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Differentiation of a function We need to find out the derivative $\frac{dy}{dx}$ of the following:
$x^m$$y^n$=$(x+y)^{m+n}$
*
*I know how to differentiate the function and on solving we get $\frac{dy}{dx}$=$\frac{y}{x}$. But we notice that $\frac{dy}{dx}$ is independent of values of $m$ and $n$. So what I did was again starting the problem from start but this time substituting any arbitrary values of $m$ and $n$, like $m$=$n$=$1$ (lets say). That is we get $xy$=$(x+y)^2$. Now opening the square bracket and solving we get
$x^2+y^2+xy=0$
So $\frac{dy}{dx}$=$\frac{-2x-y}{2y+x}$
But the answer should be $\frac{y}{x}$.Whats wrong in assuming arbitrary values of $m$ and $n$ if the answer dosen't depend upon values of $m$ and $n$.Please help.
How I got $\frac{dy}{dx}$=$\frac{y}{x}$?
$x+y$=$x^\frac{m}{m+n}$$y^\frac{m}{m+n}$
$ln(x+y)$=$\frac{m(ln(x))}{(m+n)}$+$\frac{n(ln(y))}{(m+n)}$
Differentiating we get
$\frac{1}{x+y}$.$(1+\frac{dy}{dx})$=$\frac{m}{(m+n)x}$+$\frac{n}{(m+n)y}\frac{dy}{dx}$
$\frac{dy}{dx}(\frac{1}{x+y}-\frac{n}{(m+n)y})$=$\frac{m}{(m+n)x}-\frac{1}{x+y}$
Just rearrange and you are done $\frac{dy}{dx}$=$\frac{y}{x}$
|
Take $log$ on both sides and differentiate w.r.t $x$;
$\frac{m}{x} +\frac{n}{y}. y'=\frac{(m+n)}{(x+y)}. (1+y')$
$\implies y'(\frac{n}{y}-\frac{m+n}{x+y})=\frac{m+n}{x+y}-\frac{m}{x}$
$\implies y'=\frac{y}{x}$ on simplification
|
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|
Writing columns of a matrix as linear combinations of other columns Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$
What is the best way to write the fifth and sixth columns of the matrix as linear combinations of the first and third columns?
|
First consider column 5
We can think of it as
\begin{align}a\begin{pmatrix}1\\2\\1\end{pmatrix}+b\begin{pmatrix}3\\5\\2\end{pmatrix}&=\begin{pmatrix}1\\3\\2\end{pmatrix}\\
&\Downarrow \\
a+3b&=1\\
2a+5b&=3\\
a+2b&=2
\end{align}
So $a=1-3b$
Therefore \begin{align}2(1-3b)+5b&=3\\
2-6b+5b&=3\\
2-3&=b\\
b&=-1
\end{align}
And so $a=1-(3\times-1)=4$
So we have \begin{align}4\begin{pmatrix}1\\2\\1\end{pmatrix}-\begin{pmatrix}3\\5\\2\end{pmatrix}&=\begin{pmatrix}4\\8\\4\end{pmatrix}-\begin{pmatrix}3\\5\\2\end{pmatrix}\\
&=\begin{pmatrix}1\\3\\2\end{pmatrix}
\end{align}
We can then do this again for column 6
\begin{align}c\begin{pmatrix}1\\2\\1\end{pmatrix}+d\begin{pmatrix}3\\5\\2\end{pmatrix}&=\begin{pmatrix}0\\1\\1\end{pmatrix}\\
&\Downarrow \\
c+3d&=0\\
2c+5d&=1\\
c+2d&=1\end{align}
So we have $c=-3d$
Therefore we have \begin{align}2(-3d)+5d&=1\\
-6d+5d&=1\\
-d&=1\\
d&=-1
\end{align}
And so $c=-3d= -3\times -1 = 3$
And so we have \begin{align}3\begin{pmatrix}1\\2\\1\end{pmatrix}-\begin{pmatrix}3\\5\\2\end{pmatrix}&=\begin{pmatrix}3\\6\\3\end{pmatrix}-\begin{pmatrix}3\\5\\2\end{pmatrix}\\
&=\begin{pmatrix}0\\1\\1\end{pmatrix}
\end{align}
|
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|
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$ ?
My Try :-
Total pairs of $(a,b)$ possible are $81$ .
*
*If $a=1$, then $b = 1,2,3,4,5$. Similarly for $a=2$.
*If $a=3$, then $b = 1,2,3,4,5,6$. Similarly for $a=4$.
*If $a=5$, then $b = 1,2,3,4,5,6,7$. Similarly for $a=6$.
*If $a=7$, then $b = 1,2,3,4,5,6,7,8$. Similarly for $a=8$.
*If $a=9$, then $b = 1,2,3,4,5,6,7,8,9$.
Total favourable pairs are $61$.
Hence, Total Probability = $\frac{61}{81}$
However, I don't have an answer for this. Am I right or missing something ?
|
Your solution is correct, but tedious.
A quicker way would be to note that the expression can only be non-positive if $b>5$, which gives only four cases for $b$. In these four cases of $b$, there are $2(b-5)$ values for $a$ that give non-positive results. Therefore the total probability for the expression to be positive is
$$p = 1-\frac{2+4+6+8}{81} = \frac{81 - 20}{81} = \frac{61}{81}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is $\int_0^{\infty} \frac{x^{p}}{(1+x^2)^3} dx$? We know that
$\int_0^{\infty} \frac{1}{1+x^2} dx = \frac{\pi}{2}.$
(In fact, $F(x)= \int_0^{x} \frac{1}{1+t^2}dt = [\arctan (t)]_0^{x}$, and so $\lim_{x\to \infty} F(x) = \frac{\pi}{2}.$)
Let $p \in \{2,3,4\}, n\in \mathbb N$
How should I evaluate $\int_0^{\infty} \frac{x^{p}}{(1+x^2)^3} dx$? What is $\int_0^{\infty} \frac{(x-n)^{p}}{(1+(x-n)^2)^3} dx$?
|
In the integral with $p=3$, the subsitution $y= x^2$ works out well. For $p=2$ and $p=4$ you can use the following partial fractions expansions
$$\frac{x^2}{(1+x^2)^3} = \frac{1}{(1+x^2)^2}-\frac{1}{(1+x^2)^3} $$
$$\frac{x^4}{(1+x^2)^3} = \frac{1}{1+x^2}-\frac{2}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}$$
and the problem is reduced to find a primitive for
$$ \frac{1}{(1+x^2)^2} \quad \text{and} \quad \frac{1}{(1+x^2)^3}, $$
which can be easily done integrating
$$\int 1 \cdot \frac{1}{1+x^2}dx $$
by parts.
|
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|
Showing that $m^2-n^2+1$ is a square Prove that if $m,n$ are odd integers such that $m^2-n^2+1$ divides $n^2-1$ then $m^2-n^2+1$ is a square number.
I know that a solution can be obtained from Vieta jumping, but it seems very different to any Vieta jumping problem I've seen.
To start, I chose $m=2a+1$ and and $n=2b+1$ which yields: $$ 4ka^2+4ka-4kb^2-4kb+k = b^2+b$$
Then suppose that $B$ is a solution, and $B_0$ is another solution. Then using Vieta jumping we get (with a bit of algebra) that $B+B_0 = -1$ and $B_0 = \frac {-k(2a+1)^2}{B(4k+1)}$.
But I'm not sure these final equalities are particularly helpful; I can't find any way to yield more solutions from them. How can I solve the problem? A solution without Vieta jumping is probably also possible
|
I have produced the lemma that rules out negative ratios. It is to be applied after the business of wrting the sum and difference of the pair of odd integers as double new variables.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
LEMMA
Given integers $$ m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with
$$ x^2 - Mxy + y^2 = -m. $$
PROOF
Calculus: $m+2 > \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left( \sqrt{4m+4} \right)^2 = 4m + 4.$ Therefore also
$$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 \geq 2 + M > 0. $ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$
With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get
$x < My.$
The point on the hyperbola $ x^2 - Mxy + y^2 = -m $ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and
$$ t^2 = \frac{m}{M-2}. $$
We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$
More important than first appears, that this point is inside the unit square.
We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by
$$ (x,y) \mapsto (Mx - y, x) $$
gives a smaller $x+y$ value. Or, if $2x > My,$
$$ (x,y) \mapsto (y, My - x) $$
gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$
Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the Hurwitz conditions for a fundamental solution (Grundlösung), namely
$$ 2y \leq Mx \; \; \; \; \mbox{AND} \; \; \; \; 2 x \leq My. $$
We now just fiddle with calculus type stuff, that along the hyperbola arc bounded by the Hurwitz inequalities, either $x < 1$ or $y < 1,$ so that there cannot be any integer lattice points along the arc. We have already shown that the middle point of the arc lies at $(t,t)$ with $t < 1.$ We just need to confirm that the boundary points also have either small $x$ or small $y.$ Given $y = Mx/2,$ with
$$ x^2 - Mxy + y^2 = -m $$ becomes
$$ x^2 - \frac{M^2}{2} x^2 + \frac{M^2}{4} x^2 = -m, $$
$$ x^2 \left( 1 - \frac{M^2}{4} \right) = -m $$
$$ x^2 = \frac{-m}{1 - \frac{M^2}{4}} = \frac{m}{ \frac{M^2}{4} - 1} = \frac{4m}{M^2 - 4}. $$
We already confirmed that $ M > \sqrt{4m+4}, $ so $M^2 > 4m+4$ and $M^2 - 4 > 4m.$ As a result, $ \frac{4m}{M^2 - 4} < 1.$ The intersection of the hyperbola with the Hurwitz boundary line $2y = Mx$ gives a point with $x < 1.$ Between this and the arc middle point, we always have $x < 1,$ so no integer points. Between the arc middle point and the other boundary point, we always have $y < 1.$ All together, there are no integer points in the bounded arc. There are no Hurwitz fundamental solutions. Therefore, there are no integer solutions at all.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
|
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|
$f(x,y) = 0$ when $x = y = 0$. Shouldn't $\frac{\partial^2 f}{\partial y \partial x} = 0$? I'm given the following function:
$f(x,y) =$
$
\begin{cases}
2xy \frac{x^2 - y^2}{x^2 + y^2} & x^2 + y^2 \neq 0 \\
0 & x=y=0
\end{cases}
$
And the task is to prove that $f_{xy}(0,0) = -2$ and that $f_{yx}(0,0) = 2$.
At first glance I'd say that neither of these is true since at $(0,0)$ the function is 0 and so both expressions should equal 0. I thought maybe I was wrong and tried to work with the other piece of the function but I got the following:
$$
f_{xy}=\frac{\partial^2 f}{\partial y \partial x} = \frac{2(x^6+9x^4y^2-9x^2y^4-y^6)}{(x^2 + y^2)^3}
$$
Which is clearly not $-2$ if we evaluate for $x=0, y=0$.
What am I missing? Am I going about it all wrong? Any help is appreciated.
|
Let $f$ be given by
$$f(x,y)=\begin{cases}
2xy\frac{x^2-y^2}{x^2+y^2}&,x^2+y^2>0\\\\
0&,x^2+y^2=0
\end{cases}$$
For $x^2+y^2>0$, we have
$$\begin{align}\frac{\partial f(x,y)}{\partial x}&=\frac{2y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\\\\\frac{\partial f(x,y)}{\partial y}&=-\frac{2x(y^4+4x^2y^2-x^4)}{(x^2+y^2)^2}\end{align}$$
To calculate the first partial derivaties at the origin, we revert to using the limit definition of the partial derivatives.
Proceeding, we find that
$$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=0\\\\
f_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=0$$
We proceed in a similar fashion to find the mixed partial derivatives at the origin. We find that
$$\begin{align}
f_{yx}(0,0)&=\lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h}\\\\
&=\lim_{h\to 0}\frac{-\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\
&=2
\end{align}$$
and
$$\begin{align}
f_{xy}(0,0)&=\lim_{h\to 0}\frac{f_x(0,h)-f_x(0,0)}{h}\\\\
&=\lim_{h\to 0}\frac{\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\
&=-2
\end{align}$$
as was to be shown!
|
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|
Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$
Let $A$ and $B$ be $3 \times 3$ matrices with integer entries so that $AB = BA$ and $\det(A) = \det(B) = 0$. Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$.
I thought about factorizing $A^3+B^3$. We have $$A^3+B^3 = (A+B)(A^2-AB+B^2).$$ Then we have $$\det(A^3+B^3) = \det(A+B)\det(A^2-AB+B^2).$$ How do we use the fact that $A$ and $B$ are $3 \times 3$? For $3 \times 3$ matrices is it true that $\det(A+xB) = x(a+bx)$ where $a,b$ are integers? If so, then we have since $$A^2-AB+B^2 = (A-e^{2\pi i/3}B)(A-e^{-2\pi i/3}B),$$ that $$\det(A^3+B^3) = (a+b)(a-be^{2\pi i/3})(a-be^{-2 \pi i/3}).$$
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An alternative, with the additional assumption that both $A$ and $B$ are diagonalizable: Then, since they commute we can write
$$A=PD_AP^{-1}, B=PD_BP^{-1}$$
where $D_A$ and $D_B$ are diagonal matrices. We have
$$A^{3}=PD_A^{3}P^{-1}, B^{3}=PD_B^{3}P^{-1}$$
and
$$det(A^{3}+B^{3})=det(P(D_A^{3}+D_B^{3})P^{-1})=det(D_A^{3}+D_B^{3}).$$
Now, let the diagonal of $D_A$ be $(a_1, a_2, a_3)$ and that of $D_B$ analogously. Thus we get
$$det(D_A^{3}+D_B^{3})=(a1^{3}+b_1^{3})(a_2^{3}+b_2^{3})(a_3^{3}+b_3^{3})$$
We know since $det(A)=det(B)=0$ that at least one of $a_1,a_2,a_3$ must be zero. The same holds for $B$. Now, we can identify two cases:
*
*$det(A^{3}+B^{3})=0$ This corresponds to the case when $a_i=b_i=0$ for some $i$ (i.e. the eigenvalue at the same position is zero). Then, $p=q=0$ solves the problem.
*$det(A^{3}+B^{3})\neq 0$. This means that $a_ib_i\neq 0$ for all $i=1,2,3$. W.l.o.g. assume $a_1=0, b_2=0$. Then $det(A^{3}+B^{3})=b_1^{3}a_2^{3}(a_3^{3}+b_3^{3})=(b_1a_2a_3)^{3}+(b_1a_2b_3)^{3}$ proving the second case.
|
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|
Three Distinct Points and Their Normal Lines
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $y=x^2$ such that $a\not=b\not=c$.
Find the tangent slope and the slope of their normal lines.
$$(a,a^2) \hspace{4mm}m_{tan}=2a, \hspace{4mm} m_{norm}=\frac{-1}{2a}$$
$$(b,b^2) \hspace{4mm}m_{tan}=2b, \hspace{4mm} m_{norm}=\frac{-1}{2b}$$
$$(c,c^2) \hspace{4mm}m_{tan}=2c, \hspace{4mm} m_{norm}=\frac{-1}{2c}$$
Normal Line $(a,a^2)$
$y-a^2=-\frac{1}{2a}(x-a) \implies y=-\frac{1}{2a}x+\frac{1}{2}+a^2$
Normal Line $(b,b^2)$
$y-b^2=-\frac{1}{2b}(x-b) \implies y=-\frac{1}{2b}x+\frac{1}{2}+b^2$
Normal Line $(c,c^2)$
$y-c^2=-\frac{1}{2c}(x-c) \implies y=-\frac{1}{2c}x+\frac{1}{2}+c^2$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(b,b^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2b}x+\frac{1}{2}+b^2 \implies x=-(b+a)2ab$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(c,c^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+a)2ac$
Find the $x$-interception points between the normal lines of $(b,b^2)$ and $(c,c^2)$.
$-\frac{1}{2b}x+\frac{1}{2}+b^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+b)2bc$
Show that $a+b+c=0$
$$\begin{align}
....\\
....\\
....\\
\end{align}$$
I do not know how to show that $a+b+c=0$. Any advice on how to continue? Thanks in advance!
|
$b^2a+a^2b=c^2a+a^2c=c^2b+b^2c\rightarrow b^2a+a^2b=c^2a+a^2c$
$\rightarrow b^2a-c^2a=-a^2b+a^2c\rightarrow (b^2-c^2)a=-a^2(b-c)$
|
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|
If $x^2+\lambda x+1=0, \lambda$ is part of $(-2,2)$ and $4x^3+3x+2c=0$ have common root then $c+\lambda$ can be
If $x^2+\lambda x+1=0, \lambda$ is part of $(-2,2)$ and $4x^3+3x+2c=0$ have common root then $c+\lambda$ can be
I can't figure out a way to solve this question. A hint would be great.
|
If $\lambda$ is in $(-2,2)$ (not to be confused with $[-2,2]$), then the roots of $x^2+\lambda x+1=0$ are complex because $x^2+\lambda x +1=0$ only has real solutions when $\lambda=2$ and $\lambda = -2$.
Let $a$ be a shared solution of $x^2+\lambda x+1=0$ and $4x^3+3x+2c=0$. Since $a$ is complex, $4a^3+3a+2c=0$ gives us $c=-\frac{4a^3+3a}{2}$. Thus, $c$ is complex, so $\lambda + c$ is complex.
|
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|
Prime number decomposition What is the fastest way to decompose the given number to prime numbers without using calculator?
Example : $$3575$$
What I do is :
$$3575 = 3 \times 10^3 + 5 \times 10^2 + 7 \times 10 + 5 = 3\times5^3\times2^3 + 5^2 \times 2^2 \times 5 + 7 \times 5 \times 2 + 5$$
But now I do not know how to effectively get rid of "$+$".
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By inspection, the final two digits of $75$ means $3575$ is divisible by $25=5^2$. Then the result of dividing out $25$ is $\frac{3500}{25} + \frac{75}{25}=4\times 35+3 = 143=12^2-1 = 11\cdot 13$, giving $3575 = 5^2\cdot11\cdot13$
|
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|
When is $\frac{x^2+xy+y^2}{49}$ an integer?
Find the number of distinct ordered pairs $(x, y)$ of positive integers such that $ 1\leq x, y \leq 49$ and $\frac{x^2+xy+y^2}{49}$ is an integer.
Multiplying the given equation by $(x-y)$ gives $x^3 \equiv y^3 \pmod{49}$. Thus we can't have $7 \mid x$ while $7 \nmid y$ or vice-versa. Rearranging the given equation gives $$(x+y)^2 \equiv xy \pmod{49}.$$
If $7 \mid x,y$ then there are $49$ solutions. Now suppose that $7 \nmid x,y$. Then we have $$(x+y)^2(xy)^{-1} \equiv xy^{-1}+2+xy^{-1} \equiv 1 \pmod{49}.$$ Thus, $xy^{-1}+yx^{-1} \equiv xy^{-1}(1+(yx^{-1})^2) \equiv -1 \pmod{49}$.
I didn't see how to continue from here. The answer is $$2\varphi(49)+49,$$ and $\varphi(49)$ is the number of units modulo $49$, so maybe we can use that to solve this question.
|
For any given $y$ (not dividing $7$), the congruence $x^3\equiv y^3\pmod{49}$ will have $3$ solutions:
$$x_1=y,\quad x_2=y^{\frac{42+3}3},\quad x_3=y^{\frac{2\cdot42+3}3}\,.$$
(We have $\varphi(49)=42$ possibilities for $y$.)
