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Can $10101\dots1$ be a perfect square in any base? Inpired by another question and it's answer I started to wonder if it's true in other bases as well. I've at least not found any base $b$ where $10101\dots1 = (b^{2k}-1)/(b^2-1)$ (where $k>1$) is a perfect square anytime.
From the answers we can conclude that $8|b^2$ which means that the base must be a multiple of $4$. This is because we must have $b^2/8$ to be an integer.
Also I think/guess that $b^k+1$ must be a perfect square for some $k$. This is because
$${b^{2k}-1\over b^2-1} = {(b^k-1)(b^k+1)\over (b-1)(b+1)}$$
has to be a square.
I've also found that the question is equivalent to if $111\cdots1$ can be a perfect square in base $b^2$. I've seen that it can be a perfect square in base $8$, but $8$ is no perfect square.
|
Just some ideas, too long for comments.
Summary: No such square exists if either $b$ or $k$ is even.
This can be written as $(b^k)^2-(b^2-1)y^2=1$.
The set of positive integer solutions to the Pell equation $x^2-(b^2-1)y^2=1$ are exactly the values:
$$x+y\sqrt{b^2-1}=\left(b+\sqrt{b^2-1}\right)^n; n\in\mathbb Z^{+}$$
So you need to show that $x$ is a power of $b$ only when $n=1$.
There is a recurrence: $x_0=1,x_1=b, x_{n+1}=2bx_n-x_{n-1}$.
This gives $x_2=2b^2-1, x_3=4b^3-3b=b(4b^2-3)$. There is no $b$ (other than $1$) which makes $x_3$ a power of $b$.
We can show that $x_{2n}\equiv (-1)^n\pmod {b^2}$ and $x_{2n+1}\equiv (-1)^{n}b(2n+1)\pmod{b^3}$.
So $x_{2n}$ is never a power of $b$, and if $x_{2n+1}$ is a non-trivial power of $b$, then $2n+1$ is divisible by $b^2$.
In particular, you can't have $b$ even.
You can prove inductively that $x_k>b^{k}$ for $k>1$, so the number of digits base $b$ must be at least $b$.
We also get that $x_{m}\equiv (-1)^m\pmod{b+1}$. Then if $x_{2n+1}=b^k$ then $$b^{k}+1\equiv 0\pmod{b+1}$$ and thus $k$ must be odd.
Now, from $y^2=1+b^2+b^4+\cdots +b^{2(k-1)}$ you get that $y^2\equiv k\pmod{b^2-1}$. Since $b$ is odd, you have $8\mid b^2-1$ and since $k$ is odd, we must have that $k\equiv 1\pmod{8}$.
Aside: Actually, $x_n=T_n(b)$ where $T_n$ are the Chebyshev polynomials of the first kind. This might be helpful, not sure. So we are seeking $b,m,k$ so that $T_m(b)=b^k.$
We know that $b,m$ must be odd, $b^2\mid m$, and $k$ must be odd and a square modulo $b^2-1$.
We also have $b^{m}<T_m(b)<2^mb^m$ so $m<k<m(1+\log_b 2)$.
For instance, this means that the smallest $k$ to check for $b=3$ is $k=33,41,49,57.\dots$.
For $b=5$ the smallest candidates for $k$ are: $k=33,81,97,105,\dots.$
For $b=7$, the smallest candidates are $k=57,153,169,177,\dots$.
If $f(x)=1+x+x^2+\cdots +x^{k-1}$ where $k$ is odd, then:
$$f(x)(1-x)+f(-x)(1+x)=2$$
Also, $f(x)f(-x)=\frac{x^{2k}-1}{x^2-1}$, and, again because $k$ is odd, $f(b)$ is odd for any $b$.
This means that $f(b)$ and $f(-b)$ must both be squares.
That means you need $(111\dots1)_b$ a square and $((b-1)0(b-1)0\dots (b-1)01)_b$ to be perfect squares.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\lim_{x\to 2} \sqrt{x^2+5} = 3$ As stated in the title, I need to prove that $\lim_{x\to 2} \sqrt{x^2+5} = 3$ using only the precise definition of a limit.
For any given $\varepsilon \gt 0$, there exists a $\delta = $
Such that $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} \rvert \lt \varepsilon $
I've attempted to convert $\sqrt{x^2+5}$ into the following:
$\sqrt{x^2+5} = \sqrt{(x^2-4)+9}\\
\sqrt{x^2+5} = \sqrt{(x^2-4x+4)+4x +1} = \sqrt{(x-2)^2+4(x-2)+9} $
I have a hunch that I am heading in the wrong direction. Can I get some advice on what I might be doing wrong?
|
We should prove that,for any given $ε>0$, there exists a $δ$ if $0 \lt \lvert x-2 \rvert \lt \delta \Rightarrow \lvert \sqrt{x^2+5} -3\rvert \lt \varepsilon$
$$\left| \sqrt { { x }^{ 2 }+5 } -3 \right| =\left| \frac { { x }^{ 2 }-4 }{ \sqrt { { x }^{ 2 }+5 } +3 } \right| =\left| \frac { { \left( x-2 \right) }\left( x-2+4 \right) }{ \sqrt { { x }^{ 2 }+5 } +3 } \right| <\frac { \left| { \left( x-2 \right) }^{ 2 }+4\left( x-2 \right) \right| }{ 3 } <\frac { { \left| x-2 \right| }^{ 2 }+4\left| x-2 \right| }{ 3 } <\frac { { \delta }^{ 2 }+4\delta }{ 3 } =\epsilon $$
|
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|
Evaluate $\int \frac{dx}{x(x-1)^3(x-2)^2}$ Evaluate $$I=\int \frac{dx}{x(x-1)^3(x-2)^2}$$ without using tedious partial fractions.
My Try:
we have $$1=\left((x-1)-(x-2)\right)^2$$ So
$$=\int \frac{\left((x-1)-(x-2)\right)^2dx}{x(x-1)^3(x-2)^2}$$ So
$$I=\int \frac{dx}{x(x-1)(x-2)^2}+\int \frac{dx}{x(x-1)^3}-2\int \frac{dx}{x(x-1)^2(x-2)} $$
any clue here?
|
You can reduce the labour a little by writing
$\dfrac{1}{x(x-1)^3(x-2)^2} = \dfrac{(1-x+x^2)+x(x-1)}{x(x-1)^3(x-2)^2}$
$=\dfrac{x(x-2)^2-(x-1)^3}{x(x-1)^3(x-2)^2} + \dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3} - \dfrac{1}{x(x-2)^2}+\dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3}- \dfrac{1}{x(x-2)^2}+ \dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}+2 \left[\dfrac{1}{x-1} - \dfrac{1}{x-2} \right]$
If you wish you can further decompose $\dfrac{1}{x(x-2)^2}= \dfrac{1}{(x-2)^2}+\dfrac{1}{2} \left[\dfrac{1}{x} - \dfrac{1}{x-2} \right]$
This expression can be readily integrated.
|
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How many 6-digit numbers have distinct digits but no consecutive digits both odd? How many 6-digit natural numbers exist with the
distinct digits and two arbitrary consecutive digits can not be simultaneously odd numbers?
I have tried to set up recurrence relation , by considering a valid 5 digit number satisfying the property and then appending the required digit at the last place to produce valid 6 digit number , but things are getting cumbersome. Is there a nice alternative solution? Thanks in advance.
|
One way to do it is by looking at cases, as @lulu suggested. There are $3$ basic cases: $1$ odd number, $2$ odd numbers or $3$ odd numbers. I'll go through each case below, splitting each case into scenarios starting with an odd number (O) and scenarios starting with an even number (E). The reason for this split is that scenarios starting with an even number are not allowed to start with a zero and therefore differ slightly in the computation.
Case 1
With $1$ odd number you have the following possible scenarios
(a) OEEEEE
(b) EOEEEE, EEOEEE, EEEOEE, EEEEOE, EEEEEO
The number of combinations $n_1$ for this case is
$$n_1 = \binom{5}{1} \cdot \binom{5}{5} \cdot 5! + 5 \cdot \binom{4}{1} \cdot \binom{5}{1} \cdot \binom{4}{4} \cdot 4! = 3,000$$
Case 2
With $2$ odd numbers you have the following possible scenarios
(a) OEOEEE, OEEOEE, OEEEOE, OEEEEO
(b) EOEOEE, EOEEOE, EOEEEO, EEOEOE, EEOEEO, EEEOEO
The number of combinations $n_2$ for this case is
$$n_2 = 4 \cdot \binom{5}{2} \cdot 2! \cdot \binom{5}{4} \cdot 4! + 6 \cdot \binom{4}{1} \cdot \binom{5}{2} \cdot 2! \cdot \binom{4}{3} \cdot 3! = 21,120$$
Case 3
With $3$ odd numbers you have the following possible scenarios
(a) OEOEOE, OEOEEO, OEEOEO
(b) EOEOEO
The number of combinations $n_3$ for this case is
$$n_3 = 3 \cdot \binom{5}{3} \cdot 3! \cdot \binom{5}{3} \cdot 3! + \binom{4}{1} \cdot \binom{5}{3} \cdot 3! \cdot \binom{4}{2} \cdot 2! = 13,680$$
In conclusion
The total number of combinations is then $N= n_1 + n_2 + n_3$ or
$$N = 3,000 + 21,120 + 13,680 = 37,800$$
which matches the number found by @Jeppe in the comments.
|
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|
Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.
Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{x+h}{x^2+2hx+h^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{\frac{(x+h)(x^2+1)-x(x^2+2hx+h^2+1)}{(x^2+1)(x^2+2hx+h^2+1)}}{h}\\
&= \lim_{h\to 0} \frac{\frac{hx^2-2hx-h^2+h}{(x^2+1)(x^2+2hx+h^2+1)}}{h}
\end{align}
$$
I'm currently stuck with simplify the fraction so that I can finally find the derivative. I'd really appreciate some advice on how to proceed with the problem.
|
\begin{eqnarray}
{1 \over h } ({x+h \over 1 + (x+h)^2} - {x \over 1+x^2}) &=&
{1 \over h } { (x+h)(1+x^2) - (1+(x+h)^2)x\over (x+x^2) (1 + (x+h)^2} \\
&=& {1 \over h } { h -h x^2 - h^2 x\over (x+x^2) (1 + (x+h)^2} \\
&=& { 1 - x^2 - h x\over (x+x^2) (1 + (x+h)^2}
\end{eqnarray}
The limit follows by taking $ h \to 0$.
|
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|
find the total number of possible way to reach to a particular sum suppose you have given a sum like : 5.
we have to find the total number of possible way to reach to 5.
for example
1 + 1 + 1 + 1 + 1 = 5
2 + 1 + 1 + 1 = 5
1 + 2 + 1 + 1 = 5
1 + 1 + 2 + 1 = 5
1 + 1 + 1 + 2 = 5
2 + 2 + 1 = 5
2 + 1 + 2 = 5
1 + 2 + 2 = 5
1 + 1 + 3 = 5
1 + 3 + 1 = 5
3 + 1 + 1 = 5
2 + 3 = 5
3 + 2 = 5
1 + 4 = 5
4 + 1 = 5
5 = 5
above is the total number of way to reach to a particular sum, which is nothing but : 16
|
(Related to stars and bars, but a bit more explicit).
Represent $5$ as $u\;u\;u\;u\;u$.
Between any two $u$s you can insert or not insert a separator. Once you have done this you will have uniquely determined an ordered sum to $5$.
For instance $u*u\;u\;u*u$ corresponds to $1+3+1$.
Or conversely, $2+3$ corresponds to $u\;u*u\;u\;u$.
Since there are $4$ different possible separators to insert or not insert, there are $2^4$ ways to do the problem.
These ideas should generalize to other numbers than $5$.
|
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|
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
|
The condition gives $$xyz=2+x+y+z$$ or
$$xyz=2+\frac{(x+y+z)(xy+xz+yz)}{12}$$ or
$$12xyz=24+\sum_{cyc}(x^2y+x^2z+xyz)$$ or
$$3xyz=24+\sum_{cyc}(x^2y+x^2z-2xyz),$$
which gives
$$3xyz-24=\sum_{cyc}(x^2y+x^2z-2xyz)=\sum_{cyc}z(x-y)^2\geq0.$$
Thus, $xyz\geq8$ and the equality occurs for $x=y=z$.
On the other hand, by AM-GM
$$12=xy+xz+yz\geq3\sqrt[3]{x^2y^2z^2},$$
which gives $xyz\leq8.$
Thus, $xyz=8$ and we get $x=y=z=2$.
|
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|
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1-a)(1+a)
\end{array}
\right.$$
In order to have an infinite set of solutions, I want to have $0=0$ in the last row, meaning that $a=1\Rightarrow 0=1,$ which is not what I want. Have I made any arithmetical mistake? Did this computation times now.
|
It's simpler to perform the full row reduction for the augmented matrix. We'll put the third equation at top:
\begin{gather}
\begin{bmatrix}
1&-1&1&|&2+a\\1&a&1&|&1\\
a&1&1&|&1+a\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&-1&1&|&2+a\\0&1+a&0&|&-1-a\\
0&1+a&1-a&|&1-a-a^2\end{bmatrix} \\[1ex]
\rightsquigarrow \begin{bmatrix}
1&-1&1&|&2+a\\0&1+a&0&|&-1-a\\
0&1+a&1-a&|&2-a^2\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&-1&1&|&2+a\\0&1+a&0&|&-1-a\\
0&0&1-a&|&2-a^2\end{bmatrix}\\
\end{gather}
*
*If $a\ne1,-1$, we can proceed:
\begin{align}
\begin{bmatrix}
1&-1&1&|&2+a\\0&1+a&0&|&-1-a\\
0&0&1-a&|&2-a^2\end{bmatrix}&\rightsquigarrow \begin{bmatrix}
1&-1&1&|&2+a\\0&1&0&|&-1\\
0&0&1&|&\frac{2-a^2}{1-a}\end{bmatrix}\rightsquigarrow \begin{bmatrix}
1&-1&0&|&\frac a{a-1}\\0&1&0&|& -1\\
0&0&1&|&\frac{2-a^2}{1-a}\end{bmatrix}\\[1ex]
&\rightsquigarrow \begin{bmatrix}
1&0&0&|&\frac 1{a-1}\\0&1&0&|&-1\\
0&0&1&|&\frac{2-a^2}{1-a}\end{bmatrix}
\end{align}
The (unique) solution is the last column.
*If $a=-1$, the matrix of the system becomes
$$\rightsquigarrow
\begin{bmatrix}
1&-1&1&|&1\\0&0&0&|&0\\
0&0&2&|&1\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&-1&1&|&1\\0&0&0&|&0\\
0&0&1&|&\frac12\end{bmatrix}\rightsquigarrow \begin{bmatrix}1&-1&0&|&\frac12\\0&0&0&|&0\\
0&0&1&|&\frac12\end{bmatrix} $$
and the solutions are given by $\;x=y+\frac12$, $z=\frac12$. Vectorially:
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\frac12\\0\\\frac12\end{bmatrix}+y\begin{bmatrix}1\\1\\0\end{bmatrix}$$
*If $a=1$, the system becomes
$$
\begin{bmatrix}
1&-1&1&|&3\\0&2&0&|&-2\\
0&0&0&|&1\end{bmatrix}\qquad\text{which is inconsistent.}$$
|
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|
Find the number of permutations of the word MATHEMATICS that satisfy at least one of three restrictions Find the number of permutations of letters of the word ’MATHEMATICS’ where: both letters T
are before both letters A or both letters A are before both letters M or both letters M are before
letter E.
We have 2 Ms 2 As 2 Ts, so the sum of all permutations is $$ \frac{11!}{2!\cdot 2!\cdot 2!} $$
I think i need to do it by the inclusion-exclusion principle, do i need to find that situations
A - Both T before both A $$ \frac {8 \choose2}{2!} \cdot 6!$$
B - Both M before both A $$ \text{same as } |A| $$
C - Both M before E $$ \frac {9 \choose 2}{2!\cdot2!} \cdot 7! $$
And actually I dont know is it correct, moreover I dont know how to push it forward
|
Your strategy is correct but not all of your numbers are.
I will use the same notation that you did.
Let $A$ be the event that both Ts appear before both As.
Let $B$ be the event that both Ms appear before both As.
Let $C$ be the event that both Ms appear before E.
Then, as you observed, the number of permutations of the word MATHEMATICS in which both Ts before both As or both Ms appear before both As or both Ms appear before both E is
$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
Your correctly observed that the number of distinguishable permutations of the word MATHEMATICS is
$$\binom{11}{2}\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{11!}{2!2!2!}$$
$|A|$
Within a given permutation of the word MATHEMATICS, the letters A, A, T, T can be permuted among themselves in $\binom{4}{2} = 6$ distinguishable ways. In only one of these arrangements do both Ts appear before both As. Hence, by symmetry, the number of distinguishable permutations of the word MATHEMATICS in which Ts appear before both As is
$$\frac{1}{6} \frac{11!}{2!2!2!}$$
$|B|$
Replacing T by M in the preceding argument shows that the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before both As is
also
$$\frac{1}{6} \frac{11!}{2!2!2!}$$
$|C|$
Within a given permutation of the letters of the word MATHEMATICS, the letters E, M, M can be permuted among themselves in $3$ distinguishable ways. In only one of these arrangements do both Ms appear before E. Thus, by symmetry, the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before E is
$$\frac{1}{3} \frac{11!}{2!2!2!}$$
Thus far, our answers agree. The reason for this is that the phrase "in only one of these arrangements, ..." applies. For the intersections, this is not true.
$|A \cap B|$
Within a given permutation of the word MATHEMATICS, the letters A, A, M, M, T, T can be permuted among themselves in
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
distinguishable ways. The requirements that both Ts appear before both As and both Ms appear before both As means that the two As must occupy the last two of the six positions. The two Ms and two Ts can be arranged in the first four positions in $\binom{4}{2}\binom{2}{2}$ distinguishable ways. Hence, the fraction of permissible arrangements is
$$\frac{\dbinom{4}{2}\dbinom{2}{2}}{\dbinom{6}{2}\dbinom{4}{2}\dbinom{2}{2}} = \frac{1}{\dbinom{6}{2}} = \frac{1}{15}$$
Thus, the number of distinguishable permutations of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before both As is
$$\frac{1}{15} \frac{11!}{2!2!2!}$$
$|A \cap C|$
Within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M, T, T can be permuted among themselves in
$$\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$
distinguishable ways. We require that both Ts appear before both As and both Ms appear before E. Observe that once we choose four of the seven positions for the two Ts and two As, there is only one permissible arrangement of the letters A, A, E, M, M, T, T since the first two of the four chosen positions must be occupied by Ts, the last two of those positions must be occupied by As, the first two of the remaining three positions must be occupied by Ms, and the final remaining position must be occupied by E. Hence, the fraction of permissible arrangements is
$$\frac{\dbinom{7}{4}}{\dbinom{7}{2}\dbinom{5}{1}\dbinom{4}{2}\dbinom{2}{2}} = \frac{35}{630} = \frac{1}{18}$$
Hence, the number of distinguishable permutations of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before E is
$$\frac{1}{18} \frac{11!}{2!2!2!}$$
$|B \cap C|$
Within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M can be permuted among themselves in
$$\binom{5}{2}\binom{3}{1}\binom{2}{2}$$
distinguishable ways. The requirements that both Ms appear before both As and that both Ms appear before E means the two As must appear in the first two of these five positions. The E and two Ms can be arranged in the last three positions in $\binom{3}{1}$ distinguishable ways. Hence, the fraction of permissible arrangements is
$$\frac{\dbinom{3}{1}}{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2}} = \frac{1}{\dbinom{5}{2}} = \frac{1}{10}$$
Hence, the number of distinguishable permutations of the word MATHEMATICS in which both Ms appear before both As and both Ms appear before E is
$$\frac{1}{10} \frac{11!}{2!2!2!}$$
$|A \cap B \cap C|$
We have already seen that within a given permutation of the word MATHEMATICS, the letters A, A, E, M, M, T, T can be permuted among themselves in
$$\binom{7}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2} = 630$$
distinguishable ways.
We require that both Ts appear before both As, both Ms appear before both As, and both Ms appear before E. Since two Ms and two Ts must appear before the first A, the two As must appear in the last three positions. Since the two Ms must appear before E, an M cannot appear in the last three positions. We consider cases.
An E appears in the last three positions: The last three positions can be filled with two As and an E in $\binom{3}{1}$ ways. The first four positions can be filled with two Ms and two Ts in $\binom{4}{2}$ ways. Hence, there are
$$\binom{4}{2}\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
A T appears in the last three positions and E appears in the fourth position: The last three positions can be filled with two As and a T in $\binom{3}{1}$ ways. The first three positions can be filled with two Ms and a T in $\binom{3}{1}$ ways. Hence, there are
$$\binom{3}{1}\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
A T appears in the last three positions and E appears in the third position: Since both Ms must appear before E, they must occupy the first two positions and a T must occupy the fourth position. The last three positions can be filled with two As and a T in $\binom{3}{1}$ ways. Hence, there are
$$\binom{3}{1}$$
permissible arrangements of A, A, E, M, M, T, T in this case.
Thus, there are
$$\left[\binom{4}{2} + \binom{3}{1} + 1\right]\binom{3}{1} = 30$$
permissible arrangements, so the fraction of permissible arrangements is
$$\frac{30}{630} = \frac{1}{21}$$
and the number of distinguishable arrangements of the word MATHEMATICS in which both Ts appear before both As and both Ms appear before both As and both Ms appear before both Es is
$$\frac{1}{21} \frac{11!}{2!2!2!}$$
|
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In Epsilon-Delta proof, why is $3x^2|y|/(x^2 + y^2) \Leftarrow 3|y| = 3\sqrt{y^2} \Leftarrow 3\sqrt{x^2 + y^2} $ This is part of a Epsilon-Delta proof to show that: $$\lim_{(x, y) \to 0}\frac{3x^2y}{x^2 + y^2} = 0$$ The proof began with: $$\frac {3x^2|y|}{(x^2 + y^2)} < \epsilon$$ It was then pointed out that $x^2 \Leftarrow x^2 + y^2 \because y^2 >= 0$, which I understood.
