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Is the vector space of R vector system sinx, sin(x-1), cos(x+2) linearly independent? Is the vector space of $R^R$, vector system $\sin{x}$, $\sin(x-1)$, $\cos(x+2)$ linearly independent?
|
You can use the Wronskian to verify that the functions are linearly dependent.
Alternatively, using trigonometric formulas, note that:
$\quad\quad \sin(x-1) = \cos(1) \sin(x) - \sin(1) \cos(x)$
$\quad\quad \cos(x+2) = \cos(2) \cos(x) - \sin(2) \sin(x)$
So (multiply by $\cos(2)$ and $\sin(1)$ respectively):
$\quad\quad \cos(2)\sin(x-1) = \cos(1)\cos(2) \sin(x) \color{red}{- \sin(1)\cos(2) \cos(x)}$
$\quad\quad \sin(1)\cos(x+2) = \color{red}{\sin(1)\cos(2) \cos(x)} - \sin(1)\sin(2) \sin(x)$
Add both equations:
$\quad\quad \cos(2)\sin(x-1)+\sin(1)\cos(x+2) = \left(\cos(1)\cos(2) - \sin(1)\sin(2)\right)\sin(x)$.
So it is possible to write $\sin x$ as a linear combination of $\sin(x-1)$ and $\cos(x+2)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Corollary of Fermat's Little Theorem to $a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}$ I'm not sure what to call this proposition in relation to Fermat's Little Theorem (it is not the converse, though it seems related), but I am interested to know if the following holds:
$$
a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}
$$ for $p$ prime and $a \neq 1$.
If this does hold, what is the proof? If not, are there any conditions on $x$ required in order to have $a^x \equiv 1 \pmod{p}$?
|
For each prime $p$, there exists a primitive root $r$ which has the property you want: $$r^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}.$$
For example, if we work modulo $7$, $3$ is a primitive root, because
\begin{align}
3^0 &\equiv 1 \pmod 7 \\
3^1 &\equiv 3 \pmod 7 \\
3^2 &\equiv 2 \pmod 7 \\
3^3 &\equiv 6 \pmod 7 \\
3^4 &\equiv 4 \pmod 7 \\
3^5 &\equiv 5 \pmod 7 \\
3^6 &\equiv 1 \pmod 7 \\
\end{align}
so the powers of $3$ have a period of exactly $6$. It follows that $2$ will not work mod $7$, because $$2^3 \equiv (3^2)^3 \equiv 3^6 \equiv 1 \pmod 7.$$
This MSE post gives a short proof of the existence of primitive roots modulo any prime $p$.
|
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|
Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
|
$$\int \frac{2x+3}{x^2+2x+10} dx$$
$$= \int \frac{2x+2+1}{x^2+2x+10} dx $$
$= \int \frac{2x+2}{x^2+2x+10} dx + \int \frac{1}{x^2+2x+10}dx$
Let solve them individually,
$$\int \frac{2x+2}{x^2+2x+10} dx$$
$$\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$$
$$= \log |x^2 + 2x + 10| + c$$
Other part -
$$\int \frac{1}{x^2+2x+10} dx$$
Use completing the square -
$$\int \frac{1}{x^2+2x+1+9} dx$$
$$=\int \frac{1}{(x+1)^2+(3)^2} dx$$
$$\int \frac 1{x^2+a^2} dx = \frac 1a tan^{-1}\frac xa + c$$
$$=\frac 13 tan^{-1}\frac {x+1}3+ c$$
Add both to get complete answer.
|
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|
What is the easiest way to solve the following differential equation? I am only just starting online courses on the subject of differential equations, and I passed the following differential equation
$$\frac{dy}{dx}= -2x + 3y - 5$$
I'm a bit confused by the use of two variables in the equation. What is the easiest way to solve this? In other words, in a way that a complete differential equation beginner like me can understand it.
|
We can solve equation by the Variation of parameters
$$\frac { dy }{ dx } =-2x+3y-5\\ \frac { dy }{ dx } -3y=0\\ \int { \frac { dy }{ y } } =3\int { dx } \\ \ln { y=3x+C } \\ y=C{ e }^{ 3x }\\ y=C\left( x \right) { e }^{ 3x }\\ { y }^{ \prime }={ C }^{ \prime }\left( x \right) { e }^{ 3x }+3{ C\left( x \right) e }^{ 3x }\\ { C }^{ \prime }\left( x \right) { e }^{ 3x }+3{ C\left( x \right) e }^{ 3x }=-2x+3C\left( x \right) { e }^{ 3x }-5\\ { C }^{ \prime }\left( x \right) { e }^{ 3x }=-2x-5\\ { C }^{ \prime }\left( x \right) ={ e }^{ -3x }\left( -2x-5 \right) \\ { C }\left( x \right) =-2\int { { xe }^{ -3x } } dx-5\int { { e }^{ -3x }dx } +C=-2\left[ -\frac { 1 }{ 3 } \int { xd\left( { e }^{ -3x } \right) } \right] +\frac { 5 }{ 3 } { e }^{ -3x }+C=\\ =\frac { 2 }{ 3 } \left[ x{ e }^{ -3x }-\int { { e }^{ -3x }dx } \right] +\frac { 5 }{ 3 } { e }^{ -3x }+C=\frac { 2 }{ 3 } \left[ x{ e }^{ -3x }+\frac { { e }^{ -3x } }{ 3 } \right] +\frac { 5{ e }^{ -3x } }{ 3 } ={ e }^{ -3x }\left[ \frac { 2x }{ 3 } +2 \right] +C\\ y=C{ \left( x \right) e }^{ 3x }=\frac { 2x }{ 3 } +2+C$$
so
$$ \color{blue}\ {y=\frac { 2x }{ 3 } +2} $$
|
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|
Evaluate $\sum_{n=1}^{64}(-1)^n \left\lfloor \frac{64}{n} \right\rfloor \varphi(n)$
Evaluate $$\sum_{n=1}^{64}(-1)^n \left\lfloor \dfrac{64}{n} \right\rfloor \varphi(n).$$
We have $$S = \sum_{n=1}^{64}(-1)^n \left\lfloor \dfrac{64}{n} \right\rfloor \varphi(n) = \sum_{n=1}^{32}\left\lfloor\dfrac{64}{2n}\right\rfloor \varphi(2n)-\sum_{n=1}^{32}\left\lfloor\dfrac{64}{2n-1}\right\rfloor \varphi(2n-1),$$ but this didn't seem to help. Can we relate this sum to the fact that $\displaystyle \sum_{k=1}^n\left\lfloor\dfrac{n}{k}\right\rfloor \varphi(k) = \dfrac{n(n+1)}{2}$? Also, $64$ is a power of $2$ so we may need to use that fact.
|
The identity you mention is indeed very relevant; also relevant is its proof, which rewrites the sum as $$\sum_{d=1}^n \left\lfloor \frac nd\right\rfloor \phi(d) = \sum_{k=1}^n \sum_{d \mid k} \phi(d)$$ and then uses the fact that $$\sum_{d \mid k} \phi(d) = k. \tag{1}$$ In this problem, we can similarly rewrite $$\sum_{d=1}^{64} (-1)^d \left\lfloor \frac{64}d\right\rfloor \phi(d) = \sum_{k=1}^{64} \sum_{d \mid k} (-1)^d \phi(d) \tag{2}$$ and now it remains to understand the behavior of the sum $$\sum_{d \mid k} (-1)^d \phi(d).$$
Factoring $k$ as $2^a \cdot b$, where $b$ is odd, we can group divisors of $k$ together as divisors of $b$ multiplied by a power of $2$ ranging from $2^0$ to $2^a$. This tells us $$\sum_{d \mid k} (-1)^d \phi(d) = \sum_{d \mid b} \sum_{i=0}^a (-1)^{2^i \cdot d} \phi(2^i \cdot d) = \sum_{d \mid b} \left(-\phi(d) + \sum_{i=1}^a 2^{i-1} \phi(d)\right) = (2^a-2)\sum_{d \mid b} \phi(d)$$
and knowing the identity $(1)$, we can write this as $(2^a-2)b = k - 2b$.