Since you multiplied by $x-y$, we will have to drop $x_1$ for every $y$, and thus we arrive to $2\cdot 42$ ordered pairs.
|
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}
|
Maximize $P=\frac{\sqrt{a^2-1}}{a}+\frac{\sqrt{b^2-1}}{b}+\frac{\sqrt{c^2-1}}{c} $ Let $a,b,c\geq 1$ satisfy $ 32abc=18(a+b+c)+27$. Find the maximum value $$P=\dfrac{\sqrt{a^2-1}}{a}+\dfrac{\sqrt{b^2-1}}{b}+\dfrac{\sqrt{c^2-1}}{c} $$
$\sqrt {a^2-1}=\sqrt{(a-1)(a+1)}\leq \frac{a-1+a+1}{2}=a$
$\Rightarrow \frac{\sqrt {a^2-1}}{a}\leq \frac{\sqrt {a}}{a}=\frac{1}{\sqrt {a}}$
Need prove $\sum\frac{1}{\sqrt{a}}\leq \sqrt {5}$ ?
|
If $a=b=c=\frac{3}{2}$ so $\sum\limits_{cyc}\frac{\sqrt{a^2-1}}{a}=\sqrt5.$
We'll prove that it's a maximal value.
Indeed, by C-S
$$\sum_{cyc}\frac{\sqrt{a^2-1}}{a}=\sum_{cyc}\sqrt{1-\frac{1}{a^2}}\leq\sqrt{(1+1+1)\sum_{cyc}\left(1-\frac{1}{a^2}\right)}=\sqrt{3\sum_{cyc}\left(1-\frac{1}{a^2}\right)}.$$
Thus, it remains to prove that:
$$3\sum_{cyc}\left(1-\frac{1}{a^2}\right)\leq5$$ or
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\frac{4}{3}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition does not depend on $v^2$ and we need to prove that
$$9v^4-6uw^3\geq\frac{4}{3}w^6,$$
which says that it's enough to prove the last inequality for a minimal value of $v^2$.
We know that $a$, $b$ 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 line $y=3v^2x$ and the graph of $y=-x^3+3ux^2+w^3$ have three common points
and $v^2$ gets a minimal value,
when the 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 (draw it!).
Id est, it's enough to prove the last inequality for $b=a$ and the condition gives $c=\frac{27+36a}{32a^2-18}$.
Thus, we need to prove that:
$$a^4+2a^2\left(\frac{27+36a}{32a^2-18}\right)^2\geq\frac{4}{3}a^4\left(\frac{27+36a}{32a^2-18}\right)^2$$ or
$$a^2(2a-3)^2(8a^2+12a+9)\geq0.$$
Done!
|
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|
If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$ If $\sin (\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, prove that, $\cos [2(\alpha-\beta)]-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.
My Attempt:
.$$\sin (\theta+\alpha)=a$$
$$\sin \theta. \cos \alpha+\cos \theta.\sin \alpha=a$$
Multiplying both sides by $2$
$$2\sin \theta.\cos \alpha + 2\cos \theta.\sin \alpha=2a$$
Squaring both sides,
$$4\sin^2 \theta.\cos^2 \alpha + 6\sin^2 \theta. \cos^2 \alpha + 4\cos^2 \theta.\sin^2 \alpha=4a^2$$.
How should I do further?
|
Consider a circle with radius $\frac{1}{2}$ and a triangle $ABC$ inscribed in the circle with $\angle A = 180^\circ - (\alpha+\theta), \angle B = \theta + \beta$. Then $\angle C = \alpha - \beta$. With the usual notations, we have $a = \sin(\theta+\alpha), b = \sin(\theta+\beta), c= \sin(\alpha - \beta)$. Applying the Cosine rule, $
c^2 = a^2 + b^2 - 2ab \cos C $ gives the desired result.
|
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|
Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$
$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$
Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?
|
In derivative we subtract 1.
In integration we add 1.
Your solution is fine. Except last step.
$$=\int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2}) dx $$
$$=\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1$$
$$=\left[2\cdot1^{1/2} + 2\cdot1^{3/2} + \frac{10}{7}\cdot1^{7/2}\right]-\left[2\cdot0^{1/2} + 2\cdot0^{3/2} + \frac{10}{7}\cdot0^{7/2}\right]$$
Hope now you can proceed.
|
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|
Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found the modulus to be $2\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$ units and that the argument would be $\operatorname{arg}(z+i)=\frac{\pi}{4}+\frac{\theta}{2}$.
Now, the next step that I took was that I replaced every theta with $\frac{3\pi}{8}$ in the polar form of the complex number $z+i$. So now it would look like this:
$$z+i=\left[2\cos{\left(\frac{\pi}{8}\right)}\right]\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$$
Then, I expanded the $\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$ part to become $\cos{\left(\frac{3\pi}{8}\right)}+i\sin{\left({\frac{3\pi}{8}}\right)}$. So now I've got the $\cos\left({\frac{3\pi}{8}}\right)$ part but I don't really know what to do next. I've tried to split the angle up so that there would be two angles so I can use an identity, however, it would end up with a difficult fraction instead. So if the rest of the answer or a hint would be given to finish the question, that would be great!!
Thanks!!
|
As $\frac{3\pi}{8}$ and $\frac{\pi}{8}$ are complementary angles, we get
$$\begin{align}
\cos\frac{3\pi}{8}&=\sin\frac{\pi}{8}\\
&=\sin\frac{\pi/4}{2}\\
&=\sqrt{\frac{1-\cos(\pi/4)}{2}}\\
&=\sqrt{\frac{1-(1/\sqrt{2})}{2}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}}\\
&=\sqrt{\frac{\sqrt{2}-1}{2\sqrt{2}}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}}\\
&=\sqrt{\frac{1}{4+2\sqrt{2}}}\\
&=\frac{1}{\sqrt{4+2\sqrt{2}}}
\end{align}$$
|
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|
Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$ Though we can prove this it is not something that comes up intutively.
Our ancestors must have been interested in factorising $a^3+b^3+c^3$ but why find it for $a^3+b^3+c^3-3abc$ ?
|
Well, it is
$$ \det( aI + b W + c W^2 ) $$
where
$$
W =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr}
a & b & c \\
c & a & b \\
b & c & a
\end{array}
\right)
$$
Right. It follows that the polynomial is multiplicative: as $W^3 = I$ and $W^4 = W,$ we get
$$ (aI + b W + c W^2)(pI + q W + r W^2) = xI + y W + z W^2, $$
where
$$ x = ap + br + cq, $$
$$ y = aq + bp + cr, $$
$$ z = ar + bq + cp. $$
Then, by multiplication of determinants,
$$ (a^3 + b^3 + c^3 - 3abc)(p^3 + q^3 + r^3 - 3pqr) = x^3 + y^3 + z^3 - 3xyz $$
|
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|
$a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$. Find $a_n$. Given $\{a_n\}$: $a_{n+1}=\frac{a_n}{\sqrt{a_n^2+1}}$, $a_1=1$.
Find $a_n$ like a function of $n$.
My trying.
Let $a_n=\tan\alpha_n$, where $\alpha_n\in\left(0,\frac{\pi}{2}\right)$.
Hence, $a_{n+1}=\sin\alpha_n$ and what is the rest?
Thank you!
|
Squaring gives,
$$a_{n+1}^2=\frac{a_n^2}{1+a_n^2}$$
Let $b_n=a_n^2$ then we have,
$$b_{n+1}=\frac{b_{n}}{1+b_{n}}$$
$$=\frac{1+b_n-1}{1+b_n}$$
$$=1-\frac{1}{1+b_n}$$
$$=1-\frac{b_{n+1}}{b_n}$$
$$=\frac{b_{n}-b_{n+1}}{b_{n}}$$
And hence,
$$\frac{b_{n}-b_{n+1}}{b_{n}b_{n+1}}=1$$
Thus,
$$\sum_{n=1}^{N-1} \frac{b_{n}-b_{n+1}}{b_{n}b_{n+1}}=N-1$$
$$\sum_{n=1}^{N-1} \left(\frac{1}{b_{n+1}}-\frac{1}{b_{n}} \right)=N-1$$
$$\frac{1}{b_{N}}-\frac{1}{b_1}=N-1$$
$$b_{N}=\frac{1}{N-1+1}$$
$$a_{N}=\sqrt{\frac{1}{N-1+1}}$$
$$a_{N}=\frac{1}{\sqrt{N}}$$
|
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|
How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \mathbf{0}.$$
I tried plugging it in and it looks like it is true:
Let $\vec{a} = \begin{pmatrix}2\\4\\6\end{pmatrix}$, $\vec{b} = \begin{pmatrix}4\\8\\12\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}8\\16\\24\end{pmatrix}$.
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}
=\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}\\
=
\begin{pmatrix}0\\0\\0\end{pmatrix}\\
$$
How could I prove this to be true? I am not entirely sure where to begin.
|
Vectors $\vec{A}$ and $\vec{B}$ are colinear if
$$\vec{A}=k\vec{B}$$
Then the vector product naturally comes out to be $0$.
|
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|
Finding out the range of $g(x)=\sin x + \cos x$ If $x$ is such that $$\lfloor2\sin x\rfloor+\lfloor{\cos x}\rfloor=-3$$ for some $x \in [0,25]$ then the question is to find out the range of the function $$g(x)=\sin x+\cos x$$Here the function $g(x)$ assumes those values of $x$ which satisfies the above equation in the given interval
$\lfloor.\rfloor$ represents greatest integer function.
I tried rewriting the equation as $$2 \sin x+\cos x -2\lbrace(\sin x)\rbrace-\lbrace(\cos x)\rbrace=-3$$ which can be rewritten as $$g(x)=-3+2\lbrace(\sin x)\rbrace+\lbrace(\cos x)\rbrace-\sin x$$ I again rewrote it as $$g(x)=-3-\lfloor\sin x\rfloor+\lbrace(\sin x)\rbrace+\lbrace(\cos x)\rbrace$$ $\lfloor\sin x\rfloor=\lbrace 0,-1 \rbrace$.I tried to further find out the range of this but failed.Any ideas?Thanks.
Here $\lbrace.\rbrace$ represents fractional part.
|
Note that $\lfloor \cos x \rfloor $ only takes values $-1,0,1$.
Let $S = \{x \in [0,2 \pi] | \lfloor \cos x \rfloor + \lfloor 2\sin x \rfloor = -3 \}$.
If $\lfloor \cos x \rfloor = 0$, then we would need to have $\lfloor 2\sin x \rfloor = -3$ which is impossible.
If $\lfloor \cos x \rfloor = 1$, then we would need to have $\lfloor 2\sin x \rfloor = -4$ which is impossible.
Hence we must have $\lfloor \cos x \rfloor =-1$ and so
If $\lfloor \cos x \rfloor =-1$, then we must have $\lfloor 2\sin x \rfloor = -2$, and hence
$x \in ({7 \over 6} \pi, {9 \over 6} \pi)$.
Hence $S = ({7 \over 6} \pi, {9 \over 6} \pi)$.
We note that $g$ has a (global) $\min$ at $x={5 \over 4} \pi$, and
$g({5 \over 4} \pi) = -\sqrt{2}$.
Since $g$ is unimodal in $S$, we see that $g(S) = [-\sqrt{2}, -1)$
(since $\max(g({7 \over 6} \pi), g({9 \over 6} \pi)) = -1$).
|
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|
Lucas and Fibonacci Numbers Problem:
Let
\begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
There is a unique ordered pair $(c,d)$ such that $c\phi^n + d\widehat{\phi}^n$ is the closed form for sequence $A_n$.
Find $c$ using the Fibonacci and Lucas number sequences.
My Solution:
$A_n$ is just $3L_n$+$2F_n,$ where $L_n$ is the $n^{th}$ Lucas number and $F_n$ is the $n^{th}$ Fibonacci number. Since the closed form of the Fibonacci sequence is $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),$$ and the closed form of the Lucas sequence is $$L_n = \phi^n + \widehat{\phi}^n,$$
we get the closed form of $A_n$ as $\dfrac{2}{\sqrt{5}}+3.$
What is wrong with my solution?
EDIT: My answer is actually right.
|
Although the problem is old there are two additional methods that can be considered.
First Method
Given
\begin{align*}
A_0 &= 6 \\
A_1 &= 5 \\
A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2.
\end{align*}
then, since Fibonacci and Lucas numbers satisfy the difference equation, one can take
$$A_{n} = b \, F_{n} + c \, L_{n}.$$
First note that $F_{n} = \{ 0, 1, 1, ...\}_{n\geq 0}$ and $L_{n} = \{2, 1, 3, ...\}_{n \geq 0}$. For $A_{0} = 6$ then $6 = 2 c$ or $c = 3$ and leads to $A_{n} = b F_{n} + 3 L_{n}$. For $A_{1} = 5 = b + 3$ then $b = 2$ and leads to
$$A_{n} = 2 \, F_{n} + 3 \, L_{n}.$$
In terms of Fibonacci numbers only the solution is $A_{n} = 5 \, F_{n} + 6 \, F_{n-1}$.
Second Method
The generating function for $A_{n+2} = A_{n+1} + A_{n}$ takes the form
$$\sum_{n=0}^{\infty} A_{n} \, t^{n} = \frac{A_{0} + (A_{1} - A_{0}) \, t}{1 - t - t^2}$$
and comparing to the generating function of the Fibonacci numbers,
$$\sum_{n=0}^{\infty} F_{n} \, t^{n} = \frac{t}{1 - t - t^2},$$
and yields
$$\sum_{n=0}^{\infty} A_{n} \, t^{n} = \sum_{n=0}^{\infty} (A_{0} \, F_{n+1} + (A_{1} - A_{0}) \, F_{n}) \, t^n$$
and yields the general form
$$A_{n} = A_{0} \, F_{n+1} + (A_{1} - A_{0}) \, F_{n} = A_{1} \, F_{n} + A_{0} \, F_{n-1}.$$
This result agrees with the first.
|
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|
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4 u} \, du\\
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
&= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C.
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
|
Once you get to $\tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C$, get rid of $u$. Since $\sec u = 2x = 2x/1$, that means $\tan u = \sqrt{4x^2 - 1}$ (draw a right triangle to show this), so $$\begin{aligned}[t]\text{integral} = \tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C &= \tfrac{1}{16}\Big(\sqrt{4x^2 - 1} + \tfrac{1}{3} (4x^2-1)\sqrt{4x^2-1}\,\Bigr) + C \\ &= \tfrac{1}{16}\sqrt{4x^2-1} \, \Bigl(1 + \tfrac{4}{3} x^2 - \tfrac{1}{3} \Bigr) + C \\ &= \tfrac{1}{24}\sqrt{4x^2-1} \, \Bigl(2x^2+1 \Bigr) + C.\end{aligned}$$
|
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|
Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
|
LHS:
$$
\cos (4 \theta)-4\cos(2\theta)=2\cos^2(2\theta)-1-4\cos(2\theta)
$$
RHS:
$$
8\sin^4(\theta)-3=2\left( 2\sin^2 \theta\right)^2-3=2(1-\cos(2\theta))^2-3=2+2\cos^2(2\theta)-4\cos(2\theta)-3
$$
|
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|
How do I solve this system of equations (with squares and square roots)? Consider the system of equations:
\begin{align*}
x+y+z&=6\\
x^2+y^2+z^2&=18
\\\sqrt{x}+\sqrt{y}+\sqrt{z}&=4.
\end{align*}
How do I solve this? I've tried squaring, adding equations side by side, substituting, etc., but without success, e.g.
$$x^2+y^2+z^2+2(xy+yz+xz)=36\implies xy+yz+xz=9,$$
but then I don't know what to do next. Please help me solve this.
|
\begin{align*}
\text{Let$\,$:}&&a &= \sqrt{x}\\[2pt]
&&b &= \sqrt{y}\\[2pt]
&&c &= \sqrt{z}\\[8pt]
\text{Let$\,$:}&&f(t) &= (t - a)(t - b)(t - c)\\[2pt]
&&&= t^3 - e_1t^2 + e_2t - e_3\\[8pt]
\text{where$\,$:}&&e_1 &= a + b + c\\[2pt]
&&e_2 &= ab + bc + ca\\[2pt]
&&e_3 &= abc\\[8pt]
\text{For $k \in \mathbb{Z}^{+}$, let$\,$:}&&s_k &= a^k + b^k + c^k\\[8pt]
\text{By hypothesis, we have$\,$:}&&s_1 &= 4\\[2pt]
&&s_2 &= 6\\[2pt]
&&s_4 &= 18\\[8pt]
\text{Then$\,$:}&&e_1 &= a + b + c\\[2pt]
&&&= s1\\[2pt]
&&&= 4\\[8pt]
\text{and$\,$:}&&2e_2 &= 2(ab +bc + ca)\\[2pt]
&&&= (a + b + c)^2 - (a^2 + b^2 +c^2)\\[2pt]
&&&= e_1^2 - s_2\\[2pt]
&&&= 4^2 - 6\\[2pt]
&&&= 10\\[2pt]
\implies&& e2 &= 5\\[8pt]
&&\text{Ne}&\text{xt, since $a,b,c$ are roots of $f(t)$, we have}\\[8pt]
&&a^3 &- e_1a^2 + e_2a - e_3 = 0\\[2pt]
&&b^3 &- e_1b^2 + e_2b - e_3 = 0\\[2pt]
&&c^3 &- e_1c^2 + e_2c - e_3 = 0\\[8pt]
\text{which sums to$\,$:}&&
s_3 &- e_1s_2 + e_2s_1 - 3e_3 = 0\\[8pt]
\implies&&s_3 &= e_1s_2 - e_2s_1 + 3e_3\\[2pt]
&&&=(4)(6) - (5)(4) + 3e_3\\[2pt]
&&&=4 + 3e_3\\[8pt]
&&\text{Bu}&\text{t, $a,b,c$ are also roots of $tf(t)$, hence}\\[8pt]
&&a^4 &- e_1a^3 + e_2a^2 - e_3a = 0\\[2pt]
&&b^4 &- e_1b^3 + e_2b^2 - e_3b = 0\\[2pt]
&&c^4 &- e_1c^3 + e_2c^2 - e_3c = 0\\[8pt]
\text{which sums to$\,$:}&&
s_4 &- e_1s_3 + e_2s_2 - e_3s_1 = 0\\[8pt]
\implies&&s_4 &= e_1s_3 - e_2s_2 + e_3s_1\\[2pt]
\implies&&18 &= (4)(4 + 3e_3) - (5)(6) + (e_3)(4)\\[2pt]
\implies&&e_3 &= 2\\[8pt]
\text{Then$\,$:}&&f(t) &= t^3 - e_1t^2 + e_2t - e_3\\[2pt]
&&&= t^3 - 4t^2 + 5t - 2\\[2pt]
&&&= (t-1)^2(t-2)\\[8pt]
&&\text{It}&\text{ follows that the triple}\\[8pt]
&&(a,&b,c)\;\text{is an arbitrary permutation of}\;(1,1,2)\\[2pt]
\text{Therefore$\,$:}&&(x,&y,z)\;\text{is an arbitrary permutation of}\;(1,1,4)\\[2pt]
\end{align*}
|
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How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
$2^3 = 2^2 + 2^2$
$2^5 = 4^2 + 4^2$ or $2^5 = 2^4 + 2^4$
From that, there seems to be a pattern such as $2^{k+1}=2^k + 2^k$ for some even $k \geq 0$
I've also been able to observe that $k$ has to be odd. I've tried the following:
$k = 2l + 1$ for some $l$
$2^{2l+1} = a^2 + b^2$
$2^l.2 = a^2 + b^2$ Now, if I set $a=b$
$2.2^l = 2a^2$ <=> $2^l = a^2$
But I don't know how to prove that $a$ has to be equal to $b$
Also, in the case where $k$ is even:
$k=2l$ for some $l$
$2^{2l}$ = $a^2 + b^2$ Again, setting $a=b$
$2^{2l+1} = a^2$
$2^{(2l-1)/2} = a^2$ where $(2l-1)/2$ is not an integer.
But again, I don't know how to go about it when $a \neq b$.
Am I on the right path? Any tips or suggestions on how to go forward?
|
It seems you have at least detected the pattern of the solutions and are on the right path.
A useful fact about squares is that the square of an even number is (of course) a multiple of $4$, whereas the square of an odd number is $1$ plus a multiple of $8$. To see this note that $(2n+1)^2=4\cdot n\cdot(n+1)+1$ and one of $n$, $n+1$ is even.
So if $k>1$, what can we say about $a$ and $b$?
*
*If exactly one of them is odd, then $a^2+b^2$ is $1$ plus a multiple of $4$, whereas $2^k$ is a multiple of $4$; so this is impossible.