In that way, I get why: $$\frac {3x^2|y|}{(x^2 + y^2)} \le 3|y|$$ But I don't understand how they can say it is equal to: $$ 3\sqrt{y^2} \le 3\sqrt{x^2 + y^2} $$
|
Assume $x,y\in\mathbb{R}$. First, you can start saying that $x^{2}\geq 0$, then adding $y^{2}$ in both sides of the inequality you get:
$$ x^{2}+y^{2}\geq y^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\sqrt{y^{2}}$$
because $f(t)=\sqrt{t}$ is monotone increasing on its domain. Now use that $g(t)=3t$ is also monotone increasing and then you have:
$$3\sqrt{x^{2}+y^{2}}\geq 3\sqrt{y^{2}}=3\vert{y}\vert,$$
this means
$$3\sqrt{x^{2}+y^{2}}\geq 3\vert{y}\vert,$$
as $(x,y)\to (0,0)$ you can suppose that $(x,y)\neq (0,0)$, so $$x^{2}+y^{2}\neq 0$$ and you can write:
$$3\vert{y}\vert\leq 3\sqrt{x^{2}+y^{2}}<3\delta.$$
Second, you can say that $y^{2}\geq 0$ then adding in both sides $x^{2}$ you get:
$$ y^{2}+x^{2}\geq x^{2} $$
we know that $x^{2}+y^{2}>0$ then you can divide by $x^{2}+y^{2}$ and get:
$$ 1\geq \frac{x^{2}}{x^{2}+y^{2}},$$
Now you can multiply both sides off the last inequality by $\vert{y}\vert\geq 0$:
$$ \vert{y}\vert\geq \frac{x^{2}\vert{y}\vert}{x^{2}+y^{2}} $$
Multiply by 3 both sides to get:
$$ \frac{3x^{2}\vert{y}\vert}{x^{2}+y^{2}}\leq 3\vert{y}\vert<3\sqrt{x^{2}+y^{2}}<3\delta.$$
You can put $\delta(\epsilon)=\displaystyle{\frac{\epsilon}{3}}$. Notice that you always have:
$$y^{2}\geq 0 \Rightarrow x^{2}+y^{2} \geq x^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\vert{x}\vert$$
in the same way
$$x^{2}\geq 0 \Rightarrow x^{2}+y^{2} \geq y^{2} \Rightarrow \sqrt{x^{2}+y^{2}}\geq\vert{y}\vert$$
From the inequality $x^{2}+y^{2} \leq x^{2}$ you can directly multiply both sides by $3\vert{y}\vert$ to have:
$$\frac{x^{2}}{x^{2}+y^{2}}\leq 1 \Rightarrow \frac{3\vert{y}\vert x^{2}}{x^{2}+y^{2}}\leq 3\vert{y}\vert$$
|
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Prove that if the columns of the $m\times n$ matrix $A$ are linearly independent, then $Ax=b$ has at most one solution.
Prove that if the columns of the $m\times n$ matrix $A$ are linearly independent, then $Ax=b$ has at most one solution.
I was thinking that if the $n$ columns are linearly independent, then the dimension of the column space is $n$, which implies the dimension of the row space is $n$ also. Hence $A$ is $n\times n$, but wouldn't this imply there's exactly one solution of $Ax=b$ rather than at most one?
Is this logic correct, or am I missing something?
|
Let $A = \begin{bmatrix}\mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n\end{bmatrix}$ and $\mathbf{x} = \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}b_1\\b_2\\\vdots\\b_m\end{bmatrix}$.
Now, $A\mathbf{x} = x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + \cdots + x_n\mathbf{a}_n$.
Let $\mathbf{u}$ and $\mathbf{v}$ be two solutions of $A \mathbf{x} = \mathbf{b}$. Then, we have:
$$\begin{cases}
u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + \cdots + u_n\mathbf{a}_n &=& \mathbf{b} & (1)\\
v_1 \mathbf{a}_1 + v_2 \mathbf{a}_2 + \cdots + v_n\mathbf{a}_n &=& \mathbf{b} & (2)
\end{cases}$$
Subtract $(2)$ from $(1)$ to get:
$$(u_1-v_1) \mathbf{a}_1 + (u_2-v_2) \mathbf{a}_2 + \cdots + (u_n-v_n) \mathbf{a}_n = 0$$
Using the fact that $\mathbf{a}_1$ through $\mathbf{a}_n$ are linearly independent, we get $u_1-v_1 = u_2-v_2 = \cdots = u_n-v_n = 0$, i.e. $u_i=v_i$, i.e. $\mathbf{u} = \mathbf{v}$.
|
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|
How to test for convergence for the harmonic series with irregular (binomial) sign changes? How would you test for convergence with a series such as this using the alternating series test?
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}...-\frac{1}{10}+\frac{1}{11}...+\frac{1}{15}-....$
I have attempted to find a summation equation for this by grouping in different ways in order to use the alternating series test but I can't find a way to group these in a logical way. Is there a grouping that would allow for analysis of this series using the alternating series test?
|
I hope I'm interpreting this correctly: you have one $+$ term, then two $-$, three $+$, four $-$, etc. The start and end of a run of $2k+1$ $+$ terms are $1/(2 k^2 + k + 1)$ and $1/(2 k^2 + 3 k + 1)$ and the start and end of the next run of $2k+2$ $-1$ terms are $-1/(2 k^2 + 3 k + 2)$ and $-1/(2 k^2 + 5 k + 3$.
Thus the sum of these two runs is at least
$$ \dfrac{2k+1}{2 k^2 + 3k + 1} - \dfrac{2k+2}{2k^2+3k+2} = \frac{-k}{(2 k^2 + 3k + 1)(2k^2+3k+2)}$$
and at most
$$ \dfrac{2k+1}{2 k^2 +k + 1} - \dfrac{2k+2}{2k^2+5k+3} = \frac{-k}{(2 k^2 + 3k + 1)(2k^2+3k+2)} = \frac{6k+1}{(2k+3)(2k^2+k+1)}$$
Thus in any case it is bounded in absolute value by $C/k^2$ for some $C$, and thus the series converges.
|
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|
A sequence divisible by 9 I was trying to solve this series by mathematical induction for every $n$ from $\Bbb N$ : $u_n=n4^{n+1}-(n+1)4^n+1$ is divisible by $9$.
The initiation was pretty easy, but I only managed to prove $u_{n+1}=3k$ while $k$ is an integer and I don't think if it's divisible by $3$ implies that it is divisible by $9$ ; is it ? if not how can I proceed to prove the divisibility ? by mod maybe? thanks in advance for your answer
|
Base case: $1\cdot 4^2 - 2\cdot 4^1 + 1 = 9$ is divisible by $9$.
Induction step: Assume it is true for $n = a$, say $u_a = 9k$. Now to see what we get for $n = a+1$:
$$
u_{a+1} = (a+1)4^{a+2} - (a+2)4^{a+1} + 1\\
= a4^{a+2} + 4^{a+2} - (a+1)4^{a+1} - 4^{a+1} + 1\\
= 4(a 4^{a+1} - (a+1)4^a + 1) + 4^{a+2} - 4^{a-1} - 3\\
= 4\cdot 9k + 3\cdot 4^{a+1} - 3
$$
Now what remains is showing that $3\cdot 4^{a+1} - 3$ is divisible by $9$. This can be done by induction exactly like for $u_n$, only this time it's easier. We can also show this directly by using the binomial theorem to expand $4^{a+1} = (3+1)^{a+1}$. We see that
$$
3(3+1)^{a+1} - 3= 3(3^{a+1} + (a+1)\cdot 3^a + \cdots + (a+1)\cdot 3 + 1) - 3\\
= 3^{a+2} + (a+1)\cdot 3^{a+1} + \cdots + (a+1)\cdot 3^2 + 3 - 3\\
= 9(3^{a} + (a+1)\cdot 3^{a-1} + \cdots + (a+1))
$$which is divisible by $9$.
|
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Integrating a Rational Function with a Square Root : $ \int \frac{dx}{\sqrt{x^2-3x-10}} $
Integrate $$ \int \frac{dx}{\sqrt{x^2-3x-10}}. $$
I started off with the substitution $\sqrt{x^2-3x-10} = (x-5)t$. To which I got
$$ x = \frac{2+5t^2}{t^2-1} \implies \sqrt{x^2-3x-10} = \left(\frac{2+5t^2}{t^2-1} - 5\right)t = \frac{7t}{t^2-1} $$
and
$$ dx = \frac{-14t}{(t^2-1)^2} dt. $$
Substituting back in
\begin{align*}
\int \frac{t^2-1}{7t} \frac{-14t}{(t^2-1)^2}dt &= -2 \int \frac{dt}{t^2-1} \\
&= - \ln\left| \frac{t-1}{t+1} \right| +c \\
&= - \ln\left| \frac{\frac{\sqrt{x^2-3x-10}}{x-5} - 1}{\frac{\sqrt{x^2-3x-10}}{x-5} + 1} \right| +c \\
&= - \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| + c.
\end{align*}
Is this a valid answer? Wolfram Alpha's answer is slightly different. I know this can come from a difference in method and/or the functions differ by a constant, but with an expression like this it's hard to check.
|
This is an alternative method to compute the given integral.
$$I =\int{\frac{dx}{\sqrt{x^2-3x-10}}} = \int{\frac{dx}{\sqrt{(x-3/2)^2-(7/2)^2}}}\\=\log\left|(x-3/2)+\sqrt{(x-3/2)^2-(7/2)^2}\right|+c.$$
Here $$\int{\frac{dy}{\sqrt{y^2-a^2}}}=\log\left|y+\sqrt{y^2-a^2}\right|+c.$$
|
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$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$
$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$.
Show that $(a-c)(b-c)<0$.
This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate.
Given the assumptions, it is easy to show that
$$0<(a-b)^2<a^2+b^2-ab=c^2.$$ But could not find a promising route to pursue.
Any hint or answer is welcomed.
|
we get $$c=\sqrt{a^2+b^2-ab}$$ since $c>0$ and $$a^2+b^2>ab$$ from here we get
the product
$$(a-\sqrt{a^2+b^2-ab})(b-\sqrt{a^2+b^2-ab})<0$$
we have two cases.
1) $$a>\sqrt{a^2+b^2-ab}$$ and $$b<\sqrt{a^2+b^2-ab}$$ after squaring we get $$a>b$$
2) $$a<\sqrt{a^2+b^2-ab}$$ and $$b>\sqrt{a^2+b^2-ab}$$
after squaring we get $$a<b$$ and our Statement is true.
|
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Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
$$a^3 -(b^3)i + 3(a^2)bi -3(b^2)a$$
Equating real and imaginary parts, we get:
$$ a^3 -3(b^2)a = 1 \tag{1}$$
$$ b^3 -3(a^2)b = 1 \tag{2}$$
Factoring $(1)$, $(2)$ to get some roots in the process:
$$a(a^2 - 3(b^2)) = 1 \tag{1'}$$
$$b(b^2 - 3(a^2)) = 1 \tag{2'}$$
From $(1')$, the roots are possibly given by equations below:
$$a = 1 \tag{3}$$
$$ - \text{or} - $$ // 'or' is used in logical or sense, i.e. either or both
$$ a^2 - 3(b^2) = 1 \tag{4}$$
Substituting $a = 1$ in $(4)$, we get:
$b = 0$; which cannot be a possible solution, as the right side of
question has $b = -1$. Hence, $a \ne 1$ also.
Next, trying for the possible value from $(2')$.
$$ b = 1 \tag{5} $$
$$ -\text{or}- $$ // 'or' is used in logical or sense, i.e. either or both
$$ b^2 - 3(a^2) = 1 \tag{6} $$
Substituting $b = 1$ in $(6)$, we get:
$$ a = 0; $$
which cannot be a possible solution, as the right side of question
has $a = 1$. Hence, $b \ne 1$ also.
I am unable to solve it further, as no solution emerges from the
two equations - $(1')$, $(2')$.
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Hint:
$$a^3-3b^2a=b^3-3a^2b=1 \implies a^3-b^3+3a^2b-3b^2a=0$$ which implies
$$(a-b)(a^2+ab+b^2)+3ab(a-b)=0 \implies (a-b)(a^2+4ab+b^2)=0.$$
As such, $a=b$ or $a^2 + 4ab+b^2 = a^2+4ab+4b^2-3b^2 = 0 \implies a = (-2 \pm \sqrt{3})b.$
If $a=b$ then $a^3-3b^2a = -2a^3=1 \implies \color{red}{a=b=-2^{-1/3}}$.
I leave the calculations for $a=(-2 \pm \sqrt{3})b$ to you.
An alternative way of computing the remaining roots (as also suggested by some other people) is given below.
Let $x_0 = -2^{-1/3}-2^{-1/3}i=-2^{-1/3}(1+i)$. We just found out that $x_0^3 = 1-i$. Now
$$x^3 = 1-i \implies \left(\frac{x}{x_0}\right)^3 = \frac{1-i}{x_0^3}=1.$$
Let $y = x/x_0$. We are looking for the solution of $y^3-1=0$. But
$$y^3-1 = (y-1)(y^2+y+1)=0 \implies y=1 \quad \text{ or }\quad y^2+y+1=0. $$
Now $y=1 \implies x=x_0$, and $$y^2+y+1 = 0 \implies y = \frac{x}{x_0}= \frac{-1\pm \sqrt{3}i}{2}.$$
As such, the remaining cube roots of $1-i$ are $$-2^{-1/3}(1+i)\left(\frac{-1\pm \sqrt{3}i}{2}\right).$$
|
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Help clarifying the steps to find the derivative of $y=(3x+1)^3(2x+5)^{-4}$ For the problem $y=(3x+1)^3(2x+5)^{-4}$ do I use the chain, quotient and product rules? If so how do I know what parts to break up and where the rules apply? For instance would I consider $f(x)$ to be $(3x+1)^3$ and $g(x)$ to be $(2x+5)^{-4}$? If so do I then need to take the product rule of $f(x)$ and the quotient of $g(x)$ and then combine them? Or do I just use one rule?
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Hint: You are free to apply either the quotient rule or the product rule, both of them are convenient to use.
Product rule: Set
$$f(x)=(3x+1)^3\qquad\text{and}\qquad
g(x)=(2x+5)^{-4}
$$ and consider
\begin{align*}
\left(f(x)\cdot g(x)\right)^\prime=\left((3x+1)^3\cdot(2x+5)^{-4}\right)^\prime
\end{align*}
Quotient rule: Set
\begin{align*}
f(x)=(3x+1)^3\qquad\text{and}\qquad g(x)=(2x+5)^4
\end{align*} and consider
\begin{align*}
\left(\frac{f(x)}{g(x)}\right)^\prime=\left(\frac{(3x+1)^3}{(2x+5)^4}\right)^\prime
\end{align*}
In both cases we have to apply the chain rule.
Recall:
\begin{align*}
\color{blue}{\left(f(x)(g(x))^{-1}\right)^\prime}
&=f^\prime(x)\left(g(x)\right)^{-1}+f(x)(-1)(g(x))^{-2} g^\prime(x)\\
&=\frac{f^\prime(x)}{g(x)}-\frac{f(x)g^\prime(x)}{g(x)^2}\\
&=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g^2(x)}\\
&\color{blue}{=\left(\frac{f(x)}{g(x)}\right)^\prime}
\end{align*}
Tip: Try both ways and decide which one you prefer.
|
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Proof by induction (Combinatorics) This is my first question on here so please bear with me, thank you.
Prove that for all positive integers $n$
$$2(1+2+...+n)^4 = (1^5 + 2^5 +...+ n^5) + (1^7 + 2^7 +...+ n^7)$$
After establishing the base case, I proceeded to the induction:
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ (n+1)^5) + (1^7 + 2^7 +...+ (n+1)^7)$
$2(1+2+...+(n+1))^4 = (1^5 + 2^5 +...+ n^5)+(n+1)^5 + (1^7 + 2^7 +...+n^7)+(n+1)^7$
$2(1+2+...+(n+1))^4 = 2(1+2+...+n)^4 + (n+1)^5 + (n+1)^7$
$2(1+2+...+(n+1))^4 - 2(1+2+...+n)^4 = (n+1)^5 + (n+1)^7$
$2[(1+2+...+(n+1))^4 - (1+2+...+n)^4] = (n+1)^5 + (n+1)^7$
And from here I've hit a wall and can't figure out how to continue. I tried substituting $(1+...+n) = [n(n+1)]/2$ and $(n+1)=[2(1+...+n)]/n$ but to no avail.
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You must prove $$2\left(\sum_{i=1}^{n+1}i\right)^4 = \sum_{i=1}^{n+1}i^5 + \sum_{i=1}^{n+1}i^7.$$
Let $S_n = \sum_{i=1}^{n}i = \frac{n(n+1)}{2}$
You have $$2\left(\sum_{i=1}^{n+1}i\right)^4 = 2\left(\left(\sum_{i=1}^{n}i\right)^4 + 4(n+1)S_n^3 + 6(n+1)^2S_N^2 + 4(n+1)^3S_n + (n+1)^4 \right).$$
|
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Calculate the line integral $I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$ Calculate the line integral $$I(a,b)=\int_{x^2+y^2=R^2}\limits\ln\frac{1}{\sqrt{(x-a)^2+(y-b)^2}} ds\quad(a^2+b^2\ne R^2).$$
The parametrized integral path can be given as
$$x=R\cos t,y=R\sin t,t\in[0,2\pi].$$
Then I get into trouble when computing the integral
$$-\frac{1}{2}\int_{0}^{2\pi}R\ln(a^2+b^2+R^2-2aR\cos t-2bR\sin t) dt.$$
Should I take the derivative with respect to $R$ first? Or apply integration by parts?
Update: I find the offered answer is
$$I(a,b)=-2\pi R\ln\max\{R,\sqrt{a^2+b^2}\}$$
and I have examined its correctness.
|
As noted in @Dattier answer,
\begin{align}
I&=-\frac{1}{2}\int_{0}^{2\pi}R\ln(a^2+b^2+R^2-2aR\cos t-2bR\sin t) dt\\
&=-\frac{1}{2}\int_{0}^{2\pi}R\ln(a^2+b^2+R^2-2R\sqrt{a^2+b^2}\cos t) dt\\
\end{align}
If $a^2+b^2<R^2$,
$$ \ln(a^2+b^2+R^2-2R\sqrt{a^2+b^2}\cos t)=2\ln R +\ln(1-2z\cos t +z^2)$$
with $z=\frac{\sqrt{a^2+b^2}}{R}$. We have a decomposition of the second $\ln$ using a generating function for the Chebyshev polynomials:
$$ \ln(1-2z\cos t +z^2)=-2\sum_{n=1}^\infty \frac{\cos nt}{n}z^n$$
Integral over $0,2\pi$ of each term of this series is zero, by symmetry. Thus
$$ I=-2\pi R\ln R$$
If $a^2+b^2>R^2$, calculating along the same lines gives
$$ I=-2\pi R\ln \sqrt{a^2+b^2}$$
|
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For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ =
\ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.
|
Hint $\ \ \ (d,102) = (d,\,2\cdot 3\cdot 17) = (d,2)\,(d,3)\,(d,17).\, $ Computing these gcds by Euclid & Fermat
$\qquad\quad\ \begin{align} (d,2) &= (\color{#c00}{3^{16}(2a)\! +\! 10}, \ \color{#0a0}{3^{17} a\! +\! 66},\,2) = (\color{#c00}0,\color{#0a0}a,2) = (a,2)\\[.2em]
(d,3) &= (\color{#c00}1,\ \color{#0a0}0,\ 3) = 1,\ \text{ and, finally, using $\,\color{#c00}{3^{16}}\equiv 1\!\!\!\!\pmod{\!17}\ $ we have}\\[.2em]
(d,17) &= (\color{#c00}{2a\!+\!10},\,\color{#0a0}{3a\!-\!2},17) = (2a\!+\!10,a\!+\!5,17) = (0,a\!+\!5,17)
\end{align}$
Hence $\, (d,102) = (a,2)\,(a\!+\!5,17)$
Remark $\, $ The elimination that you employed to deduce that $\,d\mid 102\,$ is a special case of the determinant criterion presented here. Namely, applying that with $\,b=3^{16}\,$ yields
$\ (ab,1)\mapsto (2ab\!+\!10,3ab\!+\!66)\,$ has $\,\det = 2(66)\!-\!10(3) = 102\ $ so $\ d\mid 102(ab,1) = 102$
|
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|
Solve nonlinear equation $-xu_x+uu_y=y$ Solve nonlinear equation
$$\left\{\begin{matrix}
-xu_x+uu_y=y & \\
u(x,2x)=0&
\end{matrix}\right.$$
using method of characteristic curves
my attempt:
for the pde we can write it as
$\frac{dx}{-x}=\frac{dy}{u}=\frac{du}{y}$
but i cant solve little please..thank you
|
The given partial differential equation is:
$$\begin{matrix}
-xu_x+uu_y=y & \\
u(x,2x)=0.&
\end{matrix}$$
Finding the general solution
Using the method of characteristics we obtain:
$$\dfrac{dx}{-x}=\dfrac{dy}{u}=\dfrac{du}{y} \qquad (1)$$
We solve the right part of the previous equality:
$$\dfrac{dy}{u}=\dfrac{du}{y}\implies ydy=udu \implies \frac{1}{2}y^2=\frac{1}{2}u^2-\frac{c_1}{2} \implies u =\pm\sqrt{c_1+y^2} $$
Now, use this in the left part of the equality (1):
$$\dfrac{dx}{-x}=\dfrac{dy}{u}\implies \dfrac{dx}{-x}=\dfrac{dy}{\pm\sqrt{c_1+y^2}} \implies -\ln x =\pm \ln(y+\sqrt{y^2+c_1})-\ln c_2 $$
$$\implies -\ln x = \pm\ln(y+u)-\ln c_2 \implies \ln c_2 = \ln(y+u)^{\pm1}+\ln x \implies c_2 = x\left(y+u \right)^{\pm 1}$$
Now, we know that $c_1=F[c_2]$, in which $F$ is an arbitrary function. Hence we obtain:
$$u =\pm\sqrt{c_1+y^2}=\pm\sqrt{F\left[x\left(y+u \right)^{\pm 1}\right]+y^2}$$
$$\implies u^2 = F\left[x\left(y+u \right)^{\pm 1}\right]+y^2$$
Determining the arbitrary function $F$ by using $u(x,2x)=0$
Now, we look as $u(x,2x)=0$:
$$\implies 0^2 = F\left[x\left(2x+0 \right)^{\pm 1}\right]+(2x)^2$$
$$\implies 0 = F\left[x\left(2x\right)^{\pm 1}\right]+4x^2.$$
We have to look at both cases $-1$ and $+1$ separately:
*
*Case $-1$:
$$0 = F\left[\frac{x}{2x}\right]+4x^2 \implies F[2]=-2[2x^2] \implies \text{contradiction as a constant cannot be a function!}$$
*Case $+1$:
$$0 = F[2x^2]+4x^2 \implies F[2x^2]=-2(2x^2) \implies F[u]=-2u,
$$hence the solution become
$$u^2=-2x(y+u)+y^2 \implies u^2+2xu+2xy-y^2.$$
Now, use the quadratic formula to solve for $u$ to obtain:
$$u_{1,2}=\frac{-2x\pm\sqrt{(2x)^2-4(2xy-y^2)}}{2}$$
$$u_{1,2}=-x\pm\sqrt{x^2-(2xy-y^2)}$$
$$u_{1,2}=-x\pm\sqrt{x^2-2xy+y^2}=-x\pm \sqrt{(x-y)^2}=-x\pm|x-y|$$
Investigating the solution
We now have to check which of the solutions do satisfy $u(x,2x)=0$.
*
*Case +1: $$u = -x +|x-y| \implies 0 = -x+|x|.$$
If $x>0$ we get $0=-x+x$, but if $x<0$ we get $0=-x-x$. So the solution $u=-x+|x-y|$ is valid for $x>0$.
*Case -1: $$u = -x -|x-y| \implies 0 = -x-|x|.$$
If $x>0$ we get $0=-x-x$, but if $x<0$ we get $0=-x+x$. So the solution $u=-x-|x-y|$ is valid for $x<0$.
|
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|
Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $
Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $
Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)
=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$
Separating the real & imaginary parts:
$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)
Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);
Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'),
dividing both sides by $b^4$, and having c = a/b, $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),
having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$.