Substituting this into $(2)$, we get $$\sum_{d=1}^{64} (-1)^d \left\lfloor \frac{64}d\right\rfloor \phi(d) = \sum_{k=1}^{64} (k - 2 \operatorname{odd}(k)) = \frac{64 \cdot 65}{2} - 2 \sum_{k=1}^{64} \operatorname{odd}(k)$$
where by $\operatorname{odd}(k)$ I denote the largest odd divisor of $k$.
Because the sum of the first $n$ odd numbers is $n^2$, we have $$\sum_{k=1}^{2n} \operatorname{odd}(k) = n^2 + \sum_{k=1}^n \operatorname{odd}(k)$$ by taking the odd and the even terms separately. By applying this rule repeatedly,
we get a final answer of $$\frac{64 \cdot 65}{2} - 2\left(32^2+16^2+8^2+4^2+2^2+1^2+1^2\right) = -652.$$
|
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What is $\lim_{n\to\infty} \frac{x_n}{y_n}$ provided $x_n=2y_{n}-y_{n-1}$ and $y_n=3y_{n-2}-y_{n-2}$ with $x_0=y_0=1$? Suppose $x_n$ and $y_n$ satisfy:
$$
\begin{pmatrix}
x_n \\
y_n \\
\end{pmatrix}=\begin{pmatrix}
2&1 \\
1&1 \\
\end{pmatrix}\begin{pmatrix}
x_{n-1} \\
y_{n-1} \\
\end{pmatrix} $$
with $x_0=y_0=1$. What is $\displaystyle\lim_{n\to\infty} \frac{x_n}{y_n}$ assuming this limit exists?
After doing some algebra, I've got $x_n=2y_{n}-y_{n-1}$ and $y_n=3y_{n-2}-y_{n-2}$, but cannot go further.
|
Hint: Let $z_{n}\equiv(x_{n},y_{n})^{\intercal}$. Then, we can rewrite
the recurrence as
$$
z_{n}=Mz_{n-1}\text{ (}n\geq1\text{)}
$$
where
$$
M\equiv\begin{pmatrix}2 & 1\\
1 & 1
\end{pmatrix}.
$$
Note that $M$ is diagonalizable. That is, we can write $M=SJS^{-1}$
where
$$
S\equiv\begin{pmatrix}\frac{1}{2}\left(1-\sqrt{5}\right) & \frac{1}{2}\left(1+\sqrt{5}\right)\\
1 & 1
\end{pmatrix}\text{ and }J\equiv\begin{pmatrix}\frac{1}{2}\left(3-\sqrt{5}\right)\\
& \frac{1}{2}\left(3+\sqrt{5}\right)
\end{pmatrix}
$$
By induction, we have $z_{n}=M^{n}z_{0}$. Note that
$$
M^{n}=\left(SJS^{-1}\right)^{n}=SJ^{n}S^{-1}.
$$
Can you figure out the rest?
|
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|
Trick to solving this Summation? I am currently following some slides I found to try and learn how to find the average case complexity for some algorithms. I got stuck when having to handle this summation though:
$$\sum_{i=1}^k i*2^i $$
According to the slides: it can be shown that the the summation is equal to:
$$(k - 1)·({2^k+1}) + 2 $$
Can somebody explain how to get to this answer?
|
$\sum_\limits{i=1}^k i*2^i = 1\cdot2 + 2\cdot2^2 + 3\cdot 2^3 +\cdots +k2^k$
Multiply by $-(1-2)$.
Since $-(1-2) = 1$ it doesn't change the value of the sum, but look what happens as you apply the distributive property.
$-(1-2)(1\cdot2 + 2\cdot2^2 + 3\cdot 2^3 +\cdots +k2^k) = -(1\cdot 2 - 1\cdot 2^2 + 2\cdot 2^2 - 2\cdot 2^3 + 3\cdot2^3 - 3\cdot2^4+\cdots - (k-1)2^k + k2^k - k2^{k+1})$
Simplify
$-(2 +2^2 + 2^3+2^4+\cdots 2^k - k2^{k+1})$
We can use the same gabit again, or you might just say that that is a more recognizable geometric series.
$-(2^{k+1} - 2 - k2^{k+1})\\
(k-1)2^{k+1} + 2$
|
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|
Prove that $\lim_{x \to2 }x^{2}+4x=12$ Please check my prove
we begin by
$$x^{2}+4x-12<\epsilon $$
$$|x+6||x-2|<\epsilon $$
I will prove by use property
$\lim_{x->n}f(x)g(x)=\lim_{x->n}f(x)\lim_{x->n}g(x)$
then we divide into to case
$|x+6|<\sqrt{\epsilon }$ and $ |x-2|<\sqrt{\epsilon }$
and let$\epsilon ,\delta _{1}\delta _{2}>0$
$$0<|x-2|<\delta _{1}\rightarrow |x+6|<\sqrt{\epsilon }$$
Choose $\delta _{1}=\sqrt{\epsilon }$
then $|x+6|<\sqrt{\epsilon }$
$$0<|x-2|<\delta _{2}\rightarrow |x-2|<\sqrt{\epsilon }$$
Choose $\delta _{1}=\sqrt{\epsilon }$
then $|x+6|<\sqrt{\epsilon }$
then use property $\lim_{x->n}f(x)g(x)=\lim_{x->n}f(x)\lim_{x->n}g(x)$
we get
$$|x+6||x-2|<\sqrt{\epsilon }\sqrt{\epsilon }$$
then $\lim_{x->2}x^{2}+4x=12$
|
Let $\epsilon>0$.
We wish to find a $\delta>0$ such that if $\vert x-2\vert<\delta$, then $\vert x^2+4x-12\vert<\epsilon$.
We see that the $x^2-4x-12=(x-2)(x+6)$. Since $x\to2$ we can assume that eventually it will be true that $\vert x-2\vert<1$ which means that
\begin{eqnarray}
-1&<&x-2<1\\
1&<&x<3\\
7&<&x+6<9
\end{eqnarray}
so we see that if $\vert x-2\vert<1$ then $\vert x+6\vert\le9$
So we let $\delta=\min\left\{1,\dfrac{\epsilon}{9}\right\}$
Now if $\vert x-2\vert<\delta$ it follows that
\begin{eqnarray}
\vert x^2+4x-12\vert &=& \vert x+6\vert\cdot\vert x-2\vert\\
&<&9\cdot\frac{\epsilon}{9}\\
&<&\epsilon
\end{eqnarray}
Note that for many limits (there are exceptions) you can find an upper bound $B$ so that
$$\left\lvert\dfrac{f(x)-L}{x-a}\right\lvert\le B$$
whenever $\vert x-a\vert<1$. In such a case you can let $\delta=\min\left\{1,\dfrac{\epsilon}{B}\right\}$.
This will not work in cases where there is a vertical tangent to $y=f(x)$ at $x=a$.
|
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|
Show that $\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$ Show that
$$\lim_{n \to \infty} \sum_{k=3}^n \frac{2k}{k^2+n^2+1} = \ln(2)$$
How many ways are there to prove it ?
Is there a standard way ?
I was thinking about making it a Riemann sum.
Or telescoping.
What is the easiest way ?
What is the shortest way ?
|
Observe
\begin{align}
\sum^n_{k=3}\frac{2k}{n^2+k^2+1} = \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+(k/n)^2}
\end{align}
then we have
\begin{align}
\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}}.
