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Probability of number formed from dice rolls being multiple of 8
A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example if the tosses give {3, 5, 4, 2, 1, 1, 6, 5}, the resultant number is $35421165$. What is the probability that the number formed is a multiple of 8.
I solved this by listing all possibilities for the last 3 digits that give multiples of 8, and found this to be $\frac{1}{8}$.
The solution key agrees with my answer, but also says that "There are quicker ways to solve the problem using a more advanced knowledge of number theory"
What would be a faster way to solve this using number theory?
|
We only care about the last three digits obviously, so the possible sums of just those three die are $100a + 10b + c$, where $a$, $b$, and $c$ are all between $1$ and $6$, inclusive.
We check the divisibility by taking this modulo $8$. If the result is equivalent to $0$, then it is divisible.
Setting up the equivalence condition:
$$
100a+10b+c \equiv 0 \pmod{8} \\
4a + 2b + c \equiv 0 \pmod{8} \\
2(2a+b) + c \equiv 0 \pmod{8} \\
2(2a+b) \equiv -c \pmod{8} \\
2a + b \equiv -\frac{c}{2} \pmod{4}
$$
The range of $2a+b$ is between $3$ and $18$, inclusive. Each of the values can be obtained in three different ways, except $\{3,4,17,18\}$ each can only be obtained one way, and $\{5,6,15,16\}$ each can be done two ways (there are $8$ integers that can be obtained $3$ ways).
Three cases: $-\frac{c}{2}$ must be $3$, $2$, or $1$ modulo $8$. Because the consecutive $8$ integers in the middle can be done $3$ ways, each case has exactly $2$ numbers that satisfy the equivalence, giving us $3\cdot{3}\cdot{2} = 18$ cases. Taking the other integers modulo $4$, only $1$ number from each of the other sets is not congruent to any of $\{3,2,1\}$, giving us $3\times{3} = 9$ more cases, totalling $18+9 = 27$.
Final probability: $\frac{27}{6^3} = \frac{1}{8}$.
|
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|
Finding a tangent to an ellipse parallel to a given line Problem: Find the lines that are tangent to the ellipse $x^2 + 4y^2 = 8$ and parallel to $x +2y = 6$.
I tried to find the derivative of $x^2 + 4y^2 = 8$ and I got: $$\frac{dx}{dy} = -\frac{x}{2y}.$$
Not quite sure if it's right, but I tried to equate it with the slope of $x +2y = 6$, which is $-\frac{1}{2}$. So I get, $$-\frac{1}{2} = -\frac{x}{2y}$$
Now, I'm stuck and I don't know what to do. How do I get the tangent lines? I have no knowledge of any point of that tangent line, only the slope. I appreciate any help. Thanks.
|
Rewrite the equation of the ellipse $x^2+4y^2=8$ as
$$
\begin{align}
x^2+4y^2&=8\\
4y^2&=8-x^2\\
y^2&=\frac{8-x^2}{4}\\
y&=\left(\frac{8-x^2}{4}\right)^{\frac12}.\tag1
\end{align}
$$
The slope of the lines that are tangent to the ellipse is
$$
\begin{align}
\frac{dy}{dx}&=\frac{d}{dx}\left(\frac{8-x^2}{4}\right)^{\frac12}\\
&=\frac12\left(\frac{8-x^2}{4}\right)^{-\frac12}\left(\frac{-2x}{4}\right)\\
&=\frac12\left(\frac{4}{8-x^2}\right)^{\frac12}\left(-\frac{1}{2}x\right)\\
&=-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}
\end{align}
$$
or the quickest way
$$
\frac d{dx}(x^2+4y^2=8)\quad\Rightarrow\quad 2x+8y \frac {dy}{dx}=0\quad\Rightarrow\quad \frac {dy}{dx}=-\frac {x}{4y}.
$$
Since it is parallel with the line $x +2y = 6$, then
$$
\begin{align}
-\frac{1}{2}x\left(\frac{1}{8-x^2}\right)^{\frac12}&=-\frac{1}{2}\\
x\left(\frac{1}{8-x^2}\right)^{\frac12}&=1.\tag2
\end{align}
$$
Square both sides of $(2)$, yield
$$
\begin{align}
\frac{x^2}{8-x^2}&=1\\
x^2&=8-x^2\\
2x^2&=8\\
x_{1,2}&=\pm2\tag3
\end{align}
$$
$y$ can be found by plugging in $(3)$ to $(1)$, yield
$$
\begin{align}
y_{1,2}&=\left(\frac{8-2^2}{4}\right)^{\frac12}\\
&=\pm1,
\end{align}
$$
where $(x_{1,2},y_{1,2})$ are the points of contact. Now, the equation of tangent line can be found by using
$$
\frac{y-y_{1,2}}{x-x_{1,2}}=-\frac12.
$$
|
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|
Generating Functions in Discrete Math a)Find the coefficient of $x^3y^4$ in $(2x + 5y)^7$. b) Find the coefficient of $x^5$ in $(3x -1)(2x +1)^8$.
I know this has to do with generating functions , but i'm not sure how to start with this problem in order to find the coefficient of part a) and b)
|
For (b), you have $(3x - 1)(2x + 1)^{8}$. So consider $(2x+1)^{8}$. You are interested in the coefficients of $x^{4}$ and $x^{5}$ in that term. When you multiply by $(3x-1)$ you can form $x^{5}$ by $3x * kx^{4}$, and $x^{5}$ is held constant by multiplying with $-1$. So $(2x + 1)^{8}$ has coefficient of $x^{5}$ as $\binom{8}{5} (2x)^{5}$ and coefficient of $x^{4}$ as $\binom{8}{4}$.
Now consider $3 * \binom{8}{4} x^{4} * x - \binom{8}{5} (2x)^{5}$.
|
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|
Recurrence relation with generating function problem I've got a recurrence problem that I'm close to solving, but having trouble with finishing up.
Solve the following recurrence relation using generating functions:
$$g_n = g_{n-1} + g_{n-2} + n$$
for $ n>=2, g_0 = 1, g_1 =2 $.
What I've done so far:
$$G(x) = \sum_{n=0}^{\infty}g_nx^n$$
$$g_n < 0 = 0$$
Changed $g_n$ to work for all $n >= 0$:
$$g_0 = 1 = g_{n-1} + g_{n-2} + n + [n=0] $$
$[n=0]$ is conditional on whether n is 0. This makes $g_n$ work for $g_0$ and $g_1$.
I want it in the form of $G(x)$, so I multiply by $x^n$ and sum over $x>=0$:
$$\sum_{n=0}^{\infty}g_nx^n = \sum_{n=0}^{\infty}g_{n-1}x^n +\sum_{n=0}^{\infty}g_{n-2}x^n + \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}[n=0]x^n$$
Factor $x$ out of the first term on the right, and $x^2$ from the second. Third is rewritten, and the final term is just 1.
$$G(x) = xG(x) +x^2G(x)+ \frac{x}{(x-1)^2}+1 $$
Solving for $G(x)$:
$$G(x) = \frac{-x^2+x-1}{(x-1)^2(x^2+x-1)}$$
This is the point at which I'm stuck. From what I've understood from lecture, I want to do a partial fraction decomposition with the parts in the form $1/(1-x)$, so they can be expressed as $\sum_{n=0}^{\infty}px^n$ so that I can extract the coefficients to create the closed formula. Once I have them in that form, I know how to proceed. I've gotten this PFD, but I can't seem to figure out how to get it in the right form:
$$\frac{-2x-4}{x^2+x-1}+\frac{2}{x-1}-\frac{1}{(x-1)^2}$$
So either I've messed up somewhere in the problem, or I'm missing something on how to get this in the right form. Thanks for any suggestions.
|
A simpler way to set this kind of problems up is to write the recurrence with no subtraction in indices:
$$
g_{n + 2} = g_{n + 1} + g_n + n + 2
$$
Multiply by $x^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} a_{n + k} x^k
&= \frac{G(x) - g_0 - g_1 z - \ldots - g_{k - 1} x^{k - 1}}{x^k} \\
\sum_{n \ge 0} x^n
&= \frac{1}{1 - x} \\
\sum_{n \ge 0} n x^n
&= x \frac{\mathrm{d}}{\mathrm{d} x} \frac{1}{1 - x}
\end{align}
and get:
$$
\frac{G(x) - 1 - 2 x}{x^2}
= \frac{G(x) - 1}{x} + G(x) + \frac{x}{(1 - x)^2} + 2 \frac{1}{1 - x}
$$
which gives:
\begin{align}
G(x)
&= \frac{1 - x + x^2}{(1 - x)^2 (1 - x - x^2)} \\
&= \frac{4 + 2 x}{1 - x - x^2} - \frac{2}{1 - x} - \frac{1}{(1 - x)^2}
\end{align}
Mow it comes handy to know that for the Fibonacci numbers $F_n$, defined by $F_0 = 0$, $F_1 = 1$, $F_{n + 2} = F_{n + 1} + F_n$ you have:
\begin{align}
\sum_{n \ge 0} F_n x^n
&= \frac{x}{1 - x - x^2} \\
\sum_{n \ge 0} F_{n + 1} x^n
&= \frac{1}{1 - x - x^2}
\end{align}
and using:
$$
\binom{-2}{n} = (-1)^n \binom{n + 2 - 1}{2 - 1}
= (-1)^n (n + 1)
$$
you get:
\begin{align}
g_n
&= 4 F_{n + 1} + 2 F_n - 2 - (n + 1) \\
&= 2 F_{n + 2} + 2 F_{n + 1} - n - 3 \\
&= 2 F_{n + 3} - n - 3
\end{align}
|
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|
Solving the non-homogeneous recurrence relation: $g_{n} = 12g_{n-2}-16g_{n-3}+6\cdot 2^n+25n$ $g_{n} = 12g_{n-2}-16g_{n-3}+6\cdot 2^n+25n$ With initial conditions $g_{0} = 23, g_{1} = 37, g_{2} = 42 $
This is a practice question I'm working on, and I'm running into absurd amounts of calculations with everything I have tried. I would really appreciate some guidance on this question, as I get the feeling there must be an easier way, or short cut to this question somewhere.
I've tried using both generating functions and the usual method of solving non-homogeneous recurrence relations.
The first method with generating functions, I let
$$A(z) = \sum^{\infty}_{n=0} a_{n}z^n$$ be a generating function. Putting in the initial conditions, I get:
$$ A(z) = 23 + 37z + 42z^2 + \sum^{\infty}_{n=3} (12g_{n-2}-16g_{n-3}+6\cdot 2^n+25n)z^n$$
after a bunch of simplifying and expressing the RHS in terms of $A(z)$, I get:
$$A(z) = 23 + 37z + 42z^2 + 12z^2(A(z)-23) - 16z^3A(Z)+ \frac{6}{1-2z}+25(\frac{z}{(1-z)^2} - z- 2z^2)$$
After rearranging and moving the $A(z)$ terms to the other side, I get:
$$A(z) = \frac{29-67z+23z^2+14z^3-24z^4}{(2z-1)^2(4z+1)(1-2z)(1-z)^2} $$
which turns into an absolutely hideous partial fraction decomposition, trying to solve for 6 constants. I pretty much went as far as I could go with it and still ran into a dead end, so i decided to try the usual method.
Doing the usual method, I try to solve the homogeneous part first, which is reasonably easy. My characteristic polynomial is $x^3-12x+16=0$, giving me roots $\lambda = 2 $(of multiplicity 2) and $\lambda = -4 $
So my solution to the homogeneous part is:
$$b_n = C_12^n+C_2n2^n+C_3(-4)^n$$
Now to get a particular solution, I try: $p_n = C_4n^2\cdot 6 \cdot 2^n + C_5n$
$$ \implies C_4n^2\cdot 6 \cdot 2^n + C_5n = 12(C_4(n-2)^2\cdot 6 \cdot 2^{(n-2)} + C_5(n-2)) - 16(C_4(n-3)^2\cdot 6 \cdot 2^{(n-3)} + C_5(n-3)) + 6\cdot2^n+25n $$
Another pretty nasty algebraic exercise (although not quite as bad as the generating function). However, I persist and after expanding the terms, I try to collect the $n^2\cdot2^n$ and $n$ terms together - to try and solve a simultaneous equation, but I run into the problem that I get terms without $n^2\cdot2^n$ nor $n$ and then some terms with $n\cdot 2^{n}$.
Would greatly appreciate some help with this practice question! Many thanks in advance.
|
Simplest way I know: Define the generating function:
$$
G(z) = \sum_{n \ge 0} g_n z^n
$$
Write the recurrence without subtraction in indices:
$$
g_{n + 3}
= 12 g_{n + 1} - 16 g_n + 48 \cdot 2^n + 25 n + 75
$$
If you multiply by $z^n$, sum over $n \ge 0$, and recognize the resulting sums:
\begin{align}
\sum_{n \ge 0} g_{n + k} z^k
&= \frac{G(z) - g_0 - g_1 z - \ldots - z_{k - 1} z^{k - 1}}{z^k} \\
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \\
&= \frac{z}{(1 - z)^2}
\end{align}
to get:
$$
\frac{G(z) - 23 - 37 z - 42 z^2}{z^3}
= 12 \frac{G(z) - 23}{z} - 16 G(z)
+ 48 \frac{1}{1 - 2 z}
+ 25 \frac{z}{(1 - z)^2}
+ 75 \frac{1}{1 - z}
$$
This gives the formidable:
\begin{align}
G(z)
&= \frac{23 - 55 z - 267 z^2 + 1198 z^3 - 1540 z^4 + 612 z^5}
{1 - 4 z + 7 z^2 - 62 z^3 + 124 z^4 - 104 z^5 + 32 z^6} \\
&= \frac{19}{1 - z}
+ \frac{5}{(1 - z)^2}
- \frac{2}{(1 - 2 z)^2}
+ \frac{2}{(1 - 2 z)^3}
- \frac{1}{1 + 4 z}
\end{align}
Using the generalized binomial theorem,
in particular for natural $m$:
$$
(1 - u)^{-m}
= \sum_{k \ge 0} \binom{m + k - 1}{m - 1} u^k
$$
and the expansion as polynomials of the binomial coefficients:
\begin{align}
g_n
&= 19
+ 5 \binom{n + 2 - 1}{2 - 1}
- 2 \binom{n + 2 - 1}{2 - 1} \cdot 2^n
+ 2 \binom{n + 3 - 1}{3 - 1} \cdot 2^n
- 4^n \\
&= 24 + 5 n + (n^2 + n) \cdot 2^n - 4^n
\end{align}
The algebra help by maxima is gratefully aknowledged.
|
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|
Which is larger $\sqrt[99]{99!}$ or $\sqrt[100]{100!}$ Which is larger
$\sqrt[99]{99!}$ or $\sqrt[100]{100!}$
I know that it is the $\sqrt[100]{100!}$
but is there a formula to figure this out instead of doing it all out by hand?
|
Let $x=\sqrt[99]{99!}$ and $y=\sqrt[100]{100!}$, then
\begin{align}
\frac{x}{y}&=\frac{\sqrt[99]{99!}}{\sqrt[100]{100!}}\\
&=\frac{(99!)^\frac{1}{99}}{(100!)^\frac{1}{100}}\\
&=\frac{(99!)^{\frac{1}{100}+\frac{1}{9900}}}{(100!)^\frac{1}{100}}\\
&=\frac{(99!)^{\frac{1}{100}}(99!)^{\frac{1}{9900}}}{(100!)^\frac{1}{100}}\\
&=(99!)^{\frac{1}{9900}}\left(\frac{99!}{100!}\right)^\frac{1}{100}\\
&=(99!)^{\frac{1}{9900}}\left(\frac{99!}{99!\cdot100}\right)^\frac{1}{100}\\
&=\frac{(99!)^{\frac{1}{9900}}}{100^\frac{1}{100}}\\
\left(\frac{x}{y}\right)^{9900}&=\left(\frac{(99!)^{\frac{1}{9900}}}{100^\frac{1}{100}}\right)^{9900}\\
&=\frac{99!}{100^{99}}\\
&=\frac{99\cdot98\cdot97\cdots1}{100\cdot100\cdot100\cdots100}\\
\frac{x}{y}&=\sqrt[9900]{\frac{99\cdot98\cdot97\cdots1}{100\cdot100\cdot100\cdots100}}
\end{align}
It is clearly $y>x$, hence $\sqrt[100]{100!}>\sqrt[99]{99!}$.
|
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|
Solving $L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ priveded $a+b+c=0$
Let $a,b,c$ be such that $a+b+c=0$ and suppose that
$$L= \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}.$$
Find the value of $L$.
I can only see the symmetry of these function but cannot solve it.
|
Since $a+b+c=0$ it follows $c=-a-b$, then
\begin{align}
2a^2+bc&=2a^2+b(-a-b)\\
&=2a^2-ab-b^2\\
&=(a-b)(2a+b)\\
&=(a-b)(a+a+b)\\
&=(a-b)(a-c)\quad\text{because $a+b=-c$}
\end{align}
In a similar way we can obtain $2b^2+ac=(b-a)(b-c)$ and $2c^2+ab=(c-a)(c-b)$. Then
\begin{align}
L&=\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} \\
&=\frac{a^2}{(a-b)(a-c)}-\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(a-c)(b-c)} \\
&=\frac{a^2(b-c)-b^2(a-c)+c^2(a-b)}{(a-b)(a-c)(b-c)} \\
&=\frac{-a(b^2-c^2)+b(a^2-c^2)-c(a^2-b^2)}{(a-b)(a-c)(b-c)} \\
&=\frac{-a(b+c)(b-c)+b(a+c)(a-c)-c(a+b)(a-b)}{(a-b)(a-c)(b-c)} \\
&=\frac{-a(b+c)(b-c)+b(a+c)(a-c)-c(a+b)[(a-c)-(b-c)]}{(a-b)(a-c)(b-c)} \\
&=\frac{[-a(b+c)+c(a+b)](b-c)+[b(a+c)-c(a+b)](a-c)}{(a-b)(a-c)(b-c)} \\
&=\frac{-b(a-c)(b-c)+a(b-c)(a-c)}{(a-b)(a-c)(b-c)} \\
&=\frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)} \\
&=1
\end{align}
|
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|
Find The Minimum Value of the quantity Find the minimum value of the quantity $$\frac{(a^2+3a+1)(b^2+3b+1)(c^2+3c+1)}{abc}$$,where $$a,b,c>0$$ and $$ a,b,c\in R $$are positive real numbers.
|
$\dfrac{a^2 + 3a + 1}{a} = a + \dfrac{1}{a} + 3 \geq 2 + 3 = 5$ by AM-GM. So $LHS \geq 5^3 = 125$ with equality at $ a = b = c = 1$
|
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|
How to find $\int \frac {dx}{(x-1)^2\sqrt{x^2+6x}}$? find the integral of $f(x)=\frac1{(x-1)^2\sqrt{x^2+6x}}$
my attempt =
$(x-1)=a$, $a=x+1$ so the integral'd be
$\int \frac {dx}{(x-1)^2\sqrt{x^2+6x}}=\int\frac{da}{a^2\sqrt{a^2+8a+7}} $
lets say $\sqrt{a^2+8a+7}=(a+1)t$
so $a=\frac{7-t}{t-1}$ and $da=\frac{-6dt}{(t-1)^2}$
$\int\frac{da}{a^2\sqrt{a^2+8a+7}}=\int\frac{\frac{-6dt}{(t-1)^2}}{(\frac{7-t}{t-1})^2\frac{6t}{t-1}}=-\int\frac{(t-1)dt}{(7-t)^2t}$
$\frac{(t-1)}{(7-t)^2t}=\frac{A}{t}+\frac{B}{7-t}+\frac{C}{(7-t)^2}$
then $A=B=\frac{-1}{7}$ and $C=6$
$-\int\frac{(t-1)dt}{(7-t)^2t}=\frac{1}{7}\int\frac{dt}{t}+\frac{1}{7}\int\frac{dt}{7-t}-6\int\frac{dt}{(7-t)^2}=\frac{ln|t|}{7}-\frac{ln|7-t|}{7}+6\frac{1}{7-t}$
finally we substitute $t=\frac{7+a}{a+1}$ and a=x+1
is my solution attempt correct? if it is, is there another simpler way to solve?
edit: a should equal to x-1 and x =a+1
|
Let $\displaystyle a=\frac{7-t^2}{t^2-1}$ instead of $\displaystyle a=\frac{7-t}{t-1} $ ??
Doing that and simplifying we get, $\displaystyle -\frac{12 t}{\left(t^2-1\right)^2}$ and integrating, we get
$$-2 \left(-\frac{3 t}{7 \left(t^2-7\right)}+\frac{2 \log \left(\sqrt{7}-t\right)}{7 \sqrt{7}}-\frac{2 \log \left(t+\sqrt{7}\right)}{7 \sqrt{7}}\right)$$
Putting back $x$ and simplifying we get
$$\frac{-7 \sqrt{x (x+6)}-4 \sqrt{7} (x-1) \log \left(\sqrt{7}-\frac{\sqrt{x (x+6)}}{x}\right)+4 \sqrt{7} (x-1) \log \left(\frac{\sqrt{7} x+\sqrt{x (x+6)}}{x}\right)}{49 (x-1)}$$
Differentiating it and simplifying we get, your original function.
|
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|
Solve $\sin x+\cos x+\tan x+\csc x+\sec x+\cot x=-2$ in the interval $0
Solve $$\sin x+\cos x+\tan x+\csc x+\sec x+\cot x=-2$$ from $0<x<2\pi$.
Could you find the most elegant solution? Does it factorize?
|
This might be considered an extended comment of lab bhattacharjee's answer above.
The roots of the factored form are
$(\cos x + \sin x)(\cos x + 1)(\sin x +1) = 0$
which gives
$\cos x = -\sin x$. The 2 solutions are $\dfrac{3\pi}{4}$ and $\dfrac{7\pi}{4}$;
$\cos x = -1$. The solution is $\pi$;
$\sin x = -1$. The solution is $\dfrac{3\pi}{2}$.
So on the interval $0 < x < 2\pi$, you have the solutions $\dfrac{3\pi}{4}, \pi, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$.
EDIT: Thanks to lab bhattacharjee for noticing which values will be undefined if evaluated in the original equation. The the only solutions that work are $\dfrac{3\pi}{4}$ and $\dfrac{7\pi}{4}$.
|
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|
If n+1 is a cube show that 504 divides n(n+1)(n+2) If $n+1$ is a cube show that 504 divides $n(n+1)(n+2)..$
I try that : $n+1=a^3$ , then $n=a^3-1$ so $n(n+1)(n+2) = (a^3-1)a^3(a^3+1) = (a^6-1)a^3$ and I try prove $(a^6-1)a^3$ divides by 504 with induction but it is not easy (or short). I just find anybody who solve this problem nice and short. And it is so important for me.. Please help me ...
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You need to show that $(a^3-1)a^3(a^3+1)$ is divisible by $9$, $8$, and $7$ (since $504=9\cdot8\cdot7$). Your idea of rewriting the product at $(a^6-1)a^3$ is a good one, since that takes care of $7$, using Fermat's little theorem.