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$
|
Let's start with equation $x^{5}-1=0$ whose one root is $x=\cos(2\pi/5)+i\sin(2\pi/5)$. The equation can be written as $$(x-1)(x^{4}+x^{3}+x^{2}+x+1)=0$$ The first factor gives the root $x=1$ and the second factor leads to the equation $$x^{2}+x^{-2}+x+x^{-1}+1=0$$ Putting $y=x+x^{-1}$ we get $$y^{2}+y-1=0$$ or $$y=\frac{-1\pm\sqrt{5}}{2}$$ Choosing the positive root we get $$2\cos(2\pi/5)=y=\frac{\sqrt{5}-1}{2}$$ Thus the value of $\sin 18^{\circ}$ is obtained as $(\sqrt{5}-1)/4$. Similarly we can find the value of $\cos 18^{\circ}$.
Observe that apart from $x=1$ there are $4$ distinct values of $x$ out of which we have to choose only one namely $x=\cos(2\pi/5)+i\sin(2\pi/5)$. The problem of choice is simplified considerably by using $y=x+x^{-1}$ which satisfies a quadratic equation and therefore has only two values. For our desired value of $x$ the expression $y>0$ and hence the positive root $y$ is chosen. And in reality we are interested in the value $\cos(2\pi/5)=(x+x^{-1})/2=y/2$ so the choice of $y$ completes our work.
The method can be generalized (thanks to Gauss) to solve higher degree equations of type $x^{n} =1$. For example we can solve $x^{17}=1$ and get the value of $\cos(2\pi/17)$ as $$\frac{-1 + \sqrt{17} + \sqrt{34 - 2 \sqrt{17}} + 2\sqrt{17 + 3 \sqrt{17} - \sqrt{34 - 2 \sqrt{17}} - 2\sqrt{34 + 2 \sqrt{17}}}}{16}$$ (see this post for more details).
|
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|
Seeking the maximal parameter value s.t. two-variable inequality still holds Consider the expression
$$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.
The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-GM, do the same with $\,ab^2$, then add and divide by $2ab\,$$\big]$,
but $\,\max\{a,b\}\,$ is not
$\big[$ choose $\,(a,b)=(3,2)\,$ for instance, the expression then evaluates to $\,2\tfrac{11}{12}\,\big]$.
Both arithmetic mean ($x=1$) and the maximum ($x=\infty$) are instances of the Hölder mean
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\quad\text{with }\; x\,\in\,\{-\infty\}\cup\mathbb R\cup\{\infty\}$$
aka Power mean, known to be strictly increasing with $\,x\,$ if $\,a\ne b$, and my question is:
What is the maximal value of $\,x\,$ such that
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\;\leq\;\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
holds for all $\,a,b>0\,$?
Let's pick up the specific "max counter-example" $\,(a,b)=(3,2)\,$.
The following plot screen-shot displays the zero, by courtesy of WolframAlpha:
Returning to the general case we can at least state that $\,x_{max}\geqslant 5\,$:
After taking the fifth power of the corresponding expression and clearing denominators (to arrive at the LHS as of below) I could find a
Certificate of positivity
$$\begin{eqnarray}
\left(a^3 + b^3\right)^5 -16a^5b^5\left(a^5 +b^5\right)\; & =\;\left(a+b\right)\left(a-b\right)^2\big[a^{12} + a^{11}b +2a^{10}b^2 +4a^9b^3 +8a^8b^4 \\
& +8a^6b^6 +8a^4b^8 +4a^3b^9 + 2a^2b^{10} +ab^{11} +b^{12}\big] \\
& + 3a^3b^3\left(a^2+b^2\right)\left(a+b\right)^3\left(a-b\right)^4
\end{eqnarray}$$
It shows also that the equality case holds iff $\,a=b$.
|
The following inequality is true.
Let $a$ and $b$ be positive numbers.
Prove that
$$\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\geq\sqrt[9]{\frac{a^9+b^9}{2}}.$$
Indeed, we need to prove that
$$(a^3+b^3)^9\geq256(a^9+b^9)a^9b^9$$ or
$$(a^3+b^3)^8\geq256(a^6-a^3b^3+b^6)a^9b^9.$$
Now, let $a^6+b^6=2ua^3b^3.$
Hence, $u\geq1$ and we need to prove that
$$(2u+2)^4\geq256(2u-1),$$ which is AM-GM:
$$(2u+2)^4=(2u-1+3)^4\geq\left(4\sqrt[4]{(2u-1)\cdot1^3}\right)^4=256(2u-1).$$
Done!
|
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|
What will be the complete area of curve $a^2x^2=y^3(2a-y)$ I want to calculate the area of $a^2x^2=y^3(2a-y)$ using double integral.
I decided to guess the curve and wrote the equation is:-
$$\pm x=\frac{y\sqrt{y(2a-y)}}{a}$$
which means the curve is symmetric about $y$ axis and $x=0$ at $y=0,2a$.
Thus, area should be:-
$$Area= 2\times\int_0^{2a}\int_0^{\frac{y\sqrt{2ay-y^2}}{a}}dx.dy$$
$$=\frac{2}{a}\times\int_0^{2a}y\sqrt{4a^2-(y-2a)^2}.dy$$
Now, how do I integrate it? Please help
|
Try to use $$\int (f(x))^a f^{(1)}(x) = \frac{f(x)^{a+1}}{(a+1)}$$
and
$$\int \sqrt{a^2-x^2}=\frac{a^2}{2} \arcsin(x/2) + \frac{x}{2}\sqrt{a^2-x^2}$$
|
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|
How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$ I have been trying to figure out how to evaluate the following sum:
$$S_n=\sum_{k=1}^{n} \frac{k}{k^4+1/4}$$
In the problem, the value of $S_{10}$ was given as $\frac{220}{221}$.
I have tried partial decomposition, no where I go. Series only seems like it telescopes, otherwise there isn't another way.
Any ideas are appreciated!
|
Try to break the denominator into product of two factors:
$$\begin{align}
4k^4 + 1 &= (2k^2)^2 + 1 + 2 (2k^2) - 2 (2k^2) \\
&= (2k^2 +1)^2 - (2k)^2 \\
&= (2k^2 +2k +1)(2k^2 -2k+1)
\end{align}$$
Using this we see the general term as:
$$T_k = \dfrac{1}{2k^2-2k+1} - \dfrac{1}{2k^2+2k+1} \\
T_{k+1} = \dfrac{1}{2k^2+2k+1} - \dfrac{1}{2(k+1)^2+2(k+1)+1} $$
Alternate terms cancel and the sum telescopes to:
$$1-\frac{1}{2n^2+2n+1}$$
|
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|
How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
1 & 3 & 1 & 1 & 1\\
1 & 1 & 4 & 1 & 1\\
1 & 1 & 1 & 5 & 1\\
1 & 1 & 1 & 1 & 6
\end{pmatrix}
I expanded it out and got $349$ but I feel there must be some trick to easily compute it.
|
Your matrix is in a very nice form. As an alternate answer, you can use the matrix determinant lemma in this case to do the calculation quite cleanly.
Let $\mathbf{M}$ be the matrix you provided. Then $\mathbf{M}$ = $\mathbf{A} + \mathbf{j}\mathbf{j^t}$,
where
$$
\mathbf{A} =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 2 & 0 & 0 & 0\\
0 & 0 & 3 & 0 & 0\\
0 & 0 & 0 & 4 & 0\\
0 & 0 & 0 & 0 & 5
\end{pmatrix}
$$ and
$$
\mathbf{j} =
\begin{pmatrix}
1\\
1\\
1\\
1\\
1
\end{pmatrix}.
$$
From applying the lemma, we have that $\text{det}(\mathbf{M}) = (1 + \mathbf{j^tA^{-1}j}) \ \text{det}(\mathbf{A})$.
Since $\mathbf{A}$ is diagonal in this case, the determinant is very easy to compute. It's just $5*4*3*2*1=120$. Moreover,
$$
\mathbf{A^{-1}} =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
0 & \frac{1}{2} & 0 & 0 & 0\\
0 & 0 & \frac{1}{3} & 0 & 0\\
0 & 0 & 0 & \frac{1}{4} & 0\\
0 & 0 & 0 & 0 & \frac{1}{5}
\end{pmatrix}
$$
and since $\mathbf{j}$ is just a vector of ones, it's easy to see that
$$\mathbf{j^tA^{-1}j} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$$.
Thus, we see that
$$ \text{det}(\mathbf{M}) = (1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}) * 120 = 394$$
after applying some simplification or a calculator.
|
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|
Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$.
A000332 has the comment:
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.
The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$):
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
|
The identity is equivalent to
\begin{align*}
n&=\binom{3}{0}\binom{n + 3}{n-1}-\binom{3}{1} \binom{n + 2}{n-2} + \binom{3}{2}\binom{n + 1}{n-3} - \binom{3}{3}\binom{n}{n-4}\\
&=\sum_{k=0}^3 (-1)^k\binom{3}{k}\binom{(n-k)+3}{(n-k)-1}=[x^n](1-x)^3\cdot \frac{x}{(1-x)^5}=[x^{n+1}]\frac{1}{(1-x)^2}
\end{align*}
which holds.
P.S. Note that by replacing $3$ with $m$, we get
$$\sum_{k=0}^m (-1)^k\binom{m}{k}\binom{(n-k)+m}{(n-k)-1}=[x^n](1-x)^m\cdot \frac{x}{(1-x)^{m+2}}=[x^{n+1}]\frac{1}{(1-x)^2}=n.$$
|
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|
How to calculate the nth term and the sum in this series if the common difference between them isn't explicit? I have this series comprised of
$$1,2,5,10,17,26,...$$,
and so on. So far i have found that they add up in intervals being odd numbers. But I don't know how to find, let's say the 16th term and the sum up to that term. What should I do?
Edit:
Although it is somewhat possible to calculate the sum up to the 16th term in the above equation, what if the series is of higher order like this?.
$$6,7,14,27,46,...$$
Is there a shortcut to calculate the sum up to the 16th term? Can this be done by hand or would be necessary to use software like Maple?.
|
As Guarang Tandon says. You can create a table of differences between consecutive terms and then repeat that process on the list you just created.
\begin{array}{|l|c|ccccccc|}
\hline
\text{index} & n & 0 & 1 & 2 & 3 & 4 & 5 & \dots\\
\hline
\text{sequence} & f_n & 1 & 2 & 5 & 10 & 17 & 26 & \dots \\
\text{first differences} & \Delta f_n && 1 & 3 & 5 & 7 & 9 & \dots \\
\text{second differences} & \Delta^2 f_n &&& 2 & 2 & 2 & 2 & \dots \\
\hline
\end{array}
If you are lucky, at some point, all of the differences will be constant. (Of course this is an unverifiable claim. All we know for sure is that the first four second differences are constant. We can only assume that this fortuitous pattern continues.)
It can be proved that, if $\Delta^d f_n$ is constant, then then $f_n$ is a $d$th degree polynomial in $n$.
If the second differences are constant, then the first differences must be an arithmetic sequence. In this case, we see that $\Delta f_n = 2n+1$. It follows that $$f_{n+1} = f_n + 2n+1 \tag 1$$.
If the first differences are an arithmetic sequence, then $f_n = an^2 + bn + c$ for some real numbers $a, b,$ and $c$.
We can now use modified form of mathematical induction to find the values of $a,b, $ and $c$.
Since $f_0=1$, then $1 = a\cdot0^2 + b\cdot 0 + c = c$. So $f_n=an^2+bn+1$.
First we note that $$f_{n+1} = a(n+1)^2 + b(n+1) + 1 = f_n + 2an + (a+b).$$
Comparing that to $f_{n+1}= f_n+ 2n+1$, we see that $2a=2$ and $a+b=1$. Hence $a=1$ and $b=0$.
We conclude that $f_n=n^2+1$.
We know that $\sum_{k=0}^n k^2=\frac 16n(n+1)(2n+1)$.
Then
\begin{align}
\sum_{k=0}^n f_k
&= \sum_{k=0}^n (k^2+1) \\
&= \sum_{k=0}^n k^2 + \sum_{k=0}^n 1 \\
&= \frac 16n(n+1)(2n+1) + (n+1) \\
&= \frac 16(n+1)(2n^2+n) + \frac 16(n+1)6 \\
&= \frac 16(n+1)(2n^2+n+6) \\
\end{align}
Check:
$\qquad \sum_{k=0}^5 f_k = 1+2+5+10+17+26=61$
$\qquad \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=5}=\frac 16(6)(61)=61$
So, since we started counting at $n=0$, the $16$th term is $f_{15} = 15^2+1=226$
and the sum of the first $16$ terms is
$\sum_{k=0}^{15} f_k = \left. \dfrac 16(n+1)(2n^2+n+6)\right|_{n=15}=1256$.
|
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|
Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $n$ is a multiple of $3$ since then $2^{3k} + 3^{3k} \equiv 1 + 6 \pmod 7$, but it doesn't work for $n = 6$
|
$2^{3}\equiv 8\equiv \color{red}1 \bmod 7$, which is why you get a cycle length of $3$ for powers of $2 \bmod 7$. The cycle only closes when you reach $a^k\equiv 1$.
However $3^3=27\equiv 6 \bmod 7$ means that the exponential cycle length for $3\bmod 7$ is more than $3$. Fermat's little theorem says that every cycle length$\bmod 7$ (of numbers coprime to $7$) must divide $6$, so the cycle length for $3$ must be $6$. (and indeed $3^6 \equiv (3^3)^2\equiv 6^2 \equiv 36 \equiv 1 \bmod 7$).
|
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|
What is $\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }$? Given that
$$f(x)=8x-f(3x)-\sin^2(2x),$$
find
$$\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }$$
|
$$f(x)=8x-f(3x)-\sin^2(2x)\Longleftrightarrow f(3x) + f(x) = 8x-\sin^2(2x)\\\Longleftrightarrow (f(3x)-3x) + (f(x) -x) = 4x-\sin^2(2x) $$
$$\Longleftrightarrow \frac{f(3x)-3x}{9x^2} + \frac{f(x) -x}{9x^2} = \frac{4x-\sin^2(2x)}{9x^2}$$
Let $$\ell = \lim_{x\to 0}\frac{f(x) -x}{x^2}$$
Then,
$$\lim_{x\to 0}\frac{4x-\sin^2(2x)}{9x^2}= \lim_{x\to 0}\frac{f(3x)-3x}{9x^2} +\frac{f(x) -x}{9x^2}=\lim_{x\to 0}\frac{f(3x)-3x}{(3x)^2} +\frac19\frac{f(x) -x}{x^2}= \frac{10}{9}\ell $$
Since $$\lim_{x\to 0}\frac{f(3x)-3x}{9x^2}=\ell$$
Hence,
$$\lim_{x\to 0}\frac{f(x) -x}{x^2}=\ell =\frac{1}{10} \lim_{x\to 0}\frac{4x-\sin^2(2x)}{x^2} \sim \frac{1}{10} \lim_{x\to 0}\frac{4x-4x^2}{x^2} $$
But this latest limit does not exists.
|
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"timestamp": "2023-03-29T00:00:00",
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|
linear algebra matrix Given the following matrix $A$, find an invertible matrix $U$ so that $UA$ is equal to the reduced row-echelon form of $A$:
$$
A=\begin{bmatrix}
0 & 2 & 2 & 2 \\
0 & 3 & 3 & 3 \\
1 & 1 & -1 & -2
\end{bmatrix}
$$
|
*
*Swap rows $1$ and $3$, i.e. pre-multiply by the matrix $\;U_1=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$. You get
$$\begin{bmatrix}1&1&-1&-2\\0&2&2&2\\0&3&3&3\end{bmatrix}. $$
*Multiply the (new) second row by $1/2$, i.e. multiply by the matrix $\;U_2=\smash{\begin{bmatrix}1&0&0\\0&\frac12&0\\0&0&1\end{bmatrix}}$ to get
$$\begin{bmatrix}1&1&-1&-2\\0&1&1&1\\0&3&3&3\end{bmatrix}. $$
*$R_3\leftarrow R_3-3R_2$, i.e. multiply on the left by $\;U_3=\begin{bmatrix}1&0&0\\0&1&0\\0&-3&1\end{bmatrix}$ to get
$\begin{bmatrix}1&1&-1&-2\\0&1&1&1\\0&0&0&0\end{bmatrix}. $
*Substract $R_2$ from $R_1$, i.e. multiply on the left by $\;U_4=\smash[b]{\begin{bmatrix}1&-1&0\\0&1&0\\0&0&1\end{bmatrix}}$ to get the reduced row echelon form:
$$\begin{bmatrix}1&0&-2&-3\\0&1&1&1\\0&0&0&0\end{bmatrix}.\qquad\qquad $$
In all you have multiplied by
$$U=U_4U_3U_2U_1.$$
|
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|
Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$b^2 = x(1-x) \;\;\implies\;\; x^2 - x + b^2 = 0,$$ which by the quadratic equation, we get $$x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right).$$
My question is: How do I know that neither solution is extraneous? These are the solutions in the back of the textbook, but I am left questioning when these are the solutions.
The alternative solution involves some more clever thinking:
$$\begin{align}2\log_bx + 2\log_b(1-x) &= 4\\\log_bx^2(1-x)^2 &= 4\\ b^4 &= x^2(x^2 - 2x + 1)\\ 0&=x^4 - 2x^3 + x^2 - b^4\\ &=x^2(x - 1)^2 - b^4\\ & = [x(x-1)]^2 - (b^2)^2 \\ &= [x(x-1) + b^2][x(x-1) - b^2],\end{align}$$ which implies that $$x = {1\over2}\left(1 \pm\sqrt{1 + 4b^2}\right) \;\;\;\text{or} \;\;\; {1\over2}\left(1 \pm\sqrt{1 - 4b^2}\right),$$
which is even worse, because now there are $4$ solutions to check.
The most I know is that since $b$ is a logarithmic base, $b>0$ and $b\ne 1$. But what happens when $b$ is something like $2$? Then you end up with a complex number under the root (of the solutions from the easier way), and that doesn't necessarily make sense if I'm trying to solve the logarithmic equation over the reals.
|
$\log_b x$ and $\log_b (1-x)$ existing mean $0 < x $ and $0 < 1-x $ so $0 < x < 1$.
When you get $b^2 = x(1-x)$ use similiarity and let $m=|\frac 12 - x|$ so $b^2= (\frac 12 + m)(\frac 12 - m^2) = \frac 14 - m^2 > \frac 14$ so $b < \frac 12$.
Therefore neither solution $x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right)$ is extraneous.
In your second method... I'm not sure why you consider that more clever. By not dividing by $2$ you are adding extraneous solutions and making it more complicated. I'd call it a lot less clever.
But as $0 < x < 1$ you know $x(x-1) + b^2 > 0$ so you can rule out the first pair of solutions. As as above $b^4 = x^2(1-x)^2$ means $b^2 = x(1-x) < \frac 14$.
|
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|
How to find the closed form of $f(n) = 9^k \times (-56) + f(n-1)$ I have to find the closed form for
$$T(n) = \begin{cases}
2 , &\text{ if } n=0 \\
9T(n-1) - 56n + 63, &\text{ if } n > 1
\end{cases}$$
I used the repeated substitution method and I found that the pattern for the coefficient of n is equal to the following:
$$f(1) = -56$$
$$f(n) = 9^{n-1} \times (-56) + f(n-1)$$
I tried to find the closed form of $f(n) = 9^{n-1} \times (-56) + f(n-1)$, but it just got more and more confusing. I believe it may be a series of some sort. Is there a way to find a closed form for this?
Thank you!
|
Usually, these problems are solved using induction. But induction requires you to already 'know' (or guess) the answer. Using generating functions we can solve the problem without having to ever guess:
$$t(n) = 9t(n-1) - 56n + 63$$
$$t(0) = 2$$
Now suppose we write:
$$T(x) = \sum_{n=0}^\infty t(n)x^n$$
We know:
$$T(x) = 2 + \sum_{n=1}^\infty (9t(n-1) - 56n + 63)x^n$$
$$T(x) = 2 +9\sum_{n=1}^\infty t(n-1)x^n -56\sum_{n=1}^\infty nx^n+ 63 \sum_{n=1}^\infty x^n$$
$$T(x) = 2 +9\sum_{n=0}^\infty t(n)x^{n+1} -56\frac{x}{(1-x)^2}+ 63 \frac{x}{1-x}$$
$$T(x) = 2 +9xT(x) -56\frac{x}{(1-x)^2}+ 63 \frac{x}{1-x}$$
$$T(x) = 2\frac{1}{1-9x} -56\frac{x}{(1-9x)(1-x)^2}+ 63 \frac{x}{(1-9x)(1-x)}$$
$$T(x) = 2\frac{1}{1-9x} +\frac{-56x + 63x(1-x)}{(1-9x)(1-x)^2}$$
$$T(x) = 2\frac{1}{1-9x} +\frac{7x(1-9x)}{(1-9x)(1-x)^2}$$
$$T(x) = 2\frac{1}{1-9x} +7\frac{x}{(1-x)^2}$$
$$T(x) = 2\sum_{n=0}^\infty 9^nx^n + 7\sum_{n=0}^\infty nx^n$$
$$T(x) = \sum_{n=0}^\infty \Big (2\cdot 9^n + 7n\Big)x^n = \sum_{n=0}^\infty t(n)x^n$$
Therefore, $t(n) = 2\cdot 9^n + 7n$.
|
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|
How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?
|
Let $x = \sin^2 \alpha,y = \cos^2 \alpha$. Note that $x+y = 1$ so $x^3 + y^3 = (x+y)(x^2+y^2-xy) = (x+y)((x+y)^2-3xy) = 1-3xy$
Also $x^2+y^2 = (x+y)^2-2xy = 1-2xy$
$R = (1-3xy) + K(1-2xy) = 1+K - (3+2K)xy$. So if $K = -3/2$ then $R$ will be independent of $\alpha$. If $K \neq -3/2$ then it will depend on $xy$ which then depends on $\alpha$.
Generally when you see an expression in the form of $\sin^{2n}x + \cos^{2n}x$ it helps to factor out $\sin^2 + \cos^2$ like what I did above
|
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|
Solve system of equations: $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$ $\sqrt2 \sin x = \sin y, \sqrt2\cos x = \sqrt5\cos y$
I tried to use tangent half-angle substitution, tried to sum these two equations and get this
$\sin(x+ \pi/4) = \sqrt3 / \sqrt2 \sin(y+\tan^{-1}(\sqrt5))$
I stuck.
Any Help is appreciated!
|
Substitute $u=\sin x;\;v=\sin y$. Then square both side of the two equations
$
\left\{
\begin{array}{l}
2 u^2=v^2 \\
2 \left(1-u^2\right)=5 \left(1-v^2\right) \\
\end{array}
\right.