\end{align}
Hence it follows
\begin{align}
\int^1_0 \frac{2x}{1+x^2} \ dx=\lim_{n\rightarrow \infty}\frac{1}{1+n}\sum^n_{k=3} \frac{2k/(1+n)}{1+k^2(1+n)^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+n^{-2}+k^2n^{-2}} \leq \lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=3} \frac{2(k/n)}{1+k^2n^{-2}} = \int^1_0 \frac{2x}{1+x^2}\ dx
\end{align}
which means
\begin{align}
\lim_{n\rightarrow \infty}\sum^n_{k=3} \frac{2k}{1+k^2+n^2} = \int^1_0 \frac{2x}{1+x^2}\ dx = \log 2.
\end{align}
|
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|
How to do this without using trigonometric substitution? $$\int \frac {\rho^2}{(\rho^2+ h^2)^\frac 32} d\rho$$
|
In addition to the pythagorean identities for circular trig functions, such as
$$ 1 + \tan^2 x = \sec^2 x $$
and for the hyperbolic trig functions, such as
$$ 1 + \sinh^2 x = \cosh^2 x $$
there is a similar identity involving rational functions:
$$ 1 + \left( \frac{y^2 - 1}{2y} \right)^2 = \left( \frac{y^2 + 1}{2y} \right)^2 $$
The various forms of this identity can be computed as needed from
$$ (2y)^2 + (y^2 - 1)^2 = (y^2 + 1)^2 $$
So, for example, when one has
$$ \sqrt{1 + x^2} \, \mathrm{d} x$$
one could make the substitution
$$x = \frac{y^2 - 1}{2y} \qquad \qquad y > 0$$
to get
$$ \left( \frac{y^2 + 1}{2y} \right) \cdot \left( \frac{1}{2} \left( 1 + \frac{1}{y^2} \right) \mathrm{d} y \right) $$
With the square root eliminated, the problem can then be solved via partial fractions. Although in this example the denominator is a power of $y$, so you have the much simpler method of just expanding everything.
Similarly, one could substitute $x = \frac{2y}{y^2 - 1}$ instead.
|
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|
If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
|
*
*Find one solution
Try to find it as $f_0(x) = ax^2 +bx + c$
$f_0(x+1) + f_0(x-1) = 2ax^2 + 2a + 2bx +2c$ => $f_0(x)=2x^2-x + 3$
*$f(x) = f_0(x) + g(x)$ => $g(x-1) + g(x+1) =0$
Find all $g(x)$
|
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|
What is the inverse element of $u=i+j+k$ with respect to the multiplication? What is the inverse element of $u=i+j+k$ with respect to the multiplication?
When I have $z = x + iy$ inverse is $\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$ but what happens in my case?
|
Try $\frac{-i-j-k}{3}$. This is a natural extension of the complex conjugate formulation to the context of having three imaginary units.
I think Don Antonio's answer gives a good indication of how to find the inverse. To give another example of this sort of thinking: in the hyperbolic numbers $z = x+jy$ where $j^2=1$ we find the multiplicative inverse of $a+bj$ by solving
$$ (a+bj)(x+jy) = ax+by+j(ay+bx) = 1 = 1+j(0)$$
for $x+jy$. In particular, we must solve real equations $ax+by=1$ and $bx+ay = 0$:
$$ \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right] \left[ \begin{array}{c}x \\ y \end{array} \right] = \left[ \begin{array}{c}1 \\ 0 \end{array} \right] \ \ \Rightarrow \ \ \left[ \begin{array}{c}x \\ y \end{array} \right] = \frac{1}{a^2-b^2}\left[ \begin{array}{cc} a & -b \\ -b & a \end{array} \right]\left[ \begin{array}{c}1 \\ 0 \end{array} \right] = \frac{1}{a^2-b^2}\left[ \begin{array}{c} a \\ -b \end{array} \right] $$
That is,
$$(a+bj)^{-1} = \frac{a-bj}{a^2-b^2}$$
Notice, the matrix
$$ \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right] $$
naturally corresponds to $a+bj$ via the map $\Psi(a+bj) = \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right]$ and you can easily show $\Psi(zw) = \Psi(z) \Psi(w)$, $\Psi(z+w) = \Psi(z)+ \Psi(w)$ and $\Psi(1) = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$. This is the regular representation of hyperbolic numbers.
Generally, given a real associative algebra, a nice way to find the inverse is given by the regular representation. Simply find the corresponding matrix to your algebra element then use standard linear algebra to find the inverse of the matrix. Furthermore, this also reveals the structure of zero divisors in the algebra. With the hyperbolic numbers, the zero divisors are those $a+bj$ for which $a^2=b^2$. In contrast, the quaternions are a skew field where every nonzero element has a multiplicative inverse.
|
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|
If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\binom{n}{1}2^2\sqrt{3}+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}+\binom{n}{n}\sqrt{3}^{n}$$
and I think we have that: $$ a_n=\binom{n}{0}2+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n}\sqrt{3}^{n}$$
$$ b_n=\binom{n}{1}2^2\sqrt{3}+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}$$
But, frankly, I do not know what to do next.
|
Hint: First observe that $a_n -b_n\sqrt{3}=(2-\sqrt{3})^n$. Combine with the assumption, we get $a_n^2 - 3b_n^2 =1$ for all $n$. Then $\frac{a_n^2}{b_n^2}-3=\frac{1}{b_n^2}$. You should prove that $b_n \to \infty$. Hence, $\frac{a_n}{b_n} \to \sqrt{3}$ since $\frac{a_n}{b_n}>0$.
|
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}
|
Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = y$
Let $x,y$ be integers. Prove that if $(x+y)^2 \equiv 0 \pmod{xy}$, then $x = \pm y$.
The given condition is equivalent to $x^2+y^2 \equiv 0 \pmod{xy}$. How do we continue from here to prove that $x = \pm y$?
|
$$ x^2 + y^2 = kxy $$
$$ x^2 - k xy + y^2 = 0. $$
Divide by $y^2,$ let $r = x/y.$
$$ r^2 - kr + 1 = 0 $$
Discriminant is
$$ k^2 - 4. $$
Thi is not a square unless $k = \pm 2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A Trigonometry + complex number Equation the question is:
Show that:
$$
[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\= 2^{n+1} \sin n \frac{(\theta - \phi)}{2} \cos x \frac{\theta + \phi - \pi}{2}$$
I gave it a try following the pattern of the previous questions: https://ibb.co/d77by5
|
I find that your result in the question is valid only for even values of $n$ (as well containing typographical errors; it should be $\sin^n$ and $\cos(n...)$). My solution is as follows:
First note that
$$\cos\theta-\cos\phi=-2\sin\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right) \\
\sin\theta-\sin\phi=2\cos\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right)$$
Then
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^n\sin^n \left(\frac{\theta-\phi}{2}\right) \cdot \left\{\left[-\sin\left(\frac{\theta+\phi}{2}\right) +i\cos\left(\frac{\theta+\phi}{2}\right)\right]^n + \left[-\sin\left(\frac{\theta+\phi}{2}\right) -i\cos\left(\frac{\theta+\phi}{2}\right)\right]^n\right\}$$
Now, let $a=(\theta+\phi)/2$ and convert the terms in the curly braces to exponential form,
$$\{\ \frac{}{}\}=\left\{\left[-\left(\frac{e^{ia}-e^{-ia}}{2i}\right) +i\left(\frac{e^{ia}+e^{-ia}}{2}\right)\right]^n + \left[-\left(\frac{e^{ia}-e^{-ia}}{2i}\right) -i\left(\frac{e^{ia}+e^{-ia}}{2}\right)\right]^n\right\}\\=\frac{i^n}{2^n} \left\{ \left[e^{ia}-e^{-ia}+e^{ia}+e^{-ia} \right]^n + \left[e^{ia}-e^{-ia}-e^{ia}-e^{-ia} \right]^n\right\} \\=\frac{i^n}{2^n} \left\{ 2^n e^{ina}+(-2)^ne^{-ina} \right\}=i^n\left\{ e^{ina}+(-1)^ne^{-ina} \right\}$$
Now,
$$i^n \left\{ e^{ina}+(-1)^ne^{-ina} \right\} = 2i^{n+1} \sin\left(n\frac{\theta+\phi}{2}\right) \ \ \text{for odd} \ n \\ i^n \left\{ e^{ina}+(-1)^ne^{-ina} \right\} = 2i^{n} \cos\left(n\frac{\theta+\phi}{2}\right) \ \ \text{for even} \ n $$
And finally,
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=(2i)^{n+1} \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \sin\left(n\frac{\theta+\phi}{2}\right) \ \ \text{for odd} \ n \\$$
and
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^{n+1} i^n \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \cos\left(n\frac{\theta+\phi}{2}\right) \ \ \text{for even} \ n \\$$
These can be combined into a single equation as follows:
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^{n+1} (-1)^n \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \cos\left(n\frac{\theta+\phi-\pi}{2}\right)$$
which is very close the answer in the original question, differing only by the factor $(-1)^n$. It's no wonder that it worked for even $n$. All the results have been verified by direct numerical simulation.