For the other two, you can factor the product at $k^3(k-1)(k+1)(k^2+k+1)(k^2-k+1)$. For divisibility by $8$, you either have $k$ is even, which gives $8\mid k^3$, or $k-1$ and $k+1$ are both even, in which case one of them is divisible by $4$ and the other by $2$. Likewise, you either have $27\mid k^3$, or you have one of $k-1$ and $k+1$ divisible by $3$ and one of $k^2+k+1$ and $k^2-k+1$ divisible by $3$.
|
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|
finding a basis for kernal and image when dealing with a matrix vectorspace
I've calculated the dimension of $M_2(\mathbb{F})$ as $4$, and calculated $T(X) = \begin{pmatrix} -(x_2+x_3) & x_1 - x_4 \\ x_1 - x_4 & x_3 + x_2 \end{pmatrix}$ and noted that $T(X) = 0$ iff $x_2 = -x_3$ and $x_1 = x_4$ but I'm not sure how I can find a basis for it, or for the image - any help please.
|
Consider arbitrary element
$$
X =\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}\in \ker T
$$
As you have calculated, $x_2 = -x_3$, $x_1=x_4$, thus
$$
X =\begin{pmatrix} x_1 & -x_3 \\ x_3 & x_1 \end{pmatrix}=
\begin{pmatrix} x_1 & 0 \\ 0 & x_1 \end{pmatrix}+
\begin{pmatrix} 0 & -x_3 \\ x_3 & 0 \end{pmatrix}=
$$
$$
=x_1\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+
x_3\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
$$
hence $\ker T$ is spanned by these two matrices. They are linearly independent, hence they form the basis of $\ker T$.
As for the image, we proceed in similar manner.
$$
T(X)=\begin{pmatrix} -x_2-x_3 & x_1 - x_4 \\ x_1 - x_4 & x_3 + x_2 \end{pmatrix}=
$$
$$
=x_1\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+
x_2\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}+
x_3\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}+
x_4\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
$$
Now, by definition $\mathrm{Im} T$ is spanned by these 4 matrices, but they are not linearly independent. Find maximal linearly independent subset and this will be our basis.
|
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|
If $\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$ then $x^2 - 44x - 36 = 0$ holds for $x=\sin 2A$ If
$$\sin A + \cos A + \tan A + \cot A + \sec A + \csc A = 7$$
then prove that
$$\sin 2A \quad\text{ is a root of }\quad x^2 - 44x - 36 = 0$$
I have no idea how to solve it. Plz help.
|
Let $$\sin{A}+\cos{A}=t$$
$$7=\sin A+\cos A+\frac{1}{\sin A\cos A}+\frac{\sin A+\cos A} {\sin A\cos A}$$
$$\sin 2A =2 \sin A \cos A$$
$$t^2=1+\sin 2A$$
|
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|
Evaluate limits by interpreting sums as integral sums Problem: Evaluate the following limits by interpreting given sums as integral sums for certain functions and by using the Fundamental Theorem of Calculus. (a) Find $\lim{S_{n}}$ as n goes to infinity where $$S_{n} = \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} $$
I do not know what this means. My "gut" guess is that I shall evaluate it is a Riemann-sum and the function to be integrated is $$\int_{0}^{x}\frac{1}{1+x} $$
|
We can rewrite our sum $S_n$ as
$$S_n=\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+\frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{n}{n}}\right).$$
Note that $\dfrac{1}{1+\frac{k}{n}}$ is the value of the function $f(x)=\frac{1}{1+x}$ at the point $x=\frac{k}{n}$.
So we have evaluated the function $f(x)$ at equally-spaced points $\frac{1}{n}$, $\frac{2}{n}$, and so on up to $\frac{n}{n}$, multiplied the value by the length $\frac{1}{n}$ of the interval from $\frac{k-1}{n}$ to $\frac{k}{n}$, and added up, $k=1$ to $n$.
So our sum $S_n$ is, for large $n$, a good approximation to the area under $y=\frac{1}{1+x}$, from $x=0$ to $x=1$. As $n\to\infty$, the sum approaches the area, which is $\int_0^1 \frac{1}{1+x}\,dx$. This area is $\ln 2$.
Remark: The Fundamental Theorem of Calculus comes into consideration because the required area can be found by finding an antiderivative $F(x)=\ln(1+x)$ of $\frac{1}{1+x}$, and calculating $F(1)-F(0)$.
We could write the answer in a different way. For example, let $g(x)=\frac{1}{x}$. We can think of our sum $S_n$ as obtained by evaluating $g(x)$ at $x=1+\frac{1}{n}$, $1+\frac{2}{n}$, and so on up to $1+\frac{n}{n}$, multiplying the result by $\frac{1}{n}$, and adding up. From that point of view, our sum is a Riemann sum for $\int_1^2\frac{1}{x}\,dx$. So $\lim_{n\to\infty}S_n=\int_1^2 \frac{1}{x}\,dx$.
|
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|
Combinatorial Problem From Book Find all the number of ways such that three elements selected from the set $\{ 1,2,3,....,4n\}$ sum of whose is divisible by $4$.
I cannot find the solution of this equation $x + y + z = 4k$. Solving this equation we can get our answer.
|
We can do the job by considering various cases. There are various ways in which the sum can be divisible by $4$.
Case (i): The $3$ chosen numbers are all congruent modulo $4$ (have the same remainder on division by $4$). Then they must all be congruent to $0$ (have remainder $0$). Since $n$ of our numbers are divisible by $4$, there are $\binom{n}{3}$ possibilities. Note that if $n\lt 3$ then $\binom{n}{3}=0$.
Case (ii): We choose $2$ numbers congruent to $1$ modulo $4$, and $1$ congruent to $2$. That gives $\binom{n}{2}\binom{n}{1}$ ways.
Case (iii): We choose $2$ numbers congruent to $3$, and $1$ congruent to $2$. We get the same number of ways as in (ii).
Case (iv): We choose $2$ numbers congruent to $2$, and $1$ congruent to $0$. Again, we get the same number as in (ii).
Case (v): all the chosen numbers are incongruent modulo $4$. Then $1$ number is congruent to $1$, another to $3$, and another to $4$. That gives $\binom{n}{1}\binom{n}{1}\binom{n}{1}$ possibilities.
Finally, add up.
|
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|
Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.
My attempt at the program was as follows:
A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$.
LONG and TEDIOUS way:
$$(8k)^2 = 0 (mod 8)$$
$$(8k+1)^2 = 1 (mod 8)$$
$$(8k+2)^2 = 4 (mod 8)$$
$$(8k+3)^2 = 1 (mod 8)$$
$$(8k+4)^2 = 0 (mod 8)$$
$$(8k+5)^2 = 1 (mod 8)$$
$$(8k+6)^2 = 4 (mod 8)$$
$$(8k+7)^2 = 1 (mod 8)$$
That is, using three of these modulo there is no way to arrive at
$$a^2 + b^2 + c^2 = 7 (mod 8)$$
|
You get a little bonus: $x^2 + y^2 + z^2$ can be even with two of the variables odd. However, $x^2 + y^2 + z^2$ cannot be divisible by $4$ unless all three of $x,y,z$ are even. So, Assume $x^2 + y^2 + z^2 \equiv 28 \pmod {32}.$ It follows that $x,y,z$ are even, and we get $(x/2)^2 + (y/2)^2 + (z/2)^2 \equiv 7 \pmod 8.$ This is a contradiction, so the sum of three squares cannot be $28 \pmod {32}.$ Do it again, the sum cannot be $112 \pmod {128}.$ And induction...the traditional way to write this is $$ x^2 + y^2 + z^2 \neq 4^k \cdot (8n+7). $$ On the other hand, every other positive integer $m$ can be written as $m=x^2 + y^2 + z^2$ with integer variables. Gauss.
|
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|
Probability of rolling 3 sixes in 6 rolls I roll a die six times. What is the probability of rolling at least 3 sixes? Please share the formula as well, so I can figure it out myself in future.
|
Let $p$ = probability of getting a $6$ for each toss, then $p = \dfrac{1}{6}$, and $q = 1 - p = \dfrac{5}{6}$ is the probability of getting a number other than $6$. Let $k$ = the number of successful trials. This is a binomial experiment consisting of $6$ independent trials. So $n = 6$. Let $x$ be the number of $6$'s in $6$ tosses of the fair die, then we find $P(x \geq 3) = 1 - P(0) - P(1) - P(2)$.
$P(x = k) = \binom n k\cdot p^k\cdot q^{n-k}$. So:
$P(0) = P(x=0) = \binom 6 0\cdot \left(\dfrac{1}{6}\right)^0\cdot \left(\dfrac{5}{6}\right)^6 = 0.335$
$P(1) = P(x=1) = \binom 6 1\cdot \left(\dfrac{1}{6}\right)^1\cdot \left(\dfrac{5}{6}\right)^5 = 0.402$
$P(2) = P(x=2) = \binom 6 2\cdot \left(\dfrac{1}{6}\right)^2\cdot \left(\dfrac{5}{6}\right)^4 = 0.201$.
So: $P(x \geq 3) = 1 - 0.335 - 0.402 - 0.201 = 0.062$
|
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|
A geometric inequality, proving $8r+2R\le AM_1+BM_2+CM_3\le 6R$
Here, $AM_1$ is the angle bisector of $\angle A$ extended to the circumcircle and so on. $R$ is the circumradius and $r$ is the inradius, respectively. I have to prove that:
$$8r+2R\le AM_1+BM_2+CM_3\le 6R$$
The second part is easy, since each of $AM_1$ is a chord, $AM_1\le 2R$, so $\sum AM_1 \le 6R$. But the first part is giving me nightmares. Applying Euler's inequality gives $8r+2R\le 6R$, which is not much helpful and I'm out of ideas.
Please help. Besides, playing GeoGebra tells that its true.
|
The quadrilaretal $ABM_1C$ is cyclic. Thus the Ptolemy Theorem says:
$AB.CM_1+AC.BM_1=BC.AM_1$, that is $AM_1=\dfrac{b+c}{a}.BM_1$
(Since, angle subtended by equal chords at the circumcircle are equal and vice-versa, $BM_1=CM_1$, as $\angle BAM_1=\angle CAM_1=\dfrac{A}{2}$).
Also, $BM_1=2R\sin\frac{A}{2}$.
Similarly, getting expressions for $BM_2$ and $CM_2$, we have:
$\displaystyle AM_1+BM_2+CM_3=\sum\limits_{cyc} \frac{b+c}{a}.\left(2R\sin\frac{A}{2}\right)=\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\left(2R\sin\frac{A}{2}\right)$
$\displaystyle =2R\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\sin\frac{A}{2}=2R\sum\limits_{cyc} \frac{2\sin \frac{B+C}{2}.\cos\frac{B-C}{2}}{2\sin \frac{A}{2}.\cos \frac{A}{2}}.\sin \frac{A}{2}$
$\displaystyle = 2R\sum\limits_{cyc}\cos\frac{B-C}{2}$.
We also have, $r=4R\prod\limits_{cyc}\sin\dfrac{A}{2}=R(-1+\sum\limits_{cyc}\cos A)$,
Therefore, $8r+2R = 2R(-3+4\sum\limits_{cyc}\cos A)$
Thus, the inequality in the question is, $\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A = 1 + \frac{4r}{R}$
Using the substitution, $a=x+y$, $b=y+z$ and $c=x+z:$
$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\sin A + \sin B}{2\cos \frac{C}{2}} \ge \sum\limits_{cyc} \frac{2(b^2+c^2-a^2)}{bc} - 3$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\frac{2\Delta}{bc} + \frac{2\Delta}{ac}}{2\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}} \ge \sum\limits_{cyc} \frac{2ab^2+2ac^2-2a^3-abc}{abc}$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{2\Delta(a+b)}{2\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}} \ge \sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{4\Delta(a+b)\sqrt{ab}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{(a+b)\sqrt{ab}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} (2ab^2+2ac^2-2a^3-abc)$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} (a+b)\sqrt{ab(a+c-b)(b+c-a)} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$
$\displaystyle \Leftrightarrow 2\sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge 2\sum\limits_{cyc} (xz(x+z) + 6xyz)$
$$\displaystyle \Leftrightarrow \sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$$
On the other hand we have the inequality $\sqrt{(x+y)(z+y)} \ge (y+\sqrt{xz})$ (By Squaring and applying AM-GM)
Thus it suffices to show $\displaystyle \sum\limits_{cyc} (x+z+2y)(y\sqrt{xz}+xz) \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$
$\displaystyle \Leftrightarrow (\sum\limits_{cyc} xz(x+z)) + 6xyz + (\sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz})\ge \sum\limits_{cyc} (x^2z + xz^2) + 18xyz$
$\displaystyle \Leftrightarrow \sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz}\ge 12xyz$
Which is AM-GM Inequality with $12$-terms.
|
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|
Last Two digits of ${14}^{{14}^{14}}$ How to calculate the last two digits of ${14}^{{14}^{14}}$? With the help of any method. I have tried and have got the last digit to be $6$. But not sure.
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Clearly, $14^{14^{14}}$ is a multiple of $4$. To compute $14^n\pmod{25}$ we should know $n\pmod {\phi(25)}$, i.e. $14^{14}\pmod{20}$. Again, $14^{14}$ is a multiple of $4$, and it is $\equiv (-1)^{14}\equiv 1\pmod 5$. Hence $14^{14}\equiv 16\pmod {20}$. Thus $14^{14^{14}}\equiv 14^{16}\pmod {25}$. This can me computed by repeated squareing:
$$ 14^{16}=(14^2)^8=196^8\equiv (-4)^8=16^4=256^2\equiv 6^2=36\pmod{25}.$$
Since $36$ is already a multiple of $4$, we have immediately that $14^{14^{14}}\equiv 36\pmod{100}$.
|
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Calculus Question: $\int_0^1\frac{(x-x^2)^4}{1+x^2}dx$ I have no idea how to find this integral:
$$\int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx\ ?$$
Any help would be appreciated. Thanks in advance.
|
HINT :
$$
\begin{align}
\int_0^1\frac{(x-x^2)^4}{1+x^2}\ dx&=\int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\ dx\tag1\\
&=\int_0^1\left(x^6-4x^5+5x^4-4x^2+4-\frac4{1+x^2}\right)\ dx.\tag2\\
\end{align}
$$
See binomial theorem for $(1)$ and polynomial long division for $(2)$. Also
$$
\begin{align}
\int_0^1\frac1{1+x^2}\ dx=\frac\pi4
\end{align}
$$
by letting $x=\tan\theta$. The answer is $\ \color{blue}{\dfrac{22}7-\pi}$.
|
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|
Intersection multiplicity
Let $f=y^2-x^3$ and $g=y^3-x^7$. Calculate the intersection multiplicity of $f$ and $g$ at $(0,0)$.
I know the general technique for this (passing to the local ring) but I having difficulty with the fact that $3,7,2$ have no common factors.
|
I think you no longer need an answer, but I’d like to write something here since it may help some other students.
We need four lemmas here(You can find them in lecture 2 of Andreas Gathmann’s notes in Plane Algebraic Curves):
Lemma (1): For any three plane curves F, G, H, $$\mu_p(F, G)=\mu_p(F, G+FH)$$
Lemma (2): For any three plane curves F, G, H, $$\mu_p(F, GH)=\mu_p(F, G)+\mu_p(F, H)$$
Lemma (3): $\mu_p(F, G)\ge 1$ iff $P\in F\cap G$.
Lemma (4): $\mu_p(F, G)=1$ iff $<F, G>=I_p$, where $I_p =\{f/g: f, g\in K[x, y], f(P)=0, g(P)\neq 0\}$.
Now we can compute $\mu_0(y^2-x^3, y^3-x^7)$.
$\begin{align}\mu_0(y^2-x^3, y^3-x^7)\overset{by (1)}{=}\mu_0(y^2-x^3, y^3-x^7-y(y^2-x^3))=\mu_0(y^2-x^3, x^3y-x^7)\overset{by (2)}{=}\mu_0(y^2-x^3, x^3)+\mu_0(y^2-x^3, y-x^4)\overset{by (1)}{=}\mu_0(y^2-x^3+x^3, x^3)+\mu_0(y^2-x^3-y(y-x^4), y-x^4))=\mu_0(y^2, x^3)+\mu_0(x^4y-x^3, y-x^4)\overset{by (2), (2), respectively}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}6\mu_0(y, x)+\mu_0(x^3, y-x^4)+\mu_0(xy-1, y-x^4)\overset{by (4), (1), (3), respectively}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}6+\mu_0(x^3, y-x^4+x(x^3))+0=6+\mu_0(x^3, y)\overset{by (2)}{=}6+3\mu_0(x, y)\overset{by (4)}{=}6+3=9\end{align}.$
|
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|
Find all real $x$ ,such $8x^3-20$ and $2x^5-2$ is perfect square
Find all real numbers $x$,such
$$8x^3-20,2x^5-2$$ is the perfect square of an integer
My idea: First we find the real number $x$ such $$8x^3-20,2x^5-2$$ is postive integer numbers,and second the real numbers such
$$8x^3-20=m^2,2x^5-2=n^5$$
How prove this all real numbers?
|
Hint. If we have
$$8x^3-20=m^2\ ,\quad 2x^5-2=n^2$$
then
$$x^3=\frac{m^2+20}{8}\ ,\quad x^5=\frac{n^2+2}{2}\ .$$
Can you see how to obtain $x$ from these two equations?
|
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Indefinite Integral $\int \frac{dx}{\sqrt {ax^4-bx^2}}$ I am trying to Integrate
$$
I=\int \frac{dx}{\sqrt {ax^4-bx^2}}, \qquad a,b\in \mathbb{R}.
$$
Thanks.
I tried to do $x=\sin \phi$
$$
\int \frac{\cos \phi\, d\phi}{\sqrt{a\sin^4 \phi-b\sin^2 \phi}}=\int \frac{\cot \phi \, b\phi}{\sqrt{a\sin^2\phi-b}}
$$
but get stuck here.
Mathematica gives a closed form result
$$
I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg).
$$
|
$$
\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt{ax^4 - bx^2}}&=\int\frac{\mathrm{d}x}{\sqrt{bx^4\left(\frac{a}{b} - \frac{1}{x^2}\right)}}\\
&=\frac{1}{\sqrt{b}}\int\frac{1}{\sqrt{\frac{a}{b} - \left(\frac{1}{x}\right)^2}}\frac{1}{x^2}\,\mathrm{d}x
\end{aligned}
$$
Now, set $u=\dfrac{1}{x}$ and $\mathrm{d}u=-\dfrac{1}{x^2}\,\mathrm{d}x$:
$$
I=\frac{1}{\sqrt{b}} \int-\frac{1}{\sqrt{\left(\sqrt{\frac{a}{b}}\right)^{\!2} - u^2}}\,\mathrm{d}u
$$
We have the integral
$$
\int-\frac{\mathrm{d}u}{\sqrt{\alpha^2 - u^2}}=\arccos\left(\frac{u}{\alpha}\right)+C
$$
Then,
$$
\begin{aligned}
I&=\frac{1}{\sqrt{b}}\arccos\left(u\sqrt{\frac{b}{a}}\right)+C\\
&=\frac{1}{\sqrt{b}}\arccos\left(\frac{1}{x}\sqrt{\frac{b}{a}}\right)+C
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/787377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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|
block matrix multiplication If $A,B$ are $2 \times 2$ matrices of real or complex numbers, then
$$AB = \left[
\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\cdot
\left[
\begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right]
=
\left[
\begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \end{array} \right]
$$
What if the entries $a_{ij}, b_{ij}$ are themselves $2 \times 2$ matrices? Does matrix multiplication hold in some sort of "block" form ?
$$AB = \left[
\begin{array}{c|c} A_{11} & A_{12} \\\hline A_{21} & A_{22} \end{array} \right]\cdot
\left[
\begin{array}{c|c} B_{11} & B_{12} \\\hline B_{21} & B_{22} \end{array} \right]
=
\left[
\begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{array} \right]
$$
This identity would be very useful in my research.
|
It depends on how you partition it, not all partitions work.
For example, if you partition these two matrices
$$\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix},
\begin{bmatrix}
a' & b' & c' \\
d' & e' & f' \\
g' & h' & i'
\end{bmatrix} $$
in this way
$$ \left[\begin{array}{c|cc}a&b&c\\ d&e&f\\ \hline g&h&i \end{array}\right],
\left[\begin{array}{c|cc}a'&b'&c'\\ d'&e'&f'\\ \hline g'&h'&i' \end{array}\right] $$
and then multiply them, it won't work. But this would
$$\left[\begin{array}{c|cc}a&b&c\\ \hline d&e&f\\ g&h&i \end{array}\right] ,\left[\begin{array}{c|cc}a'&b'&c'\\ \hline d'&e'&f'\\ g'&h'&i' \end{array}\right] $$
What's the difference? Well, in the first case, all submatrix products are not defined, like $\begin{bmatrix}
a \\ d \\ \end{bmatrix}$ cannot be multiplied with $\begin{bmatrix}
a' \\ d' \\ \end{bmatrix}$
So, what is the general rule? (Taken entirely from the Wiki page on Block matrix)
Given, an $(m \times p)$ matrix $\mathbf{A}$ with $q$ row partitions and $s$ column partitions
$$\begin{bmatrix}
\mathbf{A}_{11} & \mathbf{A}_{12} & \cdots &\mathbf{A}_{1s}\\
\mathbf{A}_{21} & \mathbf{A}_{22} & \cdots &\mathbf{A}_{2s}\\
\vdots & \vdots & \ddots &\vdots \\
\mathbf{A}_{q1} & \mathbf{A}_{q2} & \cdots &\mathbf{A}_{qs}\end{bmatrix}$$
and a $(p \times n)$ matrix $\mathbf{B}$ with $s$ row partitions and $r$ column parttions
$$\begin{bmatrix}
\mathbf{B}_{11} & \mathbf{B}_{12} & \cdots &\mathbf{B}_{1r}\\
\mathbf{B}_{21} & \mathbf{B}_{22} & \cdots &\mathbf{B}_{2r}\\
\vdots & \vdots & \ddots &\vdots \\
\mathbf{B}_{s1} & \mathbf{B}_{s2} & \cdots &\mathbf{B}_{sr}\end{bmatrix}$$
that are compatible with the partitions of $\mathbf{A}$, the matrix product
$
\mathbf{C}=\mathbf{A}\mathbf{B}
$
can be formed blockwise, yielding $\mathbf{C}$ as an $(m\times n)$ matrix with $q$ row partitions and $r$ column partitions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/787909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 3,
"answer_id": 1
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|
Evaluate $a+b+c+d$ If $a$, $b$, $c$, and $d$ are distinct integers such that
$$(x-a)(x-b)(x-c)(x-d)=4$$
has an integral root $r$, what is the value of $a+b+c+d$ in terms of $r$?
I tried to analyze graphically by shifting the graph of
$f(x)=(x-a)(x-b)(x-c)(x-d)$ four units downward but couldn't infer anything due to the random nature of $a$, $b$, $c$ and $d$.
|
The product of the four distinct integers $r-a$, $r-b$, $r-c$, and $r-d$ is $4$. Thus $r-a,r-b,r-c,r-d$ are, in some order, $-1$, $1$, $-2$, and $2$.
We may assume it is in the order we gave. So $r-a=-1$, $r-b=1$, $r-c=-2$, $r+d=2$. Solve for $a,b,c,d$ and add.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/787979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
Non-unitary representation How to prove $\pi :\mathbb R\to \mathbb C^2$, defined by $t\mapsto \begin{pmatrix} 1 & t\\ 0 & 1\end{pmatrix}$ is a non-unitary representation?