*If both are odd, then $a^2+b^2$ is $2$ plus a multiple of $8$, hence certainly not a multiple of $4$; again, this is impossible.
*Remains the case that $a$ and $b$ are both even. But then $(a/2)^2+(b/2)^2=2^{k-2}$ is a solution for a smaller $k$.
So once we have verified that $a^2+b^2=2^1$ has the only solution (in poisitive integers) $a=b=1$ and $a^2+b^2=2^2$ has no solutions, the above observatoins help us show (by induction) that
*
*If $k$ is even, then $a^2+b^2=2^k$ has no solution in positive integers
*If $k$ is odd, then the only solution in positive integers is $a=b=2^{(k-1)/2}$
|
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|
How to shorten the derivation of the Laplacian in polar coordinates? Yesterday, I computed the formula for the Laplacian in polar coordinates.
$
\newcommand{p}[2]{\frac {\partial #1} {\partial #2}}
\newcommand{s}[2]{\p {^2 #1} {#2 ^2}}
\newcommand{m}[3]{\p {^2 #1} {#2 \partial #3}}
\newcommand{po}[0]{\p {}}
\newcommand{so}[0]{\s {}}
\newcommand{mo}[0]{\m {}}
\newcommand{t}[0]{\theta}
\newcommand{w}[1]{\left( #1 \right)}
\newcommand{ct}[0]{\cos \t}
\newcommand{st}[0]{\sin \t}
\newcommand{f}[1]{\frac #1 r}
\newcommand{g}[1]{\frac #1 {r^2}}
\newcommand{jx}[2]{\ct #1 - \f \st #2}
\newcommand{jy}[2]{\st #1 + \f \ct #2}
\newcommand{px}[2]{\jx {\po r #1} {\po \t #2}}
\newcommand{py}[2]{\jy {\po r #1} {\po \t #2}}
\newcommand{pox}[0]{\px {} {}}
\newcommand{poy}[0]{\py {} {}}
\newcommand{one}[0]{\sin^2\t + \cos^2\t}
\newcommand{zero}[0]{\st\ct - \st\ct}
$
\begin{align*}
x & = r \ct \\ y & = r \st \\ r^2 & = x^2 + y^2
\end{align*}
First, I used the chain rule to relate $\po r$ and $\po \t$ to $\po x$ and $\po y$:
\begin{align*}
\po x = \p r x \po r + \p \t x \po \t = \pox \\
\po y = \p r y \po r + \p \t y \po \t = \poy \\
\end{align*}
Applying each of $\po x$ and $\po y$ to itself:
\begin{align*}
\so x + \so y
& = \po x {\w {\po x}} + \po y {\w {\po y}} \\
& = \px {\w \pox} {\w \pox} \\
& \qquad + \py {\w \poy} {\w \poy} \\
& = \ct \w {\ct \so r + \g \st \po \t - \f \st \mo r \t} \\
& \qquad+ \f \st \w {\st \po r - \ct \mo \t r + \f \ct \po \t + \f \st \so \t} \\
& \qquad + \st \w {\st \so r - \g \st \po \t + \f \ct \mo r \t} \\
& \qquad + \f \ct \w {\ct \po r + \st \mo \t r - \f \st \po \t + \f \ct \so \t} \\
& = \w \one \so r + \g \zero \po \t \\
& \qquad - \f \zero \mo r \t + \f \one \po r \\
& \qquad - \f \zero \mo \t r + \g \zero \po \t \\
& \qquad + \g \one \so \t \\
& = \so r + \f 1 \po r + \g 1 \so \t \\
\end{align*}
This was long and tedious. Is there some way to make the calculation shorter.
|
Apparently, Math.SE does not allow me to “delete for real” my previous wrong answer, so I will try to provide a non-wrong answer.
Let $\newcommand \t \theta F(x,y) = f(r,\t)$, where $(r,\t) = g(x,y)$. Of course, in our case, $g$ is not a globally well-defined function, but who cares? Differential calculus is all about working locally. By the chain rule,
$$F_{xx} = (f \circ g)_{xx} = (\nabla f \cdot g_x)_x = (\nabla f)_x \cdot g_x + \nabla f \cdot g_{xx} = g_x^T \cdot Hf \cdot g_x + \nabla f \cdot g_{xx}$$
Of course, analogously, we have
$$F_{yy} = g_y^T \cdot Hf \cdot g_y + \nabla f \cdot g_{yy}$$
Adding everything, the Laplacian expands to
$$F_{xx} + F_{yy} = g_x^T \cdot Hf \cdot g_y + g_y^T \cdot Hf \cdot g_y + \nabla f \cdot (g_{xx} + g_{yy})$$
Call the sum of the first two terms $A$ and call the last term $B$.
In our particular case, $r^2 = x^2 + y^2$ and $\tan \t = y/x$. Thus,
$$r_x = x/r, \qquad r_y = y/r, \qquad \sec^2 \t \cdot \t_x = -y/x^2, \qquad \sec^2 \t \cdot \t_y = 1/x$$
Rearranging the last two expressions, we have
$$\t_x = -y/r^2, \qquad \qquad \qquad \t_y = x/r^2$$
By the chain rule, $A$ reduces to
$$f_{rr} (r_x^2 + r_y^2) + 2f_{r\t} (r_x \t_x + r_y \t_y) + f_{\t\t} (\t_x^2 + \t_y^2) = f_{rr} + \frac 1 {r^2} f_{\t\t}$$
Differentiating once again, we have
$$r_{xx} = 1/r, \qquad r_{yy} = 1/r, \qquad \t_{xx} = 2yr_x /r^3, \qquad \t_{yy} = -2xr_y / r^3$$
Rearranging the last two expressions, we have
$$\t_{xx} = 2xy / r^4, \qquad \qquad \qquad \t_{yy} = -2xy / r^4$$
By the chain rule, $B$ reduces to
$$f_r \cdot (r_{xx} + r_{yy}) + f_\t \cdot (\t_{xx} + \t_{yy}) = \frac 1r f_r$$
Combining everything, we have
$$F_{xx} + F_{yy} = f_{rr} + \frac 1r f_r + \frac 1 {r^2} f_{\t\t}$$
|
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|
A Nasty Elliptic Integral I am trying to evaluate:
$$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|},\textrm{ }\alpha\in\mathbb{C}\backslash(-\infty,0]$$
It's simple to see that it is convergent for such $\alpha$—but that's probably the only simple thing about it! I've been lost in trying to solve this one for several days, now. Performing a change of variables yields:
$$\int_{0}^{\infty}\frac{\sqrt{q}dx}{\sqrt{x}\sqrt{x^{2}+px+q}}$$
where $p=\frac{\cos\theta}{r}$ and $q=\frac{1}{r^{2}}$, where $\alpha=re^{i\theta}$.
Performing yet more changes of variables and integrating by parts yields (assuming I didn't screw up somewhere along the way):
$$\frac{2}{r}+\frac{2}{r}\int_{\frac{p}{2}}^{\infty}\frac{x\sqrt{x-\frac{p}{2}}}{\left(x^{2}-\Delta^{2}\right)^{3/2}}dx$$
where $\Delta=\frac{i}{2r}\sqrt{4-\cos^{2}\theta}$. This version, Mathematica is able to compute, giving the formula:
$$\int_{a}^{\infty}\frac{x\sqrt{x-a}}{\left(x^{2}-b^{2}\right)^{3/2}}dx=\frac{\textrm{sgn}\left(\textrm{arg}\left(-b^{-2}\right)\right)}{\left(a^{2}-b^{2}\right)^{1/4}}K\left(\frac{1}{2}-\frac{a}{2\sqrt{a^{2}-b^{2}}}\right)$$
where $K$ is the complete elliptic integral of the first kind. However, this formula is only valid for $a,b\in\mathbb{C}$ satisfying $\textrm{Im}\left(a\right)=0$, $\textrm{Re}\left(a\right)>0$, $\textrm{Re}\left(b^{2}\right)<0$, and satisfying either: “$\textrm{Re}\left(a\right)>\textrm{Re}\left(b\right)$ and $a>\textrm{Re}\left(b\right)$” OR “$b\notin\mathbb{R}$”.
All of these conditions are satisfied for my integral, except for the $\textrm{Re}\left(a\right)>0$ condition, which makes no sense. $a=\frac{p}{2}=\frac{\textrm{Re}\left(\alpha\right)}{\left|\alpha\right|^{2}}$, and the initial integral is valid even for $\alpha$ with $a\leq0$, as long as $\textrm{Im}\left(\alpha\right)\neq0$.
So: any ideas for how to evaluate:
$$\int_{-\infty}^{\infty}\frac{dx}{\left|1+\alpha x^{2}\right|}?$$
Thanks!
|
In this answer I use a variable substitution which I cannot find in the already published answers.
Say that $\alpha \neq 0$ and $\alpha = \varrho e^{i\theta}, \, -\pi <\theta< \pi$. Then $|1+\alpha x^2| = \sqrt{\varrho^2x^4 +2\varrho\cos \theta x^2+1}$ and
\begin{gather*}
I = \int_{-\infty}^{\infty}\dfrac{dx}{|1+\alpha x^2|} = 2\int_{0}^{\infty}\dfrac{dx}{\sqrt{\varrho^2x^4 +2\varrho\cos \theta x^2+1}} =\dfrac{2}{\sqrt{\varrho}} \int_{0}^{\infty}\dfrac{dx}{\sqrt{x^4 +2\cos \theta x^2+1}} = \\[2ex]
\dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{\sqrt{x^4 +2\cos \theta x^2+1}} = \dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{\sqrt{x^4+2x^2+1-4x^2\sin^2\frac{\theta}{2}}} = \\[2ex]
\dfrac{4}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dx}{(x^2+1)\sqrt{1-\frac{4x^2}{(x^2+1)^2}\sin^2\frac{\theta}{2}}}.\tag{1}
\end{gather*}
For $0<x<1$ we put $y = \dfrac{2x}{x^2+1}, 0<y<1$.
Then
\begin{equation*}
y(x^2+1)=2x\tag{2}
\end{equation*}
and
\begin{equation*}
x= \dfrac{1-\sqrt{1-y^2}}{y}.\tag {3}
\end{equation*}
From (2) we get
\begin{equation*}
(x^2+1)dy + 2xydx=2dx \Leftrightarrow dx= \dfrac{x^2+1}{2(1-xy)}dy = \dfrac{x^2+1}{2\sqrt{1-y^2}}dy
\end{equation*}
where we have used (3) in the last step. Finally we use that in (1). Thus
\begin{equation*}
I = \dfrac{2}{\sqrt{\varrho}} \int_{0}^{1}\dfrac{dy}{\sqrt{1-y^2}\sqrt{1-y^2\sin^2\frac{\theta}{2}}} = \dfrac{2}{\sqrt{|\alpha|}}K\left(\sin^2\frac{\theta}{2}\right).
\end{equation*}
|
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|
Probability that $x^2-y^2$ is divisible by $k$ Let two numbers $x$ and $y$ be selected from the set of first $n$ natural numbers with replacement(i.e. the two numbers can be same).The question is to find out the probability that $x^2-y^2$ is divisible by $k\in \mathbb{N}$
For $k=2$
Any number can be expressed as $2p,2p+1$.Now $x^2-y^2=(x-y)(x+y)$ If both numbers are of form $2p+1$ then (x-y) would be divisible by $2$ .if two numbers are of different forms then it will not be divisible by $2$.So the probability in this case is $a^2+(1-a)^2$ where $a$ is probability that number chosen is divisible by $2$ which is $\frac{\lfloor \frac{n}{2} \rfloor}{n}$.However this gets complicated with $k=3$ onwards because numbers in different forms may be divisible.In other words if there a generalisation or way to solve for some large $k$.Thanks.
|
A generalization expressed by a set
A good way to generalize this is to use modular arithmetic, or essentially look at the remainder of $\frac{x}{k}$ and $\frac{y}{k}$. As you pointed out in your example for $k=2$, the numbers can only be expressed as $$2*p,2*p+1$$
This can be further generalized into a set of unique expressions that define every number or every $x$ and $y$ for a value $k>1$ and $p≥0$$$S={kp,kp+1...kp+(k-2),kp+(k-1)}$$
Using modular arithmetic, we know that $S\equiv \{0,1,2...k-2,k-1\}\pmod k$
Since $x$ and $y$ are any two terms from the set $S$ for any $p≥0$, we know that $x$ and $y$ are really just equal to either any value from $\{0,1,2...k-2,k-1\}$ since we only need to look at the remainder when divided by $k$ to determine divisibility.
Probability using mod
Now we know $x^2-y^2 = (x-y)(x+y)$ meaning either $(x-y)$ or $(x+y)$ have to be divisible by $k$ in order to fulfill the requirement that $x^2-y^2$ is divisible by $k$.
It should be noted that the probability of choosing a random number being expressed by any expression in set $S$ is uniform and is equal to $\frac{1}{k}$. For example, the probability of choosing a number is expressible by $3p+1$ is $\frac{1}{3}$. This can be proved by showing how every number expressed by $kp$ can always be paired with $kp+1,kp+2,...kp+(k-2),kp+(k-1)$, thus showing how there are an equal number of numbers of each type.
Onto the question....
$(x-y)$ is divisible by $k$ if $(x-y)\equiv 0\pmod k$. This simplifies to $(x$ mod $k)-(y \text{ mod}\ k) = 0$. Thus we now need to find the probability that $x$ and $y$ are both expressible by the same expression in the set S.
Now this comes down to a simple problem of asking what is the probability of choosing two of the same elements from the set $S$ without replacement, which is $\frac{k}{k} *\frac{1}{k} =$ $\frac{k}{k^2}$
$(x+y)$ is divisible by $k$ if $(x+y)\equiv 0\pmod k$. This simplifies to $(x$ mod $k)+(y \text{ mod}\ k) = 0$. Thus we now need to find the probability that $x$ and $y$ are chosen such that the $x+y =$ a multiple of $k$.
Knowing that $x$ and $y$ must come from the set $S$, we can see there is a specific pairing of elements from $S$ that causes the addition of the pair of elements to be equal to $2*kp+k$, a multiple of $k$. The only times that the pair of elements, which will be the expression that expresses $x$ and $y$, add up to $2*kp$ is when one is expressed by $kp+m$ and the other being $kp+(k-m)$ for any whole number $m$. This is because $(kp+m)+(kp+(k-m)) = 2*kp$. For example $3p+1$ can be paired up with $3p+2$ since their sum is $6p+3$, a multiple of 3.
Now we are now looking for the probability the two chosen elements from $S$ are pairs of each other (If both chosen elements are $kp$, it will be a multiple of k still) The total number of pairs that can be chosen is the total number of elements in set $S$ squared, which is $k^2$. The total number of pairs that fulfill the divisibility by $k$ is equal to $\lfloor\frac{k}{2}\rfloor + 1$. The ceiling function is applied to correct errors that occur when $k$ is odd as you obviously can't have a fractional number of pairs. The extra $+1$ is for including the case when both elements are the same.
Thus the probability that $(x+y)$ is divisible by $k$ is $\frac{\lfloor\frac{k}{2}\rfloor + 1}{k^2}$.
However now we encounter a problem of overlap, which are the cases when the two elements chosen from $S$ make $(x-y)$ and $(x+y)$ divisible by k. To get rid of the over count we need to subtract the number of times this happens for a given set $S$. We know we need to subtract once for the case when both elements are $kp$. We also need to subtract another one depending if $k$ is even or not. This is because when $k$ is even, we have the case where the element $kp+\frac{k}{2}$ can pair with itself to fulfill everything. Thus the final answer is
$$\frac{k}{k^2}+\frac{\lfloor\frac{k}{2}\rfloor + 1}{k^2} - \frac{(k+1 \text{ mod }2)+1}{k^2}$$
The $\frac{(k+1 \text{ mod }2)+1}{k^2}$ tells us to subtract one more and subtract another if $k$ is even, hence we shifted the modular part by 1 since $( \text{ even number mod }2 = 0)$
|
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|
How to convert this $k$th term to Telescopic series Given $$t_k=k^4-2k^3+2k^2-1$$ we need to find
$$S=\sum_{k=1}^{k=10}t_k$$
I factorized $t_k$ as
$$t_k=k(k-1)(k^2-k+1)$$ which further i reduced it as
$$t_k=(k^2-k)^2+(k^2-k)$$
Can i have any hint to reduce this in form $$F(k+1)-F(k)$$
|
It is more practical to convert $t_k$ into a linear combination of binomial coefficients, then exploit the hockey stick identity. We have
$$ t_k = k^4-2k^3+2k^2-1 = 24\binom{k+1}{4}+6\binom{k}{2}+\binom{k-1}{1} \tag{1}$$
hence:
$$ \sum_{k=1}^{10}t_k = 24\binom{12}{5}+6\binom{11}{3}+\binom{10}{2}=\color{red}{20043}.\tag{2}$$
As a side effect, we also get that $t_k$ can be expressed in terms of the backward differences of $\frac{k^5}{5}+\frac{k^2}{2}-\frac{7k}{10}=24\binom{k+2}{5}+6\binom{k+1}{3}+\binom{k}{2}$.
|
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|
How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\
\log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\
5^{2x^2} &= 5^\frac{5}{4}\\
(2x^2)\log5 &= \frac{5}{4}\log5\\
2x^2 &= \frac{5}{4}\\
2x^2 &= 1.25\\
x^2 &= 0.625\\
\sqrt {x^2} &= \sqrt {0.625}\\
x &= \frac{\sqrt {10}}{4}\end{align}
Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer.
|
You made a pretty big mistake because $\log a + \log b \neq \log (a+b)$. So, your third line does not follow from your second.
To actually solve the problem, here's two hints:
*
*Set $y=5^x$
*Use the fact that $a^{bc} = (a^b)^c$
|
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|
what is the relationship between the SVD of $[A;B]$ and that of $A$ and $B$? In my question, $A\in R^{m\times r}$, $B\in R^{n\times r}$, and $[A;B]\in R^{(m+n)\times r}$ results from stacking $A$ over $B$. Given $m>r$ and $n>r$, we do the SVD on $A$ and $B$ and have that $A=U_aS_aV_a^T$ and $B=U_bS_bV_b^T$, where $U_a\in R^{m\times r}$, $S_a\in R^{r\times r}$, $U_b\in R^{n\times r}$, and $S_b\in R^{r\times r}$.
Can we use the above results to get the SVD of $[A;B]$?
|
Not really, even take the simplest case of $1 \times 1$ matrices.
*
*Let $A= \begin{bmatrix}1 \end{bmatrix}$ and let $B=\begin{bmatrix} 2 \end{bmatrix}$.
*Singular value of $A$: $\sigma_a = 1$
*Singular value of $B$: $\sigma_b = 2$
*SVD of $[A ; B] = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$: $$U S V^T = \begin{bmatrix}
\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\
\end{bmatrix}
\begin{bmatrix} \sqrt{5} \\ 0 \end{bmatrix} \begin{bmatrix} 1 \end{bmatrix}$$
There is a slight relationship on the singular value, you'll note that $\sqrt{5} = \sqrt{ \sigma_a^2 + \sigma_b^2}$. However, this is really only an artifact of $\mathbf{v} = \begin{bmatrix} 1 \end{bmatrix}$ being an eigenvector for $A$, $B$, and $[A; B]$; the interplay is generally much more complicated. Note that since $U$ and $V$ are rotation matrices, the singular value decomposition doesn't play particularly well with most block structures.
