+(2n-1)\binom n{n-1}^2
&+\cdots
&+(n+2)\binom n{\frac {n+1}2}^2
\\
=(2n+2)\binom n0^2
&+(2n+2)\binom n1^2
&+\cdots
&+(2n+2)\binom n{\frac{n-1}2}^2\\
\end{array}\\
&=(2n+2)\displaystyle\sum_{r=0}^{\frac {n-1}2}\binom nr^2\\
&=(2n+2)\cdot \frac 12\displaystyle\sum_{r=0}^{n}\binom nr^2\\
&=(2n+2)\cdot \frac 12 \binom {2n}n\\
&=\color{red}{(n+1)\binom {2n}n}
\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2061461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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|
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