Is the following correct? $\pi$ is a representation because
$$\pi (t+s)=\begin{pmatrix}1 & t+s \\ 0 & 1\end{pmatrix}=\begin{pmatrix}1 & t\\ 0 &1\end{pmatrix}\begin{pmatrix}1 & s \\ 0 & 1\end{pmatrix} = \pi(t)\pi(s)$$
For non-unitary, I need to prove $U:\mathbb C^2\to \mathbb C^2$ is not unitary.
|
For $\pi$ to be a unitary representation, $\pi(x)$ must be a unitary matrix for every element $x$ of the group. Here,
$$\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}$$
is not a unitary matrix unless $t=0$. Indeed,
$$\begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}^* \begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix}
=\begin{pmatrix}1 & 0 \\ \bar t & 1\end{pmatrix} \begin{pmatrix}1 & t \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & t \\ \bar t & 1+|t|^2\end{pmatrix} $$ which is not the identity matrix.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/788743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int_0^\infty \frac{(\ln x)^2}{x^2+4} \ dx$ using complex analysis. Evaluate $\int_0^\infty \frac{(\ln x)^2}{x^2+4} \ dx$. This is the last question in our review for complex analysis. Hints were available upon request, but being the student I am, I waited until the last minute to do this. Can anyone give a detailed answer? It would greatly appreciated. This is part of our residue theorem review, which is how we were taught to evaluate such integrals.
|
Consider the contour integral
$$\oint_C dz \frac{\log^3{z}}{z^2+4} $$
where $C$ is a keyhole contour about the positive real axis, with outer radius $R$ and inner radius $\epsilon$. The contour integral is then equal to
$$\int_{\epsilon}^R dx \frac{\log^3{x}}{x^2+4} +i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^3{(R e^{i \theta})}}{R^2 e^{i 2 \theta}+4} \\ + \int_R^{\epsilon} dx \, \frac{(\log{x}+i 2 \pi)^3}{x^2+4} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^3{(\epsilon e^{i \phi})}}{\epsilon^2 e^{i 2 \phi}+4}$$
As $R \to \infty$ the second integral vanishes as $2 \pi R \log^3{R}/(R^2-4)$; as $\epsilon \to 0$, the fourth integral vanishes as $(\pi/2) \epsilon \log^3{\epsilon}$. In this limit, then, the contour integral is
$$-i 6 \pi\int_0^{\infty} dx \frac{\log^2{x}}{x^2+4} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{x^2+4} + i 8 \pi^3 \int_0^{\infty} \frac{dx}{x^2+4}$$
The third integral is familiar, and is equal to $i 8 \pi^3 (\pi/4) = i 2 \pi^4$. The second integral is evaluated in a similar manner:
$$\begin{align}\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{x^2+4} &= -i 4 \pi\int_0^{\infty} dx \frac{\log{x}}{x^2+4} + 4 \pi^2 \int_0^{\infty} \frac{dx}{x^2+4} \\ &=-i 4 \pi\int_0^{\infty} dx \frac{\log{x}}{x^2+4} +\pi^3 \\ &= \underbrace{i 2 \pi \frac14 \left ( -i (\log{2}+i \frac{\pi}{2})^2+ i (\log{2}+i \frac{3\pi}{2})^2\right )}_{\text{residue theorem}}\\ &= i \pi (-\pi \log{2} - i \pi^2)\\ &= \pi^3 - i \pi^2 \log{2}\end{align}$$
so that
$$\int_0^{\infty} dx \frac{\log{x}}{x^2+4} = \frac{\pi}{4} \log{2}$$
Therefore, by the residue theorem,
$$\begin{align}-i 6 \pi\int_0^{\infty} dx \frac{\log^2{x}}{x^2+4} + 3 \pi^3 \log{2} + i 2 \pi^4 &= i 2 \pi \frac14 \left ( -i (\log{2}+i \frac{\pi}{2})^3+ i (\log{2}+i \frac{3\pi}{2})^3\right ) \\ &= 3 \pi^3 \log{2} + i \left (\frac{13 \pi^4}{8}-\frac{3 \pi^2}{2} \log^2{2} \right ) \end{align} $$
Therefore
$$\int_0^{\infty} dx \frac{\log^2{x}}{x^2+4} = \frac{\pi^3}{16} + \frac{\pi}{4} \log^2{2} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/792497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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|
Area calculation of ellipse $x^2/2+y^2=1$ Calculate the area of the ellipse that you get when you rotate the ellipse $$\frac{x^2}{2}+y^2= 1$$ around the x-axis.
My approach has been to use the formula for rotation area from $-2$ to $2$. But this gives a complicated integral and I'm unsure about the limits $-2$ to $2$.
I would really appriciate a detailed solution to this
|
Surface
$\frac{x^2}{2}+y^2=1$
$S=2\pi\int_{-a}^{a}yds=4\pi\int_{0}^{a}yds$
$y'= -\frac{x}{2y}$
$ds=\sqrt{1+y'^2}dx=\sqrt{1+\frac{x^2}{4y^2}}dx=\frac{\sqrt{4y^2+x^2}}{2y}dx$
$S=4\pi\int_{0}^{\sqrt{2}}y\frac{\sqrt{4y^2+x^2}}{2y}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4y^2+x^2}dx=2\pi\int_{0}^{\sqrt{2}}\sqrt{4-x^2}dx$
Substitution: $x=2sint, dx=2cost\cdot dt$
$S=2\pi\int_{0}^{\pi/4}\sqrt{4-4sin^2t}\cdot 2costdt=8\pi\int_{0}^{\pi/4}cos^2tdt=$
$=4\pi\int_{0}^{\pi/4}(1+cos2t)dt=4\pi[t+\frac{sin2t}{2}]_{0}^{\pi/4}dt=\pi(\pi+2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/793036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
no. and nature of roots of $x^{\frac{3}{4}(\log_{2}{x})^2 + \log_{2}{x} - \frac{5}{4}} = \sqrt{2}$ The given equation is
$$x^{\frac{3}{4}(\log_{2}{x})^2 + \log_{2}{x} - \frac{5}{4}} = \sqrt{2}$$
I took $\log_{2}{x}$ = $t$
and then rewrote the given equation as
$$x^{3t^2 + 4t - 5} = \sqrt{2}$$
But I don't know what to do after this.
How will I find the nature and no. of roots?
|
$$x^{\frac{3}{4}(\log_{2}{x})^2 + \log_{2}{x} - \frac{5}{4}} = \sqrt{2}$$
$$\log_2 {x^{\frac{3}{4}(\log_{2}{x})^2 + \log_{2}{x} - \frac{5}{4}}} = \log_2 {\sqrt{2}}$$
$$\log_{2}{x} ({\frac{3}{4}(\log_{2}{x})^2 + \log_{2}{x} - \frac{5}{4}})=\frac12$$
$t=\log_2{x}$
$$3t^3+4t^2-5t-2=0$$
$t_1=1.$
Can you finish?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/794231",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How to show $(n-1)^3n^3(n+1)^3$ is divisible by 7 and 9? Yeah it looks like a basic, really elementary question, but i'm having hard time with it.
First i tried to show that it's divisible by 9
$$(n-1)^3n^3(n+1)^3 = ((n+1)(n-1))^3n^3 = (n^2-1)^3n^3 = (n^3-n)^3$$
and using eulers theorem we know that
$$[n^{\varphi(9)} \equiv 1 (mod \ 9)] = [n^6 \equiv 1 (mod \ 9)]$$
My doubt : can we do that? Cause $n$ and $9$ have to be coprime.
Is it right direction? I'd love some help on this, cause i never did tasks which asks for proving divisiblity of some polynomial. Cheers!
|
It's divisible by $9$ and in fact $27$, but not necessarily by $7$.
To see it is divisible by $27$, use the fact that that one of $n$, $n - 1$, $n - 2$ is divisible by $3$.
Your way
Expand out what you have:
$$
(n^3 - n)^3
= n^{9} - 3 n^{7} + 3n^5 - n^3
$$
If $n$ is divisible by $3$, you are done.
Otherwise, as you notice, we have $n^6 \equiv 1 \pmod 9$.
This implies $n^9 \equiv n^3$, so the above is
$$
\equiv n^3 - 3n^7 + 3n^5 - n^3 = -3(n^7 - n^5) \pmod 9
$$
Now since you have one factor of $3$ for sure, you consider $n^7 - n^5$ modulo $3$.
Euler's theorem gives $n^{\varphi(3)} = n^2 \equiv 1 \pmod 3$,
so $n^7 \equiv n^5$.
Therefore $n^7 - n^5 \equiv 0 \pmod 3$, and the expression in question is
$$
\equiv 0 \pmod 9.
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/794478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Method of characteristics for a system of pdes
I can do parts a) and b) as follows
$\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}\frac{\partial}{\partial{}x}\begin{pmatrix} u \\ v \\ w\end{pmatrix}+\begin{pmatrix} 1&1&0 \\ 1&2&1 \\ 0&1&1\end{pmatrix}\frac{\partial}{\partial y}\begin{pmatrix} u \\ v \\ w\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
and hence
$\bigg| \begin{pmatrix} 1-\lambda&1&0 \\ 1&2-\lambda&1 \\ 0&1&1-\lambda\end{pmatrix} \bigg| \implies \lambda = 0,1,3$
for part c) we have that $u(0,y)=1, v(0,y)=0$ and $w(0,y)=y$ on $x=0$, so I parameterise the conditions so that
$u(0,\xi)=1, v(0,\xi)=0$ and $w(0,\xi)=\xi$ on $x=0$
therefore
on $\frac{dy}{dx}=0 \implies y= constant \implies y=\xi$
now we know that $u-v+w=constant$ and so
$\implies u-v+w=1-0+\xi \implies u-v+w=1+y$
similarly I get that
on $y=x+\xi \implies u=w=1-y+x$ and on $y=3x+\xi$ I have that $u+2v+w=1+y-3x$.
However this system is not consistent with the equation at the top of the question and therefore clearly not correct. Could someone please show me how to approach these equations. Thanks
|
The solution is done in two steps.
First, diagonalise the linear system. We denote by $\mathbf{U}=[u;v;w].$ The original system becomes
$$\partial_x\mathbf{U}+M\partial_y\mathbf{U}=0,$$ where
$$
M=\begin{pmatrix}
1 & 1 & 0\\
1 & 2 & 1\\
0 & 1 & 1
\end{pmatrix}.
$$
We first diagonalise $M.$ We have (using Wolfram Alpha)
$$
M=S\,J\,S^{-1},
$$
where
$$
S = \begin{pmatrix}
1 & -1 & 1\\
-1 & 0 & 2\\
1 & 1 & 1
\end{pmatrix},
$$
$$
S^{-1} = \begin{pmatrix}
\frac{1}{3} & -\frac{1}{3} & \frac{1}{3}\\
-\frac{1}{2} & 0 & \frac{1}{2}\\
\frac{1}{6} & \frac{1}{3} & \frac{1}{6},
\end{pmatrix}
$$
$$
J = \mathrm{diag}(0,1,3).
$$
We denote by $\widetilde{\mathbf{U}}=S^{-1}\mathbf{U}.$ Then, the linear system becomes
$$
\partial_x\widetilde{\mathbf{U}} + J\partial_y\widetilde{\mathbf{U}} = 0,
$$
which consists of three independent scalar equations
$$
\begin{aligned}
\partial_x\widetilde{u} &= 0,\\
\partial_x\widetilde{v} + \partial_y\widetilde{v} &= 0,\\
\partial_x\widetilde{w} + 3\partial_y\widetilde{w} &= 0.
\end{aligned}
$$
The initial condition $\mathbf{U}(0)$ gives $\widetilde{\mathbf{U}}(0)=S^{-1}\mathbf{U}(0).$ In the case of (c),
$$
\begin{aligned}
&\left\{
\begin{aligned}
&\partial_x\tilde{u}=0,\\
&\tilde{u}(0,y)=\frac{1}{3}(1+y),
\end{aligned}
\right.
&\left\{
\begin{aligned}
&\partial_x\tilde{v}+\partial_y\tilde{v}=0,\\
&\tilde{v}(0,y)=\frac{1}{2}(-1+y),
\end{aligned}
\right.
&\left\{
\begin{aligned}
&\partial_x\tilde{w}+3\partial_y\tilde{w}=0,\\
&\tilde{w}(0,y)=\frac{1}{6}(1+y).
\end{aligned}
\right.
\end{aligned}
$$
Second step, we solve them independently,
$$
\begin{aligned}
&\tilde{u}(x,y)\equiv\tilde{u}(0,y)=\frac{1}{3}(1+y),\\
&\tilde{v}(x,y)=\tilde{v}(x,x+c)\equiv\tilde{v}(0,c)=\frac{1}{2}(-1+c)=\frac{1}{2}(-1+y-x),\\
&\tilde{w}(x,y)=\tilde{w}(x,3x+c)\equiv\tilde{w}(0,c)=\frac{1}{6}(1+c)=\frac{1}{6}(1+y-3x).
\end{aligned}
$$
Finally, we use $\mathbf{U}=S\,\tilde{\mathbf{U}}$ to obtain
$$
\begin{pmatrix}
u\\v\\w
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 & 1\\
-1 & 0 & 2\\
1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
\tilde{u}\\\tilde{v}\\\tilde{w}
\end{pmatrix}
=\begin{pmatrix}
1 \\ -x\\ y
\end{pmatrix}.
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/799360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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|
Find the area of the trapezium ABCD is trapezium AB||CD. 10 & 40 are the areas of the respective parts
How to find out the area of the trapezium?
|
$$\triangle COD \sim \triangle AOB.$$
If $\dfrac{CD}{AB}=a$, then $\dfrac{\text{Area of}\space \triangle COD}{\text{Area of} \space \triangle AOB}=a^2$.
So, we conclude: $\dfrac{CD}{AB}=\sqrt{\frac{40}{10}}=\sqrt{4}=2$.
If we denote trapezium total height $h$, then
heights of corresponding triangles are $\frac{2h}{3}$ and $\frac{h}{3}$ (all linear ratios of triangles $\triangle AOB$ and $\triangle COD$ are $\equiv 2$).
$\text{Area of} \space \triangle COD=\dfrac{CD}{2}\cdot \dfrac{2h}{3}=40$ $\implies$ $CD\cdot h=3\cdot 40=120$,
$\text{Area of} \space \triangle AOB=\dfrac{AB}{2} \cdot \dfrac{h}{3}=10$ $\implies$ $AB\cdot h = 6\cdot 10=60$;
Now Area of trapezium is
$$
\dfrac{AB+CD}{2}\cdot h = \frac{(120+60)}{2}=90.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/801328",
"timestamp": "2023-03-29T00:00:00",
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|
For which $n$ is $ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx}= \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$? I have been trying to figure out for which $n$ is $$ \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin nx} = \int_0^{\pi/2} \frac{\mathrm{d}x}{2+\sin x}=\frac{\pi}{3\sqrt{3\,}\,}$$
Using maple I got the following values below 100
$$
n = 1
,\,2
,\,12
,\,13
,\,14
,\,24
,\,25
,\,26
,\,36
,\,37
,\,38
,\,48
,\,49
,\,50
,\,60
,\,61
,\,62
,\,73
,\,74
,\,85
,\,86
,\,98
,\,110
$$
But I am having a hard time seeing any pattern. I tried calculating the integral directly.
The integral seems to converge for all $n$, but a closed form seems hard. Any help would be appreciated =)
EDIT: I ran a few more tests and can not find any more integer values that work. Are the list above exhaustive?
|
Although the problem has been solved a different method is presented here.
The problem asks to find which values of $n$ that satisfy
\begin{align}
\int_{0}^{\pi/2} \frac{dx}{2 + \sin(nx)} = \int_{0}^{\pi/2} \frac{dx}{2 + \sin(x)}
= \frac{\pi}{3 \sqrt{3}}.
\end{align}
The solution proposed is as follows.
Consider the integral
\begin{align}
I = \int_{0}^{\pi/2} \frac{dx}{a + \sin(nx)}.
\end{align}
The integral $I$ can be evaluated as
\begin{align}
I &= \left[ \frac{2}{ \sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{a \tan\left(\frac{nx}{2}\right) + 1}
{ \sqrt{a^{2}-1}} \right) \right]_{0}^{\pi/2} \\
&= \frac{2}{\sqrt{a^{2}-1} \ n } \left[ \tan^{-1} \left( \frac{a \tan\left(\frac{n\pi}{4}\right) + 1}
{ \sqrt{a^{2}-1}} \right) - \tan^{-1} \left( \frac{1}{ \sqrt{a^{2}-1}} \right) \right] \\
&= \frac{2}{\sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{ \sqrt{a^{2}-1} \ \theta_{n} }{ a + \theta_{n}}
\right) \\
I &= \frac{2}{\sqrt{a^{2}-1} \ n } \tan^{-1} \left( \frac{ \sqrt{a^{2}-1} }{ a/\theta_{n} + 1}
\right),
\end{align}
where $\theta_{n} = \tan(n\pi/4)$. Now, returning to the original integral, make the substitution
$x = \pi/2 -u$ to obtain
\begin{align}
I &= \int_{0}^{\pi/2} \frac{du}{a + \sin(\frac{n \pi}{2} - nu)} \\
&= \int_{0}^{\pi/2} \frac{du}{a + \sin\left(\frac{n \pi}{2}\right) \cos(n u) - \cos\left(
\frac{n \pi}{2}\right) \sin(n u)}
\end{align}
for which if the integral is to remain of the same form then it is required that
\begin{align}
\sin\left( \frac{n \pi}{2} \right) &= 0 \\
\cos\left( \frac{n \pi}{2} \right) &= -1.
\end{align}
From the sine equation it is seen that it is satisfied if $n = 2r$ for $r \geq 0$. The cosine
equation is satisfied if $n = 4k+2$ for $k \geq 0$. It is evident that $2,6,10,\cdots$ is a
subset of values in the set defined by $2r$ for $r \geq 0$. Since both equations are satisfied by the subset of values, this leads to the allowed values of $n$ for which the integral remains of the same form is $n = 4k+2$ for $k \geq 0$. This can be
seen as
\begin{align}
\int_{0}^{\pi/2} \frac{dx}{a + \sin((4k+2)x)} = \int_{0}^{\pi/2} \frac{dx}{a + \sin(x)},
\end{align}
for $k \geq 0$.
Now, with this $n$ value, it is seen that $\theta_{n}$ becomes
\begin{align}
\theta_{4k+2} &= \tan\left( \frac{(4k+2) \pi}{4} \right) = \tan\left( k \pi + \frac{\pi}{2} \right) \\
&= \frac{ \sin(k \pi + \pi/2) }{ \cos(k \pi + \pi/2) } = \frac{ (-1)^{k} }{ 0 } = \infty .
\end{align}
for which $1/\theta_{4k+2} = 0$. By using this result in the evaluation of the integral defined
earlier provides
\begin{align}
I = \frac{2}{\sqrt{a^{2}-1} \ (4k+2) } \tan^{-1}( \sqrt{a^{2}-1} ).
\end{align}
Hence,
\begin{align}
\int_{0}^{\pi/2} \frac{dx}{a + \sin((4k+2)x)} &= \int_{0}^{\pi/2} \frac{dx}{a + \sin(x)} \\
&= \frac{1}{\sqrt{a^{2}-1} \ (2k+1) } \tan^{-1}( \sqrt{a^{2}-1} ),
\end{align}
for $k \geq 0$.
When $a=2$ the reduction leads to
\begin{align}
\int_{0}^{\pi/2} \frac{dx}{2 + \sin((4k+2)x)} &= \frac{ \tan^{-1}( \sqrt{3} ) }{\sqrt{3} \ (2k+1) } \\
&= \frac{ \tan^{-1}( \tan(\pi/3) ) }{\sqrt{3} \ (2k+1) } = \frac{\pi}{3 \sqrt{3} \ (2k+1) }.
\end{align}
It is now evident that the only value of $k$ that satisfies the desired result is $k=0$.
Hence, in terms of the original notation, the only values of $n$ are 1 and 2.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$ How would I prove
$$
\sum\limits_{\vphantom{\large A}i\,,\,j\ \geq\ 0}{n-i \choose j} {n-j \choose i}
=F_{2n+1}
$$
where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of Fibonacci numbers?
Thank you very much! :)
|
The problem asks to show that
\begin{align}
\sum_{i=0}^{n} \sum_{j=0}^{n} \binom{n-i}{j} \binom{n-j}{i} = F_{2n+1}.
\end{align}
The problem, as stated, is incorrect. It should read $F_{2n+2}$. This will be shown
in the following.
Consider the double summation
\begin{align}
S_{n} = \sum_{i=0}^{n} \sum_{j=0}^{n} \binom{n-i}{j} \binom{n-j}{i}.
\end{align}
By reversing the summation over the index $i$ this becomes
\begin{align}
S_{n} = \sum_{i=0}^{n} \sum_{j=0}^{i} \binom{i}{j} \binom{n-j}{n-i}.
\end{align}
Now consider the generating function of $S_{n}$.
\begin{align}
\sum_{n=0}^{\infty} S_{n} t^{n} &= \sum_{n=0}^{\infty} \sum_{i=0}^{n} \sum_{j=0}^{i}
\binom{i}{j} \binom{n-j}{n-i} \ t^{n} \\
&= \sum_{n=0}^{\infty} \sum_{i=0}^{\infty} \sum_{j=0}^{i} \binom{i}{j} \binom{n+i-j}{n} \ t^{n+i} \\
&= \sum_{i=0}^{\infty} \sum_{j=0}^{i} \binom{i}{j} \ t^{i} \cdot \sum_{n=0}^{\infty} \binom{n+i-j}{n}
\ t^{n} \\
&= \sum_{i=0}^{\infty} \sum_{j=0}^{i} \binom{i}{j} \ t^{i} \ (1-t)^{-i+j-1} \\
&= \frac{1}{1-t} \ \sum_{i=0}^{\infty} \left( \frac{t}{1-t} \right)^{i} \cdot \sum_{j=0}^{i}
\binom{i}{j} \ (1-t)^{j} \\
&= \frac{1}{1-t} \ \sum_{i=0}^{\infty} \left( \frac{t}{1-t} \right)^{i} \ (2-t)^{i} \\
&= \frac{1}{1-t} \ \sum_{i=0}^{\infty} \left( \frac{2t - t^{2}}{1-t} \right)^{i} \\
&= \frac{1}{1-t} \ \frac{1-t}{1-3t+t^{2}} \\
&= \frac{1}{1-3t+t^{2}}.
\end{align}
Now,
\begin{align}
\frac{1}{1-3t+t^{2}} &= \frac{1-t}{1-3t+t^{2}} + \frac{t}{1-3t+t^{2}} \\
&= \sum_{n=0}^{\infty} F_{2n+1} \ t^{n} + \sum_{n=0}^{\infty} F_{2n} \ t^{n} \\
&= \sum_{n=0}^{\infty} F_{2n+2} \ t^{n}
\end{align}
which, when compared to the previous result, leads to
\begin{align}
\sum_{i=0}^{n} \sum_{j=0}^{n} \binom{n-i}{j} \binom{n-j}{i} = F_{2n+2}.
\end{align}
|
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|
derivative of $y=(x^2+x^3)^4$ I can't figure out where I am going wrong.
$$y=(x^2+x^3)^4$$
chain rule it first
$$4(x^2+x^3)^3* \frac d{dx}(x^2+x^3)$$
which should become:
$$4(x^2+x^3)^3(2x+3x^2)$$
factoring out should give me:
$$4*x^2*x(1+x)^3(2+3x)$$
which to me says the answer is: $4x^3(1+x)^3(2+3x)$
but the book says: $4x^7(1+x)^3(2+3x)$
where did I go wrong?
|
Your mistake is in the factoring out. When you factor $x^2$ out from the $(x^2+x^3)^3$ term, it becomes $(x^2)^3(1+x)^3=x^6(1+x)^3$ instead of $x^2(1+x)^3$
|
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|
What are the steps to derive the following inverse Fourier transformations I'm reading a text which is an introductory text on Fourier transforms. The author has two expressions:
$$ F(\omega_{o}) = \frac{1}{\sigma \sqrt {2 \pi} } e^{\Large- \frac{{\omega_{o}}^2}{2\sigma^2}} $$
and
$$G(\omega_{o}) = 2\pi \cos(\omega_{o}t).$$
Then he states that the inverse transformations of the functions are
$$f(t_{o}) = \frac{1}{2 \pi} e^{\Large-\frac{1}{2}\sigma^2 t_{o}^2}$$
and
$$g(t_{o}) = 2 \pi \left( \frac{\delta(t_{o} - t)}{2} + \frac{\delta(t_{o} + t)}{2} \right) .$$
Would someone be able to derive how $f(t_{o})$ and $g(t_{o})$ are obtained. I'd appreciate it because I'd probably learn a lot from seeing the steps. Thank you very much.
|
For the first one, we have
$$
F(\omega)=\frac{1}{\sigma\sqrt{2\pi}} e^{\Large- \frac{\omega^2}{2\sigma^2}}.
$$
It seems that you are using the textbook for physics or engineering, so the inverse Fourier transform of $F(\omega)$ using the 'convention' notation in those fields is
$$
\begin{align}
\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)\ e^{\large it\omega}\ d\omega\\
f(t)&=\frac{1}{2\pi}\cdot\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{\Large- \frac{\omega^2}{2\sigma^2}}\ e^{\large it\omega}\ d\omega\\
&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega.