For a $2 \times 2$ matrix, consider $A= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (singular values of $1$) and $B= \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ (singular values of $2$). The SVD of $[A;B]$ is
$$USV^T = \begin{bmatrix}
0 & \frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} & 0 \\
0 & \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} & 0 \\
\end{bmatrix}
\begin{bmatrix}
\sqrt{5} & 0 \\
0 & \sqrt{5} \\
0 & 0 \\
0 & 0 \end{bmatrix}
\begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}. $$
It only gets worse if you make $A$ and $B$ "messier." Again, this is an example where $A$ and $B$ share all of their eigenvectors.
|
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|
Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} \ge 9$$
$$\Leftrightarrow \frac{(a+b)^2}{(a-b)^2}+\frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}\ge 2$$
Which $$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b} =-1$$
Use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$
We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1) => xy+yz+zx=-1 $
|
your inequality is true, since it is equivalent to
$$3/4\,{\frac { \left( {a}^{2}b+{a}^{2}c+a{b}^{2}-6\,abc+a{c}^{2}+{b}^{2
}c+b{c}^{2} \right) ^{2}}{ \left( -c+a \right) ^{2} \left( b-c
\right) ^{2} \left( a-b \right) ^{2}}}
\geq 0$$ which is obviously true.
|
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|
solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There's where am stuck.
Can someone pull me out.
|
If
$$
\begin{cases}
2x^2+y^2=6\\
x^2+y^2=3+3xy \\
\end{cases}
$$
then by setting $x^2=a$, $y^2=b$
$$
\begin{cases}
2a+b=6 \\
a+b=3+3xy \\
\end{cases}
$$
Hence, by Cramer's rule (of linear algebra)
$$
\begin{cases}
a=x^2=3-3xy & (1)\\
b=y^2=6xy & (2)
\end{cases}
$$
Hence, by (2) $y=0$ or $y=6x$. Plug it into (1) yields
$$
x^2=3-3x\cdot0\qquad\text{or}\qquad
x^2=3-3x\cdot6x
$$
Hence $(x,y)=(\pm\sqrt{3},0)$ or $(x,y)=(\pm\frac{\sqrt{57}}{19},\pm\frac{6\sqrt{57}}{19})$.
|
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|
Change of coordinates in a polynomial ring I want to prove that $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x\in \mathbb{C}[x,y,z]$$ can be transformed into $$y^2z+x^3+\alpha xz^2+\beta z^3$$ by means of an invertible linear change of coordinates.
I tried to complete the square in $y$, I got $z(y^2+yz+xy)=z(y+t)^2-zt^2$ where $t=(z+x)/2$, but I don't know how to proceed.
I also tried to substitute $x+py+qz$ for $x$ and expand. In the resulting polynomial, I got the terms $3p^2xy^2$ and $3px^2y$ which must be zero, so $p=0$. But then there are also the terms $(d+3q)x^2z$ and $(bq+1)yz^2$ which also must vanish simultaneously, which is impossible. Also I got the term $bxyz$ which must vanish, but $b$ is not necessarily zero...
|
Rewrite $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x=0$$ as $$y^2z+bxyz+yz^2=x^3+dx^2z+axz^2+cz^3.$$ Make the change of coordinates $y$ goes to $y-\frac{bx+z}{2}$. This gives $$y^2z-\frac{bxz^2}{2}-\frac{b^2x^2z}{4}-\frac{z^3}{4} = x^3+dx^2z+axz^2+cz^3$$
Or equivalently, after collecting terms, $$y^2z=x^3+\left(d+\frac{b^2}{4}\right)x^2z+\left(a+\frac{b}{2}\right)xz^2+\frac{4c+1}{4}z^3$$
Setting $p=d+\frac{b^2}{4}$, $q=a+\frac{b}{2}$, and $r=c+\frac{1}{4}$, we may replace $x$ by $x-\frac{p}{3}$ to get an equation of the form
$$y^2z=x^3+\alpha xz^2+\beta z^3$$
Both coordinate changes here are instances of the following general fact: If $p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$, you can control the coefficient $a_{n-1}$ by linear changes of variables. Recall that $a_{n-1}$ is the negative of the sum of the roots: if we make the substitution $x\mapsto x-\frac{a_{n-1}}{n}$, the sum of the new roots is therefore $n\cdot\frac{-a_{n-1}}{n}=-a_{n-1}$ more than the old sum, which means it is zero.
The application of this to the RHS should be somewhat obvious. On the left, note that we're really applying this to $y^2+(bxz+z)y$, as $y^2z+bxyz+yz^2=z(y^2+(bxz+z)y)$.
|
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|
Find the variance of sum of indicator variables
Compute $\operatorname{Var} (X)$ where $X = \sum_{i=1}^n x_i$
I am aware of the formula
$$\operatorname{Var} (X) = \sum_{i=1}^9 \operatorname{Var} (X_i) + \sum_{i \ne j} \operatorname{Cov } (X_i, X_j)$$
But I cannot seem to apply it here
Edit:
We know that $\text{Var } (X_i) = 0.234$
Thus $\sum_{i=1}^{9} \text{Var} (X_i) = 9 \times 0.234 = 2.106$
$\displaystyle \sum_{i \ne j} \text{Cov} (X_i, X_j)$, we know that if $j > i + 1$ then $\text{Cov} (X_i, X_j) = 0$, thus $\sum = 9 \times 0.046875 = 0.4219$
Thus $\text{Var} (X) = 9 \times 0.2344 + 0.421875 = 2.53$
|
\begin{align}
X_i & = \begin{cases} 1 & \text{if the $i$th and $(i+1)$th characters are LW or WL}, \\ 0 & \text{if LL or WW}, \end{cases} \\[10pt]
& = \begin{cases} 1 & \text{with probability } \left(\dfrac 1 4 \times \dfrac 3 4\right) + \left(\dfrac 3 4\times \dfrac 1 4 \right) = \dfrac 3 8, \\[8pt] 0 & \text{with probability } \dfrac 5 8. \end{cases} \\[8pt]
\text{Therefore } \operatorname{var}(X_i) & = \frac 3 8 \times \frac 5 8 = \frac{15}{64}. \\[8pt] {}
\end{align}
\begin{align}
\operatorname{cov}(X_i X_{i+1}) & = \operatorname{E}(X_i X_{i+1}) - (\operatorname{E}X_i)(\operatorname{E}X_{i+1}) \\[8pt]
& = \Pr(X_i X_{i+1} = 1) - \Pr(X_i=1)\Pr(X_{i+1}=1) \\[8pt]
& = \Pr( \text{LWL or WLW)} - \Pr(\text{LW or WL})\Pr(\text{LW or WL)} \\[8pt]
& = \frac 3 {64} + \frac 9 {64} - \left( \frac 3 8 \times \frac 3 8 \right) = \frac {12} {64} - \frac 9 {64} = \frac 3 {64}, \\[10pt]
\text{and } \operatorname{cov}(X_i X_j ) & = 0 \text{ if } |i-j|>1.
\end{align}
\begin{align}
\text{So } \operatorname{var}(X_1+\cdots+X_9) = 9\operatorname{var}(X_1) + 8\times 2\operatorname{cov}(X_1 X_2) = \cdots\cdots
\end{align}
|
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Why $\frac {a}{b} = \frac {c}{d} \implies \frac {a+c}{b+d} = \frac {a}{b} = \frac {c}{d}$? In geometry class, this property was quickly introduced:
$$\dfrac {a}{b} = \dfrac {c}{d} \implies \dfrac {a+c}{b+d} = \dfrac {a}{b} = \dfrac {c}{d} $$
It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.
How to demonstrate it?
|
$$\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+c}{b+d}=\frac{a+\frac{ad}{b}}{b+d}=\frac{ab+ad}{b(b+d)}=\frac{a}{b}=\frac{c}{d}$$
|
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|
Limit of $\int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\mathrm{d}x$ without Taylor expansion $\def\d{\mathrm{d}}$Consider $$I_n(a) = \int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\d x, \quad n\ge 1,\ a \in [0, 1)$$
and evaluate
$$\lim _{n \to \infty} I_n(a). \quad a \in (0, 1)$$
One thing I have found it that
$$\int _{\frac{1}{n}}^n \frac{\arctan x}{x^2 + 2ax + 1}\,\d x = \int _{\frac{1}{n}}^n \frac{\arctan \frac{1}{x}}{x^2 + 2ax + 1}\,\d x$$
Also, I know $$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}. \quad \forall x > 0$$
I didn't make it further with that method, so I tried this:
$$\frac{1}{n} \le x \le n,\\
\arctan \frac{1}{n} \le \arctan x \le \arctan n.$$
So I deduced $$I_n(a) \le \frac{\pi}{2} \int _{\frac{1}{n}}^n \frac{\d x}{x^2 + 2at + 1}. \tag{1}$$
The discriminant of the denominator is $4 (a^2 - 1) \le 0\ (\forall a \in [0, 1))$, so $(1)$ becomes
$$I_n(a) \le \int _{\frac{1}{n}}^n \frac{\d x}{(x+a)^2 + \left(\sqrt{\frac{1-a^2}{a}}\right)^2}.$$
I have to get to the answer $$\frac{\pi}{4 \sqrt{1-a^2}} \arctan \frac{\sqrt{1-a^2}}{a}.$$
|
You are almost there. Since
$$
I_n(a)=\int_{1/n}^n\frac{\arctan x}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\arctan (1/x)}{x^2+2ax+1}\,dx
$$
and
$$
\arctan x+\arctan(1/x)=\frac\pi2
$$
you find that
$$
2I_n=\int_{1/n}^n\frac{\arctan x+\arctan(1/x)}{x^2+2ax+1}\,dx
=\int_{1/n}^n\frac{\pi/2}{x^2+2ax+1}\,dx.
$$
Hence
$$
I_n(a)=\frac\pi4\int_{1/n}^n\frac{1}{x^2+2ax+1}\,dx
=\frac{\pi}{4}\Bigl[\frac{1}{\sqrt{1-a^2}}\arctan\frac{a+x}{\sqrt{1-a^2}}\Bigr]_{1/n}^{n}.
$$
Insert the limits, take the limit $n\to+\infty$ and finally use $\arctan x+\arctan(1/x)=\pi/2$ again, and you are done.
|
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|
How to solve the differential equation $y' = \frac{x+y}{x-y}$? Solve the following differential equation:
$y' = \frac{x+y}{x-y}$
Someone please help me start this problem. This does not look like a regular first-order differential equation in the form $y' + 2xy = x$. Thank you.
|
Hint: It's homogenous of degree 1 which suggests the substitution
$$
y=xv\implies y'=xv'+v
$$
and
$$
xv'+v=\frac{x+xv}{x-xv}=\frac{1+v}{1-v}\stackrel{\text{algebra}}{\implies}
\frac{1-v}{v^2+1}\mathrm dv=\frac{1}{x}\mathrm dx\implies \int \frac{1-v}{v^2+1}\,\mathrm dv=\ln x+c\\
\implies \int\frac{1}{v^2+1}\,\mathrm dv-\int \frac{v}{v^2+1}\,\mathrm dv=\ln x+c\\
\implies\arctan(v)-\frac{1}{2}\ln (v^2+1)=\ln x+c\\
\implies\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\ln \left(\frac{y^2}{x^2}+1\right)=\ln x+c
$$
and I am afraid we will have to be satisfied with the implicit solution.
edit: Following @Artem's suggestion we can rewrite this curve in polar coordinates and it simplifies nicely, although the caveat is it makes $y$'s role difficult to
see
$$
\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\ln \left(\frac{y^2}{x^2}+1\right)=\ln x+c\\
\implies \theta-\ln \left(\sqrt{\frac{r^2}{r^2\cos^2 \theta}}\right)-\ln r\cos \theta=c\\
\implies \theta+\ln (\cos \theta)-\ln (r\cos \theta)=c\\
\implies \theta+\ln\left(\frac{1}{r}\right)=c\\
\implies \theta-c=-\ln\left(\frac{1}{r}\right)=\ln r\\
\implies Ae^{\theta}=r
$$
|
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Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where a and b are integers. Find the greatest common factor of b and 81. Let $( \sqrt{2} + 1)^{1000} = a + b \sqrt{2}$, where $a$ and $b$ are integers. Find the greatest common factor of $b$ and $81$.
How would one solve this question? I tried to use the binomial theorem but that approach did not work. How would one solve this problem if it were to appear on a Math Olympiad?
I know the answer is one of:
(A) 1 (B) 3 (C) 9 (D) 27 (E) 81.
According to one of the Stack Exchange members, the answer is 3. It was found using Python.
|
This is so long for a comment:
We say $ ( \sqrt{2} + 1)^{n} = a_n + b_n \sqrt{2}$ and $a_n,b_n$ are integers. Therefore $( 1- \sqrt{2})^{n} = a_n - b_n \sqrt{2}$. For positive odd value of $n=2k-1$, by product
$-1=( \sqrt{2} + 1)^{n} ( 1- \sqrt{2} )^{n}= (a_n + b_n \sqrt{2})(a_n - b_n \sqrt{2})$.
Hence $a_n^2 - 2b_n^2= -1$. $a^2-2b^2=-1$ is negative Pell equation and minimal solution is $(a_1,b_1)=(1,1)$. Other solutions: $a_{n+2}+b_{n+2}\sqrt 2= (a_n + b_n \sqrt 2)\cdot (1+\sqrt 2)^2$ and then
$$a_{n+2}= 3a_n + 4b_n$$ $$ b_{n+2} = 2a_n + 3b_b$$
For example $(a_3,b_3)=(7,5)$. So, $$a_{n+2} \equiv b_n \pmod{3}$$ $$b_{n+2} \equiv 2a_n \equiv 2b_{n-2}\pmod{3}$$
Since $b_1=1$ is not divisible by $3$, then for all odd $n$, $b_n$ is not divisible by $3$. Similarly, $a_n$ is not divisible by $3$.
In $ ( \sqrt{2} + 1)^{999} = a_{999} + b_{999} \sqrt{2}$, $a_{999}$ and $b_{999}$ can't divisible by $3$. Yet, I didn't say anything about $b_{1000}$. I will think about even values of $n$.
|
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|
Least possible polynomial deegree of complex roots What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$
As there are four roots so the deegree should be four but the answer is given as five . how ?
|
Let's look at the given roots. We see that one of these roots,
$2-\omega -\omega ^2$, is in fact real:
$$
2-\omega -\omega ^2 = 3, \quad\mbox{ while }\quad
(x-(2+3\omega))(x-(2+3\omega^2)) = x^2-x+7.
$$
A fourth degree polynomial with the four given roots would have complex coefficients, namely, it's the polynomial
$$
P_4(x) = (x - 3) ( x - 2\omega ) (x^2 - x + 7). \tag{1}
$$
There exists a $5$th-degree polynomial that has all four given roots and an additional fifth root:
$$
P_5(x) = (x - 3) ( x - 2\omega )( x - 2\bar\omega ) (x^2 - x + 7)
$$
$$
= x^5 - 2 x^4 + 6 x^3 - 17 x^2 - 2 x - 84. \tag{2}
$$
So the sought-for minimal degree is $5$; the polynomial $(2)$ proves this.
The additional root, $2\bar\omega$, is obtained by complex conjugation of the root $2\omega$ of $P_4(x)$, eq. $(1)$.
|
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|
Prove that $e^x>x(x+1)$
Let $x>0$,prove that $$e^x>x(x+1)$$or
$$x>\ln{x}+\ln{(x+1)}$$
we can use this Taylor some first four term,
$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3$$ prove it
But this inequality it seem very nice,maybe there exsit simple methods or Amazing way?
|
We need to prove$$e^x>1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3\ge x^2+x$$which is$$x^3+3x^2+6x+6\ge 6x^2+6x$$which is$$x^3-3x^2+6=(x-2)^2(x+1)+2>0$$which is obvious for positive $x$.
|
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|
How do you find the Taylor Series of $f(x) = x^{2}*\sin(\frac{x^{2}}{3})$ about $a = 0$? Please without getting too technical, this is calculus 2. I know that the general form of a Taylor Series about 0 (Maclaurin) is: $$\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} * x^{n}$$ and so, $$f(x) = f(0)/0! + f (0)/1!*x + f (0)/2!*x^2 + f (0)/3!*x^3 + .... $$
Just very unsure how this one would be carried out and simplified. Is it simply taking derivatives? Can it be put into general summation notation?
Please help
Thank you
|
Use Taylor Series for $\sin x$
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}\ldots$$
$$\sin {\left(\frac{x^2}{3} \right)}=x-\frac{{\left(\frac{x^2}{3} \right)}^3}{3!}+\frac{{\left(\frac{x^2}{3} \right)}^5}{5!} -\frac{{\left(\frac{x^2}{3} \right)}^7}{7!}\ldots$$
You want $x^2 \sin {\left(\frac{x^2}{3} \right)}$, so multiply the above series by $x^2$.
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = x^3- \frac{x^8}{162} +\frac{x^{12}} {29160} \ldots$$
In general :
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = \sum _{n=0}^{\infty}\frac{(-1)^n x^{4n+6}}{3^{2n+1} (2n+1)!}$$
|
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Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \times 2 = 7 \times 3^{2k+1} + 2 \times 3^{2k+1} + 2^{k-1} \times 2.$$
Where do I go from here?
|
By the binomial theorem
$$
3^{2n+1}=3\cdot 9^n=3(7+2)^n=3(7a+2^n)=21a+6\cdot2^{n-1}
$$
Therefore
$$
3^{2n+1}+2^{n-1}=21a+7\cdot2^{n-1}=7(3a+2^{n-1})
$$
|
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|
Radii of inscribed and circumscribed circles in right-angled triangle In a right angled triangle, △ ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R.
Prove that in △ABC, $a+b=2\cdot \left(r+R\right)$.
|
We have the following two formulas for $R$ and $r$:
*
*$$R=\tfrac12c=\tfrac12 \sqrt{a^2+b^2}$$
($c^2=a^2+b^2$ by Pythagoras.)
*
*$$r=\dfrac{ab}{a+b+c}$$
(proof: the area of the (right) triangle ABC, i.e., $\tfrac12ab$ is equal to the sum of the areas of triangles $IBC, ICA, IAB$, i.e., $\tfrac12ra+\tfrac12rb+\tfrac12rc$, where $I$ is the center of the inscribed circle.)
Thus we have to prove that
$$\tag{1}a+b=2\dfrac{ab}{a+b+\sqrt{a^2+b^2}}+2\tfrac12 \sqrt{a^2+b^2}$$
In fact, (1) is equivalent with:
$$\tag{2}a+b-\sqrt{a^2+b^2}=\dfrac{2ab}{a+b+\sqrt{a^2+b^2}} $$
itself equivalent to:
$$\tag{3}(a+b)^2-(\sqrt{a^2+b^2})^2=2ab$$
which is true.
|
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If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
|
Consider $\frac{a}{b}=\frac{ka}{kb}$ Then, $$\frac{a+ka}{b+kb}=\frac{(k+1)a}{(k+1)b}=\frac{a}{b}$$
which is exactly what you noticed, but with $a=1,b=2,k=2$
|
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bijection between number of non-congruent triangles with perimeter 2n and integer sides and number of partitions of n into exactly three terms. Is there any a bijection between the following problems?
Let $f(n)$ be the number of non-congruent triangles with perimeter 2n
and integer sides. Let $h(n)$ be the number of partitions of n
into exactly three terms.
I thought about it a lot, but I got nothing. I checked that for n=3,4,5 and 6 and it works.