|
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|
How can I prove that $\arcsin \left(\sqrt\frac{1 + \sin2x}{2} \right)= |x + \frac{\pi}{4}|$ the function $ f$ is defined as: $\arcsin \sqrt\frac{1 + \sin2x}{2}$
The function is defined on $\mathbb{R}$.
Prove that $f(x) = |x + \frac{\pi}{4}|$ for all $x$ in $[\frac{-3\pi}{4}, \frac{\pi}{4}]$
We have for all in $ x \in \mathbb{R} $ :
$$f'(x) = \frac{\cos2x}{\sqrt{ \frac{3 - 2\sin2x - \sin^2 2x}{4}}} $$
I don't know how to proceed to prove that $f(x) = |x + \frac{\pi}{4}|$ for all $x$ in $[\frac{-3\pi}{4}, \frac{\pi}{4}]$
Thank you.
|
Note that we can write
$$\begin{align}
\frac{1+\sin(2x)}{2}&=\frac{\cos^2(x)+2\sin(x)\cos(x)+\sin^2(x)}{2}\\\\
&=\frac{(\sin(x)+\cos(x))^2}{2}\\\\
&=\frac{2\sin^2(x+\pi/4)}{2}
\end{align}$$
Can you proceed now?
|
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|
Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
|
The above sequence is given by $a_i = \frac{i(i+5)}{2}$. The finite sum is given by
\begin{align}
S_n & = \sum_{i=1}^n a_i \\
& = \frac{1}{2} \sum_{i=1}^n i^2 + \frac{5}{2} \sum_{i=1}^n i \\
& = \frac{n(n+1)(2n+1)}{12} + \frac{5n(n+1)}{4} \\
& = \frac{n(n+1)(n+8)}{6}
\end{align}
EDIT: (In response to comments) To a certain point, finding the formula for $a_i$ is trial and error. However, it is not difficult to note here that the difference of the difference of terms is always 1 (i.e. $\Delta^2 a_i = 1$). This implies that the dependence is quadratic. Using $a_i = A i^2 + B i + C$ we can determine the formula with three terms.
|
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|
How to prove $\int_{0}^{1}{ dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}?$ Variation of Ahmed's integral
$$\int_{0}^{1}{\mathrm dx\over 1+x^2}\cdot{\tan^{-1}\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)\over \sqrt{x^2+2}}={13\pi^2\over 288}\tag1$$
Making an attempt:
I can't think of what to do. It is too complicate.
|
One may observe that, for $x \in [0,1]$,
$$
\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)=\arctan\left({1\over \sqrt{x^2+2}}\right)+\arctan\left({x\over \sqrt{x^2+2}}\right)
$$ then
$$
I:=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=I_1+I_2
$$with
$$
\begin{align}
I_1:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left({1\over \sqrt{x^2+2}}\right)dx
\\&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\left(\frac \pi2-\arctan\sqrt{x^2+2}\right)dx
\\&=\frac \pi2\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}-\int_0^1 \frac{\arctan\sqrt{x^2+2}}{(1+x^2)\sqrt{x^2+2}}\:dx
\\&=\frac{\pi^2}{12}-\frac{5\pi^2}{96}
\\&=\frac{\pi^2}{32}
\end{align}
$$ where the latter integral is Ahmed's evaluation.
On the other hand we have
$$
\begin{align}
I_2:&=\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\cdot\arctan\left({x\over \sqrt{x^2+2}}\right)dx
\\&=\int_1^\infty \frac{u}{(1+u^2)\sqrt{1+2u^2}}\cdot\arctan\left({1\over \sqrt{1+2u^2}}\right)du \quad (u=1/x)
\\&=\frac12\int_1^\infty \frac{1}{(1+v)\sqrt{1+2v}}\cdot\arctan\left({1\over \sqrt{1+2v}}\right)dv \qquad (v=u^2)
\\&=-\int_1^\infty \left[\arctan\left({1\over \sqrt{1+2v}}\right)\right]'\arctan\left({1\over \sqrt{1+2v}}\right)dv
\\&=-\left[\frac12\cdot\arctan^2\left({1\over \sqrt{1+2v}}\right)\right]_1^\infty
\\&=\frac{\pi^2}{72}.
\end{align}
$$
Finally,
$$
\int_0^1 \frac{1}{(1+x^2)\sqrt{x^2+2}}\arctan\left(\sqrt{x^2+2}\cdot{x+1\over x^2-x+2}\right)dx=\frac{\pi^2}{32}+\frac{\pi^2}{72}=\color{blue}{\frac{13\pi^2}{288}}
$$
as announced.
|
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|
Two numbers $x,y$ are randomly chosen from the segment [0,5] find the probability that $x+y\ge 5$ and $\left|x-y\right|\le 1$ Two numbers $x,y$ are randomly chosen from the segment [0,5] find the probability that $x+y\ge 5$ and $\left|x-y\right|\le 1$.
So the total area is 25 and the yellow area that has the needed requirements is given by:
$\int _2^3\:\left(x+1-\left(5-x\right)\right)dx+\int _3^4\:\left(x+1-\left(x-1\right)\right)dx+\int _4^5\:\left(5-\left(x-1\right)\right)dx=\:\frac{9}{2}$
So we devide them and get the probability, anyway is the way I calculated the yellow area correct?
|
You are correct. The area is easy to get with geometry: the area of the rectangle is $2\sqrt2\times\sqrt2=4$ and that of the triangle is $1\times1/2=1/2$ for a total of $9/2$. Divide by the total possible area ($5^2=25$) and you get a probability
$$
p=\frac{9/2}{25}=9/50=0,18.
$$
|
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|
Length of two sides in a quadrilateral with given angles I'm stuck finding the length of two sides in a quadrilateral for which I know all angles and the length of two sides.
All red objects are know ($a,b,\alpha,\beta,\gamma $ and $\delta$). I need to find the length of $c$ and $d$. I know it can't be too difficult but I can't figure it out atm. Thanks for your help!
|
Suppose our quadrilateral has vertices $A,B,C,D$. We can deduce the length of $BD$ using the law of cosines (Ignore the labels):
We get that
$$BD^2 = a^2 + b^2 - 2ab \cos \alpha$$.