$
Solutions need to be checked. Solution in $[...]$ are not solutions of the given equation
$$\sin x = -\frac{\sqrt{\frac{3}{2}}}{2},\sin y = -\frac{\sqrt{3}}{2},\left[\sin x = -\frac{\sqrt{\frac{3}{2}}}{2},\sin y = \frac{\sqrt{3}}{2}\right],\\\left[\sin x = \frac{\sqrt{\frac{3}{2}}}{2},\sin y = -\frac{\sqrt{3}}{2}\right],\sin x = \frac{\sqrt{\frac{3}{2}}}{2},\sin y = \frac{\sqrt{3}}{2}$$
Hope this helps
|
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|
Find area of triangle ABD Let ABC be a right-angled triangle $B=90$. Let in a triangle pick a point $D$ inside the triangele such that $AD= 20, DC=15, DB=10$ and $AB=2BC$. What is the area of triangle $ABD$?
Thanks!
|
i have got three equations
$$5a^2=20^2+15^2-2\cdot 20\cdot 15\cos(2\pi-\alpha-\beta)$$
$$a^2=15^2+10^2-2\cdot 15\cdot 10\cos(\beta)$$
$$4a^2=10^2+20^2-2\cdot 10\cdot20\cos(\alpha)$$
solving this we get $$\alpha=\arctan\left(\frac{1}{2}\right)$$
and $$A_{\Delta ABD}=\frac{1}{2}20\cdot 10\sin\left(\arctan\left(\frac{1}{2}\right)\right)$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor $? find the limits :
$$\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor =?$$
My try :
$$\lfloor \frac{1}{x+1}\rfloor=\frac{1}{x+1}-p_x \ \ \ : 0\leq p_x <1$$
So we have :
$$\lim_{x \to 1^-} (x+1) (\frac{1}{x+1}-p_x) \\= \lim_{x \to 1^-} -(x+1)p_x=!??$$
Now what ?
|
$$\lim_{x \to 1^-} (x+1) \left\lfloor \frac{1}{x+1}\right\rfloor $$
Let ${x = 1-y}$
So our equation is same as
$$\lim_{y \to 0^+} (2-y) \left\lfloor \frac{1}{2-y}\right\rfloor $$
Floor value is $0$. Also value outside floor funtion tends to $2$.
Thus our equation becomes
$2\cdot 0=0$
So answer is $0$.
QED
|
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|
Proving that: $ | a + b | + |a-b| \ge|a| + |b|$ I am trying to prove this for nearly an hour now:
$$
\tag{$\forall a,b \in \mathbb{R}$}| a + b | + |a-b| \ge|a| + |b|
$$
I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ?
Thanks in advance.
|
Here's the way I see this geometrically in $\mathbb{C}$: let's say we have two complex numbers $a, b$ and consider the parallelogram formed by $0, a, a + b, b$. The midpoints of the diagonals coincide at the point $\frac{a + b}{2}$. These diagonals cut the parallelogram into four triangles, on each of which we can perform the triangle inequality. We get the following inequalities:
\begin{align*}
|a - 0| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right| \\
|(a + b) - a| &\le \left| (a + b) - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - a \right| \\
|b - (a + b)| &\le \left| b - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - (a + b) \right| \\
|0 - b| &\le \left| a - \frac{a + b}{2} \right| + \left| \frac{a + b}{2} - 0 \right|
\end{align*}
Simplifying the above inequalities and summing them up yields the desired inequality.
|
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|
Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sqrt{\frac{1+x}{1-x}}}\right)'}_{=D}dx.$$
So, the integrand $D$ remains to simplify:
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2} \quad \quad (1).$$
Setting $a=\frac{1+x}{1-x}$ for notation's sake I get
$$D=x\cdot\frac{1}{1+a}\cdot\frac{1}{2\sqrt{a}}\cdot\frac{1}{(x-1)^2}=\frac{x}{(2\sqrt{a}+2a\sqrt{a})(x^2-2x+1)},$$
and I get nowhere. Any tips on how to move on from $(1)?$
NOTE: I don't want other suggestions to solutions, I need help to sort out the arithmetic to the above from equation (1).
|
Inverse trig identities are less familiar than regular trig identities. So perhaps if you invert this, so that regular trig identities would apply, you find something noteworthy:
$$\begin{align}
y&=\arctan\sqrt{\frac{1+x}{1-x}}\\
\tan(y)&=\sqrt{\frac{1+x}{1-x}}\\
\tan^2(y)&=\frac{1+x}{1-x}\\
\tan^2(y)-x\tan^2(y)&=1+x\\
-x-x\tan^2(y)&=1-\tan^2(y)\\
x&=\frac{\tan^2(y)-1}{\tan^2(y)-1}\\
&=\frac{\tan^2(y)-1}{\sec^2(y)}\\
&=\sin^2(y)-\cos^2(y)\\
&x=-\cos(2y)
\end{align}$$
So now you can invert that back, and actually $y=\frac{1}{2}\arccos(-x)$, much easier to work with if you know the antiderivative for $\arccos$.
|
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|
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
I have a solution involving a variable change ($\alpha=x+2$) which is not beautiful in my opinion! I'm looking for a more beautiful solution.
|
Adding both the equations we have $$\alpha^3-6\alpha^2+13\alpha+\beta^3-6\beta^2+13\beta=20$$
Which is equivalent to
$$(\alpha-2)^3+(\beta-2)^3+(\alpha-2)+(\beta-2)=0$$
Let $\alpha-2=a$ and $\beta -2=b$
Now we've ;
$$a^3+b^3+a+b=0$$
$$(a+b)(a^2-ab+b^2+1)=0$$
This give us
$$a+b=0 \implies \color{blue}{\alpha+\beta=4}$$
Looking the given equation, we can conclude that $\alpha,\beta >0 $ and thus another factor cannot be zero.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all real solutions of $\sqrt{x} - \sqrt{2-2x} = 1$ Squaring both sides of $\sqrt{x} - \sqrt{2-2x} = 1$ and rearranging
I arrive at the quadratic $9x^2 - 10x + 1 = 0$ which has solutions $x=1/9$ and $x=1$.
I don't understand why $x=1$ fits the original equation but $x=1/9$ doesn't (left hand side gives $-1$).
|
$$\sqrt{x} - 1 = \sqrt{2-2x}$$
$$x -2\sqrt{x} + 1 = 2-2x$$
$$3x -2\sqrt{x} - 1 = 0$$
Now if $\sqrt{x} = u$ we have:
$$3u^2 -2u - 1 = 0$$
$$u = 1 \vee u=-\frac{1}{3}$$
In the case $u = 1$ we've found the solution $x = 1$. But in the case $u = -\dfrac{1}{3}$ we get:
$$\sqrt{x} = -\frac{1}{3}$$
Do you see the issue now?
|
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|
Center of Mass Help For a question, it asked us to find the center of mass of the region between $y=\sqrt x$ and $y=x^3$. I was wondering if the way that I did it was correct. If not, what is the correct formuala for solving center of mass Center of Mass
|
The center of mass would be located by using the following formulas:
$\bar{x}=\frac{1}{A}\int_\limits{a}^bx[f(x)-g(x)]dx$ and $\bar{y}=\frac{1}{A}\int_\limits{a}^b\frac{1}{2}[(f(x))^2-(g(x))^2]dx$
$A=\int_\limits{a}^b[f(x)-g(x)]dx$
$\bar{x}=\frac{\int_\limits{0}^1x(\sqrt{x}-x^3)dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{\int_\limits{0}^1({x}^\frac{3}2-x^4)dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{[\frac{2}{5}x^\frac{5}{2}-\frac{x^5}{5}]_0^1}{[\frac{2}{3}x^\frac{3}{2}-\frac{x^4}{4}]_0^1}=\frac{12}{25}$
$\bar{y}=\frac{\int_\limits{0}^1\frac{1}{2}[(\sqrt{x})^2-(x^3)^2]dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{\int_\limits{0}^1\frac{1}{2}[x-x^6]dx}{\int_\limits{0}^1\sqrt{x}-x^3dx}=\frac{1}{2}\frac{[\frac{1}{2}x^2-\frac{x^7}{7}]_0^1}{[\frac{2}{3}x^\frac{3}{2}-\frac{x^4}{4}]_0^1}=\frac{3}{7}$
Your approach is right just need to do it for the y axis too because its a point which needs an x and y coordinate.
|
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|
Finding the value of a tricky limit I have some difficulties with my homework in mathematical analysis and I don't really have any good ideas, how to get off the mark. If you, guys, could give me any ideas, tips or solutions for the following task, I would be really thankful.
The task is as follows:
Let $a_{n}$ be a sequence with positive members so that
$$ \lim_{n} \frac{a_{n}}{n} = 0 ,$$
$$\limsup\limits_{n}\frac{a_1 + a_2 + ... + a_n }{n} \in \mathbb{R}.$$
Find the value of:
$$ \lim_{n} \frac{a_1^2 + a_2^2 + ... + a_n^2}{n^2} $$
|
Trying the problem with $a_n=c$ suggests the answer is $0$.
Since $\limsup\limits_{n}\frac{a_1 + a_2 + ... + a_n }{n} <\infty$, there exists some $M>0$ such that for all $n$, $\displaystyle \frac{a_1 + a_2 + ... + a_n }{n}\leq M$.
Let $\epsilon>0$. Since $\lim_n \frac{a_n}n=0$, there is some $N$ such that $n\geq N\implies a_n\leq \frac{\epsilon}{2M} n\implies a_n^2\leq \frac{\epsilon}{2M} na_n$.
Then for $n\geq N$, $$\begin{align} \frac{a_1^2 + a_2^2 + ... + a_n^2 }{n^2}
&\leq
\frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} + \frac{a_N^2 + a_{N+1}^2 + ... + a_n^2 }{n^2}\\&\leq \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} +\frac{\epsilon}{2M}\frac{a_N + a_{N+1} + ... + a_n }{n} \\
&\leq \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} +\frac{\epsilon}{2M}\frac{a_1 + a_2 + ... + a_n }{n} \\
&\leq \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} + \frac{\epsilon}{2}
\end{align}$$
Since $\displaystyle \lim_n \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2} = 0$, there is some $N'$ such that $n\geq N'\implies \frac{a_1^2 + a_2^2 + ... + a_N^2 }{n^2}\leq \frac{\epsilon}{2}$
For $n\geq \max(N,N')$, $$\frac{a_1^2 + a_2^2 + ... + a_n^2 }{n^2}\leq \epsilon$$
Hence convergence to $0$.
|
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|
Finding value of determinant, Where $A+B+C+P+Q+R=0$ Finding value of Determinant $$\begin{vmatrix}
\tan (A+P) & \tan(B+P) & \tan(C+P)\\\\
\tan (A+Q)& \tan (B+Q) & \tan (C+Q)\\\\
\tan (A+R)&\tan (B+R) & \tan(C+R)
\end{vmatrix}$$ for all values of $A,B,C,P,Q,R$ Where $A+B+C+P+Q+R=0$
I did not understand how to start it, Could some help me, Thanks
|
We want to find the determinant of this Matrix.
$$
\begin{pmatrix}
Tan(A+P) & Tan(B+P) & Tan(C+P)\\
Tan(A+Q) & Tan(B+Q) & Tan(C+Q)\\
Tan(A+R) & Tan(B+R) & Tan(C+R)
\end{pmatrix}
$$
The determinant of this matrix gives us \to
$$
-Tan(A+R) \cdot Tan(B+Q) \cdot Tan(C+P)\\
+Tan(A+Q) \cdot Tan(B+R) \cdot Tan(C+P)\\
+Tan(A+R) \cdot Tan(B+P) \cdot Tan(C+Q)\\
-Tan(A+P) \cdot Tan(B+R) \cdot Tan(C+Q)\\
-Tan(A+Q) \cdot Tan(B+P) \cdot Tan(C+R)\\
+Tan(A+P) \cdot Tan(B+Q) \cdot Tan(C+R)
$$
But I still don't know how to depress it after this, even though you said
$
A+B+C+P+Q+R = 0
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the sum of function series The serie is :
$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 2\,n-1
\right) {x}^{2\,n}}{{4}^{n}}}
$$
I found out the derivative $2\,{\frac {n \left( -1 \right) ^{n}{\left( 2\,n-1 \right)}{x}^{2\,n-1}
}{{4}^{n}}}
$ , which gave me this $2\,n-1$ in power and in numerator, but I have no idea what to do next. So what do I do next ? Any detailed explanation would be very appreciated.
|
$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 2\,n-1
\right) {x}^{2\,n}}{{4}^{n}}}=\sum _{n=1}^{\infty } \frac{(-1)^n (2 n) x^{2 n}}{4^n} + \sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^{2 n}}{4^n}$$
For the first series
$$\sum _{n=1}^{\infty } \frac{(-1)^n x^{2 n}}{4^n}=\sum _{n=1}^{\infty } \left(-\frac{x^2}{4}\right)^n =\frac{1}{1+x^2/4}-1=-\frac{x^2}{x^2+4}$$
Then derive and get
$$\sum _{n=1}^{\infty } \frac{(-1)^n (2 n) x^{2 n}}{4^n} =-\frac{8 x^2}{\left(x^2+4\right)^2}$$
The second series is
$$\sum _{n=1}^{\infty } \frac{(-1)^{n+1} x^{2 n}}{4^n}=-\sum _{n=1}^{\infty } \left(-\frac{x^2}{4}\right)^{n} =-\left(\frac{1}{1+\frac{x^2}{4}}-1\right)=\frac{x^2}{x^2+4}$$
so the sum is
$$-\frac{8 x^2}{\left(x^2+4\right)^2}+\frac{x^2}{x^2+4}=\frac{x^2 \left(x^2-4\right)}{\left(x^2+4\right)^2}$$
Hope this is useful
|
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|
Numbers $a^{p}a^q$ congruent to 1 modulo pq Consider numbers of the form $a^{p}a^q=a^{p+q} $ where $p,q$ are primes.
Now I'm interested when such number is congruent to $1$ modulo product $pq$.
Take for example $p=3$ and $q=5$.
Then it seems that every number $a$ coprime with $p$ and $q$ is congruent to $1$ modulo $(pq)$.
For example
$7^8-1 \equiv 1 \pmod{15} , \ \ 11^8-1 \equiv 1 \pmod{15}, \ \ 13^8-1 \equiv 1 \pmod{15} , \ \dots\ \ \ 601^8-1 \equiv 1 \pmod{15} $ etc...
If we take $p=5$ and $q=7$ we have
$7^{12}-1 \equiv 1 \pmod{35} , \ \ 11^{12}-1 \equiv 1 \pmod{35} , \ \ 13^{12}-1 \equiv 1 \pmod{35}$, etc...
I don't know whether it is true for all $a$ but for pairs $(p,q)=\{(3,5),(5,7)\}$ I have found no exceptions.
One could think that for others pairs of primes situation is similar however for pairs like $(11,13)$ or $(17,19)$ the pattern is not repeating.
For example $7^{11 + 13} \ \equiv 14 \pmod {11 \cdot 13}, \ 17^{11 + 13} \ \equiv 53 \pmod {11 \cdot 13} $.
So my question:
*
*What are conditions imposed on pairs $(p,q)$ for which $a^{p+q} \equiv 1 \pmod {pq}$ holds?
*Why it holds for $(3,5)$ and $(5,7)$, ... for others probably not?
|
This is a partial answer.
*
*Why it holds for $(3,5)$ and $(5,7)$, ... for others probably not?
Let us prove that $a^{3+5}\equiv 1\pmod{3\times 5}$ if $\gcd(a,3\times 5)=1$.
We have$$a^{3+5}-1=a^8-1=(a-1)(a+1)(a^2+1)(a^4+1)$$
*
*If $a\equiv 1\pmod 3$, then $a-1\equiv 0\pmod 3$
*If $a\equiv 2\pmod 3$, then $a+1\equiv 0\pmod 3$
*If $a\equiv 1\pmod 5$, then $a-1\equiv 0\pmod 5$
*If $a\equiv 2,3\pmod 5$, then $a^2+1\equiv 0\pmod 5$
*If $a\equiv 4\pmod 5$, then $a+1\equiv 0\pmod 5\qquad\blacksquare$
Next, let us prove that $a^{5+7}\equiv 1\pmod{5\times 7}$ if $\gcd(a,5\times 7)=1$.
We have
$$a^{5+7}-1=a^{12}-1=(a-1)(a^2+a+1)(a^3+1)(a^6+1)$$
*
*If $a\equiv 1\pmod 5$, then $a-1\equiv 0\pmod 5$
*If $a\equiv 2,3\pmod 5$, then $a^6+1\equiv 0\pmod 5$
*If $a\equiv 4\pmod 5$, then $a+1\equiv 0\pmod 5$
*If $a\equiv 1\pmod 7$, then $a-1\equiv 0\pmod 7$
*If $a\equiv 2,4\pmod 7$, then $a^2+a+1\equiv 0\pmod 7$
*If $a\equiv 3,5,6\pmod 7$, then $a^3+1\equiv 0\pmod 7\qquad\blacksquare$
As lab bhattacharjee commented, Carmichael Function $\lambda(n)$ might help.
$\lambda(n)$ is defined as the smallest integer such that $a^{\lambda(n)}\equiv 1\pmod n$ where $\gcd(a,n)=1$.
So, we can say that $$\text{$p+q\ $ is divisible by $\lambda(pq)\quad\iff\quad a^{p+q}\equiv 1\pmod{pq}$}$$
For example, $$a^{3+5}\equiv 1\pmod{3\times 5}\quad\text{and}\quad a^{5+7}\equiv 1\pmod{5\times 7}$$ follow from $$\lambda(3\times 5)=4\quad\text{and}\quad \lambda(5\times 7)=12$$
which can be seen at OEIS A002322.
Added :
If $q=p$, then
$$\lambda(pq)=\lambda(p^2)=p(p-1)$$
So, $$\lambda(pq)\mid p+q\implies p(p-1)\mid 2p\implies p=2,3$$
and it is easy to see that $a^{2+2}\equiv 1\pmod{2\times 2}$ and $a^{3+3}\equiv 1\pmod{3\times 3}$.
So, if $q=p$, $(p,q)=(2,2),(3,3)$ are the only such pairs.
If $q\not=p$, then
$$\lambda (pq)=\text{lcm}(\lambda(p),\lambda(q))=\text{lcm}(p-1,q-1)$$
So, $$\lambda(pq)\mid p+q\implies \text{lcm}(p-1,q-1)\mid p+q$$
So, if $q\not =p$, then in order to have $a^{p+q}\equiv 1\pmod{pq}$, we have to have $$\text{lcm}(p-1,q-1)\mid p+q$$
|
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|
Solve for x: $2^x+4^x=8^x$ I tried turning $4^x$ and $8^x$ into powers of $2$ and manipulating the equation but could not make progress. What would be the next steps for this problem?
|
Since $4^{x} = 2^{2x}$ and $8^{x} = 2^{3x}$ then the equation becomes
$$2^{x} + 2^{2 x} = 2^{3 x}.$$
Let $t = 2^{x}$ to obtain $t^{3} - t^{2} - t = 0$ or $t (t^{2} - t - 1) = 0$. The quadratic can be factored using $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$ for which
$$t \, (t - \alpha) \, (t - \beta) = 0.$$
This yields that $t = 2^{x}$ has the potential solutions $2^{x} \in \{0, \alpha, \beta\}$. Solving for $x$ in terms of $t$ is $2^{x} = e^{x \ln(2)} = t$,
$$x = \frac{\ln(t)}{\ln(2)}.$$
For the case $t = 0$ then $x = \ln(0)/\ln(2) = - \infty$ and should be eliminated, but is valid since it is essentially saying $0 + 0 = 0$. The other two cases result in
$$x \in \left\{ \frac{\ln(\alpha)}{\ln(2)}, - \frac{\ln(\alpha)}{\ln(2)} \right\}.$$
The second is obtained by $\alpha \cdot \beta = -1$ and $\alpha^{2} \cdot \beta^{2} =1$ and
$$\ln(\beta) = \frac{1}{2} \ln(\beta^{2}) = \frac{1}{2} \ln\left(\frac{1}{\alpha^{2}}\right) = - \frac{\ln(\alpha^{2})}{2} = - \ln(\alpha).$$
|
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|
Find Probability that even numbered face occurs odd number of times A Die is tossed $2n+1$ times. Find Probability that even numbered face occurs odd number of times
First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have
$$a+b+c+d+e+f=2n+1$$ is a linear equation where variables are non negative integers.
So number of non negative integer solutions of above is $\binom{2n+6}{5}$
But $b$, $d$ and $f$ should be odd. So let $b=2x+1$, $d=2y+1$ and $f=2z+1$ where $x,y,z$ are non negative integers.
so we get
$$a+2x+c+2y+e+2z=2n-2$$ which is possible in the following two cases
Case $1.$ $a,c,e$ are all even so let $a=2m+1$, $c=2q+1$ and $e=2r+1$ we get
$$m+x+q+y+r+z=n-1$$ so number of solutions is $\binom{n+4}{5}$
Case $2.$ Exactly two of $a,c,e$ are odd and other is even which can be possible in 3 ways.
So let $a=2m+1$, $c=2q+1$, $e=2r$ we get
$$m+x+q+y+r+z=n-2$$
Number of solutions is $3 \times \binom{n+3}{5}$
Hence Probability is
$$P(A)=\frac{\binom{n+4}{5}+3\binom{n+3}{5}}{\binom{2n+6}{5}}$$
Is my approach fine?
|
Even and odd faces both occur with a proabability of $\frac{1}{2}$. The probability of attaining an odd number of evens is
\begin{eqnarray*}
\sum_{i=0}^{n} \binom{2n+1}{2i+1} \left( \frac{1}{2} \right)^{2n+1}
\end{eqnarray*}
and an even number of odd faces is
\begin{eqnarray*}
\sum_{i=0}^{n} \binom{2n+1}{2i} \left( \frac{1}{2} \right)^{2n+1}
\end{eqnarray*}
Now note that these two sums are equal by the symmetry of the binomial coefficients so they are both $\color{red}{\frac{1}{2}}$.
|
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|
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$ let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$
then find $f(x)$
My Try :
$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$
So we have :
$$f(x)=x^2-1$$
it is right ?Is there another answer?
|
A particular solution is $f(x)=x^2-1$. The general solution of the associated homogeneous problem
$$f\left({x\over3}\right)+f\left({2\over x}\right)=0$$
is $$f_{\rm hom}(x)=u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ ,$$
whereby $u$ is an arbitrary odd function. It follows that the general solution of the original problem is given by
$$f(x)=x^2-1+ u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ .$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that: $1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$ Prove that:
$$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$
I know only this method:
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+....=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-...$
But, unfortunately, I could not a hint.
|
We have
\begin{split} &&1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{199} - \frac{1}{200} \\&=&
\left(1 + \frac{1}{3} + \frac{1}{5} +\cdots + \frac{1}{197}+\frac{1}{199} \right) - \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{200}\right)\\
&=&
\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{199} + \frac{1}{200}\right) - 2 \cdot \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{200}\right)\\
&=&\left(1 + \frac{1}{2} + \cdots \frac{1}{100}\right)+\left( \frac{1}{101}+\cdots +\frac{1}{199} + \frac{1}{200}\right) - \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{100}\right)\\[10pt]
&= &\frac{1}{101} + \frac{1}{102} + \cdots + \frac{1}{200}\end{split}
|
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|
Explicit sum of $\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$ I want to find the sum of the infinite series $$\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$$
So far I have managed to simplify the expression to $$\sum_{n=0}^{\infty} \left(\frac {1}{6^{n+1}} + \frac{1}{9\cdot2^{n+2}}\right)$$.