Any help would be appreciated.
|
The number of non congruent triangles is given here ... https://oeis.org/A005044
The first few values are $0, 0, 0, \color{red}{1}, 0, \color{red}{1}, 1, \color{red}{2}, 1, \color{red}{3}, 2, \color{red}{4}, 3, \color{red}{5}, 4,\color{red}{7}, 5, \color{red}{8}, 7, \color{red}{10}, 8, \color{red}{12}, 10, \color{red}{14},\cdots$
The number of partitions into three parts is given here ...https://oeis.org/A001399
The first few values are $1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, \cdots$
The generating function for partitions into three parts is given by
\begin{eqnarray*}
\frac{x^3}{(1-x)(1-x^2)(1-x^3)}
\end{eqnarray*}
The generating function for the number of non congruent triangles is given
\begin{eqnarray*}
\frac{x^3}{(1-x^2)(1-x^3)(1-x^4)}= \frac{x^3+x^6}{(1-x^2)(1-x^6)(1-x^4)} = \\\frac{x^3}{(1-x^2)(1-x^4)(1-x^6)}+\color{red}{\frac{x^6}{(1-x^2)(1-x^4)(1-x^6)}}
\end{eqnarray*}
but we only want every second term in this, which is shown in red above ... & so clearly enumerates to the same as the number of partitions of $n$ into three parts.
|
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|
how SVD is calculated in reality let us suppose that we have following matrix
$
A=
\left[ {\begin{array}{cc}
4 & 0 \\
3 & -5 \\
\end{array} } \right]
$
for calculation of SVD,first i have calculated $A'*A$ which is equal to
$
A'*A=
\left[ {\begin{array}{cc}
25 & -15 \\
-15 & 25 \\
\end{array} } \right]
$
eigenvalues of this matrix is equal to $40$ and $10$, eigenvector of following matrix
$
\left[ {\begin{array}{cc}
-15 & -15 \\
-15 & -15 \\
\end{array} } \right]
$
i got this matrix after subtraction of $40$ from diagonal elements, eigenvector is equal to
\begin{bmatrix}-1 \\ 1\\ \end{bmatrix}
after inserting of eigenvalues of $10$, i got following matrix
$
\left[ {\begin{array}{cc}
15 & -15 \\
-15 & 15 \\
\end{array} } \right]
$
eigenvector of this matrix is equal to
\begin{bmatrix}1 \\ 1\\ \end{bmatrix}
so normalization of these vectors and putting in one matrix $V$ will have the following form
$
\left[ {\begin{array}{cc}
-1/\sqrt{2} & 1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2} \\
\end{array} } \right]
$
now i know that
$A*V=U*E$ where $E$ is equal to
$
\left[ {\begin{array}{cc}
\sqrt{40} & 0 \\
0 & \sqrt{10} \\
\end{array} } \right]
$
we know that
$A*v1=u1*\sigma$
let us try to multiply
$
\left[ {\begin{array}{cc}
4 & 0 \\
3 & -5 \\
\end{array} } \right]
$
by
\begin{bmatrix}-1/\sqrt{2} \\ 1/\sqrt{2} \\ \end{bmatrix}
i got the following result
\begin{bmatrix}-4/\sqrt{2} \\ -8/\sqrt{2} \\ \end{bmatrix}
but i can't get equation for $\sigma$ and $u$ please help me
|
You did everything correctly. You already found $\sigma_1,\sigma_2$ - they are the square roots of the eigenvalues of $A^TA$ so $\sigma_1 = \sqrt{40},\sigma_2 = \sqrt{10}$. After you found $V$ with columns $v_1,v_2$, you must have
$$ Av_1 = \sigma_1 u_1, Av_2 = \sigma_2 u_2 $$
so $u_1$ is just $\frac{Av_1}{\sigma_1}$ and $u_2 = \frac{Av_2}{\sigma_2}$. In your case,
$$ u_1 = \begin{pmatrix} -\frac{4}{\sqrt{2}} \\ -\frac{8}{\sqrt{2}} \end{pmatrix} \frac{1}{\sqrt{40}} = \begin{pmatrix} -\frac{1}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} \end{pmatrix}, \\
u_2 = \begin{pmatrix} \frac{4}{\sqrt{2}} \\ -\frac{2}{\sqrt{2}} \end{pmatrix} \frac{1}{\sqrt{10}} = \begin{pmatrix} \frac{2}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} \end{pmatrix} $$
and indeed
$$ \begin{pmatrix} -\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 3 & -5\end{pmatrix} \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \sqrt{40} & 0 \\ 0 & \sqrt{10} \end{pmatrix}. $$
|
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|
Polynomials $P(x)$ such that $P(x-1)$ $=$ $P(-x)$ Are there an infinite number of polynomials $P(x)$ of even degree such that
$P(x-1) = P(-x)$
$P(x) = x^2+x+1$ is a good example because $P(x-1) = (x-1)^2+(x-1)+1 = x^2-x+1 = P(-x)$.
$P(x) = x^4+2x^3+4x^2+3x+1$ is another example because $P(x-1) = (x-1)^4+2(x-1)^3+4(x-1)^2+3(x-1)+1 = x^4-2x^3+4x^2-3x+1 = P(-x)$.
Is there an easy way to generate such polynomials? Thanks for help.
|
The condition $P(-x)=P(x-1)$ is equivalent to $P(x-1/2)=P(-1/2-x)$,
namely that $P(x-1/2)$ is an even function. That should be helpful.
|
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|
For how many numbers $X^2 \equiv X \mod 10^n$? I've been looking through this post:
Square of four digit number $a$
And wanted to see if there's a way to generalize the idea, in order to answer the following question:
For how many numbers $X^2 \equiv X \mod 10^n$?
Also, is $9376$ the only four digit number such that $9376^2 \equiv 9376 \mod 10^4$
Thanks
|
The number of distinct numbers modulo $10^n$ solving
$$x^2\equiv x\pmod{10^n}\tag1$$
is four.
To see this by the Chinese remainder theorem $x$ solves (1) iff both
$$x^2\equiv x\pmod{2^n}\tag2$$
and
$$x^2\equiv x\pmod{5^n}\tag3.$$
These congruences state that $p^n\mid x(x-1)$ for $p=2$ or $5$.
As $x$ and $x-1$ have no common factor then $p^n\mid x$ or $p^n
\mid(x-1)$. So the solutions are $x\equiv 0$ or $1\pmod{p^n}$.
By the Chinese remainder theorem again these can be assembled into
four solutions modulo $10^n$. Two are obvious: $0$ and $1$. The other
two are $x_n$ and $1-x_n$ where $x_n\equiv 0\pmod{2^n}$ and
$x_n\equiv 1\pmod{5^n}$. These can be found by applying the extended
Euclidean algorithm to $2^n$ and $5^n$.
For $n=4$ you have $x_4=9376$, as you say, so $1-9376\equiv 625\pmod{10^4}$. So in this case $x_4$ is the only solution
between $10^3$ and $10^4-1$ inclusive, that is "with four digits".
|
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|
Find a quadratic integer polynomial that annihilates a given $2\times2$ integer matrix Suppose $A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$. How can we find integers $b$ and $c$ such that $A^2+bA+cI_2=0$?
|
This is an application of the Cayley-Hamilton Theorem.
Every square matrix satisfies its own characteristic function.
Compute the characteristic polynomial of
$$\mathbf{A}=
\left(
\begin{array}{cc}
3 & 2 \\
2 & 3 \\
\end{array}
\right)
$$
$$ \det \mathbf{A} = 5, \qquad \text{tr }\mathbf{A} = 6$$
The characteristic polynomial is
$$
\color{blue}{p(\lambda)} = \lambda^{2} - \lambda \text{tr }\mathbf{A} + \det \mathbf{A} = \color{blue}{\lambda^{2} - 6 \lambda + 5}
$$
By the Cayley-Hamilton theorem
$$
\boxed{
\color{blue}{\mathbf{A}^{2} - 6 \mathbf{A} + 5 \mathbf{I}_{2}} = \mathbf{0}
}
$$
This is the characteristic polynomial with the target matrix as the argument: $\color{blue}{p \left( \mathbf{A} \right)} $
Confirmation:
$$
%
\begin{align}
%
\mathbf{A}^{2} - 6 \mathbf{A} + 5 \mathbf{I}_{2} =
%
\left(
\begin{array}{cc}
13 & 12 \\
12 & 13 \\
\end{array}
\right)
%
-
\left(
\begin{array}{cc}
18 & 12 \\
12 & 18 \\
\end{array}
\right)
+
\left(
\begin{array}{cc}
5 & 0 \\
0 & 5 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array}
\right)
%
\end{align}
%
$$
(Thanks to @amd for helpful suggestions.)
|
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|
Finding rank of a matrix depending on parameter a We have the matrix
$$A = \begin{bmatrix}1 & a & 1\\-4 & 2 & a \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}$$
and we want to find the rank of this matrix as a function of parameter $a$.
|
The rank of the matrix is equal to the number of nonzero rows in the matrix after reducing it to the row echelon form using elementary transformations over the rows of the matrix.
$\ \begin{bmatrix}1 & a & 1\\-4 & 2 & a \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}$ substract (-4)*row 1 from row 2 ->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}$ substract a*row 1 from row 3 ->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ 0 & -a^2-1 & -a-2 \\ 4 & 6 & 1\end{bmatrix}$ substract 4*row 1 from row 4 ->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ 0 & -a^2-1 & -a-2 \\ 0 & -4a+6 & -3\end{bmatrix}$ substract $\ (\frac{-a^2-1}{4a+2})$*row 2 from row 3, $\ 4a+2\neq0$ ->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ 0 & 0 & \frac{a^3-9a}{4a+2} \\ 0 & -4a+6 & -3\end{bmatrix}$ substract $\ (\frac{-2a+3}{2x+1})$*row 2 from row 4,$\ 2a+1\neq0$ ->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ 0 & 0 & \frac{a^3-9a}{4a+2} \\ 0 & 0 & \frac{2a^2-a-15}{2a+1}\end{bmatrix}$ substract $\ (\frac{4a+10}{a^2+3a})$*row 3 from row 4 ,$\ a^3-9a\neq0$->
$\ \begin{bmatrix}1 & a & 1\\0 & 4a+2 & a+4 \\ 0 & 0 & \frac{a^3-9a}{4a+2} \\ 0 & 0 & 0\end{bmatrix}=M =>$ rankM=3
And now to check for the values of $\ a$ that are excluded from the previous calculation
for $\ a=\frac{-1}{2} ->rankA=3,a=3-> rankA=2, a=-3->rankA=3, a=0->rankA=3$
|
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|
If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals? If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals?
I got this and similar type of question in a book and I don't really know how to exactly solve it. Any help will be appreciated.
|
Let $f(x)=4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $. Then $f(-1)=-14$ since the remainder when we divide by $x+1$ is $-14$.
So, $4+1-3-5+k-2+1+k-5-5=-14$
Thus, $2k=4$and $k=2$
|
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show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$.
|
Note that
$$\dfrac{(n+1)(n+2)\dots (2n)}{1\cdot 3 \cdot 5 \dots (2n-1)} = \dfrac{(2n)!/n!}{(2n)!/(2\cdot 4 \cdot 6 \cdot \dots \cdot (2n))} = \dfrac{(2n)!/n!}{(2n)!/(2^n n!)} = 2^n$$
Hence,
$$(n+1)(n+2)\dots (2n) = 1\cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n$$
Since the first $n$ multipliers on the right side are all odd, the maximum power of two that divides the product in question is indeed $2^n$.
Here is some intuition behind this solution.
First of all, it is clear that every natural number $m$ can be represented in the form $$m = 2^k \operatorname{maxodd}(m)$$
where $2^k$ is the maximum power of two that divides $m$ and $\operatorname{maxodd}(m)$ is the maximum odd divisor of $m$.
Note that the numbers in the range $[n+1, 2n]$ have different maximum odd divisors. Indeed, if for two different numbers in this range their maximum odd divisors were the same, one of them should be at least two times greater than the other one, which is impossible. This means that each of the number $1, 3, 5, \dots, 2n-1$ (here are exactly $n$ numbers) should be the maximum odd divisor for exactly one number from $[n+1, 2n]$ range.
For example, if $n=5$ then $[n+1, 2n] = \{6, 7, 8, 9, 10\}$ and maximum odd divisors are $\{3, 7, 1, 9, 5\} = \{1, 3, 5, 7, 9\}$.
This means that the power of two in the product is equal to
$$\dfrac{(n+1)(n+2)\dots (2n)}{1\cdot 3 \cdot 5 \dots (2n-1)} = \dfrac{(2n)!/n!}{(2n)!/(2\cdot 4 \cdot 6 \cdot \dots \cdot (2n))} = \dfrac{(2n)!/n!}{(2n)!/(2^n n!)} = 2^n$$
|
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How to find two vectors orthogonal to the gradient space of the feasible set? After finding the solution to the minimization of $f(x,y,z)=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2$ constrained by $h(x,y,z)=x^2+y^2+z^2-1=0$ by using the Lagrange multiplier method:
$$\nabla f(x,y,z) = 2 \begin{pmatrix} x-\frac{1}{2}\ \\ y-\frac{1}{2} \\ z-\frac{1}{2}\end{pmatrix}=\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}, \nabla h(x,y,z)=\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$$
such that: $\begin{pmatrix} 2x-1\ \\ 2y-1 \\ 2z-1\end{pmatrix}=\lambda \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}, x=y=z=\frac{1}{2(1-\lambda)}$
solving for $\lambda$ using constraint: $3\left(\frac{1}{2(1-\lambda)}\right)^2=1 \Rightarrow 2(1-\lambda)= \pm \sqrt{3} \Rightarrow \lambda=1 \pm \frac{\sqrt{3}}{2}$
which yields: $x=y=z=\pm \frac{1}{\sqrt{3}}$
making the final solution: $(x,y,x,\lambda)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},1 - \frac{\sqrt{3}}{2}\right)$
Using that solution, it was suggested to use Gramm-Schmidt orthogonalization to construct two independent vectors orthogonal to $\nabla h\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ which will be in the tangential space of the feasible set.
I understand that I could just take two solutions to $a\frac{1}{\sqrt{3}}+b\frac{1}{\sqrt{3}}+c\frac{1}{\sqrt{3}}=0$ where $\vec{v}=(a,b,c)$ are the orthogonal vectors, but I'm curious how to implement Gramm-Schmidt with just one vector because as soon as you get to the second iteration, it seems that you need another vector. Is the idea to construct another linearly independent vector first and then use Gramm-Schmidt?
|
I guess that the idea is that you can choose any two vectors, as long as they are independent to $\nabla h$, but this is easy to achieve by simple inspection, e.g. two vectors from the canonical basis $e_1=(1,0,0)$, $e_2=(0,1,0)$ will do the job. Then apply Gram-Schmidt to the basis $(\nabla h,e_1,e_2)$ and you obtain an orthogonal basis where the second and third vectors are a basis of the plane that is orthogonal to $\nabla h$.
|
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|
Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm {log} (x+2) . \textrm {log} (x+3) + C$$
Is this correct? Or, How do I proceed the other way?
|
Method to do -
$\frac{1}{(x+3)(x+2)} =\frac{A}{x+2}+\frac{B}{x+3}$
$1 = A(x+3)+B(x+2)$
Case 1 -
When $x+3=0$
$x=-3$
Put $x=-3$
$1 = -B$
$B = -1$
Case 2 -
When $x+2=0$
$x=-2$
Put $x=-2$
$1 = A$
$A = 1$
Now your integral becomes,
$\int (\frac{1}{x+2}-\frac{1}{x+3})\,dx$
|
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|
Finding all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ Find all real values of $a$ for which $\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$ for all $x>0$
Attempt: Let $$\sqrt{9-(x-a)^2}> \sqrt{16-x^2}$$
So $$x^2-(x-a)^2>7$$
So $$a(2x-a)>7\Rightarrow a^2-2ax+7<0$$
could some help me how to find range of $a,$ Thanks
|
The functions
$g(x;a) = \sqrt{9-(x-a)^2}$ and
$f(x)=\sqrt{16-x^2}$
are semicircles whose domains are $x \in [a-3, a+3]$ and $x \in [-4,4]$ respectively. Pictured above are $f(x), g(x;-1)$ and $g(x;1)$.
As the above picture suggests, $\sqrt{9-a^2+2ax-x^2} \le \sqrt{16-x^2}$ for all $x \in [-1,1]$ when $a \in [-1,1]$.
If $a \in [-7 -1] \cup [1,7]$, then the domains of $f$ and $g$ will overlap and, at least for some $x$ in that overlap region, we will have
$\sqrt{9-a^2+2ax-x^2}>\sqrt{16-x^2}$
|
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How does $y=\frac{2x+2}{x+2}$ contribute to demonstrating the Dedekind cut for $\sqrt{2}$? In a previous question a respondent corrected my description a Dedekind cut of $\sqrt{2}$. This Wikipedia article states: "Showing that it is a cut requires showing that for any positive rational $x$ with $x^2 < 2$, there is a rational $y$ with $x < y$ and $y^2 < 2$." I can follow this easily enough. I get lost on the next statement: "The choice $y=\frac{2x+2}{x+2}$ works." How so? I've tried squaring the expression and substituting $2$ for $x^2$, but I'm having trouble seeing it? Can anyone help me ask through this?
|
Let $x$ be a positive rational number with $x^2<2$. First note that
$$x < \frac{2x+2}{x+2}$$
is equivalent to
$$x^2+2x<2x+2$$
which in turn is equivalent to $x^2<2$, which is true by hypothesis. So $x<y$.
Now note that
$$\left( \frac{2x+2}{x+2} \right)^2< 2$$
is equivalent to
$$(2x+2)^2 < 2(x+2)^2$$
which simplifies to
$$4x^2+8x+4 < 2x^2 +8x + 8$$
which simple algebra shows is equivalent to $x^2 < 2$, again. So $y^2<2$, as required.
EDITED TO ADD:
There is nothing particularly magic about this specific choice of $y$. Here, for example, is an alternative way of constructing a (different) rational number $y$ with $x<y$ and $y^2<2$.
Let $x$ be a positive rational number with $x^2 < 2$. If we define $z = \frac{2}{x}$, then $z > 2$. Now construct the average of $x$ and $z$, and call that $w$.
Note first that $w$ lies strictly between $x$ and $z$. Moreover, we have (by the AM-GM inequality) that $2 = xz < w^2$.
Finally, define $y = \frac{2}{w}$. It's straightforward to verify that $x<y$ and $y^2 < 2$.
If you follow the trail of breadcrumbs, you can show that this choice of $y$ is given by $y = \frac{4x}{x^2 + 2}$. You can also prove directly from this formula (using methods similar to the ones at the top of this answer) that this choice of $y$ has the properties that $x<y$ and $y^2 < 2$.
|
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Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
|
By C-S we obtain:
$$v=$$
$$=\sqrt{a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a^2\sin^2x+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}=$$
$$=\sqrt{a^2+b^2+2\sqrt{(a^2\cos^2x+b^2\sin^2x)(b^2\cos^2x+a^2\sin^2x)}}\geq$$
$$\geq\sqrt{a^2+b^2+2|ab|(\cos^2x+\sin^2x)}=|a|+|b|.$$
$$v\leq\sqrt{2(a^2\cos^2x+b^2\sin^2x+b^2\cos^2x+a^2\sin^2x)}=\sqrt{2(a^2+b^2)}.$$
|
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|
Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible over the integers for all $n$? It seems $x^2-x+1$ or $x^4-x^2+1$ are irreducible over the integers.
Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible for all non-negative integer $n$? If so, how to prove?
|
Multiplying by $ (X^{2^n} + 1) $, we have that
$$ (X^{2^n} + 1)(X^{2^{n+1}} - X^{2^n} + 1) = X^{3 \cdot 2^n} + 1 $$
It is clear from this that the roots of $ X^{2^{n+1}} - X^{2^n} + 1 $ are the primitive $ 3 \cdot 2^{n+1} $th roots of unity, and the degree of the corresponding minimal polynomial is $ \varphi(3 \cdot 2^{n+1}) = 2^{n+1} $. Thus, $ X^{2^{n+1}} - X^{2^n} + 1 $ is the $ 3 \cdot 2^{n+1} $th cyclotomic polynomial, and it follows that it is irreducible.
|
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|
Integrals with fractions in exponentials I came across this integral in a research paper while trying to understand Bernstein's inequalities. I want to upper bound or evaluate this integral but it looks too complicated. The integral is:
\begin{align}
I=\int_0^\infty \exp \left ( \frac{-at^2}{bt+c} \right ) dt, \quad a,b,c >0.