Next, we find the angles $\angle ABD$ and $\angle ADB$ using law of sines to get:
$$m\angle ADB = \sin^{-1}\left(\frac{a \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)$$
$$m\angle ABD = \sin^{-1}\left(\frac{b \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)$$
Next, we deduce the measures of angles $\angle CBD$ and $\angle CDB$ using angle addition. We get
$$m\angle CBD = \beta-\sin^{-1}\left(\frac{a \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)$$
$$m\angle CBD = \alpha-\sin^{-1}\left(\frac{b \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)$$
Finally, use law of sines one last time to get:
$$c = \frac{\sin\left(\beta-\sin^{-1}\left(\frac{a \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)\right) \sqrt{a^2+b^2-2ab\cos\alpha}}{\sin \gamma}$$
$$d = \frac{\sin\left(\alpha-\sin^{-1}\left(\frac{b \sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}\right)\right) \sqrt{a^2+b^2-2ab\cos\alpha}}{\sin \gamma}$$
I may have made a mistake, someone please point it out if you find one. The general gist however, is the same. Law of cosines to find $BD$, law of sines to get the angles of triangle $BDC$, then law of sines again to get the measures of the sides of the triangle.
|
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|
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15. Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. But, apparently there are no primitive solutions to $mod 15.$ So what's another option?
|
By the Chinese remainder theorem, $a\equiv b\pmod{15}$ if and only if
\begin{cases}
a \equiv b\pmod{3}\\[4px]
a \equiv b\pmod{5}
\end{cases}
In the case $a=x^2$ and $b=1$, the first equation gives
$$
x=1+3k \qquad\text{or}\qquad x=2+3k
$$
From $x=1+3k$, we get
$$
1+6k+9k^2\equiv 1\pmod{5}
$$
that is, $k-k^2\equiv0\pmod{5}$, hence $k\equiv0\pmod{5}$ or $k\equiv1\pmod{5}$. Thus we get $x=1$ or $x=4$.
From $x=2+3k$ we similarly get
$$
4+12k+9k^2\equiv1\pmod{5}
$$
that is, $k^2-2k-3\equiv0\pmod{5}$ that yields $k=3$ or $k=4$, so
$x=14$ or $x=11$.
On the other hand, a higher level solution using the fact that the group of (multiplicative) units in $\mathbb{Z}/15\mathbb{Z}$ is the product of the groups of units in $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/5\mathbb{Z}$ is preferable.
|
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|
prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start.
Do I need contradiction?
|
$x^2 + y^2 = z^2$ and $y = x + 1 $ $\implies (3x+2z+1)^2 + (3x+2z+2)^2 = (4x+3z+2)^2$
And this series goes on forever by $\cases{x := 3x+2z+1 \\
y := 3x+2z+2 \\
z := 4x+3z+2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (here $:=$ stands for assignment)
It is not very difficult to check that $\gcd(x,y,z) = 1$ at each step, thus ensuring the primitivity of the triplets.
EDIT:
An example as to how a triplet is generated from the triplet in the previous step.
Suppose we start off from the triplet $3,4,5$ as indeed $4 = 3 + 1$.
Then the next triplet would be $(3 \times 3 + 2 \times 5 + 1, 3 \times 3 + 2 \times 5 + 2, 4 \times 3 + 3 \times 5 + 2) = (20,21,29)$.
And at the next step $(x,y,z) = (20,21,29)$.
Hence the next triplet now would be $(3 \times 20 + 2 \times 29 + 1, 3 \times 20 + 2 \times 29 + 2, 4 \times 20 + 3 \times 29 + 2) = (119,120,169)$.
And at the next step $(x,y,z) = (119,120,169)$.
And this goes on, if you will, forever.
|
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|
If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$. If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.
How does even even begin to prove this? I have proved that if $a+b$ is a multiple of $n$, then so is $a^n+b^n$, using FLT, but how to progress next eludes me.
|
When $n=2$ you have $\frac{a^n+b^n}{a+b}=a-b+\frac{2b}{a+b}$ which is not an integer in general.
So I assume $n=2k+1$ is an odd prime should be part of the statement.
In this case
$$\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+...+b^{n-1}\\=
a^{n-1}+a^{n-2}b-2a^{n-2}b-2a^{n-3}b^2+3a^{n-3}b^2+3a^{n-4}b^3-...\\
=(a+b)(a^{n-2}-2a^{n-3}b+3a^{n-4}b^-...)+nb^{n-1}$$
Therefore
$$gcd(\frac{a^n+b^n}{a+b}, a+b)=gcd(nb^{n-1}, a+b)=gcd(n, a+b)$$
with the last equality following from $gcd(a+b, b^{n-1})=1$.
|
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|
Find the interval of $c$ such that the rational function $\frac{x^2+2x+c}{x^2+4x+3c}$ takes all real values Find the interval of $c$ such that the rational function $$f(x)=\frac{x^2+2x+c}{x^2+4x+3c}$$ takes all real values $\forall$ $x\in D_f$
I tried in the following way:
Let $$y=\frac{x^2+2x+c}{x^2+4x+3c}$$ converting this to a quadratic in $x$ we get
$$(y-1)x^2+x(4y-2)+c(3y-1)=0$$ and since $x \in D_f$ this equation should have Discriminant Non negative
so
$$(4y-2)^2-4c(y-1)(3y-1) \ge 0$$ i.e.,
$$(4-3c)y^2+(4c-4)y+1-c \ge 0$$ since $\forall y \in \mathbb{R}$ the above quadratic is non negative we have two conditions
$$4-3c \gt 0$$ and
$$(4c-4)^2-4(4-3c)(1-c) \lt 0$$
$$4c^2-4c \lt 0$$ so
$c \in (0 \:\: 1)$
Is there any other approach
|
\begin{align}
f(x) &= \frac{x^2+2x+c}{x^2+4x+3c} \\
&= \frac{x^2+4x+3c - 2x - 2c}{x^2+4x+3c} \\
&= 1-\frac{2x + 2c}{x^2+4x+3c}
\end{align}
therefore $x^2+4x+3c$ cannot be zero.
$x^2+4x+3c = 0$ has no solution
from discrimant
\begin{align}
b^2 - 4ac &< 0 \\
16 & < 12c \\
3c & > 4 \\
c & > \frac{4}{3}
\end{align}
however if $2x + 2c$ is also zero, it would take the value, so if $(x + c)$ is a factor of denominator
$x^2 + 4x + 3c = (x + c)(x + 3)$ implies that $cx + 3x = 4x$ so $c = 1$
However at $c = 1$, $x = -3$ cannot be taken.
So $c > \dfrac{4}{3}$.
|
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|
Find a $3\times 3$ matrix $A\not = I_3$ such that $A^3 = I_{3}$ Use the correspondence between matrices and linear transformation to find find a $3\times 3$ matrix $A$ such that $A^3 = I_{3}$ and find an $A$ matrix that is not $I_{3}$
Where $I_{3}$ is the identity matrix:
$$I_{3}=
\left[ {\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array} } \right]$$
I was tried with the following $A$ matrix:
$$A=
\left[ {\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array} } \right]$$
and when I multiply $A \times A \times A$ I got the same matrix as $I_{3}$.
And to find a matrix $A$ that is not equal to $I_{3}$ I can take any $A$ matrix that is not:
$$A=
\left[ {\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array} } \right]$$
but I think that the exercise is expecting something else using linear transformations.
Sorry I realised that $A$ cannot be equal to $I_{3}$
|
Try
\begin{align}
A =
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
1 & 0 & 0
\end{bmatrix}
\end{align}
|
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|
The integral of ${x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}$
$$\int {x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}dx$$
Substituting $u = x^2$, I get $${1\over 4}\int {u - 1 \over u^2 \sqrt{2u^2 - 2u + 1}}du$$
Then completing the square under the square root.
$${1\over 4}\int {u - 1 \over u^2 \sqrt{\left(\sqrt{2} u - \frac1{\sqrt{2}} \right)+ 1/2} }du$$
Substituting $\tan z = \sqrt{2}\left(\sqrt{2} u - \frac1{\sqrt{2}} \right)$
$${ \sqrt{2} \over 4}\int {\tan z -3 \over (\tan z- 1)^2 } \times \sec z \ dz$$
I think I can solve the last integral if I think a bit more but I need a better way to solve this question, this was seems very tedious.
|
Hint pull out $x^2$ common from numerator and $x^4$ in the root from denominator ie $x^2$ from numerator. Then put $1-\frac {1}{x^2}=u $ thus $\frac {2}{x^3}dx=du $ also write the new denominator as $1+(1-\frac {1}{x^2})^2$ hope you can handle from here.
|
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|
Number of solutions of $x+2y+3z=2n+k$ where $1 \leq x,y,z,k \leq n$ is about $\frac 14 n^2$ Let $n \in \Bbb N$ and $k \in [n]$, where $[n]=\{1,...,n\}$. I want to show:
The number of solutions $(x,y,z)$ of the equation $x+2y+3z = 2n+k$ where $x,y,z \in [n]$ is at least $cn^2$, for some constant $c$ that is independent of $n$ and $k$.