The sum clearly converges as both expressions are equal to $0$ as $\lim_{n\to\infty}$. Any tips as to how I may find an explicit expression for the sum of this series?
thanks in advance
|
As mentioned,it is geometric series $$\sum _{ n=0 }^{ \infty } \left( \frac { 1 }{ 6^{ n+1 } } +\frac { 1 }{ 9\cdot 2^{ n+2 } } \right) =\frac { 1 }{ 6 } \sum _{ n=0 }^{ \infty } \frac { 1 }{ 6^{ n } } +\frac { 1 }{ 36 } \sum _{ n=0 }^{ \infty } \frac { 1 }{ 2^{ n } } =\\ =\frac { 1 }{ 6 } \left( 1+\frac { 1 }{ 6 } +\frac { 1 }{ { 6 }^{ 2 } } +... \right) +\frac { 1 }{ 36 } \left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ { 2 }^{ 2 } } +... \right) =\frac { 1 }{ 6 } \frac { 1 }{ 1-\frac { 1 }{ 6 } } +\frac { 1 }{ 36 } \frac { 1 }{ 1-\frac { 1 }{ 2 } } =\\ =\frac { 1 }{ 5 } +\frac { 1 }{ 18 } =\frac { 23 }{ 90 } $$
|
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|
coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ coefficient of $x^{17}$ in $(x^2+x^3+x^4+x^5+x^6+x^7)^3$ - I'm reviewing for an exam and don't understand the answer, C(11+3−1,11)−C(3,1)×C(5+3−1,5).
|
$$
\begin{align}
\left[x^{17}\right]\left(x^2+x^3+x^4+x^5+x^6+x^7\right)^3
&=\left[x^{11}\right]\left(1+x+x^2+x^3+x^4+x^5\right)^3\\
&=\left[x^{11}\right]\left(\frac{1-x^6}{1-x}\right)^3\\
&=\left[x^{11}\right]\sum_{j=0}^3(-1)^j\binom{3}{j}x^{6j}
\sum_{k=0}^\infty(-1)^k\binom{-3}{k}x^k\\
&=\left[x^{11}\right]\sum_{j=0}^3(-1)^j\binom{3}{j}x^{6j}
\sum_{k=0}^\infty\binom{k+2}{k}x^k\\
&=\sum_{j=0}^1(-1)^j\binom{3}{j}\binom{13-6j}{11-6j}\\
&=\binom{3}{0}\binom{13}{11}-\binom{3}{1}\binom{7}{5}\\[9pt]
&=15
\end{align}
$$
|
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|
Limit of a sequence by definition I need to prove by definition that the limit of sequence :
$$a_n = \frac{5n^3-3n^2+1}{4n^3+n+2}$$
is $\dfrac54$ which means I need to show that :
$$\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| < \varepsilon$$
I tried for hours to solve it but could not,
Can anyone help please?
Thanks
|
Let $\varepsilon > 0$. Pick $n_0 \in \mathbb{N}$ such that $n_0 > \frac{23}{16\varepsilon}$.
For $n \ge n_0$ we have:
\begin{align}
\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| &= \left|\frac{4(5n^3-3n^2+1)-5(4n^3+n+2)}{4(4n^3+n+2)}\right|\\
&= \left|\frac{-12n^2-5n-6}{4(4n^3+n+2)}\right|\\
&= \frac{12n^2+5n+6}{4(4n^3+n+2)}\\
&\le \frac{12n^2+5n^2+6n^2}{4(4n^3 + 0 + 0)}\\
&= \frac{23n^2}{16n^3}\\
&= \frac{23}{16}\cdot\frac1n\\
&< \varepsilon
\end{align}
Therefore $$\lim_{n\to\infty} a_n = \frac54$$
|
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|
Prove identity composed of floor functions How to prove:
$$\left\lfloor \frac{x+2}{6}\right\rfloor -\left\lfloor \frac{x+3}{6}\right\rfloor +\left\lfloor \frac{x+4}{6}\right\rfloor =\left\lfloor \frac{x}{2}\right\rfloor -\left\lfloor \frac{x}{3}\right\rfloor$$
where $x\in \mathbb{R}$.
I think it is problem made by Ramanujan, but do not have the source.
|
Hint: write $x= 6k+r$ where $o\leq r <6$ and $k\in \mathbb{Z}$. Then we have:
$$\left\lfloor \frac{x+2}{6}\right\rfloor -\left\lfloor \frac{x+3}{6}\right\rfloor +\left\lfloor \frac{x+4}{6}\right\rfloor =
k + \underbrace{\left\lfloor \frac{r+2}{6}\right\rfloor -\left\lfloor
\frac{r+3}{6}\right\rfloor +\left\lfloor
\frac{r+4}{6}\right\rfloor}_{A}$$
and
$$\left\lfloor \frac{x}{2}\right\rfloor -\left\lfloor \frac{x}{3}\right\rfloor = k +\underbrace{\left\lfloor \frac{r}{2}\right\rfloor -\left\lfloor
\frac{r}{3}\right\rfloor}_{B}$$
|
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|
Proving $4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}$ $4^n = \sum\limits_{k=0}^{n}2^k\cdot{{2n - k} \choose n}$
I tried formal power series, but failed.
|
With formal power series as requested by OP we may write
$$\sum_{k=0}^n 2^k {2n-k\choose n}
= 2^n \sum_{k=0}^n 2^{-k} {n+k\choose n}
= 2^n \sum_{k=0}^n 2^{-k} [z^n] (1+z)^{n+k}
\\ = 2^n [z^n] (1+z)^n \sum_{k=0}^n 2^{-k} (1+z)^k
= 2^n [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2}
\\ = 2^{n+1} [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-z}.$$
We get from the first piece
$$2^{n+1} [z^n] (1+z)^n \frac{1}{1-z} =
2^{n+1} \sum_{k=0}^n {n\choose k} = 2^{2n+1}.$$
The second piece yields
$$- [z^n] (1+z)^{2n+1} \frac{1}{1-z}
= - \sum_{k=0}^n {2n+1\choose k}
= - \frac{1}{2} 2^{2n+1}.$$
Joining the two pieces we find
$$2^{2n+1} - 2^{2n} = 2^{2n} = 4^n.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is there a good upper bound for $(x-1)^n-x^n$ for $x\ge 1$ and $n=2,3,4,...$? For an positive integer $n\ge 2$, is there a good upper bound for $(x-1)^n-x^n$ for $x> 1$?
Revised question: Is it less than $-\log(x-1)$ for all $x>1$?
By Binomial theorem, $(x-1)^n-x^n=\sum_{k=1}^n\binom{n}{k}x^{n-k}(-1)^k$. Or by Mean Value Theorem, it is equal to $-n\xi^{n-1}$ for some $\xi\in(x-1,x)$. I wonder if there are good estimate of this function?
|
The upper bound $-n\zeta^{n-1} \le -n(x-1)^{n-1}$ is certainly less that $-\log(x-1)$ for $x \ge 2.
You can get better upper bounds by using Taylor approximations of $f(t) = t^n$ about $t = x$ and Lagrange's remainder term. For example:
$$
(x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)\zeta^{n-2} \le
-nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} \, .
$$
and
$$
(x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3} \zeta^{n-3} \le
-nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3}(x-1)^3\, .
$$
|
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|
Muller's recurrence: limit In Kahan's account of the Muller's recurrence (p.16): $x_{n+1} = E(x_n, x_{n-1})$ for the function
$$
E(y, z) = 108 - \frac{815-\frac{1500}{z}}{y},
$$
he uses the characteristic polynomial for the recurrence to deduce the closed form:
$$
x_n = \frac{\alpha 3^{n+1} + \beta 5^{n+1} + \gamma 100^{n+1}}{\alpha 3^{n} + \beta 5^{n} + \gamma 100^{n}}.
$$
If $x_0=4, x_1=4.25$ exactly, then we can choose $\alpha=\beta=1, \gamma=0$. But, if there is some numerical round-off in a numerical calculation of the $x_n$, then the closed form becomes ("at least initially"),
$$
x_n = \frac{3^{n+1} + 5^{n+1} + \gamma_n 100^{n+1}}{3^{n} + 5^{n} + \gamma_n 100^{n}}\\
= 100 - \frac{95+97(3/5)^n}{20^n\gamma_n+1+(3/5)^n}
$$
where $\gamma_n$ is small.
I can't derive the last equality, and can't see how this changes the limit (as $n\rightarrow \infty$) to 100 instead of 5. Furthermore, what justifies the choice of $\alpha$, $\beta$, $\gamma$ above amongst the presumably infinite options?
|
$\alpha = 1, \beta = 1, \gamma = 0$ gives the exact solution from the starting values. For the round-off error setting they are slightly disturbed; we still have
$\alpha \approx 1, \beta \approx 1,$ but $\gamma$ has the order of the machine epsilon $\gamma \approx \epsilon.$ To compute Kahan's form from
$$x_n = \frac{3^{n+1} + 5^{n+1} + \gamma_n 100^{n+1}}{3^{n} + 5^{n} + \gamma_n 100^{n}}$$
first do some algebraic manipulations with the numerator
$$3^{n+1} + 5^{n+1} + \gamma_n 100^{n+1}=
3\cdot 3^{n} + 5\cdot 5^{n} + 100\gamma_n 100^{n}$$
$$=(100-97)\cdot 3^{n} + (100-95)\cdot 5^{n} + 100\gamma_n 100^{n}$$
$$=100(3^{n} + 5^{n} + \gamma_n 100^{n})-97\cdot 3^{n} -95\cdot 5^{n}
$$
hence the intermediate value for $x_n$ is
$$x_n = 100 - \frac{97\cdot 3^n + 95\cdot 5^n}{3^n + 5^n + \gamma_n 100^n}$$
now divide by $5^n$ and get
$$x_n = 100 - \frac{ 95+ 97\cdot (3/5)^n }{ 20^n \gamma_n+ 1 +(3/5)^n }$$
Now it is clear that the limit is $\lim_\limits{n\rightarrow \infty} x_n = 100$ for $\gamma_n \ne 0$ and $100-95 = 5$ for $\gamma_n = 0$
The main difference between exact (error=0) and round-off is the growing error term
$20^n \gamma_n$. For double precision calculations with a constant $\gamma_n = \epsilon = 2.22\times 10^{-16}$ you have $20^{15}\epsilon \approx 7275.9576$ and
$20^{20}\epsilon \approx 23283064365.4$
|
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|
Prove that $\cosh(x)=\sec(\theta )$ if $x=\ln(\sec \theta + \tan \theta)$ I'm trying to prove that $\cosh(x)=\sec(\theta )$ if $x=x=\ln(\sec \theta + \tan \theta)$. I've substituted the value of $x$ into $\cosh(x)$ to get $$\frac{e^{\ln(\sec \theta + \tan \theta)}+e^{-\ln(\sec \theta + \tan \theta)}}{2}$$ and simplified to get $$\frac{(\sec \theta +\tan \theta)+\frac{1}{(\sec \theta + \tan \theta)}}{2}$$. However, I do not know where to continue from here. Help would be greatly appreciated.
|
You got to
$$
\cosh x = \frac{1}{2}\left(\sec\theta + \tan\theta + \frac{1}{\sec\theta+\tan\theta}\right)
$$
Write this is as single fraction:
$$
\frac{1}{2}\left(\sec\theta + \tan\theta + \frac{1}{\sec\theta+\tan\theta}\right)
= \frac{(\sec\theta + \tan\theta)^2 + 1}{2(\sec\theta + \tan\theta)}
$$
Expand and remember that $1 + \tan^2 \theta = \sec^2\theta$:
\begin{align*}
\frac{(\sec\theta + \tan\theta)^2 + 1}{2(\sec\theta + \tan\theta)}
&= \frac{\sec^2\theta + 2\tan\theta\sec\theta + \tan^2\theta + 1}{2(\sec\theta + \tan\theta)}
\\&= \frac{2\sec^2\theta + 2 \sec\theta\tan\theta}{2(\sec\theta + \tan\theta)}
\\&= \frac{(2\sec\theta)(\sec\theta + \tan\theta)}{2(\sec\theta + \tan\theta)}
= \sec\theta
\end{align*}
|
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|
an optimisation problem folding paper
A standard $8.5$ inches by $11$ inches piece of paper is folded so that one corner touches the opposite long side and the crease runs from the adjacent short side to the other long side, as shown in the picture below. What is the minimum length of the crease?
So I put the shorter leg of the folded portion (since it's a right triangle) as $x$ and the longer leg $y$. Then the crease is $\sqrt{x^2+y^2}$ and deriving it (which is what I'm supposed to be doing for optimisation) in terms of x to minimise the crease length, I got $\frac{x+yy'}{\sqrt{x^2+y^2}} = 0$, so $x+yy' = 0$. Then I was able to find some similar triangles - one with hypotenuse $x$ and one leg $8.5-x$ and the other with hypotenuse $y$ and the not-corresponding leg $8.5$. From there I was able to get the equation $y=\sqrt{y^2-8.5^2} + \sqrt{x^2-(8.5-x)^2}$. Then I started squaring the equation but it got quite messy, and I'm not even sure if the problem is supposed to be so messy... Can someone help me solve this?
Disclaimer: I do know that there is a post with the same problem on this site, but unfortunately the answers there didn't really help me, and I don't get what I am doing wrong.
|
Perhaps, a simpler approach
would be
to express $|EG|$ in terms of
$\angle ABF=\phi$.
In this case
\begin{align}
|BF|&=\frac{w}{\cos\phi}
,\\
|BK|&=\tfrac12\,\frac{w}{\cos\phi}
,\\
|BE|&=\frac{|BK|}{\cos(\tfrac\pi2-\phi)}
=\frac{w}{2\,\cos\phi\sin\phi}
,\\
|EG|&=\frac{|BE|}{\cos\phi}
=\tfrac12\,\frac{w}{\cos^2\phi\sin\phi}
\\
&=\tfrac12\,\frac{w}{(1-\sin^2\phi)\,\sin\phi}
\\
&=\tfrac12\,\frac{w}{(1-x^2)\,x}
.
\end{align}
Now we can use $x=\sin\phi$
as a control parameter:
\begin{align}
|EG|&=f(x)=
\tfrac12\,\frac{w}{(1-x)(1+x)x}
=
\frac{w}4\left(
\frac{1}{1-x}
-\frac{1}{1+x}
+\frac{2}{x}
\right)
,\\
f'(x)&=
\frac{w}4\left(
\frac{1}{(1-x)^2}
+\frac{1}{(1+x)^2}
-\frac{2}{x^2}
\right)
\\
&=
\frac{w}2\cdot
\frac{3\,x^2-1}{(1-x)^2(1+x)^2\,x^2}
.
\end{align}
Solution to
\begin{align}
f'(x)&=0\quad
\text{gives }
x=\pm\tfrac{\sqrt3}3,
\\
|EG|_{\min}&=\tfrac34\,w\,\sqrt3
=\tfrac{51}8\,\sqrt3
\approx 11.0418
.
\end{align}
|
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|
Proving that limit (without using calculus) $(n^2 + n) ^{1/2} - n = \frac{1}{2}$ Prove, Without using Calculus.
$$\lim_{n->\infty} \left((n^2 + n)^{1/2} - n\right) = \frac{1}{2} $$
I am taking an introduction to real analysis course going through Rudin.
My work so far,
By definition, limit of a sequence $p$ has the following property:
For every $\varepsilon > 0$ There is an integer $N$ s.t. $n \geq N$ implies that $d(p_n, p) < \varepsilon $.
Let $ \varepsilon > 0$. Then we need to find an expression for $N$. So:
$$ \left|\left((N^2 + N)^{1/2} - N\right) - \frac{1}{2} \right| < \varepsilon $$
I am wondering whether this is the right approach?
|
Yep it's essentially the right approach. We want to find $N$ so that if $n > N$, then
$$ \left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| < \varepsilon $$
We know that
$\left((n^2+n)^{1/2} - (n + 1/2)\right)\left((n^2+n)^{1/2} + (n+1/2)\right) = (n^2+n)-(n+1/2)^2$ using a "difference of two squares". Hence given some $n \in \mathbb{N}$
$$\left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| = \frac{\left|(n^2+n)-(n+1/2)^2 \right| }{\left|\left((n^2+n)^{1/2} + (n+1/2)\right) \right|} = \frac{1/4}{(n^2+n)^{1/2}+n+1/2} < \frac{1}{4n}<\frac{1}{n}$$
We see that if $N = \frac{1}{\varepsilon}$, then for any $n > N$,
$$\left|\left((n^2 + n)^{1/2} - n\right) - \frac{1}{2} \right| <\frac{1}{n} < \frac{1}{N} = \varepsilon$$
|
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|
Residue of $f(z) = \frac{1}{z(1-\cos(z))}$ I tried to find the residue of the function $$f(z) = \frac{1}{z(1-\cos(z))}$$ at the $z=0$. So I did:
\begin{align}
\hbox{Res}(0)&=\lim_{x \to 0}(z)(f(z))\\
\hbox{Res}(0)&= \lim_{x \to 0} \frac{1}{1-\cos(z)}
\end{align}
and I got that the residue to be $\infty$, and it seems to be wrong.
|
$$\cos z=1-\dfrac{1}{2!}z^2+\dfrac{1}{4!}z^4-\dfrac{1}{6!}z^6+\cdots$$
then
\begin{align}
\dfrac{1}{z(1-\cos z)}
&= \dfrac{1}{z^3\left(\dfrac{1}{2!}-\dfrac{1}{4!}z^2+\dfrac{1}{6!}z^4+\cdots\right)} \\
&= \dfrac{1}{z^3}\left(2+\dfrac{1}{6}z^2+\dfrac{1}{120}z^4+\cdots\right) \\
&= \dfrac{2}{z^3}+\dfrac{1}{6z}+\dfrac{z}{120}+\cdots
\end{align}
s0 $a_{-1}=\dfrac16$.
|
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|
A problem :$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$ Hello during a problem I have this to solve :
Let $a,b,c,d,e,f$ be real positiv number such that $a\geq b\geq c\geq d\geq e\geq f\geq 1$ then :
$$(edf)^3(a+b+c+d+e+f)^3\geq(abcdef)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f})^3$$
I did not find any counter-example (with Pari-Gp and Wolfram alpha) so maybe it's true
Advice :I do not think that Am-Gm will works on it .
Thanks a lot
|
This inequality simply follows from the original inequality for $n=3$, so I guess its proof won't be easier. Indeed, dividing both sides by $def$ we transform it to
$$(def)^2(a+b+c+d+e+f)^3\geq abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\right)^3.$$
Now we note that with $a$, $b$, and $c$ fixed the left hand side is at least $(a+b+c+3)^3$, the right hand side is at most $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\right)^3$ and both sides are equal to their extremal values when $d=e=f=1$.
|
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|
Simplifying expressions. How do you simplify the following expression?
$$Q=(1-\tan^2(x)) \left(1-\tan^2 \left(\frac{x}{2}\right)\right)\cdots \left(1-\tan^2\left(\frac{x}{2^n}\right)\right)$$
I've tried the the following :
$\tan x = \frac{\sin x}{\cos x}$
$$\cos x = \frac{\sin 2x}{2\sin x}$$
$$\tan x = \frac{2\sin^2x}{\sin2x}$$
$$Q= \left( 1- \left(\frac{2\sin^2x}{\sin2x}\right)^2\right) \left(1-\left(\frac{2\sin^2\frac{x}{2}}{\sin x}\right)^2\right) \cdots \left(1-\left(\frac{2\sin^2\frac{x}{2^{n+1}}}{\sin(\frac{x}{2^n})}\right)^2\right)$$
But i have no idea how to continue.
Help appreciated!
|
Hint:
$$
1-\tan^2 t=\frac{\cos^2t-\sin^2t}{\cos^2t}=\frac{\cos(2t)}{\cos^2t}
$$
and for arguments of the form $\frac{x}{2^n}$, many things will vanish in the product.
|
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|
Maximum possible value of $P(10)$ The real numbers $a$, $b$, $c$, and $d$ are each less than or equal to $12$. The polynomial $$P(x)=ax^3+bx^2+cx+d$$ satisfies $P(2)=2$, $P(4)=4$ and $P(6)=6$. Find the maximum possible value of $P(10)$.
What I did was first I used the given information to get $3$ equation in $a, b,c,d$. Then I obtained values of $a$, $b$, $c$ in terms of $d$. Then using these values, I found $P(10)$ in terms of $d$ and substituted $d=12$. But I couldn't arrive at the answer.
Thanks in advance
|
$P(x)=x+a(x-2)(x-4)(x-6)$ because $2,4,6$ are roots of the cubic $P(x)-x$, whose leading coefficient is $a$.
Therefore, $P(x)=a x^3 - 12 a x^2 + (44 a + 1) x - 48 a$.
Since the coefficients are all at most $12$, we need
$$
a \le 12, \quad -12 a \le 12, \quad 44a+1 \le 12, \quad -48a \le 12
$$
Therefore,
$
-\frac14 \le a \le \frac14
$.
Now, $P(10)=10+192a$, whose maximum value is $58$, attained at $a=\frac14$.
|
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|
Find $\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1} n^{n^2}}$ Woflram gives $\frac{1}{e}$ as the limit, but I failed to obtain it. Please help.
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$
|
\begin{align}
\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}
&=\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2}(n+1)^{2n+1}}{n^{n^2}(n+2)^{n^2}(n+2)^{2n+1}}\\
&=\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2}(n+1)^{2n}(n+1)}{\left[n(n+2)\right]^{n^2}(n+2)^{2n}(n+2)}\\
&=\lim_{n\rightarrow\infty} \left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot \lim_{n\rightarrow\infty} \left(\frac{n+1}{n+2}\right)^{2n}\cdot\lim_{n\rightarrow\infty}\frac{n+1}{n+2}\\
&=\lim_{n\rightarrow\infty} \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n^2}\cdot
\left[\lim_{n\rightarrow\infty} \left(\frac{n+1}{n+2}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}\right)^{n}\right]^2\cdot\lim_{n\rightarrow\infty}\left(\frac{n+1}{n+2}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}\right)\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left[\lim_{n\rightarrow\infty} \left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\right)^{n}\right]^2\cdot\lim_{n\rightarrow\infty}\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left[\frac{\lim_\limits{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^{n}}{\lim_\limits{n\rightarrow\infty} \left(1+\frac{2}{n}\right)^{n}}\right]^2\cdot\lim_{n\rightarrow\infty}\frac{1+0}{1+0}\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left(\frac{e}{e^2}\right)^2\cdot 1
\end{align}
Now, as $n$ approaches infinity, the ratio of the quantities $n^2+2n$ and $n^2$ approaches $1$. What this means for us is that somewhere out there in infinity they want to become the same quantity and we can treat them as such:
$$
\lim_{x\rightarrow\infty} \left(1+\frac{1}{x}\right)^{x}\cdot\frac{1}{e^2}=
e\cdot\frac{1}{e^2}=\frac{1}{e}
$$
|
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|
Number of solutions to $a + b + c + d + f = n$ in nonnegative integers $a, b, c, d, f,$ given restrictions on $a, b,$ and $f$ Count the number of solutions to $a + b + c + d + f = n$ in nonnegative integers $a, b, c, d, f,$ such that $a$ is a multiple of 4, $b$ is at most 1, and $f$ is either 0 or 2.