\end{align}
Is there any procedure to upper bound the above integral? My attempt is to simplify the fraction so that I would get
$$
I= \int_0^\infty \exp \left ( \frac{-at}{b} + \frac{ac}{b} - \frac{ac^2}{b^3}\left(\frac{1}{t+c/b} \right) \right) dt.
$$
But this doesn't seem too useful either. Any suggestions?
|
$\int_0^\infty e^\frac{-at^2}{bt+c}~dt$
$=\int_c^\infty e^\frac{-a\left(\frac{t-c}{b}\right)^2}{t}~d\left(\dfrac{t-c}{b}\right)$
$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_c^\infty e^{-\frac{at}{b^2}-\frac{ac^2}{b^2t}}~dt$
$=\dfrac{e^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{act}{b^2}-\frac{ac^2}{b^2ct}}~d(ct)$
$=\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{act}{b^2}-\frac{ac}{b^2t}}~dt$
$=\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
Consider $\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$ ,
$\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\int_0^1e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\int_\infty^1e^{-\frac{ac}{b^2}\left(\frac{1}{t}+t\right)}~d\left(\dfrac{1}{t}\right)+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\int_1^\infty\dfrac{1}{t^2}e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\int_1^\infty\left(1+\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=2\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt-\int_1^\infty\left(1-\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$\therefore\dfrac{ce^\frac{2ac}{b^2}}{b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_0^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt+\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_1^\infty\left(1-\dfrac{1}{t^2}\right)e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~dt$
$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)+\dfrac{ce^\frac{2ac}{b^2}}{2b}\int_1^\infty e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}~d\left(t+\dfrac{1}{t}\right)$
$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)-\dfrac{ce^\frac{2ac}{b^2}}{2b}\left[\dfrac{b^2}{ac}e^{-\frac{ac}{b^2}\left(t+\frac{1}{t}\right)}\right]_1^\infty$
$=\dfrac{ce^\frac{2ac}{b^2}}{2b}K_1\left(\dfrac{2ac}{b^2}\right)+\dfrac{b}{2a}$
|
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|
Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this article it started to use mod, and the whole explanation seemed to very difficult to understand. Is there a more plain English explanation of how this is solved?
|
Find the smallest integer that leave a remainder of 3 when divided by 5, a remainder of 5 when divided by 7, and a remainder of 7 when divided by 11
We want to find a solution to:
$$
\begin{align}
x&\equiv 3 \pmod 5\\
x&\equiv 5 \pmod 7\\
x&\equiv 7 \pmod {11}
\end{align}
$$
The solution is unique $\pmod {385}$.
We can find a simple formula for this, by first finding the solutions to:
$$
\begin{align}
x&\equiv 1 \pmod 5\\
x&\equiv 0 \pmod 7\\
x&\equiv 0 \pmod {11}
\end{align}
$$
$$
\begin{align}
x&\equiv 0 \pmod 5\\
x&\equiv 1 \pmod 7\\
x&\equiv 0 \pmod {11}
\end{align}
$$
$$
\begin{align}
x&\equiv 0 \pmod 5\\
x&\equiv 0 \pmod 7\\
x&\equiv 1 \pmod {11}
\end{align}
$$
By using the Extended Euclidean Algorithm, we find:
$$
\begin{align}
x\equiv 231\pmod {385}\\
x\equiv 330\pmod {385}\\
x\equiv 210\pmod {385}\\
\end{align}
$$
Hence the solution to:
$$
\begin{align}
x&\equiv a \pmod 5\\
x&\equiv b \pmod 7\\
x&\equiv c \pmod {11}
\end{align}
$$
is:
$$231a+330b+210c\pmod{385}$$
For $a=3,b=5,c=7$, we get $693+1650+1470=3813\equiv 348\pmod{385}$.
|
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|
Find the value of $\sum_{n=1}^{\infty} \frac{2}{n}-\frac{4}{2n+1}$ Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$
My Try:we have
$$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$
$$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so
$$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$ But we know
$$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ So
$$S=2(2-\ln 2)$$
Is this correct?
|
We can use the digamma function $\psi$. It is known that
$$\psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\dfrac{z}{n(n+z)},\ \ \forall z \in \mathbb{C} \setminus \{-1,-2,...\}.$$
Thus for $z=\frac{1}{2}$ we obtain
$$\psi\left(\dfrac{3}{2}\right)=-\gamma+\sum_{n=1}^{\infty}\dfrac{1}{n(2n+1)}=-\gamma+\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{2}{(2n+1)}\right)=-\gamma+\dfrac{S}{2}.$$
From the other hand, for the half-integer values of $\psi$ we have
$$\psi\left(n+\dfrac{1}{2}\right)=-\gamma-2\ln 2+\sum_{k=1}^n\dfrac{2}{2k-1},$$
from where, for $n=1$, we conclude that
$$\psi\left(\dfrac{3}{2}\right)=-\gamma-2\ln 2+2.$$
Finally $-\gamma+\dfrac{S}{2}=-\gamma-2\ln 2+2 \Rightarrow S=4(1-\ln 2),$
as Simply Beautiful Art mentioned.
Note: S is a real number and it can be justified via comparison with the series of the Basel Problem.
|
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|
Evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$ For $a \geq 1$, evaluate $\lim _ { x \rightarrow 0} \frac { \tan ( x ^ { a } ) - ( \sin x ) ^ { a } } { x ^ { a + 2} }$
I would like to solve this without Taylor's expansion. I've tried adding and subtracting $x^a$ but I don't know if this is a correct approach. I've also tried L'Hospital and reached the result $(a^2 - a) /(6a +12). $
|
If $a$ is a positive integer then there is an easy approach via factorization. Let us first deal with $a=1$. We have $$\frac{\tan x - \sin x}{x^{3}} =\frac{\sin x} {x} \frac{1}{\cos x} \frac{1-\cos x} {x^{2}}\to \frac{1}{2}\tag{1}$$ Also note that using L'Hospital's Rule or Taylor series we can easily show that $$\frac{\tan x- x} {x^{3}}\to \frac{1}{3}\tag{2}$$ and subtracting these above two limits we see that $$\frac{x-\sin x} {x^{3}}\to\frac{1}{6}\tag{3}$$ For general integer $a>1$ we have $$\frac{\tan x^{a} - x^{a}} {x^{a+2}} = \frac{\tan x^{a} - x^{a}} {x^{3a}} \cdot x^{2(a-1)}\to\frac{1}{3}\cdot 0=0\tag{4}$$ and $$\frac{x^{a} -\sin^{a} x} {x^{a+2}}=\frac{x-\sin x} {x^{3}}\sum_{i=0}^{a-1}\frac{x^{a-1-i}}{x^{a-1-i}}\frac{\sin^{i}x}{x^{i}}\to\frac{1}{6}\cdot a=\frac{a} {6}\tag{5}$$ Adding equations $(4)$ and $(5)$ we get the desired answer as $a/6$. If $a$ is not an integer then you need to resort to powerful technique of Taylor series as mentioned in another answer.
|
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|
Convolution Integral problem I'm having a hard time to do this exercise. I'm using the definition for convolution but I'm stuck at the integral. Thanks in advance.
For $t>0$ consider $f_a(x)=\frac{1}{\sqrt{4 \pi a}}e^{-\frac{x^2}{4a}}$.
Show that $f_a*f_b=f_{a+b}$
By definition: $(f*g)(x)$= $\int_{-\infty}^{+\infty} f(u)g(x-u) du$
One computes,
$(f_a*f_b)(x)=$ $\int_{-\infty}^{+\infty} \frac{1}{\sqrt{4 \pi a}}e^{-\frac{u^2}{4a}} \frac{1}{\sqrt{4 \pi b}}e^{-\frac{(x-u)^2}{4b}} du$
$=\frac{1}{\sqrt{4 \pi a}}\frac{1}{\sqrt{4 \pi b}} \int_{-\infty}^{+\infty} e^{-\frac{u^2}{4a}}e^{-\frac{(x-u)^2}{4b}}$
I can't solve this integral. I tried using Gaussian Integral but still cant solve it.
|
One of the most useful elementary tricks to know when dealing with Gaussian-type integrals is completing the square.
On that note, we calculate
\begin{align}
f_a*f_b(x)
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{y^2}{4a}-\frac{(x-y)^2}{4b}\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(y^2+\frac{a}{b}(x-y)^2\right)\right)\,\mathrm{d}y\\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(\left(1+\frac{a}{b}\right)y^2-\frac{2a}{b}xy+\frac{a}{b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2-\frac{2a}{a+b}xy+\frac{a}{a+b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(\left(y-\frac{a}{a+b}x\right)^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}y^2\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\sqrt{\frac{4ab}{a+b}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty e^{-z^2}\,\mathrm{d}z \\
&= \frac{1}{\sqrt{4\pi(a+b)}}\exp\left(-\frac{x^2}{4(a+b)}\right) \\
&=f_{a+b}(x).
\end{align}
Of course, using the Fourier transform, we can simplify this calculation somewhat – if you haven't read about the Fourier transform (and its action on Gaussians), I highly recommend you take a look. The basic techniques used in calculating the Fourier transform of a Gaussian are in fact very closely analogous to the calculations above.
|
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|
Solve the first order differential equation, $\frac{x\mathop{dx}-y\mathop{dy}}{x\mathop{dy}-y\mathop{dx}}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}$ I need to find the general solution to the differential equation,
$$\dfrac{x\mathop{dx}-y\mathop{dy}}{x\mathop{dy}-y\mathop{dx}}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}$$
I have tried simplifying it upto
$$\mathop{d\sqrt{1+x^2-y^2}}=\dfrac{x\mathop{dy}-y\mathop{dx}}{\sqrt{x^2-y^2}}$$
but but do not know what to do with the RHS. How to solve the differential equation?
|
Substitute $x=r \sec \theta$, $y=r \tan \theta$.
$x^2 - y^2 = r^2$
Taking differentials, $xdx - y dy = r dr$
and $\frac y x=\sin\theta$
Taking differentials, $\dfrac {x dy - y dx} {x^2} = \cos \theta d\theta$
or, $x dy - y dx = r^2 \sec \theta d\theta$
Thus the differential equation becomes
$\dfrac {rdr}{r^2 \sec\theta d\theta} =\sqrt{\dfrac{1+r^2}{r^2}}$
Can you follow from here?
|
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|
Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$ Show that $x^{12}-x^9+x^4-x+1\geq0$ for all $x$.
When $x\leq0$ , this is easy.
When $x\geq1$, then also this is easy.
I need help with the case when $0\leq x\leq 1$
|
$x^{12} - x^9 +x^4 - x + 1 \ge 0$;
Rewritten:
$x^{12} + (x^4 - x^9) + (1 - x) \ge 0$.
For $0 \le x \le 1$:
We have $x^{12} \ge 0$; $(x^4 -x^9) \ge 0$ and $(1-x) \ge 0$.
The sum of the above terms is $ \ge 0$.
|
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|
Divisors of the form $4n+1$ I read a question
Number of divisors of the form $(4n+1)$
.
In the soultion
Any positive divisor of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form
$4k+1$ is a number of the form: $$3^a\cdot 5^b\cdot 7^c$$ with $0\leq
> a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $\color{red}{a+c}$ being even. There are:
$$\frac{4\cdot 4\cdot 6}{2}=\color{red}{48}$$
why a+c should be even?
Plz use easy language. I don't know the meaning of mod....
|
HINT:
As $3=4-1,7=8-1$
$$3^a7^c=(4-1)^a(8-1)^b$$
using Binomial Expansion, we can prove it to be of the form $4c+(-1)^{a+b}$ where $c$ is some integer
|
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|
To prove $0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$ Prove that $$0.4 \le \int_{0}^{1} x^{\sin x+\cos x} dx \le 0.5$$
i have proved as follows:
when $x \in (0 \: \: 1)$ $$\sin x+\cos x \in (1 \:\:\:\sin 1+\cos 1)$$
and so $$x^{\sin x+\cos x} \in (x^{\sin 1+\cos 1} \:\:\: x)$$ $\implies$
$$\int_{0}^{1}x^{\sin x+\cos x} \in \left(\int_{0}^{1}x^{\sin 1+\cos 1} \:\:\:\int_{0}^{1} x\right)$$ that is
$$\int_{0}^{1}x^{\sin x+\cos x} \in \left(\frac{1}{\sin 1+\cos 1+1} \: \: \: 0.5\right)$$
But $\frac{1}{\sin 1+\cos 1+1} $ is approximately $0.4$ , is this proof correct?
|
You may note that $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4)$.
By investigating the monotonicity of $\sin x$, we have, $\forall x \in [0,1]$, $\sin x+\cos x =\sqrt 2\sin(x+\frac \pi 4) \in [1, \sqrt 2]$.
Now can you go further starting from here?
P.S: To get the first identity, we can exploit the sine of the sum identity.
$\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$.
First, $a\sin \alpha+b\cos \alpha=\sqrt{a^2+b^2} (\sin \alpha {a\over \sqrt{a^2+b^2}}+\cos \alpha {b\over \sqrt{a^2+b^2}})$.
Obviously there must exist one $\theta$ where $\cos \theta = {a\over \sqrt{a^2+b^2}}$ and $\sin \theta={b\over \sqrt{a^2+b^2}}$, and this leads to $a\sin \alpha+b\cos \alpha=\sqrt{a^2+b^2} (\sin \alpha {a\over \sqrt{a^2+b^2}}+\cos \alpha {b\over \sqrt{a^2+b^2}})=\sqrt{a^2+b^2} (\sin \alpha \cos \theta+\cos \alpha \sin \theta)=\sqrt{a^2+b^2} \sin(\alpha+\theta)$
|
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|
For $x \ge 1$, is $2^{\sqrt{x}} \ge x$ For $x \ge 1$, is $2^{\sqrt{x}} \ge x$
The answer appears to me to be yes since:
$$\sqrt{x} \ge \log_2 x$$
Here's my reasoning:
(1) For $x > 16$, $5\sqrt{x} > 4\sqrt{x} + 4$
(2) For $x > 25$, $x > 5\sqrt{x}$
(3) So, $x > 25$, $2x > x + 4\sqrt{x} + 4$ and it follows that $\sqrt{2x} > \sqrt{x} +2$
(4) So, as $x$ doubles, $\sqrt{2x} > \sqrt{x} + 2$ but $\log_2(2x) = \log_2(x) +1$
Is my reasoning correct? Is there a simpler way to analyze this?
|
The claim is in fact false. Take $x=9$, we have
$$2^{\sqrt{9}}=2^3=8<9.$$
|
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|
If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
|
WLOG let $\theta=\dfrac\pi2-2y\implies x=\csc2y+\cot2y=\dfrac{1+\cos2y}{\sin2y}=\cot y$
$$\sin\theta=\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{\cot^2y-1}{\cot^2y+1}=?$$
|
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|
A number $x$ such that $x^n = x$ for every $n$ Is there proof fully algebraic that shows that $1$ is the only number that when put in the function $f(x) = x^n$, for $n = $ any number, the output is always $x$?
This is pretty obvious, but you seem to prove it using the trivial knowledge .
For example, $1^4 = 1, 1^{46373} = 1$, but $2^4 \neq 2$
|
For positive integer $n$, write
$$f(x)=x^n=x^n-1+1=(1+x+\cdots+x^{n-1})(x-1)+1.$$
If $x > 1$, then $(1+x+\cdots +x^{n-1})(x-1) > n(x-1) $ so $f(x) > n(x+1)+1 > 1$.
If $0<x < 1$, then $(1+x+\cdots +x^{n-1})(x-1) < \frac{1}{1-x}(x-1) = -1$. Hence $f(x) < -1+1 = 0$.
|
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|
What is the limit of this sequence? How does it relate to the exponential function?
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$
I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.
Edit:
I'm trying to prove a result that would be true if
$(1+x^2/n^2)^n -1 = x^kO(x/n).$
|
The limit can be evaluated without appeal to the exponential function. To proceed we first note that
$$\left(1+\frac{x^2}{n^2}\right)^n\left(1-\frac{x^2}{n^2}\right)^n\le 1\tag1$$
Then, rearranging $(1)$, we obtain
$$ \left(1+\frac{x^2}{n^2}\right)^n\le \frac{1}{\left(1-\frac{x^2}{n^2}\right)^n}\tag 2$$
Using Bernoulli's Inequality on the right-hand side of $(2)$ reveals for $n>|x|$
$$1\le \left(1+\frac{x^2}{n^2}\right)^n\le\frac{1}{1-\frac{x^2}{n}}\tag3$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\lim_{n\to \infty}\left(1+\frac{x^2}{n^2}\right)^n=1$$
Now, we can relate this to the exponential function as follows. First, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality, that the exponential function satisfies the inequality
$$ \left(1+\frac xn\right)^n\le e^x \le \frac{1}{1-x} \tag 4$$
for $-n<x<1$.
From $(4)$ we see that
$$\begin{align}
\left(1+\frac{x^2}{n^2}\right)^n&=\left(\left(1+\frac{x^2}{n^2}\right)^{n^2}\right)^{1/n}\\\\
&\le e^{x^2/n}\\\\
&\le \frac{1}{1-\frac {x^2}n}\tag 5
\end{align}$$
Note that $$\frac{1}{1-\frac {x^2}n}=1+\frac{x^2}{n}+O\left(\frac{x^4}{n^2}\right)$$
|
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|
$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$
I've to show that $$a^2+c^2=1$$ $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the books is $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$
|
Just a very basic brute force method,
$a^2 + b^2 = 1$ - (1)
$c^2 + d^2 = 1$ -(2)
$ac+bd=0$
On multiplying the two we get, $a^2c^2+b^2d^2+a^2d^2+b^2c^2=1$
Whereas $a^2c^2+b^2d^2+2abcd=0$ {squaring both sides $ac + bd = 0$}
Thus we get $-2abcd+a^2d^2+b^2c^2=1 \Rightarrow (ad-bc)^2=1 \Rightarrow ad-bc=\pm1$
Adding (1) and (2) we get, $a^2+b^2+c^2+d^2=2$ , Which can be re written as $a^2+b^2+c^2+d^2=2ad-2bc$.
$\Rightarrow (b+c)^2+(a-d)^2=0$. Thus we get, $b=-c$ and $a =d$.
#Q.E.D
|
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|
$\frac{'ab'}{'ba'}=\frac{'bc'}{'cb'}$ Question:
We write $'pq'$ to denote a two digit integer with tens digit $p$ and units digit $q$. For which values of $a,b$ aand $c$ are the two fractions $\frac{'ab'}{'ba'}$and$\frac{'bc'}{'cb'}$ equal and different from $1$?
My attempt:
So, write this using "normal algebra";$$\frac{10a+b}{10b+a}=\frac{10b+c}{10c+b}$$Multiply by the two denominators:$$(10a+b)(10c+b)=(10b+c)(10b+a)$$Multiply out:$$100ac+10ab+10bc+b^2=100b^2+10ab+10bc+ac$$ And we can subtract $(10ab + 10bc) $ from each side. We are left with:$$100ac+b^2=100b^2+ac$$ Which, by rearranging, we get $$99ac=99b^2$$which is equivalent to (if we divide by $99$) $$ac=b^2$$
$$[a,b,c]\neq 0$$ Since they are all parts of the tens digit, and if they are $0$, then they will not be a two digit number, which is vital. $a \neq b$, since if they were $=$, the fraction would $= 1$, and neither can this relationship occur between $b \neq c$
This was all written down on paper by me. Can anybody help me finish this off?
|
Looks good so far. Now you just have to write down all the ways that $b^2 = ac$ for $1 \leq a, b, c \leq 9$, $a \neq b$, $b \neq c$ : $$\begin{align*} 2^2 &= 1 \times 4 \\ 3^2 &= 1 \times 9 \\ 4^2 &= 2 \times 8 \\ 6^2 &= 4 \times 9 \end{align*} $$
Each of these gives two solutions; e.g. $2^2 = 1 \times 4$ gives $(a, b, c) = (1, 2, 4)$ or $(a, b, c) = (4, 2, 1)$. The criterion $b^2 = ac$ states that $a, b, c$ stand in a geometric progression; if the common ratio of this progression is $r$, then $'bc' = r \times {'ab'}$ and $'cb' = r \times {'ba'}$, so this should make intuitive sense.
|
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Generating functions ( recurrence relations ) Find $a_n$ using Generating Functions : $a_n = -a_{n-1} + 2a_{n−2}$, $n\ge2$ and $a_0 = 1$, $a_1 = 2$.
Approach : So I will form a characteristic equation $ r^2 + r - 2 = 0$ whose roots are $r_1 = -2$, $r_2 = 1$.