\end{align}
$$
In general
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x-\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx.
\end{align}
$$
Let $u=x-\frac{b}{2a}\;\rightarrow\;du=dx$, then
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\
\end{align}
$$
The last form integral is Gaussian integral that equals to $\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}$. Hence
$$
\int_{x=0}^\infty e^{-(ax^2-bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).
$$
It yields the same result for $\displaystyle\int_{x=0}^\infty e^{-(ax^2\color{red}+bx)}\,dx$. Thus
$$
\begin{align}
\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\cdot\frac{1}{2}\sqrt{2\pi\sigma^2}\exp\left(\frac{( it)^2\cdot2\sigma^2}{4}\right)\\
f(t)&=\color{blue}{\frac{1}{2\pi}\ e^{\Large- \frac{\sigma^2t^2}{2}}}.
\end{align}
$$
|
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|
Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} = 0 \\ $
It seems that x = 0 is a critical point.
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' = -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
|
A trick is to get rid of the transcendental function.
As the sine is monotonic in $(0,\pi/2)$,
$$x>\arcsin\left(x-\frac{x^3}3\right),$$ and, as the two members coincide for $x=0$, by differentiation
$$1>\frac{1-x^2}{\sqrt{1-\left(x-\dfrac{x^3}3\right)^2}}.$$
Then it suffices that for $x<1$,
$$1-\left(x-\dfrac{x^3}3\right)^2>(1-x^2)^2,$$
$$-\frac{x^6}9-\frac{x^4}3+x^2>0,$$ which is true.
|
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|
Inequality involving conjugate numerator/denominator pairs Question is to solve:
$$\frac{(x-2)(x-4)(x-7)}{(x+2)(x+4)(x+7)} > 1$$
I thought I could negate terms to make them equal (i.e. $-(x-2)$), but that does not happen. I could subtract $1$ from each side but that would be $(very)^2$ heavy work. Is there any simple way of doing this ?
|
Another way would be to multiply throughout by $(x+2)^2(x+4)^2(x+7)^2$, which is positive ($x =-2,-4,-7$ are not allowable anyway), to get
$$(x^2-4)(x^2-16)(x^2-49)>(x+2)^2(x+4)^2(x+7)^2$$
Collecting terms in the LHS and factoring, this is equivalent to
$$-2(x+7)(x+4)(x+2)(13x^2+56)>0$$
and now checking intervals where LHS can change signs gives you the answer $x \in (-\infty, -7) \cup (-4,-2)$.
|
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|
For which real numbers $c$ is $\frac{e^x+e^{-x}}{2} \le e^{cx^2}$ for all real numbers $x$? This question comes from the 1980 Putnam exam. My work is shown below.
For all integers $n \ge 1$, \begin{align}
(2n)!&=n!\cdot\underbrace{(n+1)(n+2)(n+3)\cdots(2n-2)(2n-1)(2n)}_{n \text{ terms}} \\
&\ge n! \cdot \underbrace{2 \cdot 2 \cdot 2\cdots 2\cdot2 \cdot 2}_{n \text{ terms}} \\ &=2^nn! \\ \implies \frac{1}{(2n)!} &\le \frac{1}{2^nn!}.\end{align} So we have \begin{align}
\frac{e^x+e^{-x}}{2} = \cosh(x) &= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\ &\le \sum_{n=0}^{\infty} \frac{x^{2n}}{2^nn!} \\ &= \sum_{n=0}^{\infty} \frac{\left(\frac{x^2}{2} \right)^n}{n!} \\ &= e^{\frac{1}{2}x^2}
\end{align}
Therefore, $\boxed{c \ge \frac{1}{2}}$.
How do I know that it is $\ge$, not $\le$? Other than that, how did I do on my work?
|
The method presented in the proposal of the problem is correct.
An alternate method is:
\begin{align}
\cosh(x) \approx 1 + \frac{x^{2}}{2} + \mathcal{O}(x^{4})
\end{align}
and
\begin{align}
e^{c x^{2}} \approx 1 + c x^{2} + \mathcal{O}(x^{4})
\end{align}
for which the inequality
\begin{align}
e^{c x^{2}} \geq \cosh(x)
\end{align}
leads to
\begin{align}
c \geq \frac{1}{2}.
\end{align}
|
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|
Number of Rolls of Fair Dice to get '6' and '5' A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$
|
We cannot get both 5 and 6 on the fifth trial, so $P(X=5 \mid Y=5)=0$
$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5) + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5)$
To unconditionally get a 5 on the fifth trial, we need four not 5 before a 5 on the fifth.$$\mathrm{\large P}(Y=5) = \left(\frac{5}{6}\right)^4 \frac 16$$
To get a 6 before getting a 5 on the fifth trial, we have $(x-1)$ rolls that are neither, a 6 on the $x^{th}$, $(3-x)$ rolls that are not 5, then a 5 on the fifth trial.
$$\mathrm{\large P}(X=x\mid Y=5)|_{x<5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x<5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{6}\right)^{x-1}\frac{1}{6}\left(\frac{5}{6}\right)^{3-x}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^{x-1}5^{3-x}}{6^5}\cdot\dfrac{6^5}{5^4} = \dfrac{1}{20} \cdot\dfrac{4^{x}}{5^{x}}$$
To get a 6 after getting a 5 on the fifth trial, we have $4$ rolls that are neither, a 5 on the fifth trial, $(x-6)$ rolls that are not 6, then a 6 on the $x^{th}$ trial.
$$\mathrm{\large P}(X=x\mid Y=5)|_{x>5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x>5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{5}\right)^{4}\frac{1}{6}\left(\frac{5}{6}\right)^{x-6}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^45^{x-6}}{6^x}\cdot\dfrac{6^5}{5^4} = \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}}$$
So $$P(X=x\mid Y=5) = \begin{cases} \dfrac{1}{20}\cdot\dfrac{4^{x}}{5^{x}} & 0< x < 5 \\ \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}} & x>5 \\ 0 & \text{elsewhere}\end{cases}$$
Putting it together:
$$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x<5} + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x>5} \\ = \dfrac{1}{20}\sum\limits_{x=1}^{4} x\left(\dfrac{4}{5}\right)^x + \dfrac{4^4}{5^5}\sum\limits_{x=6}^{\infty} x \left(\dfrac{5}{6}\right)^{x-5} \\ = \dfrac{14901}{3125} \approx 4.76832 $$
|
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|
Series Question: $\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$$
I tried to use partial fraction
$$\begin{align}\frac{n^2}{(4n^2-1)^3}&=\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}+\frac{1}{64(2n+1)^2}+\frac{1}{64(2n-1)^2}\\&-\frac{1}{32(2n+1)^3}+\frac{1}{32(2n-1)^3}\end{align}$$
I can compute
$$\sum_{n=1}^{\infty}\left[\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}\right]=-\frac{1}{64}$$
using telescoping series, but I cannot compute the rest. I believe there's a better way than this. Any help would be appreciated. Thanks in advance.
|
Note that:
$$\sum _{n=1}^{\infty }{\frac {{n}^{2}}{ \left( 4\,{n}^{2}-1 \right) ^{3
}}}=\frac{1}{4}\,\sum _{n=1}^{\infty }\frac{1}{\left( 4\,{n}^{2}-1 \right)^3 }+
\frac{1}{\left( 4\,{n}^{2}-1 \right)^2 }\tag{1}$$
then note that:
$$\sum _{n=1}^{\infty } \frac{1}{\left( {n}^{2}-{x}^{2} \right)}=\frac{1}{2x^2}-\frac{1}{2}\,{\frac {\pi \,\cot \left( \pi \,x \right) }{x}}\tag{2}$$
insert $(2)$ into the differential equation:
$$ \left( {\frac {d^{2}}{d{x}^{2}}}f \left( x \right) \right) a+
\left( {\frac {d}{dx}}f \left( x \right) \right) b \tag{3}$$
and evaluate $(3)$ at $x=1/2$ to get:
$$\frac{1}{4}\,\sum _{n=1}^{\infty }{\frac {a\,512}{ \left( 4\,{n}^{2}-1
\right) ^{3}}}+{\frac {128\,a+64\,b}{ \left( 4\,{n}^{2}-1 \right) ^{2
}}}=a \left( 48-4\,{\pi }^{2} \right) +b \left( -8+{\pi }^{2} \right)\tag{4} $$
then let:
$$a=\frac{1}{512},\quad b=\frac{3}{256} \tag{5} $$
to get:
$$\sum _{n=1}^{\infty }{\frac {{n}^{2}}{ \left( 4\,{n}^{2}-1 \right) ^{3
}}}=\frac{1}{4}\,\sum _{n=1}^{\infty }\frac{1}{\left( 4\,{n}^{2}-1 \right)^3 }+
\frac{1}{\left( 4\,{n}^{2}-1 \right)^2 }={\frac {1}{256}}\,{\pi }^{2} \tag{6} $$
|
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|
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$?
How to solve $\int \frac{x^4 + 1 }{x^6 + 1}$ ?
The numerator is a irreducible polynomial so I can't use partial fractions.
I tried the substitutions $t = x^2, t=x^4$ and for the formula $\int u\,dv = uv - \int v\,du$ I tried using: $u=\frac{x^4 + 1 }{x^6 + 1} , \,dv=\,dx \\ u=\frac{1}{x^6 + 1} , \,dv= (x^4 + 1) \,dx \\u=x^4 + 1 , \,dv=\frac{\,dx}{x^6 + 1}$
But I always get more complicated integrals.
Any hints are appreciated!
|
OK it is a rational function so we preform a partial fraction decomposition.
The first step is to factor the denominator into linear and quadratic factors.
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)((x+1)^2-x^2))=
(x^2+1)(x^2-x+1)(x^2+x+1)$$
Now we find
$$\frac{x^4+1}{x^6+1}=\frac{2}{3}\frac{1}{x^2+1}+\frac{1}{6}\frac{1}{x^2-x+1}+\frac{1}{6}\frac{1}{x^2+x+1}$$
So we have
$$\frac{2}{3}\int \frac{dx}{x^2+1}+\frac{1}{6}\int\frac{dx}{(x-\frac{1}{2})^2+
\frac{3}{4}}+\frac{1}{6}\int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}$$
Giving $$\frac{2}{3}\arctan x+\frac{1}{3\sqrt{3}}\arctan \frac{2}{\sqrt{3}}(x-\frac{1}{2})+\frac{1}{3\sqrt{3}}\arctan \frac{2}{\sqrt{3}}(x+\frac{1}{2})$$
in the highly unlikely event that I made no mistakes.
|
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|
Integrals of a particular form Recently, I was given a question sheet with a lot of integrals, and I could solve all of them except for a particular type of them:
$$I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=\int \frac{dx}{x^2\sqrt{x^2+1}}=\frac{-\sqrt{x^2+1}}{x}+C$$
$$I_2=\int \frac{dx}{(x-1)^2\sqrt{x^2-x+1}}$$
$$I_3=\int \frac{dx}{(2x+1)\sqrt{x^2-8}}$$
$$I_4=\int \frac{dx}{x\sqrt{x^2+x+1}}=\log x + \log\left(x+2+2\sqrt{x^2+x+1}\right)+C$$
And the like($I_2,I_3$ also are expressible in elementary functions, but they are not that cute). The method I see that wolfram is using(trial pro subscription) is to complete the square inside the roots to transform them to $\sqrt{ax^2+b}$ and then substitute $x=\sqrt{\frac{-b}{a}}\sec x$ if $b$ is negative, and $x=\sqrt{\frac{b}{a}}\tan x$ if positive. Finally, it uses the tangent half-angle subsitution. I really don't think this is the best way.
In particular, the trigonometric substitutions in $I_1,I_4$ seem to be unnecessary, since at the end we transform back the resulting composition of trigonometric functions into a function that doesn't need trigonometry at all (e.g. $-\csc(\tan^{-1}x)=-\frac{\sqrt{x^2+1}}{x}$)
So my question is: Is there some easy way to deal with the integrals:
$$\int\frac{dx}{(ax+b)\sqrt{x^2+c}}, \int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}$$
Without having to do the secant-Weiertrass substitution?
|
Usually more direct, with $\sqrt {u^2 + 1},$ to take $u = \sinh t.$ With $\sqrt {u^2 - 1},$ to take $u = \cosh t.$ Tends to save several steps.
Relevant: it will still be necessary, to integrate a rational function of $\sinh x$ and $\cosh x,$ to use a stereographic projection. I worked it out:
$$ v = \tanh \frac{x}{2} = \frac{e^x - 1}{e^x + 1}; \; \; \; \mbox{so} \; |v| < 1. $$
$$ x = 2 \operatorname{argtanh} x = \log \left( \frac{1+v}{1-v}\right) $$
$$ \sinh x = \frac{2v}{1-v^2} $$
$$ \cosh x = \frac{1+v^2}{1-v^2} $$
$$ dx = \frac{2}{1-v^2} \, dv $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/812061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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|
Showing that $3x^2+2x\sin(x) + x^2\cos(x) > 0$ for all $x\neq 0$ I got this question:
Show that for all $x\neq 0$, $3x^2+2x\sin(x) + x^2\cos(x) > 0$
I tried to show it but got stuck.
|
If $x<0$ then
$$\sin x<-x\implies 2x\sin x>-2x^2$$
and
$$\cos x\ge-1\implies x^2\cos x\ge-x^2$$
and so
$$3x^2+2x\sin x+x^2\cos x>3x^2-2x^2-x^2=0\ .$$
The case $x>0$ is similar, even a bit easier.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/812453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Compute $\int_0^{\pi/4}\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)} x\exp(\frac{x^2-1}{x^2+1}) dx$
Compute the following integral
\begin{equation}
\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx
\end{equation}
I was given two integral questions by my teacher. I can answer this one although it took a lot of time to compute it. I want to share this problem to the other users here and I would love to see how Mathematics SE users compute this monster. Thank you.
|
Rewrite
\begin{align}
&\int\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\
&=-\int\left[\frac{(1-x^2)\ln\left(\dfrac{1-x^2}{1+x^2}\right)-(1+x^2)}{(1-x^2)(1+x^2)(1+x^2)}\right] x\ \exp\left[-\frac{1-x^2}{1+x^2}\right]\ dx\\
&=-\frac14\int\left[\frac{(1-x^2)\ln\left(\dfrac{1-x^2}{1+x^2}\right)-(1+x^2)}{(1-x^2)}\right] \frac{2x}{1+x^2}\, \exp\left[-\frac{1-x^2}{1+x^2}\right]\ \frac{2\ dx}{1+x^2}\\
&=-\frac14\int\left[\ln\left(\frac{1-x^2}{1+x^2}\right)-\frac{1+x^2}{1-x^2}\right] \frac{2x}{1+x^2}\, \exp\left[-\frac{1-x^2}{1+x^2}\right]\ \frac{2\ dx}{1+x^2}.\tag1
\end{align}
Now, consider Weierstrass substitution:
$$
x=\tan\frac{t}{2}\;,\;\sin t=\frac{2x}{1+x^2}\;,\;\cos t=\frac{1-x^2}{1+x^2}\;,\;\text{ and }\;dt=\frac{2\ dx}{1+x^2}.
$$
The integral in $(1)$ turns out to be
$$
-\frac14\int\left[\ln\left(\cos t\right)-\frac{1}{\cos t}\right] \sin t\, \exp\left[-\cos t\right]\ dt.\tag2
$$
Let $y=\cos t\;\Rightarrow\;dy=-\sin t\ dt$, then $(2)$ becomes
$$
\frac14\int\left[\ln y-\frac{1}{y}\right] e^{-y}\ dy=\frac14\left[\int e^{-y}\ln y\ dy-\int\frac{e^{-y}}{y}\ dy\right].\tag3
$$
The second integral in the RHS $(3)$ can be evaluated by using IBP. Taking $u=e^{-y}\;\Rightarrow\;du=-e^{-y}\ dy$ and $dv=\dfrac1y\ dy\;\Rightarrow\;v=\ln y$, then
$$
\int\frac{e^{-y}}{y}\ dy=e^{-y}\ln y+\int e^{-y}\ln y\ dy.\tag4
$$
Substituting $(4)$ to $(3)$, we obtain
$$
\frac14\left[\int e^{-y}\ln y\ dy-e^{-y}\ln y-\int e^{-y}\ln y\ dy\right]=-\frac14e^{-y}\ln y+C.
$$
Thus
\begin{align}
&\int\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\
&=\color{blue}{-\frac14\exp\left[-\frac{1-x^2}{1+x^2}\right]\ln \left|\frac{1-x^2}{1+x^2}\right|+C}
\end{align}
and
\begin{align}
&\int_0^{\Large\frac\pi4}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\
&=\color{blue}{-\frac14\exp\left[\frac{\pi^2-16}{\pi^2+16}\right]\ln \left|\frac{16-\pi^2}{16+\pi^2}\right|}.
\end{align}
|
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"url": "https://math.stackexchange.com/questions/815863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "116",
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"answer_id": 1
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|
Creating two groups where two people can't be in the same group. If you have a group of $12$ men and $10$ women, and you need to make two groups - one with $6$ people and the other with $9$, the ways to form such groups would be
$${22 \choose 6}\cdot {16 \choose 9}$$
Since I guess the order doesn't matter.
But then, there is another constraint: Bob (male) and Hilda (female) cannot be in the same group.
For the time being, maybe I can remove them from the whole thing, and make one group with $5$ people and another with $8$ without them:
$${20 \choose 5}\cdot {15 \choose 8}$$
There are now $7$ people left, plus Bob and Hilda we got $9$. There is one spot free in the first group and another one in the second.
If Hilda is placed in the first group, then it is guaranteed Bob won't be in it (because it's full), so he would end up in the second group and viceversa.
So I guess the answer would be like this?
$${20 \choose 5}\cdot {15 \choose 8} \cdot 9 \cdot 8$$
|
In the initial case, you do not distinguish between men and women. So effectively, you have $22$ distinct people to make up two groups one of which has $6$ members and the other has $9$ members. Let's assume we have assembled all possible groups of $6$. For each group of $6$, there would be $n$ sets of 9 elements from the remaining $22 - 6 = 16$. This means $\binom{22 - 6}{9}$. Which indeed gives your initial calculation: $\binom{22}{6}\cdot\binom{22 - 6}{9} = \binom{22}{6}\cdot\binom{16}{9}$.
Now, as you say, lets take these two individuals out. They must always be on opposite teams. The fact that one appears to be male and one appears to be female is irrelevant. If we choose a set of, instead of $6$ and $9$ members, $5$ and $8$ members, from the total group of $22 - 2 = 20$ total members (the two excluded...).
This gives:
$$
\binom{20}{5}\cdot\binom{15}{8}
$$
However, for each of these groups there is $\binom{2}{1} = 2$ ways to put these two constrained members. Since one group has $6$ members and the other $9$, each of these groups is unique! Therefore switching which group Bob is in and which group Hilda is in does create two distinct groupings.
I do not believe it should be $9\cdot8$, rather:
$$
\binom{20}{5}\cdot\binom{15}{8} \cdot \binom{2}{1}
$$
In general for dividing into two regions of size $w$ and $h$ from a number of values numerating $N \geq w + h$. For the number of unique set of two groups of size $w$ and $h$:
$$
n = \binom{N}{w}\cdot\binom{N - w}{h} = \binom{N}{h}\cdot\binom{N - h}{w}
$$
If we want to exclude two from being together we simply take them out--as you prescribed:
\begin{align}
n_{\text{divorce}} =& \binom{N - 2}{w - 1}\cdot\binom{N - 2 - (w - 1)}{h - 1}\cdot\binom{2}{1} \\
=& 2*\binom{N - 2}{w - 1}\cdot\binom{N - w -1}{h - 1} \\
=& 2*\binom{N - 2}{h - 1}\cdot\binom{N - h - 1}{w - 1}
\end{align}
For your case this is:
\begin{align}
N =& 22\\
w =& 6 \\
h = & 9 \\
n =& 2*\binom{20}{5}\cdot\binom{22 - 6 - 1}{8} \\
=& 2*\binom{20}{5}\cdot\binom{16 - 1}{8} \\
=& 2*\binom{20}{5}\cdot\binom{15}{8} \\ \\
=& 2*\binom{20}{8}\cdot\binom{22 - 9 - 1}{5} \\
=& 2*\binom{20}{8}\cdot\binom{13 - 1}{5} \\
=& 2*\binom{20}{8}\cdot\binom{12}{5}
\end{align}
There is one more thing we should think about. The above constraint is actually that both Bob and Hilda will always be chosen. Assuming the only actual constraint is that they are not in the same group, then we also need to consider when exactly one of them is chosen and when neither of them is chosen.
This would mean you get:
$$
2*\binom{20}{8}\cdot\binom{12}{5} + 2*\binom{21}{9}\cdot\binom{12}{6} + \binom{20}{9}\cdot\binom{11}{6}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
A little Problem in Trigonometry (Multiple Angle) If $\tan^2 \theta = 1 + 2\tan^2 \phi$, show that $\cos 2\phi = 1 + 2\cos2\theta$.
What I have done..
$$\implies \tan^2 \theta = 1 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2 + 2\tan^2 \phi\\
\implies 1 + \tan^2 \theta = 2(1 + \tan^2 \phi)\\
\implies \sec^2 \theta = 2(\sec^2 \phi)$$
|
Your steps are fine you just need to do the following (I'm putting your steps first):
\begin{align*}
\tan^{2}\theta & = 1+2 \tan^{2}\phi\\
1+\tan^{2}\theta & = 2+2 \tan^{2}\phi\\
\sec^2 \theta & = 2 \sec^2 \phi\\
\text{Do the following to continue:}\\
\frac{1}{\cos^2 \theta} & = \frac{2}{\cos^2 \phi}\\
\cos^2 \phi & = 2 \cos^2 \theta\\
\text{using the half angle formula}\\
\frac{1+\cos 2 \phi}{2} & = 1+\cos2 \theta\\
1+\cos 2 \phi & = 2(1+\cos2 \theta)\\
\cos 2\phi & = 1 +2\cos 2\theta.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/817521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove that $\exists k>0$ such that$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}
Consider a positive sequence $\{a_{n}\}$ such that $a_{n+1}>a_{n}$, and $\{a_n\}$ is unbounded.
Show that there exits a positive integer $k$ such that, when $n>k$
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<n-2014$$
This problem is from china test 2014.