Sidenotes:
*
*The $2n$ term is not important to me and can be replaced by any term that depends only on $n$, if such a $c>0$ exists. for $2n$ it seems to exist.
*My aim is to show the same for general $r$ with the equation $x_1+2x_2+3x_3+...+rx_r=k+(r-1)n$ and $c$ a function of $r$ only.
In the case $r=2$ we see that $x+2y=n+k$ has at least $\lfloor\frac n2 \rfloor$ solutions, so we can take $c=\frac 13$. This is since if we start increasing $y$ then at $y=\lceil \frac k2 \rceil$ we will have a solution and then we will have a solution for $x$ in the next $\lfloor \frac n 2 \rfloor $ values of $y$. Geometrically this is a line that has a lattice point once in every two points.
This reasoning is less clear in the case $r=3$, looking at the possible solutions of $x,y$ as we increase $z$.
UPDATE
Experimenting using Python shows that for $r=3$ the number of solutions converges to $\frac 14 n^2$.
It is enough to prove that there is such a constant $c$ for all large enough $n$, since then we can make $c$ smaller so that it will work for the small values of $n$ as well.
The natural guesses that the number of solutions is at least $\lfloor \frac {n^2} 4 \rfloor$ or at least $\lfloor\frac n2\rfloor ^2$ are both sometimes false, though.
The script, in case anyone wants to experiment too:
from itertools import product
def number_of_solutions(n, k, r):
return sum(1 for variables in product(range(1, n + 1), repeat=r-1) if
(r - 1) * n + k - sum((k+2)*variables[k] for k in range(len(variables)))
in range(1, n + 1))
def minimal_number_of_solutions_over_all_k(n, r):
return min([number_of_solutions(n, k, r) for k in range(1, n + 1)])
def find_approximated_c(r):
for n in range(2, 1000):
approximated_c = minimal_number_of_solutions_over_all_k(n,r) / (n ** (r - 1))
print(n, approximated_c) # stabilizes on 0.25 when r=3
|
Since you are looking for "at least" then its obvious that $2n+k$ have less solutions when $k=1$ than when $k$ is very big or $k=n$ (because the partitions of a number grows exponentially), so we are left with $x+2y+3z = 2n+1$ in order to give the "at least" constant $c$.
Now $z$ can run from $1$ up to $\lfloor \frac{2n+1}{3} \rfloor$.
Now we are left with $x+2y = 2n+1-3z$ such that $z$ runs form $1$ up to $\lfloor \frac{2n+1}{3} \rfloor$.
Let $m=2n+1-3z$ so $x+2y=m$ , in how many ways we can write some integer $m$ as sum of $x+2y$ , its easy to see that when we give $y$ a fixed value, $x$ is forced into a fixed value also, so we have $\lfloor \frac{m}{2} \rfloor$ solutions.
So we arrive at this sum $\sum \limits_{z=1}^{\lfloor \frac{2n+1}{3} \rfloor} \lfloor \frac{m}{2} \rfloor = \sum \limits_{z=1}^{\lfloor \frac{2n+1}{3} \rfloor} \lfloor \frac{2n+1-3z}{2} \rfloor $.
But there is a catch in these solutions, in these solutions i made sure that $1 \leq y,z \leq n$ but not for $x$ so for some solutions $x$ could equal $0$ or $x >n$.
To correct this let $x =2n+1-3z-2y \leq n$ which means $3z+2y \geq n+1$ which means that $y \geq \frac{n+1-3z}{2}$ which means that $y \geq \lceil \frac{n+1-3z}{2} \rceil$ (solution which will be subtracted each time). until we reach that $\lceil \frac{n+1-3z}{2} \rceil < 1$
So Approximately we get this summation $\sum \limits_{z=1}^{\lfloor \frac{n+1}{3} \rfloor} \lfloor \frac{2n+1-3z}{2} \rfloor - \lceil \frac{n+1-3z}{2} \rceil + \sum \limits_{z=1+\lfloor \frac{n+1}{3} \rfloor}^{\lfloor \frac{2n+1}{3} \rfloor} \lfloor \frac{2n+1-3z}{2} \rfloor $ and because we are trying to find the "at least" we can use the fact that $x-1 \leq \lfloor x \rfloor \leq x$ and $x \leq \lceil x \rceil \leq x+1$ and with simple arithmetic we arrive at :
$$ \sum _{z=1}^{\frac{1}{3} (n+1-3)} \frac{1}{2} (-4+n)+\sum
_{z=1+\frac{n+1}{3}}^{\frac{1}{3} (2 n+1-3)} \frac{1}{2} (2 n+1-3 z-2) = \frac{1}{12}(28-19n+3n^2) \approx \frac{n^2}{4}.$$
And thus concluding the proof by setting $c=\frac{1}{4}$(actually it have to be a little bit less than $1/4$).
(Note) your first side note is correct we can take for any value $c>0$ , $c n$ and apply the proof above to come up with approximation, also i think the second side note (the Generalization) is also true but proving it might be hard.
|
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|
How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$? How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$?
I tried show that difference between terms of sums $\rightarrow 0$
$|\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}} - \frac{2^{\frac{k}{n}}}{n}| = 2^\frac{k}{n}(\frac{1}{n+\frac{1}{k}} - \frac{1}{n})= 2^\frac{k}{n}(\frac{n-n-\frac{1}{k}}{n^2+\frac{n}{k}}) = -2^\frac{k}{n} \frac{1}{n^2k+n}\rightarrow 0$ if $n\rightarrow \infty$
But i'm not sure that is right method, because difference of terms can accumulete and it would be $\neq 0$
Which correct method i must use to prove it?
|
If you're trying to show the difference goes to $0$, here's a big hint:
$$
\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}} = \frac{1}{n} \sum\limits_{k=1}^n \frac{2^{k/n}}{1 + \frac{1}{kn}}
$$
and hence
$$ 0 \leq
\sum\limits_{k=1}^n \frac{2^{k/n}}{n} - \sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}= \frac{1}{n}\sum\limits_{k=1}^n\frac{(1+\frac{1}{kn})2^{k/n} - 2^{k/n}}{1 + \frac{1}{kn}} = \frac{1}{n}\sum\limits_{k=1}^n\frac{2^{k/n}}{kn\left(1 + \frac{1}{kn}\right)} < \frac{1}{n} \sum\limits_{k=1}^n \frac{2}{k^2}
$$
Or, perhaps the following will serve you better:
$$
\sum_{k=1}^n \frac{2^{k/n}}{n + 1} \leq \sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}} < \sum_{k=1}^n \frac{2^{k/n}}{n}
$$
and hence
$$
\frac{n}{n+1} \leq \frac{\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}}{\sum\limits_{k=1}^n \frac{2^{k/n}}{n}} < 1
$$
|
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|
Find the values of $k$ such that the equation $x^3+x^2-x+2=k$ has three distinct real solutions Find the values of $k$ such that the equation
$$x^3+x^2-x+2=k$$
has three distinct real solutions.