So my general solution is $a_n = α_1r_1^n + α_2r_2^n$.
$a_n = α_1(-2)^n + α_2(1)^n$
When $a_0 = 1$, then $1 = α_1(-2)^0 + α_2(1)^0$, then $α_2 = 1 - α_1 $.
When $a_1 = 2$, then $2 = α_1(-2)^1 + α_2(1)^1$, then $-2α_1 + 1 - α_1 = 2$.
$α_1 = -1/3$ and $α_2 = 4/3 $
So $a_n = -1/3r_1^n + 4/3r_2^n$.
Can anyone tell me if it is correct or not and any help will be appreciated :) .
Also, if I have $a_{n+2}=a_{n+1}+2a_n$. Can someone tell me if its char. equation should be like $r^2-r-2 = 0$? Just asking because of the addition symbol rather than subtraction.
|
If you really want to use generating functions to get $a_n$. In general with a recurrence relation with initial conditions $a_0$ and $a_1$ and \begin{equation}
a_n = r_1a_{n-1}+r_2a_{n-2}
\end{equation}
you can write the generating function as \begin{equation}
G(x) = \frac{-a_0-a_1x+a_0r_1x}{r_2x^2+r_1x-1}
\end{equation}
so your question will have \begin{equation}
G(x) = \frac{1+3x}{1+x-2x^2} = 1 + 2x +0x^2 + 4x^3 -4x^4 + \cdots
\end{equation}
where the coefficients of the expansion are the sequence $a_n$. To extract the coefficients we can see that \begin{equation}
\frac{1}{n!}\frac{d^nG(x)}{dx^n}\bigg|_{x=0} =a_n
\end{equation}
however working out the $n^{th}$ derivative if a function can be tricky.
|
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|
Intersection of a hypersphere (4d) and a hyperplane I have a problem to solve and I really am in over my head here. Just to get me started, if I have a four dimensional plane and a 4 dimensional sphere
$ax + by + cz + dw = k$ and $x^2+y^2+z^2+w^2 = R$, and they intersect, am I expecting a three dimensional sphere.
In particular, what if I simply have x = r as the hyperplane, where r is less than R?
|
Multiply the $4d$ sphere by $(a^2+b^2+c^2+d^2)$ & rearrange it to give
\begin{eqnarray*}
(ax+by+cz+dw)^2+(bx-ay+dz-cw)^2+(cx-dy-az+bw)^2+(dx+cy-bz-aw)^2=R^2(a^2+b^2+c^2+d^2)
\end{eqnarray*}
Now substitute the equation for the plane & let
\begin{eqnarray*}
X=bx-ay+dz-cw \\
Y=cx-dy-az+bw\\
Z=dx+cy-bz-aw
\end{eqnarray*}
and $\rho^2 =R^2(a^2+b^2+c^2+d^2)-k^2$
\begin{eqnarray*}
X^2+Y^2+Z^2= \rho^2
\end{eqnarray*}
So the manifold of intersetion is indeed a $3d$ sphere.
|
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Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using 3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$
Is everything right here?
But how to check it for convergence?
|
Write your sum like:
$$S = {S}_1 + {S}_2 + {S}_3$$
Where,
$${S}_1 = 1 + \frac{1}{\sqrt{10}\cdot 10} + \frac{1}{10^{3}} + ... = \frac{1}{1 - \frac{1}{10 \cdot\sqrt(10)}} = \frac{10\sqrt{10}}{10\sqrt{10} - 1}$$
$${S}_2 = -\frac{1}{\sqrt{10}} - \frac{1}{10^{2}} - ... = -\frac{1}{\sqrt{10}} \cdot{S}_1$$
$${S}_3 = -\frac{1}{10} - \frac{1}{10^{2}\sqrt{10}} - ... = -\frac{1}{10}\cdot{S}_1$$
And finally:
$$S = \frac{9\sqrt{10} - 10}{10\sqrt{10} - 1}$$
|
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|
Number of values for a such that the following polynomial has only integral roots: $P(x) =X^3-X+a^2+a$? How do I solve this problem? I thought of solving it with Viete but I only have 2 relations from which I can find the roots.
|
Let $(X-\alpha)(X-\beta)(X-\gamma) = X^3 - X + a^2 + a$ where $\alpha,\beta,\gamma \in \mathbb{Z}$. Then
$$\alpha + \beta + \gamma = 0, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = -1, \qquad \alpha\beta\gamma = -a^2 -a.$$
From the first equation, $\gamma = -(\alpha + \beta)$. Then from the second equation
$$\alpha^2 + \alpha\beta + \beta^2 - 1 = 0$$
so
$$\beta = \dfrac{-\alpha \pm \sqrt{4 -3\alpha^2}}{2}$$
but since $\alpha,\beta\in \mathbb{Z}$, then $\alpha$ must be either $0$ or $\pm 1$. This gives the solutions
$$(\alpha,\beta,\gamma) \in \{(-1,0,1), (-1,1,0),(0,-1,1),(0,1,-1),(1,-1,0),(1-0,-1)\}$$
Which means that $X^3 - X + a^2 + a = X(X-1)(X+1) = X^3 - X$, i.e. $a^2 + a = 0$ which can only happen if $a = 0$ or if $a = -1$ so there are only two possible values of $a$ for which the equation only have integral roots.
|
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Find the total number of partitions of $12$ having unequal positive parts. $P(k, n) = P(k - 1, n - 1) + P(k - n, n)$
Let $P^\star(k,n)$ denote the number of partitions of $k$ having exactly $n$ positive parts, all of which are unequal. For example, $P(8, 3)$ implies $8$ can be partitioned as $8 = 2 + 2 + 4,$ but $P^\star(8,3)$ does not.
$P^\star(k, n) = P(k - \binom n2, n)$
We need the $\sum_{i=1}^4P^\star(12,i)$ because $P(k, n)$ is not defined for $P^\star(12,n)$ where $5 \le n \le 12.$
Now,
$P^\star(12,1) = P(12, 1)$
$P^\star(12,2) = P(11, 2)$
$P^\star(12,3) = P(9, 3)$
$P^\star(12,4) = P(6, 4)$
Now we calculate the four $P(k, n)$ above.
$P(12, 1) = 1.$
$P(11, 2) \\ = P(10, 1) + P(9, 2) \\ = 1 + P(8, 1) + P(7, 2) \\ = 1 + 1 + P(6, 1) + P(5, 2) \\ = 1 + 1 + 1 + P(4, 1) + P(3, 2) \\ = 1 + 1 + 1 + 1 + P(2, 1) + P(1, 2) \\ = 1 + 1 + 1 + 1 + 1 \\ = 5.$
$P(9, 3) \\ = P(8, 2) + P(6, 3) \\ = P(7, 1) + P(6, 2) + P(5, 2) + P(3, 3) \\ = 1 + P(5, 1) + P(4, 2) + P(4, 1) + P(3, 2) + 1 \\ =1 + 1 + P(3, 1) + P(2, 2) + 1 + P(2, 1) + P(1, 2) + 1 \\ = 1 + 1 + 1 + 1 + 1 + 1 + 1 \\ = 7.$
$P(6, 4) \\ = P(5, 3) + P(2, 4) \\ = P(4, 2) + P(2, 3) \\ = P(3, 1) + P(2, 2) \\ = 1 + 1 \\ = 2.$
So, the answer to this problem is $1 + 5 + 7 + 2 = 15.$
Does the calculation above make sense to you? Thanks.
|
Your answer is correct. As a check, here is a list of the $15$ partitions of $12$ into unequal positive parts:
\begin{align*}
12 & = 12\\
& = 1 + 11\\
& = 2 + 10\\
& = 3 + 9\\
& = 4 + 8\\
& = 5 + 7\\
& = 1 + 2 + 9\\
& = 1 + 3 + 8\\
& = 1 + 4 + 7\\
& = 1 + 5 + 6\\
& = 2 + 3 + 7\\
& = 2 + 4 + 6\\
& = 3 + 4 + 5\\
& = 1 + 2 + 3 + 6\\
& = 1 + 2 + 4 + 5
\end{align*}
|
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|
Solve system in $a,b,c$ What's the best way to solve the system of equations ?:
$$
\left\{\begin{array}{rcccccr}
a & + & b & + & c & = & 1
\\
a^{2} & + & b^{2} & + & c^{2} & = & 7
\\
a^{3} & + & b^{3} & + & c^{3} & = & 13
\end{array}\right.
$$
If I square the first one I get $2\left(\,ab + bc + ac\,\right)=-6$, and if I multiply the second by $a + b + c$ I get some other expression that equals $-6$ but I'm not sure how to proceed.
|
By using Newton's identities, we can find
that
$$s_2:=ab+bc+ca=\frac{1}{2}(a+b+c)^2-\frac{1}{2}(a^2+b^2+c^2)=-3$$
and
$$s_3:=abc=\frac{1}{3}(a^3+b^3+c^3)-\frac{1}{3}(a+b+c)^3+(ab+bc+ca)(a+b+c)=1.$$
Then we obtain $(a,b,c)$ by solving the polynomial equation
$$x^3-s_1x^2+s_2x-s_3=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\\=(x+1)(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))=0$$
where $s_1=a+b+c$.
|
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|
Compute $\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}$ Compute
$$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}.$$
Today I have participated in Olympiad which had 8 different tasks. This was one of them.
I've started by substituting concrete values and looking for a pattern:
$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!} = 1 - \frac{2^2}{1!} + \frac{3^2}{2!} - \frac{4^2}{3!} + \frac{5^2}{4!} + \dots$
This gave me insight that it may be connected to the following series:
$e = 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$
After that for a few hours I was trying to transform original series to the series for $e$. My intuition tells me that the result is something similar to $e \times constant$. But I wasn't able to prove it or disprove it.
|
HINT:
As $(k+1)^2=k(k-1)+3k+1,$
For $k\ge2,$
$$\dfrac{(k+1)^2}{k!}=\dfrac1{(k-2)!}+3\cdot\dfrac1{(k-1)!}+\dfrac1{k!}$$
$$\dfrac{(k+1)^2}{k!}x^k=x^2\cdot\dfrac{x^{k-2}}{(k-2)!}+3x\cdot\dfrac{x^{k-1}}{(k-1)!}+\dfrac{x^k}{k!}$$
For $k=0,$ $$\dfrac{(k+1)^2}{k!}x^k=1$$
For $k=1,$ $$\dfrac{(k+1)^2}{k!}x^k=4x=3x\cdot\dfrac1{0!}=\dfrac{x^1}{1!}$$
Now use $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
|
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|
Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and I don't know what am I doing wrong. What is the correct way to prove this equality?
|
$$\frac{c^2}{c-s}-\frac{cs}{c+s}=\frac{\cos x}{2\cos^2x-1}$$
$$\frac{c^3+sc^2-c^2s+cs^2}{c^2-s^2} =\frac{\cos x}{2\cos^2x-1}$$
$$\frac{ c }{c^2-s^2}=\frac{\cos x}{2\cos^2x-1}$$
$$\frac{ c }{2c^2-1}=\frac{\cos x}{2\cos^2x-1}$$
|
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|
Is the transformation possible? I came across this problem which asks to transform $x^2-x-2$ to $x^2-x-1$ ,if possible, using the following rules:
Given a quadratic equation $ax^2+bx+c$ you can :
1)Interchange $a$ and $c$
2)Replace $x$ by $x+t$ where $t$ is a real number.
My approach:
I wrote the equations as
${(x-\frac12)}^2-\frac94$ and ${(x-\frac12)}^2-\frac54$
Replacing $x$ by $x+t$ in first equation
${(x-\frac12+t)}^2-\frac94$
And setting this equal to ${(x-\frac12)}^2-\frac54$
Simplifying and solving gives a quadratic with variable $t$
Which on applying quadratic formula gives
$$t=\frac{1-2x\pm \sqrt{4x^2-4x+5}}{2}$$
So, is this correct? If wrong, could somebody help me in the right direction?
|
We start with a quadratic $ax^2+bx+c$ and use one of the steps to transform it to $a'x^2+b'x+c'$.
In the first case, we have $a'=c,b'=b,c'=a$, in the second case, we have $a'=a, b'=b+2at, c'=c+bt+t^2$.
Note that in the first case, we have $b'^2-4a'c'=b^2-4ac$ and in the second case $b'^2-4a'c'=b^2+4abt+4a^2t^2-4a(c+bt+t^2)=b^2-4ac$.
In other words, neithre of the allowed steps changes the discriminat of our quadratic.
Hence by starting from $x^2-x-2$ with discriminant $9$, we cannot reach $x^2-x-1$ with discriminant $5$.
|
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|
How to solve $p^{2a+1}+p^a+1=b^2$ in integers? For prime number $p$ and positive integers $a$ and $b$ solve the following equation
$$p^{2a+1}+p^a+1=b^2$$
Can you give me a hint for starting this problem?
|
Hint
$$p^a(p^{a+1}+1)=b^2-1=(b-1)(b+1)$$
Hint 2 $gcd(b-1,b+1)=1$ or $2$. If $p \neq 2$ then $p^a$ must divide either $b-1$ or $b+1$.
Hint 3
If $b+1=kp^{a}$ then $b-1=\frac{p^{a+1}+1}{k}$. Then
$$kp^a-\frac{p^{a+1}+1}{k}=2 \\
k^2p^a-p^{a+1}-1=2k \\
k^2p^a-p^{a+1}=2k+1 \\
$$
This means that $p |2k+1$.
Also, $b+1=kp^{a}$ and $b-1=\frac{p^{a+1}+1}{k}$ imply that $1 < k<p$.
There is only one possibility.
The case $b-1=kp^{a}$ then $b+1=\frac{p^{a+1}+1}{k}$, is almost identic, you just get a sign changed.
Lats, the case $p=2$ is easy to tackle separately.
|
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|
How to calculate product $\prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)$? How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right) $$....Can I take a geometric series and compare it with that?
|
As a hint, consider $n = 3$:
$\displaystyle\prod_{k = 0}^{2}\left(1+\dfrac{1}{2^{2^k}}\right) = \left(1+\dfrac{1}{2^1}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^4}\right)$
$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^2} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} \cdot \dfrac{1}{2^4}$
$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + \dfrac{1}{2^5} + \dfrac{1}{2^6} + \dfrac{1}{2^7}$
$= 2-\dfrac{1}{2^7}$
Can you generalize this for larger values of $n$?
|
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|
Evaluate the closed form of $\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)$ Last question of this form
I am very curious to what is the closed form of:
Assume where $a,b,c > 0$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b\sin^2x+c\cos^2x}\mathrm dx=f(a,b,c)\tag1$$
$$b\sin^2x+c\cos^2x$$
$$=b(1-\cos^2 x)+c\cos^2 x$$
$$=b-(b-c)\cos^2 x$$
$$\int_{0}^{\pi/2}{\arctan(a\tan^2x)\over b-(b-c)\cos^2x}\mathrm dx\tag2$$
$u=a\tan^2x\implies du=2a\tan x\sec^2 x dx$
${u-a\over a}=\sec^2 x$
$${\sqrt{a}\over 2}\int_{0}^{\infty}{\arctan(u)\over \sqrt{u}(bu+ac-2ab)}\mathrm du\tag3$$
$u=y^2\implies du=2ydy$
$$\sqrt{a}\int_{0}^{\infty}{\arctan(y^2)\over by^2+ac-2ab}\mathrm dy\tag4$$
$b=P$ and $ac-2ab=Q$
Take the form of: $$\int_{0}^{\infty}{\arctan(y^2)\over Py^2+Q}\mathrm dy\tag5$$
Apply $arctan(y)$ series
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\int_{0}^{\infty}{y^{4n+2}\over Py^2+Q}\mathrm dy\tag6$$
|
We have
$$ f(a,b,c) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan(a \tan^2\theta)}{b \sin^2\theta + c\cos\theta^2} \, d\theta = \frac{\pi}{\sqrt{bc}} \left( \arctan\left(1 + \sqrt{\frac{2ac}{b}}\right) - \frac{\pi}{4} \right). $$
Here is a sketch of computation.
Step 1. Reduction of the problem (not essential)
Consider the curve $\gamma(\theta) = (\sqrt{c}\cos\theta, \sqrt{b}\sin\theta)$ for $\theta \in [0,\pi/2]$. This is part of an ellipse in the first quadrant. By noting that $ \sqrt{bc} d\theta = x dy - y dx$, it follows that
$$ f(a,b,c) = \int_{\gamma} \omega, \quad \text{where} \quad \omega = \frac{1}{\sqrt{bc}} \arctan\left(\frac{ac}{b}\cdot\frac{y^2}{x^2}\right) \frac{x dy - y dx}{x^2 + y^2}. $$
Write $g = ac/b$ for simplicity. Then the crucial observation is that
$$ d\omega = \frac{1}{\sqrt{bc}} \left( \frac{\partial}{\partial x} \frac{x \arctan(gy^2/x^2)}{x^2 + y^2} + \frac{\partial}{\partial y} \frac{y \arctan(gy^2/x^2)}{x^2 + y^2} \right) dx \wedge dy = 0 $$
in the first quadrant. So by the Green's theorem, we can replace $\gamma$ by any piecewise $C^1$-curve in the first quadrant that joins from $P = (\sqrt{c}, 0)$ to $Q = (0, \sqrt{b})$. Since the integral of $\omega$ along any line segment on each axis is zero, we can also replace $P$ by any point on the positive $x$-axis and likewise for $Q$. Putting altogether, we can replace $\gamma$ by the circular arc $(x,y) = (\cos\theta, \sin\theta)$ to obtain
$$ f(a,b,c)
= \frac{1}{\sqrt{bc}} \int_{0}^{\frac{\pi}{2}} \arctan(g\tan^2\theta) \, d\theta
= \frac{1}{2\sqrt{bc}} \int_{0}^{\infty} \frac{\arctan(g u)}{\sqrt{u}(1+u)} \, du. $$
Step 2. Calculus
Now the rest of computation can be done with the aid of residue computation. Indeed, consider the last integral as a function of $g$. Then the residue computation shows that
$$ \frac{\partial f}{\partial g}
= \frac{1}{2\sqrt{bc}} \int_{0}^{\infty} \frac{\sqrt{u}}{(1+u)(1+ g^2 u^2)} \, du
= \frac{\pi}{2\sqrt{bc}} \cdot \frac{1}{\sqrt{2g}(1 + \sqrt{2g} + g)}. $$
Integrating both sides and using the initial value $f|_{g=0} = 0$ gives the desired answer.