I only prove follow: there exsit $k$,such $n>k$,have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<(n-1)-1$$
because
\begin{align*}
&(n-1)-\left(\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}\right)\\
&=\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{3}-a_{2}}+\cdots+\dfrac{a_{n}-a_{n-1}}{a_{n}}\\
&>\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{3}-a_{2}}{a_{n}}+\cdots+\dfrac{a_{n}-a_{n-1}}{a_{n}}
&=\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{n}-a_{2}}{a_{n}}\\
&=1+a_{2}\left(\dfrac{a_{2}-a_{1}}{a^2_{2}}-\dfrac{1}{a_{n}}\right)
\end{align*}
since $a_{n}$ is unbounded,so we choose $\{a_{n}\}$ such
$$a_{n}>\dfrac{a^2_{2}}{a_{2}-a_{1}}$$
so
$$(n-1)-\left(\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}\right)>1$$
|
Let $(1-\epsilon_n) = a_{n} / a_{n+1}$, Then $a_n = a_1 / \prod_{i=1}^{n-1} (1-\epsilon_i)$. Since $a_n$ is unbounded, it follows that $\prod_{i=1}^\infty (1-\epsilon_i) = 0$. It is well known that this product equals zero if and only if $\sum_{i=1}^\infty \epsilon_i$ diverges.
But $\frac{a_1}{a_2} + \cdots \frac{a_{n-1}}{a_n} = n - \sum_{i=1}^{n-1} \epsilon_i$, and since $\sum_{i=1}^\infty \epsilon_i$ diverges, for some integer $k$ we have that $\sum_{i=1}^{n-1} \epsilon_i > 2014$ for $n > k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/818315",
"timestamp": "2023-03-29T00:00:00",
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|
Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$
Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$
But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.
|
In your step 4, you forgot to divide (3z+2) by 2
|
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"timestamp": "2023-03-29T00:00:00",
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|
Primes as a difference of powers Find the smallest prime that cannot be written as
$$|3^a - 2^b|$$
EDIT: I forgot to mention that $a$ and $b$ are whole numbers.
I tried to expand $3^a$ as $(2+1)^a$ using binomial theorem but I couldn't infer much. Please help. Thanks in advance!
|
$41$ is the answer.
For $41=3^a-2^b$, clearly $b \not =0, 1, a \not =0$. $\pmod{3}$ gives $b$ even, $\pmod{4}$ gives $a$ even. Then $41=(3^{\frac{a}{2}}-2^{\frac{b}{2}})(3^{\frac{a}{2}}+2^{\frac{b}{2}})$, so $3^{\frac{a}{2}}-2^{\frac{b}{2}}=1, 3^{\frac{a}{2}}+2^{\frac{b}{2}}=41$, which is clearly impossible.
For $41=2^b-3^a$, clearly $b \geq 3$. $\pmod{8}$ gives a contradiction.
Note $2=3^1-2^0, 3=2^2-3^0, 5=2^3-3^1, 7=3^2-2^1, 11=3^3-2^4, 13=2^4-3^1, 17=3^4-2^6, 19=3^3-2^3, 23=3^3-2^2, 29=2^5-3^1, 31=2^5-3^0, 37=2^6-3^3$.
P.S. Try this problem instead:
Determine whether there are infinitely many primes $p$ s.t. $p$ cannot be expressed as the sum or difference of a power of $2$ and a power of $3$, i.e. $p \not =|3^a \pm 2^b|$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/821346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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|
Stuck on finding the value of $\sum_{n=0}^\infty {n(n+1) \over 3^n}$ I am trying to find the value of the series
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}
$$
Here's what I have done so far:
$$
\sum_{n=0}^\infty {n(n+1) \over 3^n}=\sum_{n=0}^\infty {n^2 \over 3^n}+\sum_{n=0}^\infty {n \over 3^n}
$$
Determining the values of the both terms separately:
\begin{eqnarray}
S_1 = \sum_{n=0}^\infty {n \over 3^n} = {1 \over 3^1}+{2 \over 3^2}+{3 \over 3^3}+ \dots \\
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^2} + {1 \over 3^3} + \dots \\ +{1 \over 3^3} + \dots \\
= {1 \over 3^1} + {1 \over 3^2} + {1 \over 3^3} + \dots \\
+{1 \over 3^1}\left( {1 \over 3^1}+{1 \over 3^2}+\dots\right) \\
+ {1 \over 3^2}\left( {1 \over 3^1}+\dots\right)\\
+ \dots\\
= \sum_{n=1}^\infty{1 \over 3^n} \cdot \sum_{n=0}^\infty {1 \over 3^n}
\end{eqnarray}
Next evaluating $\sum_{n=1}^\infty{1 \over 3^n}$:
\begin{align*}
3S_1 - S_1 = \sum_{n=1}^\infty{1 \over 3^{n-1}} - \sum_{n=1}^\infty{1 \over 3^n} = \\
1 - \require{enclose}\enclose{updiagonalstrike}{1 \over 3^1} + \enclose{updiagonalstrike}{1 \over 3^1} - \enclose{updiagonalstrike}{1 \over 3^2} + \enclose{updiagonalstrike}{1 \over 3^2} - \enclose{updiagonalstrike}{1 \over 3^3} + \enclose{updiagonalstrike}{1 \over 3^3} - \dots - {1 \over 3^n} \\
= 1-{1 \over 3^n} = 2S_1;
\end{align*}
$$
S_1=\sum_{n=1}^\infty{1 \over 3^n} = \frac 12 \lim_{n \to \infty} \left( 1-{1 \over 3^n} \right) = \frac 12
$$
And thus we receive
$$
\sum_{n=1}^\infty{n \over 3^n} = \frac 14;
$$
I was trying to evaluate the second term $\sum_{n=0}^\infty {n^2 \over 3^n}$, however it wasn't a success. Could you give me a hint, how I can proceed with this series?
|
Write it as:
$$
\frac{1}{3} \sum_{n \ge 0} (n + 2) (n + 1) \left( \frac{1}{3} \right)^n
$$
This looks suspicious...
$$
\frac{\mathrm{d}^2}{\mathrm{d} z^2} \frac{1}{1 - z}
= \sum_{n \ge 0} (n + 2) (n + 1) z^n
$$
and you are all set.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/822059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
Formula of signed distance from hyperplane to point Let $H$ be a hyperplane defined by the points $p_1, p_2, ..., p_n$ and single point $x$ generally out of the hyperplane. Is there any formula to calculate the signed distance between $x$ and $H$?
I found, that following value is the orientation (i.e. its sign matches the sign of the required signed distance):
$\displaystyle
\Delta_{H}(x)=
\left\vert
\begin{array}{ccccc}
p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\
p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \\
x_{1} & x_{2} & \cdots & x_{n} & 1
\end{array}
\right\vert$
But I suspect, that $\Delta_H(x)$ have a bound with value of signed distance.
Am I right? What is the real formula of the signed distance from hyperplane to point?
|
Finally, I found the solution (in book Andrew J. Hanson: Geometry for N-Dimensional Graphics):
$
\displaystyle
V=\frac{1}{n!}
\left\vert
\begin{array}{ccccc}
p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\
p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \\
x_{1} & x_{2} & \cdots & x_{n} & 1 \notag
\end{array}
\right\vert
\\
S=\frac{1}{(n - 1)!}
\cdot
\sqrt{
\left\vert
\left (
\begin{array}{ccccc}
p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\
p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \notag
\end{array}
\right )
\cdot
\left (
\begin{array}{ccccc}
p_{n,1} & p_{2,1} & \cdots & p_{1,1} \\
p_{n,2} & p_{2,2} & \cdots & p_{1,2} \\
\vdots & \vdots & \ddots & \vdots \\
p_{n,n} & p_{2,n} & \cdots & p_{1,n} \\
1 & 1 & \cdots &1 \notag
\end{array}
\right )
\right\vert
}
\\
D=\frac{n\cdot V}{S}
$
|
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"url": "https://math.stackexchange.com/questions/822741",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$ I have started doing integration by parts:
$$\int_{0}^{2\pi}{\sin^8(x)}{dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{\sin(x)dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{d(-\cos(x))}
= \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi}
+ \int_{0}^{2\pi}{\cos(x)}{d(\sin^7(x))}
= \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi}
+ \int_{0}^{2\pi}{\cos^2(x) \cdot 7 \cdot \sin^6(x)}{dx} = \dots$$
Is there a better way to do it?
|
$$
\begin{align}
\int_0^{2\pi}\sin^8xdx
&=\int_0^{2\pi}\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^8dx\\
&=\int_0^{2\pi}2^{-8}(e^{ix}-e^{-ix})^8dx\\
&=2^{-8}\int_0^{2\pi}\sum_{k=0}^8\binom{8}{k}(e^{ix})^k(-e^{-ix})^{8-k}dx\\\
&=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{8-k}\int_0^{2\pi}e^{i(2k-8)x}dx\\
&=2^{-8}\sum_{k=0}^8\binom{8}{k}(-1)^{k}2\pi\delta_{2k-8,0}\\
&=2^{-8}\binom{8}{4}(-1)^{4}2\pi\\
&=\frac{\binom{8}{4}}{2^7}\pi
\end{align}
$$
|
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|
Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac {x^5}{5!}$ so $\frac {f(x)}{g(x)}<1$ and both are continuous and differentiable on $x\in (0,\frac{\pi}2)$ so
$$\large \frac {f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac {f'(c)}{g'(c)}\Rightarrow
\frac{\sin x -x +\frac {x^3}{3!}} {\frac {x^5}{5!}}=\frac {\cos y -1 +\frac{y^2}{2}}{\frac{y^4}{24}}$$
Such that $y\in (0,x)$.
Now what ? I need to show that $\frac {f'(y)}{g'(y)}<1$ but I tried to input $\frac{\pi} 2$ as a value to $f'/g'$ but it was larger than 1.
The other way: With Taylor expansion.
$\large \sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}+R_{5,0}(x)\Rightarrow\\
\large \sin x-x+\frac {x^3}{3!}-\frac {x^5}{5!}=R_{5,0}(x)$
Now we're left with showing that $R_{5,0}(x)<0$
So $\large R_{5,0}(x)=\frac{-\sin(c)c^7}{7!}$ for $c\in (0,x)$, now can I say that because it's negative and $(\sin(c)c^7>0)$ for all $c\in(0,x)$, $R_{5,0}(x)<0$ ?
Also, would it be correct to say that: $\large R_{5,0}(x)=\frac{cos(c)c^6}{6!}$?
|
Observe first that $f_0(x):=1-\cos x>0$ for $0<x\leqslant\pi/2$. Then
$$f_1(x):=\int_0^xf_0(t)\,\mathrm dt=x-\sin x>0$$
for $0<x\leqslant\pi/2$. Continue in this way, with
$$f_{n+1}(x):=\int_0^xf_n(t)\,\mathrm dt>0$$
for $0<x\leqslant\pi/2$ and $n=0,1,...$, and we arrive at your result at the stage $n=4$.
It is easy to generalize this procedure to show by induction that
$$(-1)^n\left(\sum_{k=0}^n(-1)^k\frac{x^{2k}}{(2k)!}-\cos x\right)>0$$
and
$$(-1)^n\left(\sum_{k=0}^n(-1)^k\frac{x^{2k+1}}{(2k+1)!}-\sin x\right)>0$$
for $0<x\leqslant\pi/2$ and $n=0,1,...$.
|
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|
Need algebra tip about $a^4 + b^4 + c^4 - 2b^2c^2 - 2a^2b^2 - 2a^2c^2$ for sides of a triangle I just got a long expression:
$$a^4 + b^4 + c^4 - 2b^2c^2 - 2a^2b^2 - 2a^2c^2$$
and I need to prove its less than zero for every $a$, $b$, and $c$ which are triangle sides
I really need tips how to handle such large expressions so they can be more useful for me.
I know that I should probably get something sort of $(a+b+c)^2$ but I cant really find a way to do it , I tried some ways but a general rule that helps would be useful ...
|
Others have mentioned it came from Heron's formula. That expresses the area of a triangle as a function of the lengths $a$, $b$, and $c$ of the three sides. Notice that since "a straight line is the shortest distance between two points", we must have $a+b\ge c$, $b+c\ge a$, and $c+a\ge b$. If $a+b$ happens to be equal to $c$, so that the distance along one side plus the distance along the next side makes the trip in as little distance as going along the third side, then the three vertices of the triangle must be on a straight line, and the area is $0$. That means an expression for the area must have $a+b-c$ as a factor. For the same reason, it has $b+c-a$ and $c+a-b$ as factors. And since the area of a triangle is $0$ when all three sides have length $0$, it also has $a+b+c$ as a factor.
But $(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ is proportional to the fourth power of the lengths, whereas an area is proportional to the second power. Hence $\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$ is proportional to the area. Application to any particular triangle whose area and side lengths are known shows that the constant of proportionality must be $1/4$. So
$$
\text{area of triangle} = \frac 1 4 \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}.
$$
That's Heron's formula. If you multiply out the polynomial under the radical, you get
$$
-a^2-b^2-c^2+2ab+2bc+2ca.
$$
Anyone who factors that polynomial instantly is probably someone who learned that by thinking about Heron's formula.
Heron's formula is also written in the form
$$
\text{area}=\sqrt{s(s-a)(s-b)(s-c)} \text{ where }s=\frac{a+b+c}{2}.
$$
The quantity $s$ is called the semiperimeter, obviously because it's half the perimeter.
|
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|
How many 10-bit strings with more 0’s than 1’s? I have to pick the answer from:
a.512
b.386
c.256
d.252
e.none of these
The number of bit strings of length 10 with n 0's (or n 1's in fact):
is C(10,n) , where C(a,b) = a! / [(a-b)!b!] is the combinitorial function.
So the first answer is C(10,3) = 120
In the second, there could be 6,7,8,9, or 10 zeros, so:
C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 210 + 120 + 45 + 10 + 1 = 386
In the third, C(10,7) + C(10,8) + C(10,9) + C(10,10)
120 + 45 + 10 + 1 = 176
In the fourth,
C(10,3) + C(10,4) + C(10,5) + C(10,6) + C(10,7) + C(10,8) + C(10,9) + C(10,10)
= 2^10 - C(10,2) - C(10,1) - C(10,0)
= 1024 - 45 - 10 - 1
= 968
So should i go with e. none of the above?
Thank you.
|
Exactly half of the strings that don't have $5$ of each. How many have $5$ of each? $\binom{10}{5}=252$ how many are there in total? $2^{10}=1024$.
Therefore the answer is $\frac{1024-252}{2}=386$
|
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|
Probability depending on x being the difference between two numbers An urn contains 10 balls, labeled 1 – 10 . A set of two balls is drawn from the urn, and let X record the absolute value of the difference. Compute Prob( X < 2 ).
My understanding is that this is without replacement and so the probability of x<2 is just the probability of x=1 plus x=0. Therefore, the probability of the difference being 0 is 10/100 and the probability of the difference being 1 is 18/100. So x<2 is just (10/100)+(18/100). Is this correct?
|
Let us draw the balls, and then look at them one at a time.
There are two possibilities: (i) The first ball is $1$ or $10$ or (ii) The first ball is $2$ to $9$.
In Case (i), the probability that the second ball is a neighbour is $\frac{1}{9}$. In Case (ii), the probability the second ball is a neighbour is $\frac{2}{9}$.
Note that Case (i) has probability $\frac{2}{10}$, and Case (ii) has probability $\frac{8}{10}$. Thus the required probability is $\frac{2}{10}\cdot \frac{1}{9}+\frac{8}{10}\cdot \frac{2}{9}$. This simplifies to $\frac{1}{5}$.
Another way: Think of the numbers as $10$ chairs in a row. There are $\binom{10}{2}$ equally likely ways to pick a set of $2$ chairs. There are $9$ ways to pick $2$ consecutive chairs. Thus our probability is $\frac{9}{\binom{10}{2}}$. This is $\frac{1}{5}$.
|
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|
Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$
\int \sqrt{\tan x} ~ \mathrm{d}{x}.
$$
|
As already mentioned in some answers, let $t^2=\tan x \implies 2tdt=\sec^2x dx\implies dx=\frac{2tdt}{t^4+1}$. Now, We can easily reach to the final answer as follows $$I=\int \frac{2t^2 dt}{t^4+1}=\int \frac{2 dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-2}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+(\sqrt{2})^2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-(\sqrt{2})^2}$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \left(\frac{\left(t+\frac{1}{t}\right)-\sqrt{2}}{\left(t+\frac{1}{t}\right)+\sqrt{2}}\right)+C$$ Now, substituting the value of $t$, we get $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$
|
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|
Let $f(x)$ denote the sum of the infinite trigonometric series, $f(x) =\sum^{\infty}_{n=1} \sin\frac{2x}{3^n}\sin\frac{x}{3^n}$ Let $f(x)$ denote the sum of the infinite trigonometric series,
$$f(x) =\sum^{\infty}_{n=1} \sin\frac{2x}{3^n} \sin\frac{x}{3^n}$$
I am not getting any clue how to solve it only thing I can put the value of $n = 1$, $2$, $3$, $\infty$. Please help thanks.
|
Hints:
$$\sin\alpha\sin\beta=\frac12\left(\cos(\alpha-\beta)-\cos(\alpha+\beta)\right)\implies$$
$$\sin\frac{2x}{3^n}\sin\frac x{3^n}=\frac12\left(\cos\frac x{3^n}-\cos\frac x{3^{n-1}}\right)\implies$$
$$\sum_{n=1}^\infty\sin\frac{2x}{3^n}\sin\frac x{3^n}=\frac12\left(\cos\frac x3-\cos x+\cos\frac x9-\cos\frac x3+\cos\frac x{27}-\cos\frac x9+\ldots\right)$$
Can you see the telescopic series...?
|
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|
Matrix Power Formula
Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$:
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$
How would we prove this? Right now we are doing work with proofs by induction... but how would I prove this?
|
Verify that it is true for $n = 1$. This is the induction basis.
Then assume the result holds for $n$, so
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\ 0 & a^n\end{pmatrix}$$
This is the induction hypothesis.
Finally, show (with the help of the induction hypothesis) that the result holds for $n + 1$, so
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} = \begin{pmatrix}a^{n+1} & (n+1)a^n\\ 0 & a^{n+1}\end{pmatrix}$$
This can be easily proved:
\begin{align}
\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} & = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}\times\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n\\
& = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} \begin{pmatrix}a^n & na^{n-1}\\ 0 & a^n\end{pmatrix}\\
& = \begin{pmatrix}a^{n+1} & na^n + a^n\\0 & a^{n + 1}\end{pmatrix}\\
& = \begin{pmatrix}a^{n+1} & (n+1)a^n\\0 & a^{n + 1}\end{pmatrix}
\end{align}
as required.
|
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|
Prove $\sum\limits_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity) Let $n$ be a nonnegative integer, and $k$ a positive integer. Could someone explain to me why the identity
$$
\sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}
$$
holds?
|
Generating function can do the job quite easily:
\begin{align*}
\frac{1}{(1-x)^k} &= \sum_{i\ge 0} \binom{i+k-1}{k-1}\, x^i
\end{align*}
Using convolution of generating functions,
\begin{align*}
\frac{1}{(1-x)}\cdot \frac{1}{(1-x)^k} &= \sum_{n\ge 0} \left(\sum_{i=0}^n \binom{i+k-1}{k-1}\right) x^n \\
\frac{1}{(1-x)^{k+1}} &= \sum_{n\ge 0} \binom{n+k}{k} x^n \\
\implies \sum_{i=0}^n \binom{i+k-1}{k-1} &= \binom{n+k}{k}
\end{align*}
|
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|
Locate my error for this initial value separable differential equation? The problem is to solve $ sin\,2x\,dx + cos\,3y\,dy = 0, \;\;\;\;y({\pi\over 2}) = {\pi\over 3}$
Here are my steps:
$$ cos \,3y \,dy = -sin \,2x \,dx $$
$$\int cos\,3y\,dy = \int -sin\,2x\,dx$$
$$ {1\over 3}\,sin\,3y = {1\over 2}\,cos\,2x\,+C $$
$$ 2\,sin\,3y = 3\,cos\,2x+C$$
Using the initial value then.. $$ 2\,sin({3\pi\over 3}) = 3\,cos({2\pi\over 2}) + C $$
$\qquad\qquad\qquad\qquad\qquad\qquad 0 = -3+C $ and it follows that $ C = 3 $$$$$Then,
$$ 2\,sin\,3y = 3\,cos\,2x+3 $$
$$ sin\,3y = {3\over 2}cos\,2x+{3\over 2} = 3({1\over 2}cos\,2x+{1\over 2})$$
Using the trig. identity, $$ sin\,3y = 3\,cos^2x$$
$$ arcsin(sin\,3y) = arcsin(3\,cos^2x) $$
$$ 3y = arcsin(3\,cos^2x)$$
$$ y = {arcsin(3\,cos^2x)\over 3} $$
However, the solution is stated as: $$ y = {\pi - arcsin(3\,cos^2x)\over 3} $$
Could someone point out what I missed?
|
The equation
\begin{align}
- \sin(2x) \ dx = \cos(3y) \ dy
\end{align}
can be seen to have the solution
\begin{align}
\frac{1}{2} \cos(2x) = \frac{1}{3} \sin(3y) + c.
\end{align}
This can be seen in the form
\begin{align}
\sin(3y) = \frac{3}{2} \left( \cos(2x) - 2 c \right)
\end{align}
or
\begin{align}
y(x) = \frac{1}{3} \sin^{-1}\left( \frac{3}{2}\left[ \cos(2x) - 2c \right]\right).
\end{align}
From the "general solutions" section of the Wiki page this can be placed in the form
\begin{align}
y(x) = \frac{1}{3}\left\{\pi - \sin^{-1}\left( \frac{3}{2}\left[ \cos(2x) - 2c \right]\right) \right\}.
\end{align}
Now for the condition $y(\pi/2) = \pi/3$ this becomes
\begin{align}
\frac{\pi}{3} =\frac{\pi}{3} - \frac{1}{3} \sin^{-1}\left( \frac{3}{2}\left[ \cos(\pi) - 2c \right]\right)
\end{align}
or $c = -1/2$. It can now be seen that
\begin{align}
y(x) = \frac{1}{3}\left\{\pi - \sin^{-1}\left(3 \cos^{2}(x)\right) \right\}.
\end{align}
|
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|
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
|
Using Taylor expansion:
$$\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$$
$$\sqrt{x + \sqrt{x}+\dfrac{1}{2} + O(\frac{1}{\sqrt{x}})} - \sqrt{x}$$
$$\sqrt{x}+\dfrac{1}{2\sqrt{x}}(\sqrt{x}+\dfrac{1}{2}+ O(\frac{1}{\sqrt{x}})) + O(\frac{1}{\sqrt{x}})- \sqrt{x}$$
$$\dfrac{1}{2} + \dfrac{1}{4\sqrt{x}}$$
Which in the limit case tends to ${\dfrac{1}{2}}$.