Please explain how to find the solution! Thanks
|
We can sketch the graph $y=x^3+x^2-x+2$ as below (I used WolframAlpha):
Now we imagine drawing horizontal lines across this graph (setting $y=k$ in the original equation)
We can see that the only points where the line crosses the equation exactly three times are any points where $$y\text{ value of minimum point}<k<y\text{ value of maximum point}$$
Therefore, to solve the problem, we must calculate the turning points of the graph $y=x^3+x^2-x+2$
We do this by computing the derivative first:
$$y'=3x^2+2x-1$$
We then set this equal to zero to find turning points of the graph
\begin{align}3x^2+2x-1&=0\\
(x+1)(3x-1)&=0\end{align}
Therefore we have turning points at $x=-1$ and $x=\frac 13$
We can then find the $y$ co-ordinates of these points by substiuting them back into the equation:
When $x=-1$, \begin{align}y&=(-1)^3+(-1)^2-(-1)+2\\
&=-1+1+1+2\\
&=3\end{align}
When $x=\frac 13$, \begin{align}y&=\left(\frac13\right)^3+\left(\frac13\right)^2-\frac13+2\\
&=\frac 1{27}+\frac 19-\frac 13+2\\
&=\frac{49}{27}\end{align}
We know that $\frac{49}{27}<3$ and therefore we can say that $$\frac{49}{27}<k<3$$
|
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|
If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
|
Or, with less variables, you can treat your fraction as reducible to some number $a$ (e.g. a decimal), which you can write as $\frac{a}{1}$. Then:
$\frac{a+a}{1+1} = \frac{2a}{2} = \frac{a}{1} = a$
|
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|
Calculate sum of coefficients of polynomial Let $$(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5) = \sum_{k=0}^{9} (A_k \cdot x^k)$$
Compute:
*
*$$\displaystyle \sum_{k=0}^{9} A_k$$
*$$\displaystyle \sum_{k=0}^{4} A_{2k}$$
I tried to figure out from Viete's Sums how to rewrite this but I can't find the coefficients for all powers of $x$. All I know is $A_9 = 1, A_0 = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5$.
|
Let $P$ be your polynomial :
$$P(x)=(x+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)=\sum_{k=0}^9 A_kx^k$$
Then
$$\sum_{k=0}^9 A_k = \sum_{k=0}^9 A_k.1^k = P(1) = 2\times3\times4\times5\times6 = 720$$
And, as
$$\sum_{k=0}^9 (-1)^kA_k = P(-1) = 0$$
you can, by adding, find :
$$2\sum_{k=0}^4 A_{2k} = P(1)+P(-1)=720$$
so that
$$\sum_{k=0}^4 A_{2k} = 360$$
|
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|
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$
My attempt:I've tried it by considering the sum
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$
along with
$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $ 100$ as resultant but failed to separate the sum of
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.
I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.
I've invested a great amount of time in the so it would be better if someone please come up with an answer.
|
HINT
Use the fact that $$\cos^2(x) = \dfrac{1}{2}+\dfrac{\cos(2x)}{2}$$
Thus, the result is equal to
$$50 + \dfrac{1}{2}\sum\limits_{k=1}^{100} \cos\left(\dfrac{2\pi k}{101}\right)$$
And the last sum can be proven to be $-1$ using this result: How to prove $\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$ for any n>1? .
So the answer is $49.5$.
|
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|
The sum of areas of triangles omitted infinitely many times within an equilateral triangle ABC
Triangle $ABC$ is an equilateral triangle. Four new equilateral triangles are formed by joining the midpoints of the sides $A, B, C$ in $\triangle ABC$ Triangle, with the white triangle in İmage 1, and it is taken out.
Each triangle now formed in $\triangle ABC$ has an area of $\frac 14$ of the area of $\triangle ABC$, as seen in Image 1.
The same process of joining the midpoints in each triangle and taking out the newly formed white triangles is repeated infinitely many times as exemplified in Image 2 and Image 3.
If $\triangle ABC$ has an area of $1\, m^2,\,$ what is the sum of the areas of white triangles taken out?
I know the answer, which I've posted below.
|
$$\begin{align} S &=\sum_{n=1}^\infty a\cdot r^{n-1}\\ \\ & = \frac{1}4+\frac{3}{16}+\frac{9}{64}+\dots \\ \\
&= \frac{1}4\cdot\frac{3}{4}^{0}+\frac{1}4\cdot\frac{3}{4}^{1}+\frac{1}4\cdot\frac{3}{4}^{2}+\dots \\ \\ \end{align}$$
$$ a=\frac{1}4, r=\frac{3}4\implies S= 1 $$
|
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|
Have I correctly taken this function $f(x) = \frac{x^{2}}{1+2x}$ and $f ' (x) $and turned them into power series? $f(x) = \dfrac{x^{2}}{1+2x}$
To turn this into a power series I recall the similar looking geometric series,
$f(x) = \sum\limits_{n=1}^{\infty} x^{n} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^5 +\cdots$
And not I can mutate my original function to fit the form the geometric series:
$f(x) = \dfrac{x^{2}}{1-(-2x)}$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 + (-2x) + (-2x)^{2} + (-2x)^{3} + (-2x)^{4} +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2( 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots)$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2 - 2x^3 + 4x^4 -8x^5 + 16x^{6} +\cdots)$
And for $'(x)$
$f'(x) = 2x - 6x^2 + 16x^3 -40x^4 + 16*6x^{5} +\cdots$
Is this right or wrong?
Thank you
|
Suggestion:
$$
\frac{1}{2x+1} \approx 1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4 - 32 x^5 +\mathcal{O}\left( x^{6} \right)
$$
Now multiply by $x^{2}$.
$$
f(x) = \frac{x^2}{2 x+1} \approx x^2 -2 x^3+4 x^4-8 x^5+ \mathcal{O}\left( x^{6} \right)
$$
Suggestion:
$$
\frac{1}{(2 x+1)^2} \approx 1 - 4 x + 12 x^2 - 32 x^3 + 80 x^4 - 192 x^5 + \mathcal{O}\left( x^{6} \right)
$$
Multiply by $2 x (1 + x) = 2 x^2+2 x$:
$$
f'(x) = \frac{2 x (x+1)}{(2 x+1)^2} \approx 2 x - 6 x^2 + 16 x^3 - 40 x^4 + 96 x^5+ \mathcal{O}\left( x^{6} \right)
$$
@law-of-fives suggests partial fraction decomposition:
$$
f(x) =\frac{x^2}{2 x+1} = \frac{x}{2}+\frac{1}{4 (2 x+1)}-\frac{1}{4}
$$
A more elegant solution!
|
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|
Inequality with absolute value and variable on both sides I'm not able to understand how to get through this
$$\dfrac{|x+6|}{x+1} \leq x-2$$
$x$ can't, of course, be equal to $-1$.
I'm at a point where I'm not sure how to deal with the absolute value after reaching
$$|x+6| \leq (x-2)(x+1)$$
Thanks!
|
$$\dfrac{|x+6|}{x+1} \leq x-2$$
Let's start with $\dfrac{|x+6|}{x+1} = x-2$.
\begin{array}{rclcrcl}
|x+6| &= &(x+1)(x-2) &| \\
x^2-x-2 &= &|x+6| &|\\
x^2-x-2 &= &x+6 &\text{or} & x^2-x-2 &= &-x-6 \\
x^2-2x-8 &= &0 &\text{or} & x^2+4 &= &0 \\
x &\in &\{-2,4\} &\text{or} & x &\in &\emptyset \\
\end{array}
So we have three critical points: zeros at $x=-2$ and $x=4$ and a pole at $x=-1$. The critical points divide the real number line into four regions:
$(-\infty, -2), (-2,-1), (-1, 4)$, and $(4, \infty).$ Each region is either all solutions or all no solutions. We test each region.
\begin{array}{c}
\text{region} & \text{sample point} & \text{Is solution?} \\
\hline
(-\infty, -2) & -10 & \text{no} \\
(-2,-1) & -1.5 & \text{yes} \\
(-1, 4) & 0 & \text{no}\\
(4, \infty) & 12 & \text{yes} \\
\hline
\end{array}
So the solution set is $[-2,-1) \cup [4, \infty)$.