My attempt:
As far as I know, you can solve these types of problems using generating functions or multisets. For this problem, it makes sense to use generating functions. Here's what I have so far:
For $a$: $(x^4 + x^8 + x^{12} + ... ) = \frac{1}{1-x^4}$
For $b$: $(x^0 + x^1) = \frac{1-x^2}{1-x}$
I'm stuck at what to do for $f$.
|
The generating function is
\begin{eqnarray*}
\underbrace{\frac{1}{(1-x^4)}}_{a} \underbrace{(1+x)}_{b} \underbrace{\frac{1}{(1-x)^2}}_{c \text{ and } d } \underbrace{(1+x^2)}_{f}.
\end{eqnarray*}
Which can be simplified to
\begin{eqnarray*}
\color{blue}{\frac{1}{(1-x)^3}}.
\end{eqnarray*}
|
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|
Find the$A:=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{20^2}$
Find the :
$$A:=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{20^2}$$
My Try :
$$\frac{1}{2^2}+\frac{1}{3^2}=\frac{3^2+2^2}{2^2\cdot 3^2}$$
$$\frac{3^2+2^2}{2^2\cdot 3^2}+\frac{1}{4^2}=\frac{(3^2+2^2)(4^2+1)}{4^2\cdot 3^2\cdot 2^2}$$
$$\frac{(3^2+2^2)(4^2+1)}{4^2\cdot 3^2\cdot 2^2}+\frac{1}{5^2}=\frac{(5^2)(3^2+2^2)(4^2+1)+(4^2)(3^2)(2^2)}{5^2 \cdot4^2\cdot3^2\cdot 2^2}$$
Now what ?
|
We know that the sum of the reciprocal of the squares add up to $\pi^2/6.$ So your sum equals
$$\frac{\pi^2}{6} -1 -\sum_{n=21}^{\infty} \frac{1}{n^2}.$$
The infinite sum is less than, but very close to
$$\int_{20}^\infty \frac{1}{x^2} \; dx = \frac{1}{20}.$$
So your sum is about
$$\frac{\pi^2}{6} -1 -\frac{1}{20} = 0.594934068\ldots.$$
Maple says the exact answer is $0.5961632439.$
|
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|
How to evaulate $\int_0^{\infty}\frac{\ln x}{\sqrt{x}(x^2+a^2)^2}$ using contour integration I'm asked to calculate the following integral for which $0 \neq a \in \mathbb{R}$:
$$\int_0^{\infty}\frac{\ln x}{\sqrt{x}(x^2+a^2)^2}$$
I'm confused about which contour I should use, whether it should be a semi-circle deformed to avoid the origin, or a keyhole contour based on a similar question I found here: Calculating $\int_0^{\infty } \frac{\ln (x)}{\sqrt{x} \left(a^2+x^2\right)^2} \, \mathrm{d}x$ using contour integration (but this question did not go into detail about how the integration was carried out using the keyhole contour)
Also, after deciding on which contour to use, how do I proceed from there to evaluate this integral? I know that I would eventually have to use the Residue Theorem but how do I isolate the part of the contour only going from $0$ to $\infty$?
|
We integrate with $a$ a positive real
$$f(z) = \frac{\mathrm{Log}(z)}{(z-ai)^2 (z+ai)^2}
\exp(-1/2\times \mathrm{Log}(z))$$
around a keyhole contour with the slot on the positive real axis,
which is also where the branch cut of the logarithm is located
(argument of the logarithm is between $0$ and $2\pi$). Now for the
large circle we get $\lim_{R\to\infty} 2\pi R \log R / \sqrt{R} /R^4 =
0$ so there is no contribution in the limit. For the small circle
around the origin we find $\lim_{\epsilon\to\ 0} 2\pi \epsilon \log
\epsilon / \sqrt\epsilon /a^4 = 0$ so there is no contribution here
either.
We get for the upper line segment
$$\int_0^\infty \frac{\log x}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx$$
which is our target integral, call it $J$. The lower line segment
contributes
$$-\int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \exp(-1/2\times 2\pi i) \; dx
\\ = \int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx
\\ = J + 2\pi i \int_0^\infty \frac{1}{(x^2+a^2)^2}
\exp(-1/2\times \log x) \; dx = J + 2\pi i K$$
where $J$ and $K$ are real numbers. We thus have
$$2J + 2\pi i K = 2\pi i \mathrm{Res}_{z=ai} f(z)
+ 2\pi i \mathrm{Res}_{z=-ai} f(z)$$
or
$$J + \pi i K = \pi i \mathrm{Res}_{z=ai} f(z)
+ \pi i \mathrm{Res}_{z=-ai} f(z)$$
With this setup we do not actually need to compute $K$ as it must
correspond to the imaginary part of the contribution from the two
residues. We get for the first residue at $z=ai$ the derivative
$$\frac{1}{z} \frac{1}{(z+ai)^2} \exp(-1/2\times \mathrm{Log}(z))
-2 \mathrm{Log}(z)
\frac{1}{(z+ai)^3} \exp(-1/2\times \mathrm{Log}(z))
\\ + \mathrm{Log}(z) \frac{1}{(z+ai)^2}
\exp(-1/2\times \mathrm{Log}(z)) \times -\frac{1}{2} \frac{1}{z}.$$
With the branch of the logarithm we find $\mathrm{Log}(ai) =
\log a + \pi i/2$, getting
$$\frac{1}{\sqrt{a}} \exp(-\pi i/4)
\\ \times \left(\frac{1}{ai} \times - \frac{1}{4 a^2}
+ (2\log a + \pi i) \frac{1}{8 i a^3}
+ (\log a + \pi i/2) \times - \frac{1}{4 a^2}
\times -\frac{1}{2 ai}\right)
\\ = \frac{1}{\sqrt{a}} \exp(-\pi i/4) \frac{1}{8i a^3}
(3\log a + 3\pi i/2 - 2).$$
We also have $\mathrm{Log}(-ai) = \log a + 3 \pi i/2$, getting
for the second residue at $z=-ai$
$$\frac{1}{\sqrt{a}} \exp(-3\pi i/4)
\\ \times \left(- \frac{1}{ai} \times - \frac{1}{4 a^2}
- (2\log a + 3\pi i) \frac{1}{8 i a^3}
+ (\log a + 3\pi i/2) \times - \frac{1}{4 a^2}
\times \frac{1}{2 ai}\right)
\\ = \frac{1}{\sqrt{a}} \exp(-3\pi i/4) \frac{1}{8i a^3}
(- 3\log a - 9\pi i/2 + 2).$$
With $\exp(-\pi i/4) =
\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
and $\exp(-3\pi i/4) =
-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$
and factoring out $\frac{1}{\sqrt{a}} \frac{1}{8i a^3}$
we get three contributions, which are
$$\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)$$
Combine these and multiply by $\pi i$ to get
$$\frac{\pi}{8\sqrt{a} a^3}
(\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)).$$
We extract the real part as promised and obtain
$$\frac{\sqrt{2}\pi}{8\sqrt{a} a^3}
(3\log a - 2 - 3\pi/2)$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{4\sqrt{2} a^{7/2}}
\left(3\log a - 2 - \frac{3}{2}\pi\right).}$$
matching the result by @Jack D'Aurizio.
As a bonus we have shown that
$$\int_0^\infty \frac{1}{\sqrt{x} (x^2+a^2)^2} dx
= \frac{1}{\pi} \frac{\pi}{8\sqrt{a} a^3} 3\sqrt{2}\pi
= \frac{3\pi}{4\sqrt{2} a^{7/2}}.$$
|
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|
Solve $X^2=A$, where X is a 2 by 2 matrix and A is a known matrix A = $
\begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$
. I wrote X = $
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
$. So $X^2$=$
\begin{pmatrix}
a^2+bc & ab+bd \\
ac+bc & bc+d^2 \\
\end{pmatrix}
$ = $
\begin{pmatrix}
1 & 1 \\
-1 & 1 \\
\end{pmatrix}
$
Unfortunatelly I don't know how to continue from here and maybe someone can help be find the matrix X
|
You get:
$a^2 +bc$ = 1
$ac + bc$ = -1
$ab + bd$ = 1
$bc + d^2$ = 1
Hence use substitution to find a, b, c and d.
|
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|
Taylor inside an integral I know the following integral should be:
$$ \int_{0}^{1} \frac{dx}{\sqrt{1-x^2-\epsilon(1-x^3)}} \approx \pi/2 + \epsilon$$
for $\epsilon$ small. What I do is set $-\epsilon(1-x^3)=y$ and then since $y$ it's always greatly smaller than $(1-x^2)$ I do Taylor expansion around $y=0$:
$$ \frac{dx}{\sqrt{1-x^2+y}} \approx \frac{dx}{\sqrt{1-x^2}}-\frac{y \cdot dx}{2 \cdot (1-x^3)^{3/2}} $$
Then I recover $y=-\epsilon(1-x^3)$ and I get the correct answer. The problem is that I don't know if that's mathematically correct because $y$ depends on $x$.
|
You can be authorized in doing that because $\epsilon$ is small (we suppose small enough to give you the "license" of dong that).
Also notice that the range of the integral is $[0, 1]$, hence $x$ too, in a certain way, is small, and terms like $x^3$ are then even smaller.
There are many approaches for dealing with such an integral. Yours is one. Then you can take that function and make a Tylor series in $x$ or in $\epsilon$, obtaining respectively:
$$\star : \frac{1}{\sqrt{1-\epsilon }}+\frac{x^2}{2 (1-\epsilon )^{3/2}}-\frac{x^3 \epsilon }{2 (1-\epsilon )^{3/2}}+O\left(x^4\right)$$
$$\star : \frac{1}{\sqrt{1-x^2}}-\frac{\left(x^3-1\right) \epsilon }{2 \left(1-x^2\right)^{3/2}}+\frac{3 \left(x^3-1\right)^2 \epsilon ^2}{8 \left(1-x^2\right)^{5/2}}-\frac{5 \left(x^3-1\right)^3 \epsilon ^3}{16 \left(1-x^2\right)^{7/2}}+O\left(\epsilon ^4\right)$$
Those are clearly Taylor series of the whole function, that is, of
$$\frac{1}{\sqrt{1 - \epsilon - x^2 + \epsilon x^3}}$$
Another way is to think about the term $\epsilon x^3$ and rawly say "let's get rid of this", remaining with the easy integral
$$\int_0^1 \frac{\text{d}x}{\sqrt{1 - \epsilon - x^2}} = -\tan ^{-1}\left(\frac{\epsilon }{(-\epsilon )^{3/2}}\right)$$
Numerica other methods are available, but this wouldn't be the topic of the question.
EDIT
Notice that in my last result we have
$$\frac{\epsilon}{(-\epsilon)^{3/2}} = \frac{1}{\sqrt{-\epsilon}}$$
hence as $\epsilon \to 0$
$$-\arctan\left(\frac{1}{\sqrt{-\epsilon}}\right) = \frac{\pi}{2}$$
|
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|
How to show the action of $SL_2(\mathbb R)$ on the complex upper half plane is transitive? I've seen other answers here and here but I still am not understanding how to show that this action is transitive. I want to show that for all $z \in \mathbb H$ there exists a matrix $g \in SL_2(\mathbb R)$ such that $gi=z$.
If I start with $\frac{ai+b}{ci+d}=z=x+iy$, then I get
$$\frac{ai+b}{ci+d}=z \implies ai+b=z(ci+d) \implies i=\frac{dz-b}{a-cz}$$
I don't know what to do with this or what this tells me.
If I do $\frac{az+b}{cz+d}=i$ I don't get anywhere either.
How can I go about this?
|
ANSWER:
If $A \in SL(2,\mathbb R)$, then
\begin{align*}
A\cdot i&=\frac{ai+b}{ci+d}\\
&=\frac{ai+b}{ci+d}\cdot \frac{(-ci+d)}{(-ci+d)}\\
&=\frac{ac+bd+i(ad-bc)}{c^2+d^2}.
\end{align*}
We're in $SL(2, \mathbb R$), so $ad-bc=1$, and
$$\frac{ac+bd+i(ad-bc)}{c^2+d^2}=\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}.$$
If $c=0$ then $a \ne 0$, $d\ne 0$ and $ad-bc=1$ becomes $ad=1$. So, $d=1/a$. Also,
$$\frac{ac+bd}{c^2+d^2}+i\frac{1}{c^2+d^2}=\frac{bd}{d^2}+i\frac{1}{d^2}=\frac{b}{d}+i\frac{1}{d^2}=ab+a^2i.$$
Since $z \in \mathbb H$, then $z=x+iy$ where $y>0$. Now $a^2=y \implies a=\sqrt y$. Therefore, $ab=x \implies b=\frac{x}{\sqrt y}$.
So, $\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \in SL(2,\mathbb R)$ and
$$\begin{pmatrix} \sqrt y & \frac{x}{\sqrt y} \\ 0 & \frac{1}{\sqrt y} \end{pmatrix} \cdot i =z$$
implies transitive.
|
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|
How to prove this Fibonacci identity? $\sum_{k=0}^{n} F_{k} F_{n-k} = \frac{1}{5}\left(n L_{n} - F_{n}\right)$ How to prove this Fibonacci identity?
$$\sum_{k=0}^{n-3} F_{k} F_{n-k-3} = \frac{(n-3)L_{n-3} - F_{n-3}}{5}$$
i tried to used the generating function and partial decomposition but i got confused?
|
Using
\begin{align}
F_{n} &= \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta} \\
L_{n} &= \alpha^{n} + \beta^{n} \\
2 \alpha &= 1 + \sqrt{5} \\
2 \beta &= 1 - \sqrt{5}
\end{align}
then
\begin{align}
(\alpha - \beta)^{2} \, F_{n} F_{n-k} &= L_{n} - \beta^{n} \, \left(\frac{\alpha}{\beta}\right)^{k} - \alpha^{n} \, \left(\frac{\beta}{\alpha}\right)^{k} \\
&= L_{n} - \alpha^{n} (-\beta^{2})^{k} - \beta^{n} (- \alpha^{2})^{k}.
\end{align}
Now,
\begin{align}
5 \, \sum_{k=1}^{n} F_{k} F_{n-k} &= n \, L_{n} - \alpha^{n} \cdot \frac{(- \beta^{2}) ( 1- (-1)^{n} \beta^{2n})}{1 + \beta^{2}} - \beta^{n} \cdot \frac{(- \alpha^{2}) ( 1- (-1)^{n} \alpha^{2n})}{1 + \alpha^{2}} \\
&= n \, L_{n} - \beta \, F_{n} - \alpha \, F_{n} \\
&= n \, L_{n} - F_{n}.
\end{align}
Since $F_{0} = 0$ then
$$\sum_{k=0}^{n} F_{k} \, F_{n-k} = \frac{n \, L_{n} - F_{n}}{5}.$$
|
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|
Integration contour for improper integral $\int^\infty_{-\infty} \frac{dx}{(e^x + 1)(1 + e^{-x-z})}$. A while ago, I read through a book in solid state physics (Ziman's Electrons and Photons) and, as an algebraic step in the derivation of a formula, the author uses
$$
\int_{-\infty}^{\infty}\frac{d\eta}{\{\exp\eta + 1\}\{1 + \exp[-(\eta + z)]\}} = \frac{z}{1 - \exp(-z)}
$$
without proof, merely stating the integral is "elementary". Today I used this equation to derive an analytical result in a paper, so I would like to find proof that this evaluation of the integral is correct. By the form of the integrand and the infinite limits of the integral, I guessed that the result should be obtainable with contour integration, but I am struggling to find an appropriate integration contour.
Any suggestions for which contour of integration I should take or even if contour integration is the correct idea to evaluate this integral?
|
The integral is indeed elementary.
For initial notational ease, let $a = e^{-z}$ and we will consider the indefinite integral
$$I = \int \frac{dx}{(e^x + 1)(1 + a e^{-x})}.$$
Observing that
$$\frac{1}{(e^x + 1)(1 + a e^{-x})} = \frac{1}{1 -a} \left [\frac{1}{e^x + 1} - \frac{a e^{-x}}{1 + a e^{-x}} \right ],$$
the integral can be rewritten as
\begin{align*}
I &= \frac{1}{1 - a} \int \frac{dx}{e^x + 1} - \frac{1}{1 - a} \int \frac{ae^{-x}}{1 + a e^{-x}} \, dx\\
&= \frac{1}{1 - a} \int \frac{e^{-x}}{1 + e^{-x}} - \frac{1}{1 - a} \int \frac{ae^{-x}}{1 + a e^{-x}} \, dx.
\end{align*}
Setting $u = e^{-x}, du = - e^{-x} \, dx$ yields
\begin{align*}
I &= -\frac{1}{1 - a} \int \frac{du}{1 + u} + \frac{1}{1 - a} \int \frac{a}{1 + au} \, du\\
&= -\frac{1}{1 - a} \ln (1 + u) + \frac{1}{1 - a} \ln (1 + au) + C\\
&= -\frac{1}{1 - a} \ln (1 + e^{-x}) + \frac{1}{1 - a} \ln (1 + ae^{-x}) + C.
\end{align*}
Rearranging, after setting $a = e^{-z}$, the expression for the integral can be brought into the following form
$$I = \frac{1}{1 - e^{-z}} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) + C.$$
Now for your improper integral, if we note that
$$\lim_{x \to \infty} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) = z,$$
and
$$\lim_{x \to -\infty} \ln \left (\frac{1 + e^{x + z}}{1 + e^x} \right ) = 0,$$
then
$$\int^\infty_{-\infty} \frac{dx}{(e^x + 1)(1 + e^{-x-z})} = \frac{z}{1 - \exp(-z)},$$
as required.
|
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|
Determine $\lim_{n \to \infty} (2^{n+1} \sin(\frac{\pi}{2^{n+1}}))$ I was able to calculate
$$\lim_{n \to \infty} (2^{n+1} \sin(\frac{\pi}{2^{n+1}})) = \pi$$
using L'Hopital Rule, but how to get $\pi$ without it?
|
since $\sin \left( \frac { \pi }{ 2^{ n+1 } } \right) \overset { n\rightarrow \infty }{ \longrightarrow } 0$ we can apply famous limit $\lim _{ n\rightarrow 0 }{ \frac { \sin { n } }{ n } } =1$
$$\lim _{ n\to \infty } \left( 2^{ n+1 }\sin \left( \frac { \pi }{ 2^{ n+1 } } \right) \right) =\lim _{ n\to \infty } \frac { \sin \left( \frac { \pi }{ 2^{ n+1 } } \right) }{ \frac { \pi }{ 2^{ n+1 } } } \pi =\pi $$
|
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|
Triple integration of a solid bounded by 2 curves $z=1-x^2-2y^2$ and $z=x^2$.
I need to find the Vol bounded by $z=1-x^2-2y^2$ and $z=x^2$.
I have gotten
$$V=∫∫∫rdrdθdz$$
where the limits are $0$ to $2\pi$, $0$ to $\sqrt{1/2}$, and $0$ to $1$.
The ans is $\pi$ but my ans is $\pi/2$. Can anyone tell me where my mistake is?
|
$ z = 1 - x^2 - 2y^2$ and $z = x^2 $ meet along the circle $x^2+y^2=\frac{1}{2} $
So, limits of $ x$ would be $ \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$ and
$y$ would be $ -\sqrt{\frac{1}{2}-x^2}<y<\sqrt{\frac{1}{2}-x^2}$ and
$z$ would be $x^2<z<1-x^2-2y^2$
So,
$V=\int_{x^2}^{1-x^2-2y^2} \int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}dxdydz$
$ \implies V=\int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}(1-2x^2-2y^2)dxdy$
Now, convert to r-plane.
$A=\int \int rdrd{\theta }$
Put $x=r \cos \theta , y=r \sin \theta $
$V=\int_{0}^{\frac{1}{\sqrt {2}}} \int_{0}^{2\pi}(1-2r^2)rdrd{\theta}$
$ \implies V=4\int_{0}^{{\frac{1}{\sqrt {2}}}} \int_{0}^{\frac{\pi}{2}}(1-2r^2)rdrd{\theta}$
|
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|
Trigonometric sum of a ratio of two sine functions
Evaluate $$\sum^{13}_{k=1}\frac{\sin (30^\circ k +45^\circ)}{\sin(30^\circ(k-1)+45^\circ)}.$$
Put $30^\circ = \alpha, 45^\circ = \beta$. Then
$$\begin{align}
S:=&\sum^{13}_{k=1}\frac{\sin (\alpha k+\beta)}{\sin(\alpha(k-1)+\beta)} = \sum^{13}_{k=1}\frac{\sin(\alpha(k-1)+\beta+\alpha)}{\sin(\alpha(k-1)+\beta)}\\
&= \sum^{13}_{k=1}\frac{\sin(\alpha(k-1)+\beta)\cos \alpha+\cos(\alpha(k-1)+\beta)\sin \alpha}{\sin(\alpha(k-1)+\beta)}\\
&= \cos \alpha\sum^{13}_{k=1}1+\sin \alpha\sum^{13}_{k=1}\cot(\alpha(k-1)+\beta) \\
&= \frac{\sqrt{3}}{2}\cdot 13+\frac{1}{2}\sum^{13}_{k=1}\cot(\alpha(k-1)+\beta).
\end{align}
$$
Could some help me to solve it, thanks.
|
You are on the right track. Note that $\cot(x)=\tan(90^{\circ}-x)$ and $\tan(x)=\tan(x+180^{\circ})$. Hence
$$\begin{align}\sum_{k=1}^{13}&\cot((k-1)30^\circ+45^\circ)=\sum_{k=0}^{12}\tan(45^\circ-k30^\circ)\\
&=1+\sum_{k=1}^{6}\tan(45^\circ-k30^\circ)+\sum_{k=1}^{6}\tan(45^\circ-(k+6)30^\circ)\\
&=1+2\sum_{k=1}^{6}\tan(45^\circ-k30^\circ)\\
&=1+2\left(\tan(15^\circ)+\tan(-15^\circ)+\tan(-45^\circ) +\tan(-75^\circ) +\tan(-105^\circ) +\tan(-135^\circ)\right)\\
&=1+2\left(-\tan(45^\circ) -\tan(75^\circ) +\tan(-180^\circ+75^\circ)+\tan(-180^\circ+45^\circ)\right)\\
&=1.
\end{align}$$
|
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|
Find min of $\int_{-1}^1| x^3-a-bx-cx^2|^2 dx$ and max of $\int_{-1}^1 x^3 g(x) dx$
1) Apply best approximation theorem to compute min$\int_{-1}^1| x^3-a-bx-cx^2|^2 dx$.