The expansion I used was $(a+b)^{1/2} = \sqrt{a}+\frac{b}{2\sqrt{a}} + \cdots$
|
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|
Stuck on Indefinite Integral Please help me. I have been stuck on this for ages :(
$$\int \frac{1}{13\cos x+ 12}\,\mathrm{d}x$$
I appreciate any and all help. Thank you.
|
$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{1}{13\cos x+12}dx = \int\frac{1}{13(1+\cos x)-1}dx$$
Now Using $$\displaystyle 1+\cos x = 2\cos^2 \frac{x}{2}\;,$$ we get
$$\displaystyle \int\frac{1}{26\cos^2 \frac{x}{2}-1}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\displaystyle \cos^2 \frac{x}{2}$
$$\displaystyle \int\frac{\sec^2 \frac{x}{2}}{26-1-\tan^2 \frac{x}{2}}dx = \int\frac{\sec^2 \frac{x}{2}}{5^2-\tan^2\frac{x}{2}}dx$$
Now Let $\displaystyle \tan \frac{x}{2} = t\;,$ Then $\displaystyle \sec^2\frac{x}{2}dx = 2dt$
So Integral is $$\displaystyle 2\int\frac{1}{5^2-t^2}dt = \frac{1}{10}\ln \left|\frac{5+t}{5-t}\right|+\mathbb{C}$$
Where $\displaystyle t = \cos^2 \frac{x}{2}$
|
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|
What is the probability that the difference of squares of two positive integers up to $30$ is divisible by $3$ or $7$? If we choose any two numbers $a$ and $b$ from the integers $1$ through $30$, what is the probability of $a^2-b^2$ of being divisible by $3$ or $7$?
|
Note that $a^2-b^2$ is divisible by $c$ if and only if
$$a\equiv b\equiv0\pmod{c}\qquad\text{ or }\qquad a\equiv\pm b\not\equiv0\pmod{c}.$$
Taking $c=3$ we find that the sum of the probabilities of these cases is
$$\left(\frac{10}{30}\right)^2+\left(\frac{20}{30}\right)^2=\frac{500}{900}.$$
For $c=7$ there are $9$ integers congruent to $\pm1\pmod{7}$ and $9$ integers congruent to $\pm2\pmod{7}$, but only $8$ integers congruent to $\pm3\pmod{7}$, so the sum becomes
$$\left(\frac{4}{30}\right)^2+2\cdot\left(\frac{9}{30}\right)^2+\left(\frac{8}{30}\right)^2=\frac{242}{900}.$$
Similarly the probability that $a^2-b^2$ is divisible by $21$ equals
$$\left(\frac{1}{30}\right)^2+9\cdot\left(\frac{3}{30}\right)^2+\left(\frac{2}{30}\right)^2=\frac{85}{900}.$$
So the probability that $a^2-b^2$ is divisible by either $3$ or $7$ equals
$$\frac{500}{900}+\frac{242}{900}-\frac{85}{900}=\frac{662}{900}=\frac{331}{450}.$$
|
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|
Help understanding this approximation In a paper that I'm reading, the authors write:-
$$N_e \approx \frac{3}{4} (e^{-y}+y)-1.04. \tag{4.31}$$
Now, an analytic approximation can be obtained by using the expansion
with respect to the inverse number of "e-foldings" ($N_e$ is the
number of "e-foldings"). For instance, eq. $(4.31)$ yields:-
$$e^y = \dfrac{3}{4N_e} - \dfrac{9\ln(N_e)}{16(N_e)^2} -
\dfrac{0.94}{(N_e)^2} + O\left(\dfrac{\ln^2(N_e)}{(N_e)^3}\right)$$
Can anyone tell me how this approximation is done? I'm not familiar with the $O$ notation either. What does it mean? How do the authors arrive at that expression?
If anyone should require it, the original paper can be found here: https://arxiv.org/pdf/1001.5118.pdf?origin=publication_detail
|
The assumption seems to be that $y \ll -1$ and $N_\epsilon \gg 1$. For brevity let's set $x = e^y$ and $N_\epsilon = N$ so that the equation becomes
$$
N = \frac{3}{4 x} + \frac{3}{4}\ln x - 1.04. \tag{1}
$$
The condition $y \ll -1$ now corresponds to $0 < x \ll 1$ for the new variable. In this case the term involving $1/x$ is dominant on the right-hand side, and since the other two terms are negative we have
$$
N < \frac{3}{4 x} \quad \Longleftrightarrow \quad 0 < x < \frac{3}{4 N}.
$$
This is our initial estimate. Since $N \gg 1$ this implies that
$$
\frac{3}{4}\ln x - 1.04 = O(\ln N),
$$
and hence from $(1)$ that
$$
\frac{3}{4x} = N + O(\ln N).
$$
Solving for $x$ yields
$$
\begin{align}
x &= \frac{3}{4(N+O(\ln N))} \\
&= \frac{3}{4N} \cdot \frac{1}{1+O\left(\frac{\ln N}{N}\right)} \\
&= \frac{3}{4N} \left(1+O\left(\frac{\ln N}{N}\right)\right) \tag{2.1} \\
&= \frac{3}{4N} + O\left(\frac{\ln N}{N^2}\right). \tag{2.2}
\end{align}
$$
This matches the first term of the desired estimate. Achieving more terms is simply a matter of repeating this process.
Substituting $(2.1)$ for $x$ into the sub-dominant terms of $(1)$, namely $\frac{3}{4}\ln x - 1.04$, yields
$$
\begin{align}
\frac{3}{4}\ln x - 1.04 &= \frac{3}{4} \ln\left[\frac{3}{4N} \left(1+O\left(\frac{\ln N}{N}\right)\right)\right] - 1.04 \\
&= \frac{3}{4}\ln \frac{3}{4} - \frac{3}{4}\ln N + \frac{3}{4}\ln\left(1+O\left(\frac{\ln N}{N}\right)\right) - 1.04 \\
&= -\frac{3}{4}\ln N + \frac{3}{4}\ln \frac{3}{4} - 1.04 + O\left(\frac{\ln N}{N}\right),
\end{align}
$$
so that $(1)$ becomes
$$
N = \frac{3}{4x} - \frac{3}{4}\ln N + \frac{3}{4}\ln \frac{3}{4} - 1.04 + O\left(\frac{\ln N}{N}\right).
$$
Solving for $x$ as above (though with a bit more algebra this time) yields the approximation
$$
x = \frac{3}{4N} - \frac{9}{16} \frac{\ln N}{N^2} + \frac{C}{N^2} + \frac{27}{64} \frac{(\ln N)^2}{N^3} + O\left(\frac{\ln N}{N^3}\right), \tag{3}
$$
where
$$
C = \frac{9}{16} \ln \frac{3}{4} - \frac{3}{4} \cdot 1.04 \approx -0.94.
$$
Using the original variables $x = e^y$ and $N = N_\epsilon$, $(3)$ is
$$
e^y = \frac{3}{4N_\epsilon} - \frac{9}{16} \frac{\ln N_\epsilon}{N_\epsilon^2} + \frac{C}{N_\epsilon^2} + \frac{27}{64} \frac{(\ln N_\epsilon)^2}{N_\epsilon^3} + O\left(\frac{\ln N_\epsilon}{N_\epsilon^3}\right), \tag{4}
$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Solutions of a system of linear equations with a parameter I've come across a linear algebra problem that I'm not sure how to solve. It's a generic problem - I have to find the solutions of a system of equations dependent on a parameter. So, my first though was to check for which values of the parameter the matrix on the left is not invertible (1 and -2). I think that for these values there are no solutions.
Is that enough, or should I check the rank of the augmented matrix and use the Rouché–Capelli theorem (for which values of the parameter the planes meet at a point, a line or a plane)?
Thanks,
$$\begin{pmatrix}
\lambda & 1 & 1\\
1 & \lambda & 1\\
1 & 1 & \lambda
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}=
\begin{pmatrix}
1 \\ \lambda \\ \lambda^2
\end{pmatrix}
$$
|
$$\left(\begin{array}{ccc|c}
\lambda & 1 & 1 & 1 \\
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^2
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
\lambda+2 & \lambda+2 & \lambda+2 & 1+\lambda+\lambda^2 \\
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^2
\end{array}\right)\overset{(1)}\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
1 & \lambda & 1 & \lambda \\
1 & 1 & \lambda & \lambda^2
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
0 & \lambda-1 & 0 & \frac{\lambda-1}{\lambda+2} \\
0 & 0 & \lambda-1 & \frac{\lambda^3+\lambda^2-\lambda-1}{\lambda+2}
\end{array}\right)\overset{(2)}\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
0 & 1 & 0 & \frac{1}{\lambda+2} \\
0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & \frac{-\lambda-1}{\lambda+2} \\
0 & 1 & 0 & \frac{1}{\lambda+2} \\
0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2}
\end{array}\right)\sim
$$
The step (1) is only valid for $\lambda\ne -2$.
The step (2) is only valid for $\lambda\ne 1$.
For values $\lambda\ne1,-2$ we get that there is only one solution $x_1=\frac{-\lambda-1}{\lambda+2}$, $x_2=\frac{1}{\lambda+2}$, $x_3=\frac{(\lambda+1)^2}{\lambda+2}$.
We can do the sanity check.
$\lambda x_1 + x_2 + x_3 = \frac{\lambda(-\lambda-1)+1+(\lambda+1)^2}{\lambda+2} = \frac{-\lambda^2-\lambda+1+\lambda^2+2\lambda+1}{\lambda+2} = \frac{\lambda+2}{\lambda+2} = 1$
$x_1 + \lambda x_2 + x_3 = \frac{(-\lambda-1)+\lambda+(\lambda+1)^2}{\lambda+2} = \frac{\lambda^2+2\lambda}{\lambda+2} = \lambda$
$x_1 + x_2 + \lambda x_3 = \frac{(-\lambda-1)+1+\lambda(\lambda+1)^2}{\lambda+2} = \frac{-\lambda+\lambda^3+2\lambda^2+\lambda}{\lambda+2} = \frac{\lambda^3+2\lambda^2}{\lambda+2}=\lambda^2$
Notice that the check works for $\lambda=1$. So we know that there is at least one solution for $\lambda=1$.
For $\lambda=1$
We get the system
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1
\end{array}\right)$$
which has solutions $x_1=1-s-t$, $x_2=s$, $x_3=t$ (where $s$, $t$ are arbitrary).
For $\lambda=-2$
$\left(\begin{array}{ccc|c}
-2 & 1 & 1 & 1 \\
1 & -2 & 1 & -2 \\
1 & 1 & -2 & 4
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
-2 & 1 & 1 & 1 \\
1 & -2 & 1 & -2 \\
0 & 0 & 0 & 3
\end{array}\right)$.
So in this case there is no solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving that $\gamma = \int_{0}^{1} \!\!\int_{0}^{1} \!\frac{x - 1}{(1 - x y) \log(x y)} \, \mathrm{d}{x} \, \mathrm{d}{y} $. In 2005, J. Sondow found a surprising formula for the Euler-Mascheroni constant $ \gamma $. The formula is
$$
\gamma
= \int_{0}^{1} \int_{0}^{1} \frac{x - 1}{(1 - x y) \log(x y)}
~ \mathrm{d}{x} ~ \mathrm{d}{y}.
$$
Now, the definition of $ \gamma $ is
$$
\gamma
\stackrel{\text{def}}{=}
\lim_{n \to \infty} \left[ \sum_{k = 1}^{n} \frac{1}{k} - \log(n) \right].
$$
I have tried using the geometric series
$$
\frac{1}{1 - x y} = \sum_{n = 0}^{\infty} x^{n} y^{n}
$$
to obtain a proof, but it would not work. Thanks for any help.
|
We want use the transformation formula.
\begin{align*}
&U:= \left\lbrace (t,r) \mid t \in (0,1), r \in (t,1) \right\rbrace , V:= (0,1)^2,
\\ &f: V \rightarrow \mathbb{R}, \left( \begin{array}{r}
x\\
y\\
\end{array} \right) \mapsto \frac{x - 1}{(1-xy) \cdot ln(xy)},
\\ &\varphi: U \rightarrow V, \left( \begin{array}{r}
t\\
r\\
\end{array}\right) \mapsto \left( \begin{array}{r}
t/r\\
r\\
\end{array} \right),
\end{align*}
\begin{align*}
\varphi ^{-1}: V \rightarrow U, \left( \begin{array}{r}
t\\
r\\
\end{array}\right) \mapsto \left( \begin{array}{r}
t \cdot r\\
r\\
\end{array} \right)
\end{align*}
\begin{align*}
D\varphi = \left( \begin{array}{rr}
1/r & -t/r^2\\
0 & 1\\
\end{array}\right),
Det(D\varphi) = \frac{1}{r}.
\end{align*}
So we can use the transformation.
\begin{align*}
\int_0^1 \int_0^1 \frac{x-1}{(1-xy)\cdot ln(xy)} dx \, dy &= \int_0^1 \int_t^1 \frac{(t/r) - 1}{(1-t) \cdot ln(t)} \cdot \frac{1}{r} \, dr \, dt
\\ &= \int_0^1 \int_t^1 \left( \frac{t}{(1-t) \cdot ln(t)} \cdot \frac{1}{r^2} - \frac{1}{r} \cdot \frac{1}{(1-t) \cdot ln(t)} \right) \, dr \, dt
\\ &= \int_0^1 \left[ -\frac{1}{r} \cdot \frac{t}{(1-t) \cdot ln(t)} - \frac{ln(r)}{(1-t)\cdot ln(t)} \right]_t^1 \, dt
\\ &= \int_0^1 \left( \frac{1-t}{(1-t)\cdot ln(t)} + \frac{ln(t)}{(1-t) \cdot ln(t)} \right) \, dt
\\ &= \int_0^1 \left( \frac{1}{ln(t)} + \frac{1}{1-t} \right) \, dt
\end{align*}
The last integral represented $\gamma$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $x^8 + x^5 + x^3 + x^2 + 1$ an irreducible polynomial or not in GF $2^8$ Is $x^8 + x^5 + x^3 + x^2 + 1$ an irreducible polynomial or not in Galois Field $2^8$? Thanks in advance.
|
The question is a bit unclear. If you are truly asking whether this polynomial
$$
p(x)=x^8+x^5+x^3+x^2+1
$$ is irreducible in the ring $GF(256)[x]$, then the answer is trivially NO! This is because:
*
*Either it is reducible over $GF(2)$ in which case it is reducible over the extension field also,
*Or it is irreducible over $GF(2)$ in which case $GF(2)[x]/\langle p(x)\rangle$ is a field of 256 elements, and thus isomorphic (or equal to) $GF(256)$. In this case $p(x)$ has 8 distinct zeros in the field, namely the cosets $x^{2^i}+\langle p(x)\rangle$, $i=0,1,2,\ldots,7$. Consequently $p(x)$ would split into linear factors over $GF(256)$.
But this was a bit naughty of me, because I took your phrasing of the question literally. Presumably you would not ask things like: "Is $x^2+1$ irreducible over the complex numbers?", because you know that it factors as $(x+i)(x-i)$ in $\Bbb{C}[x]$. The only sensible guess is that you are inquiring whether $p(x)$ is irreducible over $GF(2)$.
As pointed out in the comments you need to show that $p(x)$ has no factors of degree 4 or lower. The usual suspects are $x,x+1,x^2+x+1,x^3+x+1,x^3+x^2+1,x^4+x+1,x^4+x^3+1$ and
$x^4+x^3+x^2+x+1$. If you were given this as an exercise, your teacher/book must have produced this at some earlier point (otherwise it would a bit naughty of them - barring introduction of a general algorithm like Berlekamp's).
*
*Because $p(x)$ has five terms including $1$ we can immediately rule out the linear factors.
*The quadratic factor is a factor of $x^3+1$. But $x^8+x^5=(x^3+1)x^5$, so
$$p(x)\equiv x^2\pmod{x^2+x+1}.$$ Thus no irreducible quadratic factor.
*Both the cubic irreducible polynomials are factors of $x^7+1$ (the multiplicative group of the field $GF(8)$ is cyclic of order seven). We see that $x^8\equiv x\pmod{x^7+1}$, so $p(x)\equiv x^5+x^3+x^2+x+1\pmod {x^7+1}$. If this were divisible by $x^3+x+1$ then so would $x^5+x^2=x^2(x^3+1)=x^2(x+1)(x^2+x+1)$ which is absurd. Likewise, if it were divisible by $x^3+x^2+1$, then so would $x^5+x=x(x^4+1)=x(x+1)^4$ again violating uniqueness of factorization.
This leaves the possibility that $p(x)$ could be a product of two irreducible quartics, i.e. a product of two of $f_1(x)=x^4+x+1$, $f_2(x)=x^4+x^3+1$, $f_3(x)=x^4+x^3+x^2+x+1$.
Could $f_1(x)$ be a factor? We have $f_1(x)^2=x^8+x^2+1$, so
$$
p(x)\equiv x^5+x^3\pmod{f_1(x)}.
$$
As in the cubic case this leads to a contradiction as $x^5+x^3=x^3(x+1)^2$ is not divisible by $f_1(x)$.
Because $p(x)$ contains odd degree terms, it is not a square of a polynomial. The only remaining possibility is
$$
p(x)=f_2(x)f_3(x).
$$
Leaving it to you to rule out this possibility. Either by brute force or by using one of the techniques outlined here.
|
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|
How to solve the recurrence $T(n) = T(n-1) +\sqrt{n}$? While solving the recurrence of the title I come to the series
$$T(n) = \sqrt1 + \sqrt2 + \sqrt{3} + \cdots + \sqrt n.$$
Please somebody help me how to solve this.
|
We can use left-hand and right-hand approximations of integrals via Riemann sums to obtain tight bounds on our desired sum. Since $f(x) = \sqrt{x}$ is a strictly increasing function, observe that:
\begin{align*}
\int_0^n \sqrt{x} \, dx \leq \sum_{k=1}^n \sqrt{k} &\leq \int_1^{n+1} \sqrt{x} \, dx \\
\left[ \frac{x^{3/2}}{3/2} \right]_0^n \leq T(n) &\leq \left[ \frac{x^{3/2}}{3/2} \right]_1^{n+1} \\
\frac{2}{3}n^{3/2} \leq T(n) &\leq \frac{2}{3}[(n+1)^{3/2} - 1] \\
\end{align*}
Thus, we conclude that $T(n) = \Theta(n^{3/2})$.
|
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|
Range of the function $f(x) = \frac{x^2+14x+9}{x^2+2x+3}\;,$ where $x\in \mathbb{R}$ Calculation of Range of the function $\displaystyle f(x) = \frac{x^2+14x+9}{x^2+2x+3}\;,$ where $x\in \mathbb{R}$
(Can we solve it Using $\bf{A.M\geq G.M}$) Inequality.
$\bf{My\; Try::}$ Let $\displaystyle y = f(x) = \frac{x^2+14x+9}{x^2+2x+3} = \frac{(x+1)^2+12(x+1)-4}{(x+1)^2+1}$
Now Let $(x+1) = t\;,$ where $t\in \mathbb{R}$
So $\displaystyle y=\frac{t^2+12t-9}{t^2+1} = 1+\frac{12t-10}{t^2+1}$
Now I did not understand How can I solve after that
Help me
Thanks
|
HINT:
Let $\displaystyle y=\frac{x^2+14x+9}{x^2+2x+3}$
Rearrange to form a Quadratic Equation in $x$
As $x$ is real, the discriminant must be $\ge0$
|
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|
Working with groups. Finding the inverse of some $S_9$ I want to compute the inverse of:
$\begin{pmatrix} 1&2&3&4&5&6&7&8&9\\3&2&1&6&5&9&4&8&7 \end{pmatrix}$
Sorry about alignment(they are all one digit though)
Now it would seem that the inverse here is itself. But I feel like maybe I am just confusing myself now, and I want some confirmation. Thank you and sorry again.
Also can someone explain why my matrix above isn't formatting properly despite having two & between each element
|
Each element in $S_9$ is a permutation (or bijection) $\{1, 2, \dots, 9\} \to \{1, 2, \dots, 9\}$. In particular, your permutation I can denote as a function $f$, where:
$$\begin{align}
f(1) &= 3\\
f(2) &= 2 \\
f(3) &= 1 \\
f(4) &= 6 \\
f(5) &= 5 \\
f(6) &= 9 \\
f(7) &= 4 \\
f(8) &= 8 \\
f(9) &= 7\text{.}
\end{align}$$
To compute the inverse of your permutation, notice that the inverse of $f$ exists because it is a bijection, and thus
$$\begin{align}
f^{-1}(3) &= 1\\
f^{-1}(2) &= 2 \\
f^{-1}(1) &= 3 \\
f^{-1}(6) &= 4 \\
f^{-1}(5) &= 5 \\
f^{-1}(9) &= 6 \\
f^{-1}(4) &= 7 \\
f^{-1}(8) &= 8 \\
f^{-1}(7) &= 9\text{.}
\end{align}$$
Hence $$\begin{pmatrix} 1&2&3&4&5&6&7&8&9\\3&2&1&6&5&9&4&8&7 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5& 6 & 7 & 8 & 9 \\ 3 & 2 & 1 & 7 & 5 & 4 & 9 & 8 & 6\end{pmatrix}\text{.}$$
|
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|
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want.
|
Using repeated differences we get
$$
\begin{array}{llll}
1 & 28 & 153 & 496 & 1225 & 2556 & \\
27 & 125 & 343 & 729 & 1331 & \\
98 & 218 & 386 & 602 & \\
120 & 168 & 216 & \\
48 & 48 & \\
0 & \\
\end{array}
$$
Newton's interpolation formula then gives us
$$
1 \binom{n-1}{0} + 27 \binom{n-1}{1} + 98 \binom{n-1}{2} + 120 \binom{n-1}{3} + 48\binom{n-1}{4}
=
2 n^4-n^2
$$
This is a technique well worth knowing. (The simplification was done with WA.)
|
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|
Evaluation of $ \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$ Evaluation of $\displaystyle \int\frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$
$\bf{My \; Try::}$ Let $\displaystyle I = \int \frac{\sin (x+\alpha)}{\cos^3 x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx$
So $\displaystyle I = \int\frac{\sin (x+\alpha)}{\cos x}\cdot \sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}dx\cdot \sec^2 x$
$\displaystyle I = \int \frac{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha}{\cos x}\cdot \sqrt{\frac{\cos +\sin x}{\cos x-\sin x}}\cdot \sec^2 x dx$
$\displaystyle I = \int\left(\cos \alpha\cdot \tan x+\sin \alpha\right)\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$
Now Let $\tan x= t\;,$ Then $\sec^2 xdx = dt$
$\displaystyle I = \int (\cos \alpha\cdot t + \sin \alpha)\sqrt{\frac{1+t}{1-t}}dt$
Now How can I solve after that
Help me
Thanks
|
Assuming what you have done up till the last part is correct, these hints I think will help you go further:
(1) Multiplying and dividing by $\sqrt{1+t}$
(2) Knowing the derivative of $\sinh^{-1} (\sqrt{t-1}) $ and $\sqrt{t+1}\sqrt{t-1}$
(3) Using partial integration
|
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|
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
|
Here is another nice proof.
We have to prove;
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}\\\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\geq \frac{3}{2}+3\\(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq \frac{9}{2}\\([b+c]+[a+c]+[a+b])\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9\\$$Now we use this well known inequality for $x>0\to x+\frac{1}{x}\geq2$;
(quick proof: $\left (\sqrt x-\frac{1}{\sqrt x}\right )^2\geq0\to x+\frac1x\geq2$)
$$1+1+1+\left[\frac{a+b}{a+c}+\frac{a+c}{a+b}\right]+\left[\frac{a+b}{b+c}+\frac{b+c}{a+b}\right]+\left[\frac{a+c}{b+c}+\frac{b+c}{a+c}\right]\geq 9\\q.e.d.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Proof of $\sin2x+x\sin^2x \lt\dfrac{1}{4}x^2+2$ How can be proven the following inequality?
$$\forall{x\in\mathbb{R}},\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$
Thanks
|
$$\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$
$$\sin(x)(2\cos(x)+x\sin(x))\lt\dfrac{1}{4}x^2+2$$
while the right side is positive for all $x \in \mathbb{R}$ it is enoght that we show:
$$|\sin(x)|\cdot|(2\cos(x)+x\sin(x))|\lt\dfrac{1}{4}x^2+2$$
by the Cauchy–Schwarz inequality we have:
$$|\sin(x)|\cdot\sqrt{|4+x^2|.|\sin(x)^2+\cos(x)^2|}\lt\dfrac{x^2+4}{4}+1$$
$$|\sin(x)|\lt\dfrac{\sqrt{|4+x^2|}}{4}+\dfrac{1}{\sqrt{|4+x^2|}}$$
and the second side is greater than 1 by arithmetic and geometric inequality and equality.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
|
Rewriting, we have $3\cdot2^{n+1}=(n+1)(4+2^n)$
First, note that whenever $n\geq 5$, we have $(n+1)(4+2^n)>6\cdot 2^n=3\cdot2^{n+1}$
Next note that whenever $n<-1$, $(n+1)(4+2^n)<0<3\cdot2^{n+1}$
Note that $n=-1$ isn't possible because in the first equation we would divide by $0$.