Remark. I realized that we can skip Step 1 and apply techniques in Step 2 directly to
$$ f(a,b,c) = \frac{1}{2} \int_{0}^{\infty} \frac{\arctan (a u)}{\sqrt{u}(c + bu)} \, du. $$
Still, Step 1 looks pleasing to me as it sanitizes parameters and demonstrates some unusual technique.
|
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|
shortcut method for calculating a determinant \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c& a+b \end{vmatrix} = 4abc
Is there any shortcut method to compute this determinant without breaking by definition. Need some help please.
|
$$\begin{vmatrix}
b+c & a & a \\ b & c+a & b \\ c & c& a+b
\end{vmatrix} \xrightarrow{-R_1+R_2} \begin{vmatrix}
b+c & a & a \\ -c & c & -a+b \\ c & c& a+b
\end{vmatrix} \xrightarrow{-R_2+R_3} \begin{vmatrix}
b+c & a & a \\ -c & c & -a+b \\ 2c & 0 & 2a
\end{vmatrix} \xrightarrow{-0.5R_3+R_1} \begin{vmatrix}
b & a & 0 \\ -c & c & -a+b \\ 2c & 0 & 2a
\end{vmatrix} \xrightarrow{-0.5R_3+R_2} \begin{vmatrix}
b & a & 0 \\ 0 & c & b \\ 2c & 0 & 2a
\end{vmatrix} = b(2ac) - a (-2bc) = 4abc$$
|
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|
Radius of converge of Laurent series for $\frac{1}{\sin z}$ I want to show that the radius of convergence of Laurent series for $\frac{1}{\sin z}$ is $\pi$
I showed that:
\begin{align}
\frac 1 {\sin z} & = \frac 1 {z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots } = \frac 1 {z\big(1 - (\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots )\big)} \\[10pt]
& = \frac{1}{z} \bigg( 1 + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg) + \bigg(\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg)^2 + \cdots \bigg) \\[10pt]
& = \frac{1}{z} +\frac{z}{3!} + \bigg( \frac{1}{3!}- \frac{1}{5!}\bigg)z^3 + \cdots
\end{align}
This equality holds if and only if:
$$\bigg|\frac{z^2}{3!} - \frac{z^4}{5!} + \cdots \bigg| < 1
\iff \bigg|\frac{\sin z}{z} - 1\bigg| < 1$$
How can I show that the last inequality implies $|z| < \pi$ ?
|
After writing my comment, I realize that it is way simpler...apparently:
The function $\;\cfrac1{\sin z}\;$ is analytic at $\;\left\{\,z\in\Bbb C\;/\;0<|z|<\pi\,\right\}\;$ and thus the radius of convergence of (the non principal part of) the Laurent series of the function around zero is at least $\;\pi\;$ . But as already commented, it can't be $\;\ge\pi\;$ as the function has a pole of order one at $\;z=\pi\;$ . Thus , the radius is exactly $\;\pi\;$ .
|
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|
Arclength, finding definite integral Find the length of the segment between $x=5$ and $x=2$ for the following curve.
$$f(x)= \frac{1}{2}x^2-\frac{1}{4}\log\left(x\right)$$
My working so far.
$$f'\left(x\right)=x-\frac{1}{4x}.$$
Then
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)\right)^2}\,dx
=\int _2^5\:\sqrt{x^2+\frac{1}{2}+\frac{1}{16x^2}}\,dx.$$
However I have no idea how to find the solution to the definite integral.
|
$$L = \int _2^5\:\sqrt{1+\left(f'\left(x\right)^2\right)}
\\
\\=\int _2^5\:\sqrt{1+\left(x-\frac{1}{4x}\right)^2}\\
\\=\int _2^5\:\sqrt{1+x^2-\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
=\int _2^5\:\sqrt{x^2+\dfrac{1}{2}+\dfrac{1}{(4x)^2}}dx\\
\\=\int _2^5\:\sqrt{\left(x+\frac{1}{4x}\right)^2}\\
=\int _2^5\:|x+\frac{1}{4x}|dx \to (2<x<5) \\
\int _2^5\:(x+\frac{1}{4x})dx$$
|
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|
Prove that the product is never a perfect square
Prove that for nonnegative integers $x_1,\ldots,x_{2011}$ and $y_1,\ldots,y_{2011}$ the product $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_{2011}^2+3y_{2011}^2)$$ is never a positive perfect square.
I thought about generalizing this question to any odd subscript $n$ instead of $2011$. Thus, $$(2x_1^2+3y_1^2)(2x_2^2+3y_2^2) \cdots (2x_n^2+3y_n^2)$$ is never a perfect square. For $n = 1$ we have $2x^2+3y^2 = z^2$ and I want to show the only solution is $x = y = z = 0$. If $x$ is even, then $y$ must be even by taking modulo $4$. If $x$ is odd, then $y$ must be odd. I didn't see how to continue from here.
|
Another viewpoint is by using the composition of binary quadratic forms, to reduce the case $ n $ odd to the case $ n = 1 $, in a `systematic' way. We have the following identities writing a product
$$
(2a^2 + 3b^2)(2c^2 + 3d^2) = (3 b c + 2 a d)^2 + 6 (-a c + b d)^2
$$
and
$$
(2a^2+3b^2)(c^2 + 6y^2) = 2 (3 b c + a d)^2 + 3 (-2 a c + b d)^2 \, .
$$
So the $ n = 2m+1 $-fold product of
$$
N = (2x_1^2 + 3y_1^2) \cdots (2x_n^2 + 3y_n^2)
$$
can be written as
$$
N = 2X^2 + 3Y^2 \,
$$
for some integer polynomials $ X = f(x_0,\ldots,x_n,y_0,\ldots,y_n) $ and $ Y = g(x_0,\ldots,x_n,y_0,\ldots,y_n) $.
So if this product is ever a positive square, we can obtain values $ X, Y $ to make $ 2X^2 + 3Y^2 $ a positive square. Following other answers, we see this is impossible.
If $ 2X^2 + 3Y^2 = Z^2 $ is a positive square, we can assume $ \gcd(X,Y,Z) = 1 $ by dividing through if necessary. Then reducing modulo 3 shows that $ -X^2 \equiv Z^2 \pmod{3} $. If $ X \not\equiv 0 \pmod{3} $, we obtain $ -1 \equiv \Box \pmod{3} $. Contradiction, as the squares modulo 3 are 0 and 1. So $ 3 \mid X $ and $ 3 \mid Z $. Hence $ 9 \mid 3Y^2 $, so $ 3 \mid Y $, contradicting $ \gcd(X,Y,Z) = 1 $. So $ Z = 0 $ is the only possible solution, which immediately gives $ X = Y = 0 $ too.
|
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|
Show which of $6-2\sqrt{3}$ and $3\sqrt{2}-2$ is greater without using calculator How do you compare $6-2\sqrt{3}$ and $3\sqrt{2}-2$? (no calculator)
Look simple but I have tried many ways and fail miserably.
Both are positive, so we cannot find which one is bigger than $0$ and the other smaller than $0$.
Taking the first minus the second in order to see the result positive or negative get me no where (perhaps I am too dense to see through).
|
$6-2√3 \approx2(1.732) = 6 - 3.464 = 2.536$
$3√2-2 \approx 3(1.414) - 2 = 4.242 - 2 = 2.242$
This implies that $\dfrac 52$ is between the two quantities:
\begin{align}
\dfrac{49}{4} &> 12 \\
\dfrac 72 &> 2\sqrt 3 \\
\dfrac{12}{2} &> 2\sqrt 3 + \dfrac 52 \\
6 - 2\sqrt 3 &> \dfrac 52
\end{align}
and
\begin{align}
\dfrac{81}{4} &> 18 \\
\dfrac 92 &> 3\sqrt 2 \\
\dfrac 52 &> 3\sqrt 2 - 2
\end{align}
It follows that $6-2√3 > 3\sqrt 2 - 2$.
|
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|
Find all real coefficient polynomial such $f(x)|f(x^2-2)$
Find all real coefficient polynomial with the degree is $3$,and such
$$f(x)|f(x^2-2),\forall x\in Z $$
I try: let $f(x)=ax^3+bx^2+cx+d$,then $$f(x^2-2)=ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d$$,then such
$$(ax^3+bx^2+cx+d)|ax^6-6ax^4+12ax^2-8a+bx^4-4bx^2+4b+cx^2-2c+d,\forall x\in Z $$following I can't it
|
Hint: Write $$f(x^2-2)=g(x)f(x)$$ If $x$ is a root of $f$ then $x^2-2$ is a root too. Then $(x^2-2)^2-2$ is a root of $f$ and etc. But $f$ cannot have infinitely many roots.
|
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|
What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)? I have the following combinatoric problem to solve; I have searched and couldn't find an answer already existing.
In the main lottery in Germany, 6 numbers are to be drawn out of 49 {1,2,…49} without replacement. Let's assume I can only choose 4 numbers between 1 and 49. What is the probability that at least 2 of the four numbers are among the six numbers that have been drawn in the lottery?
I drew a sample of numbers and therefore have a rough idea that it should be around 6.2-7% if I did everything correctly and the sample was large enough. However, I am interested in the exact probability and its derivation.
Any help would be highly appreciated.
|
We can consider three cases:
*
*Draw four numbers out of six. All four balls must belong to the six chosen balls, so the probability of this happening equals:
$$\frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{3}{46}$$
*Draw three numbers out of six. One way to do this, is by first choosing three balls belonging to the six chosen balls, followed by one ball which does not belong to these balls. This probability equals:
$$\frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{43}{46}$$
Since there are four different turns in which we can choose a ball not belonging to the six chosen balls, the probability of selecting three balls out of six equals:
$${4 \choose 1} \cdot \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{43}{46}$$
*Draw two numbers out of six. This time there are ${4 \choose 2} = 6$ combinations of turns in which we can choose a ball not belonging to the six chosen balls, so this happens with probability:
$${4 \choose 2} \cdot \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{43}{47} \cdot \frac{42}{46}$$
Adding this all up, the probability of choosing at least two correct numbers equals:
$$\frac{6 \cdot 5 \cdot 4 \cdot 3 + 4 \cdot 6 \cdot 5 \cdot 4 \cdot 43 + 6 \cdot 6 \cdot 5 \cdot 43 \cdot 42}{49 \cdot 48 \cdot 47 \cdot 46} \approx 0.0681$$
|
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|
A polynomial is divisible with another one For what $m$ and $n$ is the polynomial $2X^{19}+X^{13}+mX^{11}+X^8+2X^6+nX^2+2$ divisible by $X^4+X^3+X^2+X+1$.
I tried to find the real solutions for g but couldn't
|
HINT:
$$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0$$
$$\implies x^{19}=x^4,x^{13}=x^3$$
|
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|
Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$ Let $a^4+b^4+c^4=3$. Prove that $a^2+b^2+c^2\geq a^2b+b^2c+c^2a$
My proof :D
We have the inequality
$\sum_{cyc}^{ }a^2.\sum_{cyc}^{ }a\geq \sum_{cyc}^{ }a^2b $
which is equivalent to $\sum_{cyc}^{ }b(a-b)^2$ (true)(1)
In another side, by AM-GM, we have: $\sum_{cyc}^{ }a\leq \sum_{cyc}^{ }\dfrac{a^4+1+1+1}{4}=3$ (2)
Thus, from (1) and (2), we've completed our proof.
|
It's enough to prove our inequality for non-negative variables.
We'll prove a stronger inequality:
$$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3,$$
which solves our problem.
Indeed, by C-S $$a^2+b^2+c^2=\sqrt{(a^2+b^2+c^2)^2}\leq\sqrt{(1+1+1)(a^4+b^4+c^4)}=3.$$
Thus,
$$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3=(a^2+b^2+c^2)\sqrt{\frac{a^2+b^2+c^2}{3}}\leq a^2+b^2+c^2.$$
Since the inequality $$a^2b+b^2c+c^2a\leq3\left(\sqrt{\frac{a^2+b^2+c^2}{3}}\right)^3$$
is homogeneous, we can assume that $a^2+b^2+c^2=3$ and we need tho prove that:
$$a^2b+b^2c+c^2a\leq3.$$
Now, let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, be Rearrangement and AM-GM we obtain:
$$a^2b+b^2c+c^2a=a\cdot ab+b\cdot bc+c\cdot ca\leq x\cdot xy+y\cdot xz+z\cdot yz=$$
$$=y(x^2+xz+z^2)\leq y\left(x^2+z^2+\frac{x^2+z^2}{2}\right)=\frac{3}{2}y(3-y^2)\leq3$$ and we are done!
|
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|
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
|
Letting
$c = \cos(x)$,
$\cos x\cos2x\cos3x
=c(2c^2-1)(4c^3-3c)
=c^2(2c^2-1)(4c^2-3)
$
so
$\dfrac{1-\cos x\cos2x\cos3x}{\sin^2x}
=\dfrac{1-c^2(2c^2-1)(4c^2-3)}{1-c^2}
=8 c^4 - 2 c^2 + 1
$
(according to Wolfy).
Putting
$c^2 = 1-s^2$
(s = sin),
this becomes
$8(1-s^2)^2-2(1-s^2)+1
=8(1-2s^2+s^4)-2+2s^2+1
=8s^4-14s^2+7
$.
As $x \to \infty$,
the limit of this
does not exist
since it oscillates
from 1 to 7.
As $x \to 0$,
the limit is 7.
|
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|
Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it equal?
What's next?
From Wolfram Mathematica I know that $\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}= \frac{1}{1624}$.
|
HINT: How about rewriting the sum as
$$S=\sum_{1\leq a<b<c\\a,b,c\in\mathbb{N}}\frac{1}{2^a 3^b 5^c}
=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\sum_{c=b+1}^{\infty}\frac{1}{2^a 3^b 5^c}=\sum_{a=1}^{\infty}\frac{1}{2^a}\sum_{b=a+1}^{\infty}\frac{1}{3^b}\sum_{c=b+1}^{\infty}\frac{1}{5^c}.$$
|
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|
prove that : if $a, b \in \mathbb{R}^+$ : then : $a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$ prove that : if $a, b \in \mathbb{R}^+$ : then :
$$a^2b^2(a^2+b^2-2)\geq (a^2+b^2)(ab-1)$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2b+b^2a-a^2$$
$$a^4b^2+b^4a^2-2a^2b^2 \geq a^3b-a^2+b^3a-b^2$$
now what ?
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Let $ab=x$, $a^2+b^2=y$, then $0\leq 2x\leq y$ and we want
$$x^2(y-2)\geq y(x-1)$$
$$x^2y-2x^2\geq xy-y$$
$$y\geq\frac{2x^2}{x^2-x+1}$$
It is equivalent to show
$$2x\geq\frac{2x^2}{x^2-x+1}$$
$$x^2-x+1\geq x$$
$$(x-1)^2\geq0$$
Equality holds iff $a=b=1$.
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Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to Wolfram is $(x,y)=(\frac{1}{3},\frac{2}{3}).$
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You can use the Cauchy-Schwarz Inequality:
$$(1+1+1)\big(x^2+(1-y)^2+(y-x)^2\big)\geq \big(1\cdot x+1\cdot(1-y)+1\cdot(y-x)\big)^2=1^2=1\,.$$
Thus,
$$x^2+(1-y)^2+(y-x)^2\geq\frac{1}{3}\,.$$
The inequality becomes an equality if and only if $$\frac{x}{1}=\frac{1-y}{1}=\frac{y-x}{1}\,,$$ or equivalently $$(x,y)=\left(\frac{1}{3},\frac{2}{3}\right)\,.$$
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number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$
Total number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$
$\bf{Attempt}$ writting $1,3,3^2,3^3$ as $6k+1$ or $6k+3$
same way $1,5,5^2,5^3,5^4,5^5$ as $6k+1$ or $6k+4$
same way $1,7,7^2,7^3,7^4$ as $6k+1$
same way $1,11,11^2,11^3$ as $6k+1$ or $6k+5$
could some help me how can i solve my question, thanks
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Well, we can get $1 = 6k+1$ for $k=0$ off the bat, so that's nice. We can also ignore the $2$s and $3$s (we may do so by arguing that no divisor of $2^{2}3^{3}5^{5}7^{4}11^{3}$ with $2$ or $3$ as a factor will be congruent to $1$ mod $6$). Fix $a \in \mathbb{N}$. Since $7 = 6k+1$ for $k=1$, we clearly have $7^{a} \equiv 1 \pmod{6}$ for any $a$. For odd $a$, we can show that $$5^{a} \equiv 5 \pmod{6} \hspace{0.4cm} \text{and} \hspace{0.4cm} 11^{a} \equiv 5 \pmod{6}.$$
For even $a$, we may also show that
$$5^{a} \equiv 1 \pmod{6} \hspace{0.4cm} \text{and} \hspace{0.4cm} 11^{a} \equiv 1 \pmod{6}.$$
We should now make use of what we know about modular multiplication and the fact that $5^{2} \equiv 1 \pmod{6}$. This boils down to the question: how many combinations of $5^{i}7^{j}11^{k}$ (for $i,j,k \in \mathbb{N} \cup \lbrace 0 \rbrace$ and $i \leq 5$, $j \leq 4$, $k \leq 3$) can you make such that both $i$ and $k$ are even or odd?
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Integrate Airy function from 0 to $\infty$ From the integral representation of Airy function
$$\mathrm{Ai}(x)=\int_{-\infty}^{\infty} \frac{\mathrm{d} \tau}{2\pi} \exp(-\mathrm{i}\tau x)\exp(-\mathrm{i}\frac{\tau^3}{3}),$$
It is easy to see that $\int_{-\infty}^{\infty} \mathrm{d} x\mathrm{Ai}(x) =1$. However, I am wondering how to find
$$\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x).$$
From this website, the result of the above integral is $\frac{1}{3}.$ I could not follow the method in the reference given by that website. Could anyone give an alternative (more straightforward) derivation of $\int_{0}^{\infty} \mathrm{d}x \mathrm{Ai}(x)=\frac{1}{3}$?
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EDIT:
After posting my answer, I realized that the same approach can be used to find the Mellin transform of $\operatorname{Ai}(x)$. So I decided to modify my answer.
I don't know if this is the approach discussed in the reference, but the Airy function $\operatorname{Ai}(x)$ can be expressed in terms of the modified Bessel function of the second kind of order $\frac{1}{3}$.
Specifically, $$\operatorname{Ai}(x)= \frac{1}{\pi} \sqrt{\frac{x}{3}} K_{1/3} \left(\frac{2}{3} x^{3/2} \right), \quad x>0 .$$
And an integral representation of the modified Bessel function of the second kind is $$K_{\nu}(x) = \frac{1}{2} \left(\frac{x}{2} \right)^{\nu} \int_{0}^{\infty}\exp\left(-t-\frac{x^{2}}{4t} \right) \, \frac{dt}{t^{\nu+1}}, \quad x>0, $$
which can be derived from the integral representation$$K_{\nu}(x) =\int_{0}^{\infty} \exp(-x\cosh t) \cosh(\nu t) \, dt = \frac{1}{2} \int_{-\infty}^{\infty} \exp\left(-x \cosh t\right) e^{-\nu t} \, dt $$ by making the substitution $e^{t}= \frac{2}{x}u$.
Using this representation, and assuming that $a>0$, we get
$$ \begin{align}I(a) &= \int_{0}^{\infty} x^{a-1} \operatorname{Ai}(x) \, dx \\ &= \frac{1}{\pi \sqrt{3}}\int_{0}^{\infty} x^{a-1/2} \, K_{1/3}\left(\frac{2}{3}x^{3/2} \right) \, dx \\ &=\frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3-2/3}} \int_{0}^{\infty} u^{2a/3-2/3} K_{1/3}(u) \, du \\ &= \frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3-2/3}}\int_{0}^{\infty} u^{2a/3-2/3} \, \frac{1}{2} \left(\frac{u}{2} \right)^{1/3} \int_{0}^{\infty} \exp \left(-t - \frac{u^{2}}{4t}\right) \, \frac{dt}{t^{4/3}} \, du \\ &= \frac{1}{\pi} \, \frac{3^{2a/3-7/6}}{ 2^{2a/3+2/3}}\int_{0}^{\infty}\frac{e^{-t}}{t^{4/3}} \int_{0}^{\infty}u^{2a/3-1/3} \exp \left(- \frac{u^{2}}{4t} \right) \, du \, dt \tag{1}\\ &= \frac{3^{2a/3-7/6}}{2 \pi}\int_{0}^{\infty} t^{a/3-1} e^{-t} \int_{0}^{\infty} w^{a/3-2/3} e^{-w} \, dw \, dt \\ &= \frac{3^{2a/3-7/6}}{2 \pi}\Gamma\left(\frac{a+1}{3}\right) \int_{0}^{\infty} t^{a/3-1} e^{-t} \, \, dt \\&= \frac{3^{2a/3-7/6}}{2 \pi} \, \Gamma \left(\frac{a+1}{3} \right) \Gamma \left(\frac{a}{3} \right). \end{align}$$
$(1)$ Since the integrand is nonnegative, Tonelli's theorem allows us to change the order of integration.
Therefore, $$\begin{align} \int_{0}^{\infty} \operatorname{Ai}(x) \, dx&= I(1) = \frac{1}{2\pi \sqrt{3}} \, \Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{1}{3} \right)\\ &= \frac{1}{2 \sqrt{3}} \, \pi \csc \left(\frac{\pi }{3} \right) \tag{2} \\ &=\frac{1}{2 \sqrt{3}} \left(\frac{2}{\sqrt{3}} \right) \\ &= \frac{1}{3}. \end{align}$$
$(2)$ Euler's reflection formula
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Subsets and Splits
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