2) Find max $\int_{-1}^1 x^3 g(x) dx$, where $g$ is subject to the constraints
$$\int_{-1}^1 g(x)dx = \int_{-1}^1xg(x)dx=\int_{-1}^1 x^2 g(x) dx=0,\quad\text{and}\quad \int_{-1}^1|g(x)|^2 dx=1.$$
I don't know how to start to solve this problem, anyone can help me.
|
Hint for 1). Since $\int_{-1}^1x^d dx=0$ when $d$ is odd, and $\int_{-1}^1x^d dx=\frac{2}{d+1}$ when $d$ is even, by expanding the square we get
$$\int_{-1}^1(x^3-a-bx-cx^2)^2 dx
=\frac{2}{3}b^2-\frac{4}{5}b+\frac{2}{7}+2a^2+\frac{4}{3}ac+\frac{2}{5}c^2\\
=\frac{2}{3}\left(b-\frac{3}{5}\right)^2+\frac{8}{175}+2\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right).
$$
Moreover $(1/3)^2-1/5<0$ implies that $\left(a^2+\frac{2}{3}ac+\frac{1}{5}c^2\right)\geq 0$.
Hint for 2) (see Professor Vector's comment). By the given constraints and the Cauchy-Schwarz inequality,
$$\begin{align}
\int_{-1}^1 x^3 g(x) dx&=\int_{-1}^1 (x^3-a-bx-cx^2)g(x) dx\\
&\leq
\left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2}
\left(\int_{-1}^1 (g(x))^2dx\right)^{1/2}\\
&\leq
\left(\int_{-1}^1 (x^3-a-bx-cx^2)^2dx\right)^{1/2}.
\end{align}$$
Hence the maximum of the LHS is less or equal to the minimum of the RHS (see 1)).
|
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|
Solve the integral $\int_0^1 x^{\frac{1}{3}}(1-x)^{\frac{2}{3}}dx$ I would like to consider two ways to compute the (real) integral $\int_{0}^{1}x^{\frac{1}{3}}(1-x)^{\frac{2}{3}}dx$ using complex analysis:
(i) By residues
(ii) By Beta function
My computations:
(i) First of all, my intuition is that the value of the integral should be either $0$ or another real value.
$x^{\frac{1}{3}}(1-x)^{\frac{2}{3}}$ has two singularities in 0 and 1. Therefore, $\sqrt[3]{z(1-z)^2}$ is well defined on $\mathbb{C}\setminus [0,1]$ and $\int_{|z|=\varepsilon}\sqrt[3]{z(1-z)^2}dz \leq 2\pi \varepsilon k \rightarrow 0$, as $\varepsilon \rightarrow 0$, for some constant $k$.
$\int_{\gamma_\varepsilon}\sqrt[3]{z(1-z)^2}dz\rightarrow 3\int_0^1 x(1-x)^2dx$, as $\varepsilon \rightarrow 0$.
Now we need to find the Laurent expansions at the singularities.
For $z=0$
$\sqrt[3]{z(1-z)^2} = \sqrt[3]{z^3-2z^2+z}=z\sqrt[3]{1-\frac{2}{z}+\frac{1}{z^2}}$ and fixing $w=\frac{2}{z}-\frac{1}{z^2}$. Hence, using the binomial formula for $(1-w)^{1/3}$, we get
$(1-w)^{1/3}=1+\frac{w}{3}-\frac{w^2}{9}+\frac{5w^3}{3}+\dots$
Plugging back the value of $w$:
$z(1-w)^{1/3}=z(1+\frac{\frac{1}{z}(2-\frac{1}{z})}{3}-\frac{\frac{1}{z^2}(2-\frac{1}{z})^2}{9}+\dots)=z+\frac{2-1/z}{3}-\frac{(2-1/z)^2}{9z}+\dots$
As a result, the residue is $(-\frac{1}{3}-\frac{4}{9})\frac{1}{z}=-\frac{7}{9}\frac{1}{z}$
For $z=1$
$\sqrt[3]{z(1-z)^2}=\sqrt[3]{(z-1)^2(z-1+1)}=\sqrt[3]{(z-1)^3+(z-1)^2}=(z-1)\sqrt[3]{1-\frac{-1}{z-1}}=(z-1)(1-\frac{1}{3(z-1)}-\frac{1}{9(z-1)^2}+\dots)$
Hence, the residue for $z=1$ is $-\frac{1}{9}$
By the Cauchy-Riemann integral formula, we have
$\int_{|z|=\varepsilon}\sqrt[3]{z(1-z)^2}dz=2\pi i(-\frac{8}{9})$ which is complex-valued.
(ii) Using the Beta function $f(m,n)=\int_0^1x^{m-1}(1-x)^{n-1}=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$, for $m-1=1/3$ and $n-1=2/3$, we get:
$\int_0^1 x^{\frac{1}{3}}(1-x)^{\frac{2}{3}}dx=\frac{\Gamma(4/3)\Gamma(5/3)}{\Gamma(3)}=\frac{\frac{1}{3}\Gamma(\frac{1}{3})\frac{2}{3}\Gamma(\frac{2}{3})}{2!}=\frac{1}{9}\Gamma(\frac{1}{3})\Gamma(1-\frac{1}{3})=\frac{1}{9}\frac{\pi}{\sin\frac{\pi}{3}}=\frac{1}{9}\pi\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}\pi}{27}$
Could someone please help me to understand where the mistake is?
|
You seem to want to use Cauchy's residue theorem (not to be confused with the Cauchy-Riemann equations, BTW) to evaluate this integral. To use the residue theorem, you need to integrate over a closed contour in the complex plane. As far as I can tell, you don't have one; you're just adding up all the residues that you can find. This is not a valid approach.
It is probably possible to find a closed contour in the complex plane that allows you to evaluate this integral. Note, however, that the integrands has branch points at 0 and 1, which means that you have to be careful in what counts as a "closed contour" for the purposes of the residue theorem.
|
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|
Find the result of the root expression, is my answer correct or not? Suppose $a < 0 < b$. Then what is the result of:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = ?$
I have a solution but I can't be sure if I did a mistake, because I usually do! My solution:
Call $a = -c$ for some $0 < c $. Then,
$=\sqrt{(-c-b)^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(- (c+b))^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(c+b)^2} + \sqrt[6]{ b^6 }$
$=(c+b) + b$
$= -a +2b$
|
i would write $$\sqrt{(a-b)^2}+\sqrt[6]{b^6}=\sqrt{(b-a)^2}+\sqrt[6]{b^6}=b-a+b$$
this is the same as you
|
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|
Can this definite integral involving series be solved without a calculator? I got this question today but I can't see if there is any good way to solve it by hand.
Evaluate the definite integral
$$\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}\,\mathrm{d}x$$
where the series in the numerator and denominator continue infinitely.
If you let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$, solving for $y$ we get $y=\frac{1\pm\sqrt{1+4x}}{2}$. And similarly for the denominator we have $z=\sqrt{xz}$. So $z=x$. So the integral simplifies to
$$\int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$
Now my problems are
*
*I don't know what to with the $\pm$.
*I tried to solve the integral by separating it as a sum of two fractions. But I can't solve $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$
|
For 2. notice that
$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx=\int_{2}^{12}\frac{1+4x}{2x\sqrt{1+4x}}=\frac{1}{2}\left(\int_{2}^{12}\frac{1}{x\sqrt{1+4x}}dx+4\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}\right).$$
Now the second integral is pretty easy
$$2\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}=\left[\sqrt{1+4x}\right]_{2}^{12}=4,$$
for the first one instead
$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{2}^{12}\frac{dx}{x\sqrt{1+\left(2\sqrt{x}\right)^2}},$$
let $2\sqrt{x}=\sinh y$, so $x=\frac{\sinh^2 y}{4}$ and $dx=\frac{\cosh y \sinh y}{2}\ dy$. Hence
$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{\frac{\cosh y \sinh y}{2}}{\frac{\sinh^2}{4} y\sqrt{1+\sinh^2 y}}dy=\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{dy}{\sinh y}=$$
$$=2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy.$$
Now let $e^y=z$, so $e^y\ dy=dz$
$$2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy=2\int_{\eta}^{\psi}\frac{dz}{z^2-1},$$
where $\eta=e^{\sinh^{-1}{2\sqrt2}}$ and $\psi={e^{\sinh^{-1}{4\sqrt3}}}.$
$$2\int_{\eta}^{\psi}\frac{dz}{z^2-1}=\ln\left|\frac{z-1}{z+1}\right|_{\eta}^{\psi}=\ln\left|\frac{e^y-1}{e^y+1}\right|_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}=$$
$$=\ln\left|\frac{e^{\sinh^{-1}{2\sqrt{x}}}-1}{e^{\sinh^{-1}{2\sqrt{x}}}+1}\right|_{2}^{12},$$
so
$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$
|
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|
Help in solving linear recurrence relation I need to solve the following recurrence relation: $a_{n+2} + 2a_{n+1} + a_n = 1 + n$
My solution:
Associated homogeneous recurrence relation is:
$a_{n+2} + 2a_{n+1} + a_n = 0$
Characteristic equation: $r^2 + 2r + 1 = 0$
Solving the characteristic equation, we get: $r = -1$ with multiplicity $m = 2$
Therefore, solution of the homogeneous recurrence relation is:
$a_n^{(h)} = (c_1 + c_2n)(-1)^n$
Let the particular solution of the given equation be
$a_n = c_3 + c_4n$
since, $(n + 1)$ is polynomial of degree 1.
Substituting in the given equation, we get:
$c_3 + c_4(n + 2) + 2(c_3 + c_4(n + 1)) + c_3 + c_4n = n + 1$
Comparing the corresponding coefficients, we get: $c_4 = 1/4$ and $c_3 = 0$.
Therefore, $a_n^{(p)} = n / 4$
Hence, the solution, would be:
$a_n = (c_1 + c_2n)(-1)^n + n / 4$
But the solution in textbook is
$a_n = (c_1 + c_2n)(-1)^n + 1/6(2n - 1)$
Please explain me where I am going wrong.
Thanks!
|
General technique uses generating functions. Define:
$\begin{equation*}
A(z) = \sum_{n \ge 0} a_n z^n
\end{equation*}$
Multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
+ 2 \sum_{n \ge 0} a_{n + 1} z^n
+ \sum_{n \ge 0} a_n z^n
&= \sum_{n \ge 1} z^n
+ \sum_{n \ge 0} n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
+ 2 \frac{A(z) - a_0}{z}
+ A(z)
&= \frac{1}{1 - z}
+ z \frac{d}{d z} \frac{1}{1 - z} \\
&= \frac{1}{1 - z} + \frac{z}{(1 - z)^2}
\end{align*}$
Solve for $A(z)$, split into partial fractions:
$\begin{equation*}
A(z)
= -\frac{4 a_1 + 4 a_0 - 1}{(1 + z)^2}
+ \frac{\frac{4 a_1 + 8 a_0 - 1}{(1 + z)}
+ \frac{1}{4 (1 - z)^2}
- \frac{1}{4 (1 - z)}
\end{equation*}$
Note that:
$\begin{align*}
(1 - \alpha z)^{-m}
&= \sum_{n \ge 0}
(-1)^n \binom{-m}{n} \alpha^n \zeta^n \\
&= \sum_{n \ge 0}
\binom{m + n - 1}{m - 1} \alpha^n z^n
\end{align*}$
Also:
$\begin{align*}
\binom{n + 1}{1}
&= n + 1 \\
\binom{n + 0}{0}
&= 1
\end{align*}$
and you can read off the coefficients. The solution here has the form:
$\begin{equation*}
a_n
= (-1)^n c_1 n + (-1)^n c_2 + c_3 n + c_4
\end{equation*}$
|
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|
If $p$ is prime, $p\ne3$ then $p^2+2$ is composite I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+2$. The first one will be even when $k$ is odd, and the second one will be even when $k$ is even. So we will see what happens for each form in the case that $k$ is even (for the first form) and $k$ is odd (for the second form):
*
*$p=3k+1$, $k$ is even
Since $k$ is even, we can write it as $k=2q$ for some $q$. Then
$p^2+2=(3(2q)+1)^2+2=(6q+1)^2+2=6^2q^2+12q+1+2=3(12q^2+4q+1)$
So $p^2+2$ is composite.
*$p=3k+2$, $k$ is odd
Then, $k$ can be written as $k=2q+1$, for some $q$. Then
$p^2+2=(3(2q+1)+2)^2+2=(6q+5)^2+2=6^2q^2+60q+25+2=3(12q^2+10q+9)$
And again, $p^2+2$ is composite.
QED
Is that a correct proof? Is not the same that comes in the answer books.
|
Yes, it is correct. However, there's a much shorter proof (which is essentially the same as yours): If $p$ is not divisible by $3$, then $p=3k+1$ or $p=3k+2$; then $p^2$ leaves a remainder of $1$ when divided by $3$ ($p^2\equiv 1\mod{3}$), so that $p^2+2$ is divisible by $3$.
Note by the way that you don't really need $p$ to be prime. All you need is that $p>1$ is not divisible by $3$.
|
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|
Determine a probability of coin is fair in repeated trial The probability of getting head in a fair coin is 1/2 and the probability of getting tail in a fair coin is 1/2. While the probability of getting tail in a biased coin is 3/4 and probability of getting head in a biased coin is 1/4.
Suppose we don't know which one is a fair coin, which one is not.
What is the probability that the coin flipped was the fair coin, if 8 heads were observed in 10 trials?
Is it something like P(coin is fair | 8 heads appear in 10 trials)?
|
$$P(\text{8 heads in 10 tosses}|\text{fair coin})={10\choose{8}}\frac{1}{2}^{10}$$
$$P(\text{8 heads in 10 tosses}|\text{unfair coin})={10\choose{8}}\frac{1}{4}^{8}\frac{3}{4}^2$$
Thus, the probability that the coin was fair would be
$$\begin{align*}
P(\text{coin is fair})
&= \frac{P(\text{8 heads in 10 tosses}|\text{fair coin})}{P(\text{8 heads in 10 tosses}|\text{fair coin})+P(\text{8 heads in 10 tosses}|\text{unfair coin})}\\\\
&=\frac{{10\choose{8}}\frac{1}{2}^{10}}{{10\choose{8}}\frac{1}{2}^{10}+{10\choose{8}}\frac{1}{4}^{8}\frac{3}{4}^2}\\\\
&\approx 0.9913
\end{align*}$$
Alternatively, using Bayes' Theorem, assuming that the selected coin was random, we have
$$\begin{align*}
P(\text{coin is fair}|\text{8 heads in 10 tosses})
&= \frac{P(\text{coin is fair}\cap\text{8 heads in 10 tosses})}{P(\text{8 heads in 10 tosses})}\\\\
&= \frac{0.5\cdot{10\choose{8}}\frac{1}{2}^{10}}{0.5\cdot{10\choose{8}}\frac{1}{2}^{10}+0.5\cdot{10\choose{8}}\frac{1}{4}^{8}\frac{3}{4}^2}\\\\
&\approx 0.9913
\end{align*}$$
|
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|
How to prove complex identity? I proved that $z^6+z^3+1 = (z^2-2z\cos(\frac{2π}{9})+1)(z^2-2z\cos(\frac{4π}{9})+1)(z^2-2z\cos(\frac{8π}{9})+1)$
Using this, how to show that $$2\cos(3 \theta)+1=8\left(\cos \theta-\cos\left(\frac{2π}{9}\right)\right)\left(\cos \theta-\cos\left(\frac{4π}{9}\right)\right)\left(\cos\theta-\cos\left(\frac{8π}{9}\right)\right)$$
I know it involves some trig substitution but I'm not sure how. Help appreciated, thanks.
|
Divide by $z^3$:
$$z^3+z^{-3}+1=(z+z^{-1}-2\cos(2\pi/9))(z+z^{-1}-2\cos(4\pi/9))(z+z^{-1}-2\cos(8\pi/9))$$
Put in $z=\exp(i\phi)$, noting that $z+z^{-1}=2\cos\phi$ and
$z^3+z^{-3}=2\cos3\phi$:
$$2\cos3\phi+1=(2\cos\phi-2\cos(2\pi/9))(2\cos\phi-2\cos(4\pi/9))(2\cos\phi-2\cos(8\pi/9)).$$
|
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Prove that $\sum\limits_{r=0}^\infty\,\dfrac{(-1)^r\,n}{r!\,(r+1)!}\,\prod\limits_{k=1}^r\,\left(n^2-k^2\right)=(-1)^{n+1}$
If n is a positive integer, prove that:
$$\begin{align}n - {\frac{n(n^2-1^2)}{2!}}+{\frac{n(n^2-1^2)(n^2-2^2)}{2!3!}} + \ldots\phantom{aaaaaaaaaa} &&\\
+(-1)^r{\frac{n(n^2-1^2)(n^2-2^2)\cdots(n^2-r^2)}{r!(r+1)!}}+\ldots &=& (-1)^{n+1}\end{align}$$
I tried by simplifying it to $\binom{n}{1}\binom{n}{0}-\binom{n}{2}\binom{n+1}{1}+\binom{n}3\binom{n+2}{2}-\ldots$ , or $\sum(-1)^r\binom{n}{r+1}\binom{n+r}{r}$. But I don't know what to do from here.
|
The product is:
$$\prod (n^2-k^2)=(n-r)(n-r+1)\cdots(n-1)(n+1)\cdots(n+r-1)(n+r)$$
$$=\frac{(2r+1)!}{n}{n+r \choose 2r+1}$$
The $n$ is because we are missing that term, the binomial symbol is because we recognize a product of $2r+1$ going down from $n+r$, and the factorial is because the binomial symbol is missing its denominator.
You get
$$\sum_{r=0}^\infty (-1)^r\frac{(2r+1)!}{r!(r+1)!}{n+r \choose 2r+1}$$
Here you recognize another choose symbol
$$\sum_{r=0}^\infty (-1)^r{2r+1\choose r}{n+r \choose 2r+1}=\sum_{r=0}^{n-1} (-1)^r{2r+1\choose r}{n+r \choose 2r+1}$$
where the last term told us the largest reasonable value for $r$ (if bottom is larger than top, it's zero).
You already got to here. Now you can just calculate the recursion. Assuming you have term $a_n$, let's compute the next one:
$$a_{n+1}=\sum_{r=0}^{n} (-1)^r{2r+1\choose r}{n+1+r \choose 2r+1}$$
We use the summation property of the Pascal's triangle:
$$a_{n+1}=\sum_{r=0}^{n} (-1)^r{2r+1\choose r}\left({n+r \choose 2r+1}+{n+r \choose 2r}\right)$$
Can you continue expressing this with $a_n$?
|
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|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
|
As $3^n$ is an odd number, $4\cdot 3^n\equiv 4~(mod~8)$, also $12\equiv 4~(mod~8)$.
|
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|
Calculate the probability density function of two independent random variables: (R^3 + r^3)^(1/3) Random variables R and r are independent, both of them are uniform distributed and greater than zero. R distributes in (a,b), r in (c,d). I tried to solve the probability density function of (R^3 + r^3)^(1/3), but I couldn't figure it out. so any one can help me out?
Question Update: c > a > 0, a + t = b , c + t = d, t > 0, c - a => t
|
I recommend doing this one step at a time.
First let's find the CDFs and densities of $R^3$ and $r^3$:
$$F_{R^3}(x)=P(R^3 \leq x) = P(R \leq x^{1/3}) = \int_{-\infty}^{x^{1/3}}\frac{1}{b-a}1_{a \leq t^{1/3} \leq b} \ dt = \begin{cases}0 & x < a^3 \\ \frac{x^{1/3}-a}{b-a} & a^3\leq x\leq b^3 \\ 1 & x>b^3\end{cases}$$
so
$$f_{R^3}(x) = \frac{d}{dx}F_{R^3}(x) = \begin{cases} 0 & x<a^3, x>b^3 \\ \frac{1}{3(b-a)}x^{-2/3}& a^3 \leq x \leq b^3 \end{cases}.$$
Similarly you can find $F_{r^3}(x)$ and $f_{r^3}(x)$.
Then you can find the density of the sum $R^3+r^3$ using the fact that $R^3$ and $r^3$ are independent:
$$f_{R^3+r^3}(x) = \int_{-\infty}^{\infty} f_{R^3}(x-t)f_{r^3}(t) \ dt.$$
From that you can find $F_{R^3+r^3}(x) = \int_{-\infty}^x f_{R^3+r^3}(t) \ dt$. Finally, from that you can find
$$F_{(R^3+r^3)^{1/3}}(x) = P((R^3+r^3)^{1/3} \leq x) = P(R^3+r^3 \leq x^3)=F_{R^3+r^3}(x^3)$$
and
$$f_{(R^3+r^3)^{1/3}}(x) = \frac{d}{dx}F_{(R^3+r^3)^{1/3}}(x).$$
|
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|
How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no idea.
|
Let $x = \tan u \implies dx = \sec^2u\ du$
$$\begin{align}\int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx =& \int_0^{\pi/2} \dfrac{\arctan(\tan u)\sqrt{\sec u-1}\sec u}{\tan u }\,du &\\=& \int_0^{\pi/2} \dfrac{u\sqrt{\sec u-1}}{\sin u }\,du&\\=& \int_0^{\pi/2} \dfrac{u\sqrt{1-\cos u}}{\sin u \sqrt{\cos u}}\,du&\\=& \sqrt{2}\int_0^{\pi/2} \dfrac{u\sin u/2}{\sin u \sqrt{\cos u}}\,du &\\=& \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} \dfrac{u}{\cos (u/2) \sqrt{\cos u}}\,du&\\=& \dfrac{1}{\sqrt{2}}\left(\left[2u \sin^{-1}\left(\tan \dfrac u2\right)\right]_0^{\pi/2} - 2\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du\right)&\\=& \dfrac{\pi^2}{2\sqrt{2}} - \dfrac{2}{\sqrt{2}}\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du \end{align} $$
Let $z = u/2 \implies 2dz = du$
$$\begin{align}\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du&=& 2\int_0^{\pi/4}\sin^{-1}(\tan z )\, dz\end{align}$$
Let $t = \tan z$
$$\begin{align} 2\int_0^{\pi/4}\sin^{-1}(\tan z )\, dz=& 2\int_0^1\dfrac{\sin^{-1} t}{t^2 + 1} dt&\\=& 2[\tan^{-1} x \sin^{-1} x]_0^1 -
2\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt&\\=& 2\dfrac{\pi^2}{8} -
2\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt\end{align}$$
The integral $$\int_0^1 \dfrac{\tan^{-1}(t)}{\sqrt{1-t^2}}\, dt = -\frac{\ln^{2}(1+\sqrt{2})}{2} + \frac{\pi^{2}}{8} \tag 1$$
So,
$$\int_0^{\pi/2}\sin^{-1}\left(\tan \dfrac u2\right)\, du = \ln^2(1+ \sqrt{2})$$
The final answer is then
$$\dfrac{\pi^2}{2\sqrt{2}} - \sqrt{2}\ln^2(1+\sqrt{2})$$.