Hence any solution must have $n\in[0,4]$
Now we just check our $5$ cases and find that the only solutions are $n=1$, $n=2$, and $n=3$.
|
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|
Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ Question: Prove that if
$$a_n=\left\{
\begin{array}{ll}
a_1=1\\
a_{n+1}=1+\frac{1}{1+a_n}
\end{array}
\right.$$
then $a_n$ converges, and then find $\lim_{n \to \infty}a_n$.
I found that $\lim_{n \to \infty}a_n=\sqrt{2}$, and that $1\leq a_n<2$, but the convergence seems difficult.
My idea is to show that $a_{2n}$ is decreasing, and $a_{2n-1}$ is increasing, but I don’t see how.
I’d be thankful for any hint.
|
Given
$$
a_{n+1} = 1 + \frac{ 1 }{ 1 + a_n }.
$$
Which can be written as
$$
a_{n+1} = \frac{ 2 + a_n }{ 1 + a_n }.
$$
Write
$$
a_n = \frac{ P_n }{ Q_n },
$$
then
$$
a_{n+1} = \frac{ 2 + a_n }{ 1 + a_n } =
\frac{ \displaystyle 2 + \frac{ P_n }{ Q_n } }{ \displaystyle 1 + \frac{ P_n }{ Q_n } }
= \frac{ P_n + 2Q_n }{ P_n + Q_n }
= \frac{ P_{n+1} }{ Q_{n+1} }.
$$
Thus we have the recursion
$$
\begin{eqnarray}
P_{n+1} &=& P_n + 2 Q_n,\\
Q_{n+1} &=& P_n + Q_n.
\end{eqnarray}
$$
Note that
$$
\begin{eqnarray}
P_{n+1} + \sqrt{2} Q_{n+1} &=& \Big( 1 + \sqrt{2} \Big)
\Big( P_{n} + \sqrt{2} Q_{n} \Big),\\
P_{n+1} - \sqrt{2} Q_{n+1} &=& \Big( 1 - \sqrt{2} \Big)
\Big( P_{n} - \sqrt{2} Q_{n} \Big),\\
\end{eqnarray}
$$
and as $P_1=1$ and $Q_1=1$, we obtain
$$
\begin{eqnarray}
P_{n} + \sqrt{2} Q_{n} &=& \Big( 1 + \sqrt{2} \Big)^n,\\
P_{n} - \sqrt{2} Q_{n} &=& \Big( 1 - \sqrt{2} \Big)^n,
\end{eqnarray}
$$
whence
\begin{eqnarray}
P_{n} &=& \frac{\Big( 1 + \sqrt{2} \Big)^n + \Big( 1 - \sqrt{2} \Big)^n}{2},\\
Q_{n} &=& \frac{\Big( 1 + \sqrt{2} \Big)^n - \Big( 1 - \sqrt{2} \Big)^n}{2\sqrt{2}},
\end{eqnarray}
therefore
$$
a_n = \frac{\Big( 1 + \sqrt{2} \Big)^n + \Big( 1 - \sqrt{2} \Big)^n}
{\Big( 1 + \sqrt{2} \Big)^n - \Big( 1 - \sqrt{2} \Big)^n} \sqrt{2}.
$$
Note that
$$
\begin{eqnarray}
a_1 &=& \frac{\Big( 1 + \sqrt{2} \Big) + \Big( 1 - \sqrt{2} \Big)}
{\Big( 1 + \sqrt{2} \Big) - \Big( 1 - \sqrt{2} \Big)} \sqrt{2} = 1,\\
a_2 &=& \frac{\Big( 1 + \sqrt{2} \Big)^2 + \Big( 1 - \sqrt{2} \Big)^2}
{\Big( 1 + \sqrt{2} \Big)^2 - \Big( 1 - \sqrt{2} \Big)^2} \sqrt{2} = \frac{3}{2},\\
a_3 &=& \frac{\Big( 1 + \sqrt{2} \Big)^3 + \Big( 1 - \sqrt{2} \Big)^3}
{\Big( 1 + \sqrt{2} \Big)^3 - \Big( 1 - \sqrt{2} \Big)^3} \sqrt{2} = \frac{7}{5},\\
\vdots
\end{eqnarray}
$$
And we obtain
$$
\begin{eqnarray}
\lim_{n \rightarrow \infty} a_n &=& \lim_{n \rightarrow \infty}
\frac{\Big( 1 + \sqrt{2} \Big)^n + \Big( 1 - \sqrt{2} \Big)^n}
{\Big( 1 + \sqrt{2} \Big)^n - \Big( 1 - \sqrt{2} \Big)^n} \sqrt{2}\\
&=& \lim_{n \rightarrow \infty}
\frac{ 1 + \Big( \displaystyle \frac{1 - \sqrt{2}}{1 + \sqrt{2}} \Big)^n}
{\ 1 - \Big( \displaystyle \frac{1 - \sqrt{2}}{1 + \sqrt{2}} \Big)^n} \sqrt{2}\\
&=& \sqrt{2}
\end{eqnarray}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Fair Coin Flips
Michele flips a fair coin nonstop. Two students, Thomas and George,
decide to
make a bet about whose sequence of flips will occur first from the moment they
begin observing the results of Michele's flips. Thomas picks the sequence 'HTT'.
Find, with proof, a sequence George can pick that gives him an edge over Thomas
in their bet.
I set up a diagram for the different cases and used it to compute the probabilities of all of the cases.
The sequence we pick has to be of length $3$. Otherwise, this question is trivial.
|
First, let us see the expected number of tosses required for each sequence of three:
\begin{align*}
\begin{array}{|c|c|}\hline
\mathrm{HHH}, \mathrm{TTT} & 14\\
\mathrm{HHT}, \mathrm{THH}, \mathrm{TTH}, \mathrm{HTT} & 8\\
\mathrm{HTH}, \mathrm{THT} & 10 \\ \hline
\end{array}
\end{align*}
Since $\mathrm{HTT}$ was chosen, let's pick $\mathrm{HHT}$ to compare the probability of getting one before the other.
We can use the following absorbing markov chain to find that:
\begin{align*}
\left(\begin{array}{rrrrrrrr}
& \mathrm{I} & \mathrm{H} & \mathrm{T} & \mathrm{HT} & \mathrm{HH} & \mathrm{HTT} & \mathrm{HHT}\\\\
\mathrm{I} & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 & 0 & 0 \\\\
\mathrm{H} & 0 & 0 & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 \\\\
\mathrm{T} & 0 & \dfrac{1}{2} & \dfrac{1}{2} & 0 & 0 & 0 & 0 \\\\
\mathrm{HT} & 0 & \dfrac{1}{2} & 0 & 0 & 0 & \dfrac{1}{2} & 0 \\\\
\mathrm{HH} & 0 & 0 & 0 & 0 & \dfrac{1}{2} & 0 & \dfrac{1}{2} \\\\
\mathrm{HTT} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
\mathrm{HHT} & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
\end{align*}
'I' is the initial state.
From the matrix, we find the probability of absorption from non-absorbing states to be:
\begin{align*}
\left(\begin{array}{rr}
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{1}{3} & \dfrac{2}{3} \\\\
\dfrac{2}{3} & \dfrac{1}{3} \\\\
0 & 1
\end{array}\right)
\end{align*}
Hence, we see that from the initial state, the probability of getting $\mathrm{HHT}$ first is $\dfrac{2}{3}$
Also, see IV. 6.3. p. 271 in Analytic Combinatorics, which has an example about patterns.
Update
We can calculate the probabilities by solving the following set of equations, after tossing one head:
\begin{align*}
p_h &= \frac{1}{2}\left(p_{hh}+p_{ht}\right) \\
p_{hh} &= \frac{1}{2}\left(p_{hh}+p_{hht}\right) \\
p_{ht} &= \frac{1}{2}\left(p_{h}+p_{htt}\right) \\
p_{hht} &= 0 \\
p_{htt} &= 1
\end{align*}
which gives the probability of ${\rm HTT}$ occuring first, and is same as calculated from the markov chain.
If we solve the above equations by taking $p_{htt}=0$ and $p_{hht}=1$, we get the probability of ${\rm HHT}$ occuring first.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Is $7^{8}+8^{9}+9^{7}+1$ a prime? (no computer usage allowed)
Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.
I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.
|
That $\,47\,$ is a factor can be verified by very easy mental arithmetic:
$\begin{eqnarray}
{\rm mod}\ 47\!:\quad && 1+ \color{#c00}{7^8} +\, 8^9 +\ \color{#0a0}{9^7}\\
\equiv && 1+ \color{#c00}{2^4} + 2^{27} + \color{#0a0}{7\cdot 7^6\cdot 8^7}\quad {\rm by}\ \ \color{#0a0}{9\equiv 7\cdot 8},\ \ \color{#c00}{7^2\equiv 2}\\
\equiv&& 1+ 2^4 + 2^{27} + 7\cdot \color{#c00}{2^3}\cdot 2^{21}\\
\equiv&& 1+ 2^4 + 2^{4} +\ 7\cdot 2\,\equiv\, 0\quad {\rm by}\ \ \color{#c00}2^{23}\equiv (\color{#c00}{7^2})^{23}\equiv 7^{46}\equiv 1,\ \ \rm by\ little \ Fermat\\
\end{eqnarray}$
|
{
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"url": "https://math.stackexchange.com/questions/868663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Definite trigonometric integral This question is motivated by Iterative Mean, Covariance Algorithm Convergence:
Is there a closed form for the integral
$$
\int_0^{2 \pi} \frac{\sin^n(\theta)\cos^m(\theta)}{A\sin^2(\theta)+B\cos^2(\theta)+2C\sin(\theta)\cos(\theta)} d\theta\,?
$$
where $AB \geq C^2$ and $m,n \in \{0,1,2,3,4\}$
|
Dividing the numerator and denominator by $\cos^2 \theta$ we get
$$I = \int_0^{2\pi } {\frac{{{{\sin }^n}\theta {{\cos }^{m - 2}}\theta }}{{A{{\tan }^2}\theta + 2C\tan \theta + B}}d\theta }.$$
Now let us solve the equation $$A{{\tan }^2}\theta + 2C\tan \theta + B=0$$ for simplicity we let $u=\tan \theta$, so that the equation become $A{u^2} + 2Cu + B = 0$. Using the general law we get
\begin{align}
{u_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} &= \frac{{ - 2C \pm \sqrt {4{C^2} - 4AB} }}{{2A}}
\\
&= \frac{{ - C \pm \sqrt {{C^2} - AB} }}{A}
\\
&= \frac{{ - C \pm i\sqrt {AB - {C^2}} }}{A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{since} \,\,\,AB \ge {C^2}} \right)
\end{align}
Thus
$${A{{\tan }^2}\theta + 2C\tan \theta + B}
=\left( {\tan \theta - \frac{{-C + i\sqrt {AB - {C^2}} }}{A}} \right)\left( {\tan \theta - \frac{{-C - i\sqrt {AB - {C^2}} }}{A}} \right)$$
Substituting
$$\cos \theta = \frac{{z + {z^{ - 1}}}}{2},\sin \theta = \frac{{z - {z^{ - 1}}}}{{2i}},\tan \theta = - i\frac{{z - {z^{ - 1}}}}{{z + {z^{ - 1}}}},d\theta = \frac{{dz}}{{iz}}$$
in $I$ and simplify to get
\begin{align}
I&=\int_C {\frac{{{{\left( {\frac{{z - {z^{ - 1}}}}{{2i}}} \right)}^n}{{\left( {\frac{{z + {z^{ - 1}}}}{2}} \right)}^{m - 2}}}}{{\left( { - i\frac{{z - {z^{ - 1}}}}{{z + {z^{ - 1}}}} - \frac{{ - C + i\sqrt {AB - {C^2}} }}{A}} \right)\left( { - i\frac{{z - {z^{ - 1}}}}{{z + {z^{ - 1}}}} - \frac{{ - C - i\sqrt {AB - {C^2}} }}{A}} \right)}}\frac{{dz}}{{iz}}}
\\
&= \int_C {\frac{{{{\left( {{z^2} - 1} \right)}^n}}}{{{{\left( {2i} \right)}^n}{z^n}}}\frac{{{{\left( {{z^2} + 1} \right)}^{m - 2}}}}{{{2^{m - 2}}{z^{m - 2}}}}\frac{{{{\left( {{z^2} + 1} \right)}^2}}}{{\left( {i - {u_1} - \left( {i + {u_1}} \right){z^2}} \right)\left( {i - {u_2} - \left( {i + {u_2}} \right){z^2}} \right)}}\frac{{dz}}{{iz}}}
\\
&= \frac{1}{{{2^{m - 2}}i{{\left( {2i} \right)}^n}}}\int_C {\frac{{{{\left( {{z^2} - 1} \right)}^n}{{\left( {{z^2} + 1} \right)}^m}}}{{{z^{n + m - 1}}\left( {i - {u_1} - \left( {i + {u_1}} \right){z^2}} \right)\left( {i - {u_2} - \left( {i + {u_2}} \right){z^2}} \right)}}dz}
\end{align}
now factor the denominator and find the singularities ... i think you can proceed,
where $C$ is the unit circle.
\begin{align}
\int_0^{2\pi } {f\left( {\cos \theta ,\sin \theta } \right)d\theta } = 2\pi i\sum {\text{residues of $f\left( z \right)$ inside the unit circle}}
\end{align}
Do not forget that $z^{m+n-1}$ have three cases $m+n=1$, $m=n=0$, $m+n>1$.
|
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|
Find odd numbers $(o_1,o_2,o_3,o_4)$ such that $o_1^2-o_2^2=2(o_3^2-o_4^2)$ such that $o_1>o_2$ and $o_3>o_4$ I am working on a graph labeling problem and am stuck at the following problem on odd numbers.
Find (all) odd numbers $(o_1,o_2,o_3,o_4)$ such that $o_1^2-o_2^2=2(o_3^2-o_4^2)$ such that $o_1>o_2$ and $o_3>o_4$?
Ideally I would like to prove that no such 4-tupule $(o_1,o_2,o_3,o_4)$ exists. However if they exist, I want to count, for given $n$, how many such 4-tupules exists such that the largest odd number i.e. $o_1 \leq n$.
My approach: Say the 4-tupule is $(2p+1,2q+1,2r+1,2s+1)$. The above equation simplifies to $(p-q)(p+q+1) = 2(r-s)(r+s+1)$. I am stuck here..
Thanks for going through this.
|
Given the condition, let use write
\begin{eqnarray}
o_1 &=& o,\\
o_2 &=& o - 4p,\\
o_3 &=& o - 2q,\\
o_4 &=& o - 2 r.
\end{eqnarray}
Then we get
$$
o\Big(p+q-r\Big)=2p^2+q^2-r^2.
$$
Case 1
When
$$
p+q-r = 0,
$$
we obtain
$$
p \Big( p -2q \Big) = 0,
$$
so we obtain
\begin{eqnarray}
o_1 &=& 1 + 2 v + 8 w,\\
o_2 &=& 1 + 2 v,\\
o_3 &=& 1 + 2 v + 6 w,\\
o_4 &=& 1 + 2 v + 2 w.
\end{eqnarray}
This gives for example
$$
\begin{array}{cc|cccc}
v & w & o_1 & o_2 & o_3 & o_4\\
\hline
0 & 1 & 9 & 1 & 7 & 3\\
0 & 2 & 17 & 1 & 13 & 5\\
0 & 3 & 25 & 1 & 19 & 7\\
1 & 1 & 11 & 3 & 9 & 5\\
1 & 2 & 19 & 3 & 15 & 7\\
1 & 3 & 27 & 3 & 21 & 9\\
2 & 1 & 13 & 5 & 11 & 7\\
2 & 2 & 21 & 5 & 17 & 9\\
2 & 2 & 29 & 5 & 23 & 11\\
\end{array}
$$
Case 2
When
$$
p+q-r \ne 0,
$$
we can write
$$
r = p + q - a,
$$
so we obtain
$$
a o = 2 p^2 + q^2 - \Big( p + q - a \Big)^2
= p \Big( p - 2q \Big) - a^2 + 2ap + 2aq,
$$
whence
$$
p = ak,
$$
thus
$$
o = a \Big( k^2 + 2k - 1 \Big) - 2 \Big( k - 1 \Big) q.
$$
However, $o$ is odd and therefore $a$ is odd and $k$ is even, whence
$$
o = \Big(2 u + 1 \Big) \Big( 4 v^2 + 4 v - 1 \Big) - 2 \Big( 2 v - 1 \Big) w.
$$
So we obtain
\begin{eqnarray}
o_1 &=& \Big(2 u + 1 \Big) \Big( 4 v^2 + 4 v - 1 \Big) - 2 \Big( 2 v - 1 \Big) w,\\
o_2 &=& \Big(2 u + 1 \Big) \Big( 4 v^2 - 4 v - 1 \Big) - 2 \Big( 2 v - 1 \Big) w,\\
o_3 &=& \Big(2 u + 1 \Big) \Big( 4 v^2 + 4 v - 1 \Big) - 2 \Big( 2 v + 1 \Big) w,\\
o_4 &=& \Big(2 u + 1 \Big) \Big( 4 v^2 - 4 v - 1 \Big) - 2 \Big( 2 v + 1 \Big) w.
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the value of $ \int_{x}^{1} \arcsin \left( \frac{2t}{t^2+1} \right) \text{d}t $? Is this result true? Wolfram doesn't seem to be able to evaluate the definite integral in the allowed time. $$ \int_{x}^{1} \arcsin \left( \dfrac{2t}{t^2+1} \right) \text{d}t = \dfrac{\pi}{2} - 2x\arctan x - \log(2) + \log(1+x^2) $$
|
$$
\frac{2t}{1+t^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin (2\theta).
$$
So
$$
\arcsin\left(\frac{2t}{1+t^2}\right) = 2\theta.
$$
$$
dt = \sec^2\theta\,d\theta
$$
As $t$ goes from $x$ to $1$, then $\theta=\arctan t$ goes from $\arctan x$ to $\pi/4$.
$$
\int \theta\Big(\sec^2\theta\,d\theta\Big) = \int \theta\,dv = \theta v- \int v\,d\theta
$$
and so $v=\tan\theta$, etc.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the equation of the linear transformation of an orthogonal projection on the line y=mx. Let $T : \mathbb R^2 → \mathbb R^2$ the orthogonal projection on the line $y = mx$. Prove that for all $a, b \in \mathbb R$,
$$\begin{align}T((a,b)) = {\frac{1}{m^2 + 1}}(a+mb, ma + m^2b)\end{align}$$
The mark scheme talks about using an equation $$\begin{align}y = {\frac{-1}{m}x + b + \frac{a}{m}}\end{align}$$
First of all where did they get that equation from? I've never heard of it and it hasnt been given to us in our lecture notes. And where do i go from there to find the answer above.
Then it says let : $$\begin{align}S:\mathbb R^2 → \mathbb R^2 \end{align}$$ be the orthogonal symmetry with respect to the line $y = mx$. Prove that for all $a, b \in \mathbb R$,
$$\begin{align} S((a, b)) = −(a, b) + 2T ((a, b)) \end{align}$$
So I can visualise it but I dont understand how the mark scheme can write: $v \in \mathbb R^2$ and let $x \in \mathbb R^2$ such that $T(v) + x = v$. then $S(v) = T (v) − x = T (v) − (v − T (v)) = 2T (v) − v.$
Because I dont get how $T(v) + x = v$? it wouldnt it equal $x$ not $v$? I am sure the mark scheme is right but i cant seem to understand that part. All i know is it would be $2$ times the projection minus the vector $(a,b)$.
|
Here's how I solve this problem:
Notice I am writing vectors in columnar form; thus, the OP's $(a, b)$ is my
$\begin{pmatrix} a \\ b \end{pmatrix} \tag{0}$
etc.; these things being said , we have:
The line $y = mx$ is in fact the set of points (vectors) $L = \{(x, mx)^T = x(1, m)^T \mid x \in \Bbb R \}$; this is the one-dimensional subspace of $\Bbb R^2$ generated by the vector $\vec v = (1, m)^T$. $L$ is also generated by the unit vector $\mathbf u_{\vec v}$ in the direction of $\vec v$, and since $\Vert \vec v \Vert = \sqrt{m^2 + 1}$ we have
$\mathbf u_{\vec v} = \begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix}. \tag{1}$
The scalar component of the projection of any vector $(a, b)^T \in \Bbb R^2$ onto the subspace $L = \langle \mathbf u_{\vec v} \rangle$ generated by the unit vector $\mathbf u_{\vec v}$ is in fact
$\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v} = \begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix} = \dfrac{a + mb}{\sqrt{m^2 + 1}}, \tag{2}$
and the projected vector itself is this scalar component times $\mathbf u_{\vec v}$, or
$T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v} = \dfrac{a + mb}{\sqrt{m^2 + 1}}\begin{pmatrix} \dfrac{1}{\sqrt{m^2 + 1}} \\ \dfrac{m}{\sqrt{m^2 + 1}} \end{pmatrix} = \begin{pmatrix} \dfrac{a + mb}{m^2 + 1} \\ \dfrac{ma + m^2 b}{m^2 + 1}\end{pmatrix}; \tag{3}$
thus is the OP's first equation established.
As for the equation for $S((a, b)^T$,
$S((a, b)^T) = -(a, b)^T + 2T((a, b)^T) = -\begin{pmatrix} a \\ b \end{pmatrix} + 2T(\begin{pmatrix} a \\ b \end{pmatrix}), \tag{4}$
note that we have
$\begin{pmatrix} a \\ b \end{pmatrix} = T(\begin{pmatrix} a \\ b \end{pmatrix}) + (\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})); \tag{5}$
furthermore, we see that
$T(\begin{pmatrix} a \\ b \end{pmatrix}) \cdot (\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix}) = 0. \tag{6}$
(6) is easily seen using
$T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v}, \tag{7}$
the leftmost equation in (3) and the definition of $T((a, b)^T)$; indeed
$T((a, b)^T) \cdot T((a, b)^T) = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \Vert \mathbf u_{\vec v} \Vert^2 = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \tag{8}$
since $\Vert \mathbf u_{\vec v} \Vert^2 = 1$, and
$T((a, b)^T) \cdot \begin{pmatrix} a \\ b \end{pmatrix} = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v}) \mathbf u_{\vec v} \cdot \begin{pmatrix} a \\ b \end{pmatrix} = (\begin{pmatrix} a \\ b \end{pmatrix} \cdot \mathbf u_{\vec v})^2 \tag{9}$
as well, so that (6) indeed binds, showing that $(a, b)^T - T((a, b)^T)$ is orthogonal to the subspace $L = \langle \mathbf u_{\vec v} \rangle$. Now the reflection $S$ about/across the line $L$ leaves elements of $L$ itself invariant, and reverses vectors in $\Bbb R^2$ which are orthogonal to $L$; thus we must have
$ST(\begin{pmatrix} a \\ b \end{pmatrix}) = T(\begin{pmatrix} a \\ b \end{pmatrix}), \tag{10}$
but
$S(\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})) = T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix}; \tag{11}$
applying (10) and (11) to (5) yields
$S(\begin{pmatrix} a \\ b \end{pmatrix}) = ST(\begin{pmatrix} a \\ b \end{pmatrix}) + S(\begin{pmatrix} a \\ b \end{pmatrix} - T(\begin{pmatrix} a \\ b \end{pmatrix})) = T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix} + T(\begin{pmatrix} a \\ b \end{pmatrix})$
$= 2T(\begin{pmatrix} a \\ b \end{pmatrix}) - \begin{pmatrix} a \\ b \end{pmatrix}, \tag{12}$
as was to be shown. QED.