For the proof of $(1)$ see here.
|
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|
Prove that $\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$ I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum?
$$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
|
Starting from
$$\sum_{n=0}^N {2N-n\choose N} 2^n (n+1)$$
we write
$$\sum_{n=0}^N {2N-n\choose N-n} 2^n (n+1)
= \sum_{n=0}^N 2^n (n+1) [z^{N-n}] (1+z)^{2N-n}
\\ = [z^N] \sum_{n=0}^N 2^n (n+1) z^n (1+z)^{2N-n}.$$
We may extend $n$ to infinity beyond $N$ because the sum term does not
contribute to the coefficient extractor in that case, getting
$$[z^N] (1+z)^{2N} \sum_{n\ge 0} 2^n (n+1) z^n (1+z)^{-n}
= [z^N] (1+z)^{2N} \frac{1}{(1-2z/(1+z))^2}
\\ = [z^N] (1+z)^{2N+2} \frac{1}{(1-z)^2}.$$
Extracting the coefficient we find
$$\sum_{q=0}^N {2N+2\choose q} (N+1-q).$$
The first piece here is
$$(N+1) \sum_{q=0}^N {2N+2\choose q}
= (N+1) \frac{1}{2} \left(2^{2N+2} - {2N+2\choose N+1}\right).$$
The second piece is
$$\sum_{q=1}^N {2N+2\choose q} q
= (2N+2) \sum_{q=1}^N {2N+1\choose q-1}
\\ = (2N+2) \sum_{q=0}^{N-1} {2N+1\choose q}
= (2N+2) \frac{1}{2} \left(2^{2N+1}
- {2N+1\choose N} - {2N+1\choose N+1} \right).$$
Joining the two pieces the powers of two cancel and we are left with
$$(2N+2) {2N+1\choose N} - \frac{1}{2} (N+1) {2N+2\choose N+1}
\\ = (2N+2) \frac{2N+1}{N+1} {2N\choose N}
- \frac{1}{2} (N+1) \frac{2N+2}{N+1} {2N+1\choose N}
\\ = 2 (2N+1) {2N\choose N}
- (N+1) \frac{2N+1}{N+1} {2N\choose N}
\\ = (2N+1) {2N\choose N}$$
as claimed.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Determinant of hermitian matrix Let $M=A+iB$ be a complex $n \times n$ Hermitian matrix. First of all we know that $$(\det M)^2=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}.$$ Also $\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$ is a polynomial in $n^2$ variables of degree $2n$. Is it true that $\det M$ is a polynomial, say D, of degree $n$ in this $n^2$ variables such that $D^2(M)=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$? The explicit calculations for $n=1,2,3$ suggest so, yet I can't find the information if this is true neither proof this.
|
The matrix is similar to
$$
S = T \hat{M} T^{-1}=
\begin{pmatrix} I & iI \\ I & -iI \end{pmatrix}
\begin{pmatrix} A & -B \\ B & A \end{pmatrix}
\begin{pmatrix} \frac12I & \frac12 I\\-\frac i2I& \frac i2I\end{pmatrix} =
\begin{pmatrix} A+Bi &\\ &A-Bi \end{pmatrix}=
\begin{pmatrix} M &\\ &M^* \end{pmatrix}=\begin{pmatrix} M &\\ &M \end{pmatrix}
$$
hence $S$ and $\hat M$ have the same characteristic polynomial.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove Big Theta on polynomial function? Im working on the Big-O notation and struggle to understand whether I have done enough to prove the following:
$5n^3+4n^2+4 \in \Theta (n^3)$
So based on the definition of $Θ(g(n)):$
Step $1$: $0 ≤ c_1 n^3 \leq 5n^3+4n^2+4 \leq c_2 n^3$
Divide the inequality by the largest order n-term
Step $2$: $0 ≤ c_1 \leq 5+(4/n+4/n^3) \leq c_2$
Find constant $c_2$ that will satisfy:
Step $3$: $0 \leq 5+(4/n+4/n^3) \leq c_2$
For $n=1$, $0 \leq 5+(4/1+4/1^3)=13$
For $n=2$, $0 \leq 5+(4/2+4/2^3)=7,5$
For $n=3$, $0 \leq 5+(4/3+4/3^3)=6,7$
For $n=4$, $0 \leq 5+(4/4+4/4^3)=6,06$
Since $c_2$ approaches $5$ when $n \to \infty$, I pick 5 for $c_2$ as it satisfies the equation in Step $3$
Step $4$: Is to find $c_1$
Since $c_1$ can be $\leq 5$
I can say that $5n^3+4n^2+4 \in \Theta (n^3)$ for $c_1$ $\leq 5$ and $c_2 \geq 5$. However not sure how to find $n_0$ in that regard.
Appreciate any insights!
|
Hint: $1 \le n^2 \le n^3$ for $n \ge 1\,$, so $5n^3 \;\le\; 5n^3+4n^2+4 \;\le\; 5n^3+4n^3+4n^3 = 13 n^3\,$.
|
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|
Parsing a question about the curl of a vector field on the surface of a sphere Given the following question and temptative answers:
I am not sure what the answer is for the $V(x,y)$ box.
Normally I would have converted the problem to spherical coordinates and solved it that way.
My guess is that when one is converting $\iint_M(\nabla\times F)\cdot dS$ to $\iint_DV(x,y)dydx$, there is a term that appears from the change in the differential that needs to be multiplied to the curl in order to obtain the correct term, but I am not sure what it should be.
|
We have that $\nabla \times \mathbf{F}=\langle2yzx^3,-3y^2zx^2,-2\rangle$ and we have $z=-\sqrt{\frac{1-x^2-y^2}{6}}$ since $z\leq0$. Then we have that
$$\begin{align}
\iint_M(\nabla\times F)\cdot dS & = \iint_M \langle2yx^3z,-3y^2x^2z,-2\rangle \cdot\langle-z_{x},-z_{y},1\rangle\,dA\\
& = \iint_M\langle-2yx^3\sqrt{\frac{1-x^2-y^2}{6}}, 3y^2x^2\sqrt{\frac{1-x^2-y^2}{6}}, -2\rangle \cdot\langle-z_{x},-z_{y},1\rangle\,dA\\
& = \frac{1}{6}\iint_M -2yx^4+3y^3x^2-12\,dA.
\end{align}$$
Then $$V(x,y)= \frac{-2yx^4+3y^3x^2-12}{6}.$$
Now we will convert to polar coordinates. Since $x^2+y^2=1$, we have that $r$ will range from $0$ to $1$ and since our figure is an ellipsoid we have that $\theta$ will range from $0$ to $2\pi$. Then our equation becomes
$$\begin{align}
\iint_DV(x,y)\,dy\,dx & = \frac{1}{6}\int_{0}^{1}\int_0^{2\pi}(-2r^5\cos^4(\theta)\sin(\theta)+3r^5\cos^2(\theta)sin^3(\theta)-12)r\,d\theta\,dr \\
& = \frac{1}{6}\int_{0}^{1}(0+0-12(2\pi))r\,dr \\
& = -2\pi\,r^2 \big|_{0}^{1} \\
& = -2\pi.
\end{align}$$
|
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|
joint pdf of two statistically independent random variables Let A and B be statistically independent, identically distributed (iid)
random variables having chi-square distribution with four degrees of freedom.
*
*calculate the joint pdf for $X=\frac{A-B}{A+B}$ and $Y=A+B$
*calculate the marginal pdfs of X and Y
So I did the following:
The chi-square distribution with four degrees of freedom is given by $f_X(x) = \frac{1}{4}x e^{-x/2}$
Since A and B are statistically independent $f_{A,B} = f_A(a) \cdot f_B(b)=\frac{1}{16}a b e^{-\frac{a+b}{2}}$
So since $X=\frac{A-B}{A+B}$ and $Y=A+B$ I expressed that as $A=\frac{1}{2}(Y+XY)$ and $B=\frac{1}{2}(Y-XY)$
Now for the transformation I have to calculate the det. of the Jacobian matrix
$\det\begin{pmatrix} \frac{\partial \frac{1}{2}(Y+XY)}{\partial X}
& \frac{\partial \frac{1}{2}(Y+XY)}{\partial Y} \\ \frac{\partial \frac{1}{2}(Y-XY)}{\partial X}
& \frac{\partial \frac{1}{2}(Y-XY)}{\partial Y}\end{pmatrix} = \det\begin{pmatrix} 1/2 Y
& 1/2[1+X] \\ -1/2 Y
& 1/2[1-X]\end{pmatrix} = \frac{1}{2} Y$
So the joint pdf will be $f_{X,Y}(x,y) = f_{A,B}\left(\frac{1}{2}[y+xy], \frac{1}{2} [y-xy]\right) \cdot \frac{1}{2} Y = \frac{1}{128}y^3(1-x^2)e^{-y/2}$
When I now try to calculate the marginal density $f_Y(y) = \int_0^\infty \frac{1}{128}y^3(1-x^2)e^{-y/2} \, dx$ i have the problem that it doesn't converge
I would be thankful if someone could show me where my mistake is .
|
Thanks to @Henry I found out my mistakes, I used the wrong integration borders,
just for completeness i will answer my own question:
$f_X(x) = \int\limits_0^{\infty} \frac{1}{128}y^3(1-x^2)e^{-y/2}\, dy = \frac{3}{4} (1-x^2)$
$f_Y(y) = \int\limits_{-1}^{1} \frac{1}{128}y^3(1-x^2)e^{-y/2}\, dx = \frac{1}{96}y^3e^{-y/2}$
So we see $X$ and $Y$ are again statistically indepent since $f_{X,Y}(x,y) = f_X(x) \cdot F_Y(y)$
Thanks again Henry
|
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|
Solving the limit $\lim_{x\rightarrow+\infty}(\sqrt[5]{x^5-x^4}-x)$ Recently, I am struggling to solve the limit: $$\lim_{x\rightarrow+\infty}(\sqrt[5]{x^5-x^4}-x)$$
If I try to make some fraction with nominator $-x^4$ and some irrational denominator by multiplying, it becomes more complex. Can anyone help about this with more easier way?
|
Let $a=\sqrt[5]{x^5-x^4}$ and $b=x$. Then\begin{align}\sqrt[5]{x^5-x^4}-x&=a-b\\&=\frac{(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)}{a^4+a^3b+a^2b^2+ab^3+b^4}\\&=\frac{a^5-b^5}{a^4+a^3b+a^2b^2+ab^3+b^4}\\&=\frac{-x^4}{\sqrt[5]{x^5-x^4}^4+\sqrt[5]{x^5-x^4}^3\,x+\sqrt[5]{x^5-x^4}^2\,x^2+\sqrt[5]{x^5-x^4}\,x^3+x^4}\\&=\frac{-1}{\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^4+\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^3+\left(\frac{\sqrt[5]{x^5-x^4}}x\right)^2+\frac{\sqrt[5]{x^5-x^4}}x+1}\\&=\frac{-1}{\left(\sqrt[5]{1-\frac1x}\,\right)^4+\left(\sqrt[5]{1-\frac1x}\,\right)^3+\left(\sqrt[5]{1-\frac1x}\,\right)^2+\sqrt[5]{1-\frac1x}+1}\end{align}and therefore\begin{align}\lim_{x\to+\infty}\sqrt[5]{x^5-x^4}-x&=\lim_{x\to+\infty}\frac{-1}{\left(\sqrt[5]{1-\frac1x}\,\right)^4+\left(\sqrt[5]{1-\frac1x}\,\right)^3+\left(\sqrt[5]{1-\frac1x}\,\right)^2+\sqrt[5]{1-\frac1x}+1}\\&=-\frac15.\end{align}
|
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|
Mathematical induction proof that $f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$
The function $f(n)$ for $n=0,1...$ has the recursive definition $$f(n)= \begin{cases} 2 & \text {for n=0} \\ -f(n-1)+1 & \text{for n=1,2...} \end{cases}$$
Prove by induction that the following equation holds: $$f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$$
So, I begin by checking that the basic step holds
*
*$f(0)=\frac{1}{2}+\frac{3}{2}(-1)^0=2$ OK
*Assume that the equation holds for a given $n$
*Show that n+1 holds: $f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n+1} \Rightarrow f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n} \cdot (-1) = -f(n)-\frac{1}{2}$
I get kind of stuck here. Any advice on how I should approach this?
|
$f(n)=1-f(n-1)$. We want to show that
$$f(n+1)=1-f(n)=1-[1-f(n-1)]=f(n-1)$$
$f(n+1)=\frac{1}{2} - \frac{3}{2}(-1)^n $
$f(n-1)= \frac{1}{2} + \frac{3}{2}(-1)^{n-1}$. Multiply this equality $-1$,then
$-f(n-1)= [\frac{1}{2} + \frac{3}{2}(-1)^{n-1}](-1)=-\frac{1}{2}+\frac{3}{2}(-1)^n$. Hence we obtain
$f(n-1)=\frac{1}{2} - \frac{3}{2}(-1)^n=f(n+1)$
|
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|
Proof by induction in trigonometry.
Prove that $\cos x +\cos 2x + \cos 3x + ...+ \cos nx =\cos \left(\dfrac{n+1}{2}x\right) \sin \left(\dfrac{nx}{2}\right)\csc \dfrac{x}{2}$
Attempt:
Clearly, $P(1)$ is true.
Assume $P(m)$ is true.
Thus, $P(m+1) = (\cos x +\cos 2x + \cos 3x + ...+ \cos mx)+ \cos((m+1)x)$
$= \cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)\csc \dfrac{x}{2} + \cos((m+1)x)
\\= \csc (\dfrac x 2)\left(\cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)+ (\cos(m+1)x)\sin (\dfrac x 2)\right)$
What do I do next?
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Formula to be used:
$\sin A- \sin B = \cos\left(\dfrac{A+B}{2}\right)\sin \left(\dfrac{A-B}{2}\right)$
Thus,
$\csc \left(\dfrac x 2 \right)\left(\cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)+ (\cos(m+1)x)\sin (\dfrac x 2)\right)$
$= \csc \left(\dfrac x 2 \right)\left(\dfrac 1 2 \left(\sin \dfrac{2mx+x}{2} - \sin \dfrac x 2 \right)+ \dfrac 1 2 \left(\sin \dfrac{2mx+3x}{2} - \sin \dfrac{2mx + x}{2 } \right) \right)$
Now, again use the formula on the left out terms.
$= \csc \left(\dfrac x 2 \right)\left(\dfrac 1 2 \left(\sin \dfrac{2mx+3x}{2} - \sin \dfrac x 2 \right) \right)$
$= \cos \left(\dfrac{m+2}{2}x\right) \sin \left(\dfrac{(m+1)x}{2}\right)\csc \dfrac{x}{2} $
Thus, $P(m+1)$ is also true.
Q.E.D.
|
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Transform $\iint_D \sin(x^2+y^2)~dA$ to polar coordinates and evaluate the polar integral.
Transform the given integral in Cartesian coordinates to polar coordinates and evaluate the polar integral:
(b) $\iint_D \sin(x^2+y^2)~dA$, where $D$ is the region in the first quadrant bounded by the lines $x=0$ and $y=\sqrt{3}\cdot x$ and the circles $x^2+y^2=2\pi$ and $x^2+y^2=3\pi$.
In Cartesian coordinates I know $0\le y\le \sqrt{3 \pi}, \space0\le x \le \frac{\sqrt{3\pi}}{4}$
I also know $x=r\cos\theta$, $y=r\sin\theta$
How should I proceed from here?
$r$ remains the same $r= \sqrt{3\pi}$?
So $0\le r\le\sqrt{3\pi}$, $0\le\theta\le \frac{\pi}{2}$. Is this correct?
|
No, it is not correct. The polar representation you give yields a quarter circle of radius $\sqrt{3\pi}$ on the first quadrant, which is obviously not the region described by the question.
The first thing I always do to solve these types of problems is to sketch the region of integration. In this case, it turns out to be extremely helpful:
The region $D$ you are concerned about I highlighted in yellow. It is easy to see that $x^2+y^2=2\pi$ and $x^2+y^2=3\pi$ are just circles of radius $\sqrt{2\pi}$ and $\sqrt{3\pi}$ respectively. Similarly, it can be seen that $y=\sqrt{3}\cdot x$ and $x=0$ correspond to $\theta=60^{\circ}=\pi/3$ (Since $\arctan(\sqrt{3})=\pi/3$) and $\theta=90^{\circ}=\pi/2$ respectively. Hence, the region can be described in polar coordinates as:
$$D=\{(r,\theta)\in \mathbb{R}^2\mid \sqrt{2\pi}\leq r\leq \sqrt{3\pi}, \pi/3\leq \theta\leq \pi/2\}$$
Hence, converting the double integral to polar coordinates gives (Don't forget the Jacobian):
$$\iint_D \sin(x^2+y^2)~dA=\int_{\pi/3}^{\pi/2} \int_{\sqrt{2\pi}}^{\sqrt{3\pi}} \sin(r^2)\cdot r~dr~d\theta$$
This should be easy to evaluate.
|
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|
Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$
|
The curve has polar equation
$r=7 \sin t$
multiply both sides by $r$
$r^2=7r\sin t$
in rectangular coordinates becomes
$x^2+y^2=7y$
Equation of a circle having center $C(0,\;7/2)$ and radius $R=7/2$
The point of the circle when $t=\pi/6$ can be found intersecting the line, in polar form, $a:t=\pi/6$ with the given circle. The line $a$ in rectangular coordinates has equation
$$y=x\tan(\pi/6)\to x=y\sqrt 3$$
plugging this in the equation of the circle we get
$3y^2+y^2=7y\to y=0;\;y=7/4$
Therefore coordinates of $P$ are $(7/4\sqrt 3,\;7/4$)
To find the slope of the tangent in $P$ we write the equation of the polar line to the circle in $P$. The polar line of a conic in a point lying on the conic is the tangent to the conic in that point.
To find the polar line substitute in the equation of the conic$^{(*)}$
$x^2\to x_Px;\;y^2\to y_Py;\;y\to \dfrac{y+y_P}{2}$
and get
$\dfrac{7}{4}x\sqrt 3+\dfrac{7}{4}y =7\,\dfrac{y+\frac{7}{4}}{2}$
$x\sqrt 3+y=2\left(y+\frac{7}{4}\right)$
$y=x\sqrt{3} -\dfrac{7}{2}$
and finally the slope $\color{red}{m=\sqrt 3}$
Ehm... maybe not so much simpler :)
$(*)$ to complete the formulas
$x\to \dfrac{x+x_p}{2};\;xy\to \dfrac{x_Py+y_Px}{2}$
$$...$$
Hope this can be useful
|
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If $A+B = I_n$ and $A^2 +B^2 = O_n$ then $A$ and $B$ are invertible and $(A^{-1}+B^{-1})^n = 2^n I_n$ For $A$ and $B$ square matrices of size $n$, show that:
If $A+B = I_n$ and $A^2 +B^2 = O_n$ then $A$ and $B$ are invertible and $(A^{-1}+B^{-1})^n = 2^n I_n$
Where $I_n$ is the identity matrix of order $n$, and $O_n$ is the square matrix of size $n$ with all the entries equal to zero.
|
First, let's show that A is invertible:
$$\begin{align} A+B=I\\B=I-A\\A^2+(I-A)^2=0\\I-2A+2A^2=0\\A(2(I-A))=I\\A^{-1}=2(I-A)\end{align}$$
Similarly, $B^{-1}=2(I-B)$, so $A^{-1}+B^{-1}=2(2I-A-B)=2I$ (because $A+B=I$), so:
$$(A^{-1}+B^{-1})^n=(2I)^n=2^nI$$.
|
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On the evaluation of $\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4x^2} \right ) \, dx$ When attempting to find an alternative solution to this question here, which called for the evaluation of the integral
$$\int_0^\infty \frac{x \sin x}{1 + x^4} \, dx$$
using real methods, I ran up against the following improper integral
$$\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx. \tag1$$
A value for this improper integral can be found. It is
$$\frac{\sqrt{\pi}}{2} \exp \left (-\frac{1}{\sqrt{2}} \right ) \sin \left (\frac{1}{\sqrt{2}} \right ),$$
and is what I am having trouble in finding.
I have tried a number of the various tricks one typically employs when attempting to find such integrals such as Feynman trick of differentiating under the integral sign, series solution, and so on, all to no avail (perhaps I missed something here).
One method that looked promising was to use properties for the (inverse) Laplace transform, namely
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty (\mathcal{L} f)(s) \cdot (\mathcal{L}^{-1} g)(s) \, ds.$$
After enforcing a change of variable $x \mapsto \dfrac{1}{2 \sqrt{x}}$ we have
\begin{align*}
\int_0^\infty e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx &= \frac{1}{4} \int_0^\infty \frac{e^{-1/(4x)}}{x^{3/2}} \cdot \sin x \, dx\\
&= \int_0^\infty \mathcal{L} \{\sin x\} \cdot \mathcal{L}^{-1} \left \{\frac{e^{-1/(4x)}}{x^{3/2}} \right \} \, ds\\
&= \frac{1}{4} \int_0^\infty \frac{1}{s^2 + 1} \cdot \frac{2 \sin (\sqrt{s})}{\sqrt{\pi}} \, ds\\
&= \frac{1}{\sqrt{\pi}} \int_0^\infty \frac{s \sin s}{1 + s^4} \, ds, \tag2
\end{align*}
where in the last line a substitution of $s \mapsto s^2$ has been made.
While this is a perfectly valid approach the only problem is the integral one ends up with in (2) is exactly the integral one started out with.
So my question is
Is it possible to evaluate the integral given in (1) using real methods that does not depend on evaluating the integral given in (2)?
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Considering $$ I=\int e^{-x^2} \cos \left (\frac{1}{4 x^2} \right ) \, dx\qquad \qquad J=\int e^{-x^2} \sin \left (\frac{1}{4 x^2} \right ) \, dx$$
$$K=I+iJ=\int e^{-x^2+\frac{i}{4 x^2}}\,dx\qquad \qquad \qquad \qquad L=I-iJ=e^{-x^2-\frac{i}{4 x^2}}\,dx$$
$$-x^2+\frac{i}{4 x^2}=-\left(x^2-\frac{i}{4 x^2} \right)=-\left(x+\frac{\sqrt{-i}}{2 x} \right)^2+\sqrt{-i}$$
$$-x^2-\frac{i}{4 x^2}=-\left(x^2+\frac{i}{4 x^2} \right)=-\left(x+\frac{\sqrt{i}}{2 x} \right)^2-\sqrt{i}$$
All of that makes
$$K=\frac{ \sqrt{\pi }}{4} \left(e^{-(-1)^{3/4}}
\left(-\text{erf}\left(\frac{(-1)^{3/4}}{2 x}-x\right)-1\right)+e^{(-1)^{3/4}}
\left(\text{erf}\left(x+\frac{(-1)^{3/4}}{2 x}\right)+1\right)\right)$$
$$L=\frac{\sqrt{\pi }}{4} \left(e^{-\sqrt[4]{-1}}
\left(1-\text{erf}\left(\frac{\sqrt[4]{-1}}{2 x}-x\right)\right)+e^{\sqrt[4]{-1}}
\left(\text{erf}\left(x+\frac{\sqrt[4]{-1}}{2 x}\right)-1\right)\right)$$ and then $I$ and $J$ and finally your result since
$J=\frac{1}{2} i (L-K)$
Edit
Using Wolfram Alpha
$$\int_0^\infty e^{-x^2+\frac{a}{x^2}}\,dx=\frac{\sqrt{\pi }}{2} e^{-2 \sqrt{-a}}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/2585053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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