Additional Remarks added Thursday 24 July 2014 10:58 PM PST: I don't know exactly what the "mark scheme" is, but I think that introduction of the line
$y = -\dfrac{x}{m} + b + \dfrac{a}{m} \tag{13}$
is meant to provide a slightly different argument, though one geometrically equivalent to that given above. For we see that the slope of line (13) is $-(1/m)$; as such, it is perpendicular to $L$. (Recall that two lines are perpendicular if their slopes $m_1$, $m_2$ satisfy $m_1 m_2 = -1$.) Furthermore, the line (13) passes through the point $(a, b)^T$ as may be seen by setting $x = a$, $y = b$; thus (13) is the equation of the line normal to $L$ containing the point $(a, b)^T$. As such, it is related to the OP's equation $T(v) + x = v$ or my equation (5); $x = v - T(v)$ is normal to $T(v)$, which lies along the line $L: y = mx$; it ($v - T(v)$) may be thought of as running from $L$ to $v$. Though I can't, being unfamiliar with "mark scheme"(s), say exactly what the referenced one suggests, it appears to be focused the relationship between the geometry of this problem, based on the lines $y = mx$ and (13), and it's linear algebra, which uses vectors such as my $\mathbf u_{\vec v}$. Perhaps these observations can help our OP user135688 piece things together along these "lines", if I may so say; he certainly is on the inside track. But the best I can do is provide the preceding mere suggestions of an idea. End of Remarks.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
|
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|
Solve $\sin^2 x − \cos^2 x = \sin x$, when $x\in -\pi ≤ x ≤\pi$ I have to solve for $x$ using the domain of $-\pi ≤ x ≤\pi$
$$\sin^2 x − \cos^2 x = \sin x $$
I tried changing $\cos^2 x$ to $1 - \sin^2 x$, and then getting
$$\sin^2 x - 1 + \sin^2 x = \sin x \to 2\sin^2 x - 1 = \sin x$$
Then I have no clue where to go from there. Please help!
|
From
$\sin^2 x − \cos^2 x = \sin x$,
since
$\cos^2 x = 1-\sin^2 x$,
$\sin x
= \sin^2 x - (1-\sin^2 x)
=2\sin^2 x -1
$.
Now we only have $\sin x$,
so let $\sin x = y$.
This becomes
$y = 2y^2-1$.
You can solve this for $y$.
From the possible values of $y$,
you can then get $x$.
I will leave it at that.
|
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|
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$
*$\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$
If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$
Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.
If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$
Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$
Is there also an other way to show the equality or is it the only one?
|
For even $n$, $\left\lfloor\frac n2\right\rfloor=\left\lceil\frac n2\right\rceil=\frac n2$.
For odd $n$, $\left\lfloor\frac n2\right\rfloor=\frac{n-1}2$, and $\left\lceil\frac n2\right\rceil=\frac{n+1}2$.
|
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|
Matrix exponential of a skew-symmetric matrix without series expansion I have the following skew-symmetric matrix
$$C = \begin{bmatrix} 0 & -a_3 & a_2 \\
a_3 & 0 & -a_1 \\
-a_2 & a_1 & 0 \end{bmatrix}$$
How do I compute $e^{C}$ without resorting to the series expansion of $e^{C}$? Shall I get a finite expression for it?
NB: Values of $a_i$ s can't be changed.
|
Let $x = \sqrt{a_1^2+a_2^2+a_3^2}$. You can verify that $C^3 = -(a_1^2+a_2^2+a_3^2)C = -x^2C$.
Hence, $C^{2m+1} = (-1)^mx^{2m}C$ and $C^{2m} = (-1)^{m-1}x^{2m-2}C^2$.
Therefore, $e^C = \displaystyle\sum_{n=0}^{\infty}\dfrac{1}{n!}C^n = I + \sum_{m=0}^{\infty}\dfrac{1}{(2m+1)!}C^{2m+1} + \sum_{m=1}^{\infty}\dfrac{1}{(2m)!}C^{2m}$
$= \displaystyle I + \sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m}}{(2m+1)!}C + \sum_{m=1}^{\infty}\dfrac{(-1)^{m-1}x^{2m-2}}{(2m)!}C^{2}$
$= \displaystyle I + \dfrac{1}{x}\sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m+1}}{(2m+1)!}C - \dfrac{1}{x^2}\sum_{m=1}^{\infty}\dfrac{(-1)^{m}x^{2m}}{(2m)!}C^{2}$
$= I + \dfrac{\sin x}{x}C + \dfrac{1-\cos x}{x^2}C^2$
|
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|
Indefinite integral of trignometric function What is the trick to integrate the following
$$\int \frac{1-\cos x}{(1+\cos x)\cos x}\ dx$$
|
Using $$\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$$
$$\int\frac{1-\cos x}{\cos x(1+\cos x)}dx$$
$$=\int\frac{2\sin^2\dfrac x2}{\cos^2\dfrac x2-\sin^2\dfrac x2}\frac12\sec^2\dfrac x2dx$$
$$=\int\frac{2\tan^2\dfrac x2}{1-\tan^2\dfrac x2}\frac12\sec^2\dfrac x2dx$$
$$=\int\frac{2u^2-2+2}{1-u^2}du=-2u+2\int\frac{du}{1-u^2}du$$
|
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|
Trignometry-Prove that $(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$ Prove that $$(\csc\theta - \sec\theta )(\cot \theta -\tan\theta )=(\csc\theta +\sec\theta )(\sec\theta ·\csc\theta -2)$$
I tried solving the LHS and RHS seperately but they were not coming out to be equal.
Please help me answer this question.
And also how does one go about proving such questions?
Thanks in advance
|
$$\begin{array}{lll}
(\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&(\csc\theta - \sec\theta)\bigg(\frac{\csc\theta}{\sec\theta} - \frac{\sec\theta}{\csc\theta}\bigg)\\
&=&(\csc\theta - \sec\theta)\bigg(\frac{\csc^2\theta-\sec^2\theta}{\sec\theta\csc\theta}\bigg)\\
&=&\frac{(\csc\theta-\sec\theta)(\csc\theta-\sec\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\
&=&\frac{(\color{blue}{\csc^2\theta}-2\sec\theta\csc\theta\color{blue}{+\sec^2\theta})(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}
\end{array}$$
using the 4th Pythagorean identity
$$\sin^2 \theta+\cos^2 \theta = 1$$
$$\frac{\sin^2 \theta+\cos^2}{\cos^2\theta\sin^2\theta} = \frac{1}{\cos^2\theta\sin^2\theta}$$
$$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$
we have
$$\begin{array}{lll}
(\csc\theta - \sec\theta)(\cot\theta - \tan\theta)&=&\frac{(\color{blue}{\sec^2\theta\csc^2\theta}-2\sec\theta\csc\theta)(\csc\theta+\sec\theta)}{\sec\theta\csc\theta}\\
&=&(\csc\theta + \sec\theta)(\sec\theta\csc\theta-2)
\end{array}$$
|
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|
prove $\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$
If $x$,$y$,$z$ are positive real numbers,Prove:$$\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$$
Using this two inequality:
$\sum ^n_{i=1} \sqrt{a_ib_i}\leq\sqrt {ab} $ (we call it $A$ inequality)
$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.
Things I have tried so far:
Using inequality $A$ i can re write question inequality as$$\sum \limits_{cyc} \frac{x}{2x+\sqrt{xy}}\leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.
|
By your work and by C-S we obtain:
$$\sum_{cyc}\frac{x}{x+\sqrt{(x+y)(x+z)}}\leq\sum_{cyc}\frac{x}{2x+\sqrt{yz}}=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{x}{2x+\sqrt{yz}}\right)=$$
$$=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{\sqrt{yz}}{2x+\sqrt{yz}}=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{yz}{2x\sqrt{yz}+yz}\leq$$
$$\leq\frac{3}{2}-\frac{1}{2}\cdot\frac{\left(\sum\limits_{cyc}\sqrt{yz}\right)^2}{\sum\limits_{cyc}(2x\sqrt{yz}+yz)}=\frac{3}{2}-\frac{1}{2}=1.$$
|
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|
Find an equation for a moving rod The two endpoints of a 1-metre long rod have an initial position at $(0,0),(0,1).$ The rod slides continuously to the position $(1,0),(0,0)$ sweeping out a region in the positive quadrant. Determine the equation for the boundary of this region.
My attempt:
At any position, let $a$ denote the distance from the lower endpoint to the origin, $b$ the distance from the upper endpoint to the origin. Then we have
$$a^2+b^2=1,$$
$$\frac{x}{a}+\frac{y}{b}=1.$$
I tried to fix $x$ and use the method of Lagrange multiplier to determine the maximal $y.$ But I couldn't solve the equations. Any ideas?
|
Firstly obtain the straight-line equation of the rod at any point in the slide. If $x_0$ and $y_0$ are the $x$-intercept and $y$-intercept, then by Pythagoras, $x_0^2 + y_0^2 = 1$.
So we have the equation of this line:
\begin{eqnarray*}
y &=& -\dfrac{y_0}{x_0}x + y_0 \\
&=& -\dfrac{y_0}{\sqrt{1 - y_0^2}}x + y_0
\end{eqnarray*}
With this equation we treat $x$ as a constant and try to maximise $y$ with respect to $y_0$. That is, for each $x$ we want the maximum $y$ these equations achieve. To do this we differentiate:
\begin{eqnarray*}
\dfrac{dy}{dy_0} &=& \dfrac{-x}{\sqrt{1 - y_0^2}} - \dfrac{y_0^2x}{\left(1 - y_0^2\right)^{\frac{3}{2}}} + 1 \\
&=& 1 - \dfrac{x}{\left(1 - y_0^2\right)^{\frac{3}{2}}} \\
\end{eqnarray*}
Setting this to $0$ gives $y_0 = \dfrac{1}{\sqrt{1 - x\frac{2}{3}}}$ and $x_0 = x^\frac{1}{3}$.
Substitute these back to find $y$:
\begin{eqnarray*}
y &=& \dfrac{-\sqrt{1 - x^\frac{2}{3}}x}{x^\frac{1}{3}} + \sqrt{1 - x^\frac{2}{3}} \\
&=& \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}
\end{eqnarray*}
So the region you want is the region bounded by this curve and the $x$ and $y$ axes.
|
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|
Find X location using 3 known (X,Y) location using trilateration I post this question in stackoverflow here and was advised it was best suited for here.
I am trying to understand the maths behind trilateration, we have 3 access points $(\text{AP's: } 1,2,3)$ and we know the centre coordinates of theses $3$ (AP's). Lost device ($X$) is the location where all $3$ circle intersect.
example GRID $\Rightarrow x=30, y=30.$
Known Locations: $\text{AP}_1 = (5,5), \text{AP}_2 = (5,15), \text{AP}_3 = (20,10).$
Unknown Location: $X$
Lets say $X = (9,11)$
how do we work out the intersection of all $3$ circles by the maths.
your help is greatly appreciated.
|
Use the distance equation. If your unknown point is $(x,y)$, your known points are $(x_i, y_i)$ which are distances $r_i$ from your unknown point, then you get three equations:
$$(x-x_1)^2 + (y-y_1)^2 = r_{1}^2 \\ (x-x_2)^2 + (y-y_2)^2 = r_{2}^2 \\ (x-x_3)^2 + (y-y_3)^2 = r_{3}^2$$
We can expand out the squares in each one:
$$x^2 - 2x_1x + x_1^2 + y^2 - 2y_1y + y_1^2 = r_1^2 \\ x^2 - 2x_2x + x_2^2 + y^2 - 2y_2y + y_2^2 = r_2^2 \\ x^2 - 2x_3x + x_3^2 + y^2 - 2y_3y + y_3^2 = r_3^2$$
If we subtract the second equation from the first, we get
$$(-2x_1 + 2x_2)x + (-2y_1 + 2y_2)y = r_1^2 - r_2^2 - x_1^2 + x_2^2 - y_1^2 + y_2^2 .$$
Likewise, subtracting the third equation from the second,
$$(-2x_2 + 2x_3)x + (-2y_2 + 2y_3)y = r_2^2 - r_3^2 - x_2^2 + x_3^2 - y_2^2 + y_3^2 .$$
This is a system of two equations in two unknowns:
$$Ax + By = C \\ Dx + Ey = F$$
which has the solution:
$$x = \frac{CE-FB}{EA-BD} \\ y = \frac{CD-AF}{BD-AE}$$
|
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|
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear manner? Thanks in advance for the help!
|
Since the residue in $x=i$ of $\frac{x^2-3}{x^2+1}$ is given by:
$$\lim_{x\to i}\frac{x-i}{x^2+1}(x^2-3)=(i^2-3)\lim_{x\to i}\frac{1}{2x}=2i$$
due to De l'Hospital theorem, and the residue in $x=-i$ is just the opposite due to the Schwarz reflection principle, we have that:
$$f(x)=\frac{x^2-3}{x^2+1}-\left(\frac{2i}{x-i}-\frac{2i}{x+i}\right)=\frac{x^2-3}{x^2+1}+\frac{4}{x^2+1}$$
is an entire function. That function is bounded when $|x|$ approaches $+\infty$, hence $f(x)$ is constant due to the Liouville's theorem, and $f(x)\equiv f(0)=1$, hence we have:
$$\frac{x^2-3}{x^2+1}=1-\frac{4}{x^2+1}$$
by exploiting just two or three quite advanced results.
|
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|
Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I get stuck here but I am assuming I need to get that 4 to a 2 somehow so I can combine them... so:
$ \large \frac {(2^2)^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I feel like I am not on the right track here as I have no idea where to go now. Could anyone help please? Thank you!
|
$$\large \frac {12^{x-2}.4^{x}} {6^{x-2}}$$
$$\begin{align}
& \implies \large \frac {(3\times4)^{x-2}.2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}\times2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}2^{2x}} {(3)^{x-2}\times(2)^{x-2}} \\
& \implies \large (2)^{2(x-2)+2x-(x-2)} \\
& \implies \large (2)^{2x-4+2x-x+2}\\
& \implies \large (2)^{3x-2} \\
& \end{align}$$
|
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|
Evaluation of limit How to evaluate the value of this limit?
$$\lim_{x\to 2} \frac{\sqrt{x-2} + \sqrt x - \sqrt2}{\sqrt{x^2 - 4}}$$
Actually I'm struck at algebraic part. Please guide..
|
Deepak's answer is better, but you can also do as the followings :
$$\begin{align}\frac{(\sqrt{x-2}+\sqrt x)-\sqrt 2}{\sqrt{x^2-4}}\cdot \frac{(\sqrt{x-2}+\sqrt x)+\sqrt 2}{(\sqrt{x-2}+\sqrt x)+\sqrt 2}\end{align}$$
$$=\frac{x-2+2\sqrt{x(x-2)}+x-2}{\sqrt{(x-2)(x+2)}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$
$$=\frac{\color{red}{\sqrt{x-2}}\ (\sqrt{x-2}+2\sqrt x+\sqrt{x-2})}{\color{red}{\sqrt{x-2}}\sqrt{x+2}\ (\sqrt{x-2}+\sqrt x+\sqrt 2)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written in the same form: $49$ is already shown, and $7$ is $2^2+3\cdot 1^2$.
I'm looking for a general proof of this using infinite descent. I tried to let $a^2+3b^2=xy$, where $x,y$ are positive integer factors of the LHS, but couldn't continue from here I'm afraid. However, I have managed to prove that integers of the form $a^2+3b^2$ are closed under multiplication, i.e. the product of two of them is of the same form. I was hoping this would help.
I also had another question if you have time: Is the representation in the form $a^2+3b^2$ unique? For example, is it possible for an integer, say $N$, to be $N=a^2+3b^2=x^2+3y^2$ with $a\ne x$ and $b\ne y$?
EDIT: Sorry, here is the "closed under multiplication" proof I found:
Using Diophantus' Identity we have $(a^2+3b^2)(x^2+3y^2)=(a^2+(\sqrt{3}b)^2)(x^2+(\sqrt{3}y)^2=(ax+\sqrt{3}\cdot \sqrt{3} y)^2)+(a\cdot \sqrt{3} y - \sqrt{3} b\cdot x)^2=(ax+3by)^2+3(ay-bx)^2$ as required.
EDIT (2): I am very sorry. The question should be:
If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form.
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Note that:
$$(ax-3by)^2 + 3(ay+bx)^2 = (a^2+3b^2)(x^2+3y^2) = (ax+3by)^2 + 3(ay-bx)^2$$
So if all divisors of numbers of that form are also of that form, then only numbers of that form that are primes have unique representation.
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Solve $(1+z)^8=(1-z)^8$ My guess is to write this as $$\left(\frac{1+z}{1-z}\right)^8=1.$$ We can then find 8 possibilities for $\frac{1+z}{1-z}$, namely $\cos(k\pi/4)+i\sin(k\pi/4)$, $k=1,\ldots,8$. For each $k$ we can then deduce 2 equations by putting $z=x+iy$, for example for $k=1$ we get: $$\frac{1+x+iy}{1-x-iy}=\frac12 (1+i).$$ Now we find two equations for $x$ and $y$, by noting both the real and imaginary parts of the equations should be equal, and can thus find $z$.
However, doing this for $k=1,\ldots,8$ seems somewhat cumbersome. Does anyone know of a faster way to find the $z$? Thans in advance.
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HINT:
$\left(\dfrac{1+z}{1-z}\right)^8=1 \implies$
$\left(\dfrac{1+z}{1-z}\right)^4=\pm{1} \implies$
$\left(\dfrac{1+z}{1-z}\right)^2=\pm{1},\pm{i} \implies$
$\left(\dfrac{1+z}{1-z}\right)^1=\pm{1},\pm{i},\pm{\dfrac{1+i}{\sqrt{2}}},\pm{\dfrac{1-i}{\sqrt{2}}}$
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Interesting combinatorial identities Let $n$ be a strictly positive integer and let $j=0,\dots,n-1$.
By using Mathematica I managed to guess the following identities:
\begin{eqnarray}
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m}{j} &=& \frac{1}{2} \binom{2 n}{2j + 1} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+1}{j+1} &=& \frac{1}{2} \binom{2 n+1}{2j + 2} + \binom{n-1}{j} \binom{n+1}{j+1} \frac{n(n-j)}{2 (n+1)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+2}{j+2} &=& \frac{1}{2} \binom{2 n+2}{2j + 3} + \binom{n-1}{j} \binom{n+2}{j+2} \frac{n(n-j)}{(n+2)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+3}{j+3} &=& \frac{1}{2} \binom{2 n+3}{2j + 4} + \binom{n-1}{j} \binom{n+3}{j+3} \frac{n(n-j)(11+5 j+(7+3 j) n)}{(n+2)(n+3)(4+ 2j)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+4}{j+4} &=& \frac{1}{2} \binom{2 n+4}{2j + 5} + \binom{n-1}{j} \binom{n+4}{j+4} \frac{2 n(n-j)(5+2 j+(3+ j) n)}{(n+3)(n+4)(2+ j)(j+1)} \\
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+5}{j+5} &=& \frac{1}{2} \binom{2 n+5}{2j + 6} + \binom{n-1}{j} \binom{n+5}{j+5} \cdot \\
&&\frac{n(n-j)((137+93 j+16 j^2)+(264+161j+25 j^2)n/2+(62+35 j+ 5 j^2)n^2/2)}{(n+3)(n+4)(n+5)(j+1)(j+2)(j+3)}
\end{eqnarray}
The temptation is strong to write a general conjecture:
\begin{equation}
\sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m+a}{j+a} = \frac{1}{2} \binom{2 n+a}{2j + 1+a} + \binom{n-1}{j} \binom{n+a}{j+a} \cdot\left(\cdots\right)
\end{equation}
Can anyone advise me how to tackle such a problem?. I think that in this particular case the method of generating functions does not lead to a closed form result.
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Hint: The first identity can be explained as follows.
One wants to count how many sequences $a_1<\cdots<a_{2j+1}$ can be formed from $\{1,\dots,2n\}$. Of course the standard answer is ${2n}\choose{2j+1}$.
Let's count differently by looking at the middle number $a_{j+1}$. It can only take one of the values $j+1,\dots, 2n-j$.
For each $k=j+1,\dots,2n-j$, there are ${k-1}\choose j$ choices of $a_1<\cdots<a_j$ and ${2n-k}\choose{j}$ choices of $a_{j+2}<\cdots<a_{2j+1}$. That gives
$${{2n}\choose{2j+1}}=\sum^{2n-j}_{k=j+1} {{k-1}\choose{j}} {{2n-k}\choose{j}}.$$
Reindexing, we get
$$\begin{aligned}{{2n}\choose{2j+1}}&=\sum^{2n-j-1}_{k=j} {{k}\choose{j}} {{2n-k-1}\choose{j}}\\&=\sum_{k=j}^{n-1}\binom{k}{j}\binom{2n-k-1}{j}+\sum^{2n-j-1}_{k=n}\binom{k}{j}\binom{2n-k-1}{j}.
\end{aligned}$$
Now each of the last sums is equal to the LHS by another reindexing.
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Globally Lipschitz condition
Show that the function $f(x)=\frac1{1+x^2}$ satisfies a global
Lipschitz condition on $\mathbb R$.
By the definition we want to whow that $|f(x)-f(y)|\leqslant L|x-y|$. Here is my work so far.
$$\left|\frac1{1+x^2}-\frac1{1+y^2}\right|=\left|\frac{y^2-x^2}{(1+y^2)(1+x^2)}\right|\leqslant L|x-y|$$
$$\implies\left|\frac{y+x}{(1+y^2)(1+x^2)}\right|\leqslant L.$$
It appears to be bounded, since the denominator becomes much larger than the numerator for large values of $x$ and $y$. Is this good enough or can I find a value for $L$? Also, is it also the same as showing that $f'(x)$ is globally bounded?
Thank you in advance for any help or constructive criticism. I am forever grateful.
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$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right|=\frac{|x+y|}{(1+x^2)(1+y^2)}$$
By triangle inequality:
$$\frac{|x+y|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{(1+x^2)(1+y^2)}+\frac{|y|}{(1+x^2)(1+y^2)}$$
Now $x^2+1 \geq 1$, so:
$$ \frac{|x|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{1+x^2}$$
Next $\displaystyle \frac{|x|}{1+x^2} \leq 2$ be AM-GM inequality. The same with $\displaystyle \frac{|y|}{(1+x^2)(1+y^2)}$, so finally:
$$\left| \frac{x+y}{(1+x^2)(1+y^2)}\right| \leq 4$$
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|
Find the value of $27\csc^2\theta+8\sec^2\theta$ $10\sin^4\theta+15\cos^4\theta=6$, then find the value of $27\csc^2\theta+8\sec^2\theta$
I don't know how to do it have just tried by converting sin and cos into csc and sec. But can't get the answer.
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Since $\sin^2\theta+\cos^2\theta=1$,
$$10\sin^4\theta+15\cos^4\theta=6(\sin^2\theta+\cos^2\theta)^2.$$
It follows that
$$4\sin^4\theta-12\sin^2\theta\cos^2\theta+9\cos^4\theta=0,$$
or
$$(2\sin^2\theta-3\cos^2\theta)^2=0.$$
Thus we have $2\sin^2\theta=3\cos^2\theta$. Can you continue from here?
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.