|
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|
show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$.
|
We can write as $$(n+1)(n+2)...(2n)=\frac{2n!}{n!}\tag1$$ Now let us calculate the power of $2$ contained in $2n!$ $$\text{Power of }2 \text{ in } 2!=\frac{2n}{2}+\frac{2n}{2^2}+\frac{2n}{2^3}+\frac{2n}{2^4}\ldots\\=2n\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right)\\=2n\left(\frac{\frac{1}{2}}{1-\frac12}\right)\\=2n$$ which means $$2n!=2^{2n}\times a$$
Similarly power of $2$ contained in $n!$ $$\text{Power of }2 \text{ in } 2!=\frac{n}{2}+\frac{n}{2^2}+\frac{n}{2^3}+\frac{n}{2^4}\ldots\\=n\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots\right)\\=n\left(\frac{\frac{1}{2}}{1-\frac12}\right)\\=n$$ which means $n!=2^{n}\times b\tag{Note that $b|a$}$ $\therefore$ equation $(1)$ can be written as $$(n+1)(n+2)...(2n)=\frac{2^{2n}\times a}{2^{n}\times b}$$ $$(n+1)(n+2)...(2n)=\frac{2^{n}\times a}{b}$$
|
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|
show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k+1} - \frac{1}{k+2} + ... + \frac{1}{2k-1} - \frac{1}{2k}$
$k$ is even.
Every negative term $d$ in $X_2$ is of the form $\frac{1}{2^my}$ where $y$ is odd and could be $1$. Also $y \le k$
For every such $d$, we will have the following terms
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y}$ in $X_1$.
Adding $d$ to the above, we get
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y} - \frac{1}{2^my} = \frac{1}{2^my}$
So every negative term $d$ in $X_2$ is now replaced with a term of the same magnitude but with a positive sign.
Hence $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$
|
We can use harmonic numbers
$H_k=\sum_{i=1}^k\frac{1}{i}$.
We obtain
\begin{align*}
\sum_{i=k+1}^{2k}\frac{1}{i}=\color{blue}{H_{2k}-H_k}
\end{align*}
on the other hand we get
\begin{align*}
\sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}-H_{2k}
&=-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2k}\right)=-H_k\\
\end{align*}
which implies
\begin{align*}
\sum_{i=1}^{2k}\frac{(-1)^{i+1}}{i}
&=\color{blue}{H_{2k}-H_k}
\end{align*}
and the claim follows.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Extremely ugly integral $\int_{-\pi}^{\pi} \frac{\operatorname{sign}(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$
Evaluate: $\DeclareMathOperator{\sign}{sign}$ $$\int_{-\pi}^{\pi} \dfrac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$$
My idea:
\begin{align*}I=\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)| }\,dx &=\int_{-\pi}^{\pi} \dfrac{\sign(-x)\arctan((-x)^2)-|-x|}{\sin^2(x)+|\cos(x)|}\,dx\\
&=\int_{-\pi}^{\pi} \frac{-\sign(x)\arctan(x^2)-|x|}{ \sin^2(x)+|\cos(x)| }\,dx.
\end{align*}
So, it means that the integral $$\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)}{\sin^2(x)+|\cos(x)| }\,dx=0.$$
Therefore, my integral now looks like
$$I=\int_{-\pi}^{\pi} \frac{-|x|}{\sin^2(x)+|\cos(x)|}\,dx,$$
and since my integrand is even function, I have:
$$I=-2\int_{0}^{\pi} \frac{|x|}{ \sin^2(x)+|\cos(x)| }\,dx=-2\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx.$$
And so,
\begin{align*}
\frac{I}{2}&=-\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\int_{0}^{\pi} \frac{\pi-x}{ \sin^2(\pi-x)+|\cos(\pi-x)|}\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)|}+\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)| }-\frac{I}{2}.
\end{align*}
Hence,
$$I=-\pi\int_{0}^{\pi} \dfrac{dx}{ \sin^2(x)+|\cos(x)| }.$$
That is the end of the road. I tried to eliminate absolute value over the cosine but always get some divergent integral. Is this computation correct and if it is, what next? Bonus question: Where can I find more problems like this?
|
I can't take full credit for this, Wolfram can be very helpful when you know how to coax it along.
For the integral
$$\int{dx\over\sin^2x+\cos x},$$
let
\begin{align*}
u &= \tan\left(\frac{x}{2}\right)\\
du &= \frac{1}{2}\sec^{2}\left(\frac{x}{2}\right)\,dx.
\end{align*}
Then we have
\begin{align*}
\sin x &= \frac{2\sin x}{1 +\cos x}\cdot \frac{1+\cos x}{2} = \frac{2\tan^{2}\left(\frac{x}{2}\right)}{\sec^{2}\left(\frac{x}{2}\right)} = \frac{2\tan^{2}\left(\frac{x}{2}\right)}{\tan^{2}\left(\frac{x}{2}\right)+1}=\frac{2u^{2}}{u^{2}+1}\\
\cos x&= \frac{2\cos x}{1+\cos x}\cdot\frac{1+\cos x}{2}=\frac{1+\cos x - 1 + \cos x}{(1+\cos x)\sec^{2}\left(\frac{x}{2}\right)}=\frac{1-\frac{1-\cos x}{1+\cos x}}{\sec^{2} \left(\frac{x}{2}\right)}= \frac{1-u^{2}}{u^{2}+1}\\
dx &= \frac{2\,du}{u^{2}+1}.
\end{align*}
The integral then becomes
\begin{align*}
2\int \frac{du}{(u^{2}+1)\left[\frac{4u^{2}}{(u^{2}+1)^{2}}+\frac{1-u^{2}}{u^{2}+1}\right]} &=-2\int\frac{u^{2}+1}{u^{4}-4u^{2}-1}\,du\\
&=-2\int\frac{u^{2}+1}{\left(u^{2}-\sqrt{5}-2\right)\left(u^{2}+\sqrt{5}-2\right)}\,du.
\end{align*}
We can then use a partial fraction decomposition, which I will omit, to change the integral to
\begin{align*}
-\int\frac{\sqrt{5}-3}{\sqrt{5}\left(u^{2}+\sqrt{5}-2\right)}\,du -\int\frac{-3-\sqrt{5}}{\sqrt{5}\left(-u^{2}+\sqrt{5}+2\right)}\,du,
\end{align*}
which becomes
\begin{align*}\frac{3-\sqrt{5}}{\sqrt{5}}\int\frac{du}{u^{2}+\sqrt{5}-2} + \frac{3+\sqrt{5}}{\sqrt{5}}\int\frac{du}{-u^{2} + \sqrt{5} +2}.
\end{align*}
We can rewrite this as
$$\frac{3-\sqrt{5}}{\sqrt{5}(\sqrt{5}-2)}\int\frac{du}{1+\left(\frac{u}{\sqrt{\sqrt{5}-2}}\right)^{2}}+\frac{3+\sqrt{5}}{\sqrt{5}(\sqrt{5}+2)}\int\frac{du}{1-\left(\frac{u}{\sqrt{\sqrt{5}+2}}\right)^{2}}.
$$
This then becomes
$$\frac{3-\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}-2}}\tan^{-1}\left(\frac{u}{\sqrt{\sqrt{5}-2}}\right)+\frac{3+\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}+2}}\tanh^{-1}\left(\frac{u}{\sqrt{\sqrt{5}+2}}\right)+C,
$$
or substituting back
$$\frac{3-\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}-2}}\tan^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{\sqrt{\sqrt{5}-2}}\right)+\frac{3+\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}+2}}\tanh^{-1}\left(\frac{\tan\left(\frac{x}{2}\right)}{\sqrt{\sqrt{5}+2}}\right)+C.$$
Since this expression vanishes at $x=0$, your integral is given by
$$-2\pi\left(\frac{3-\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}-2}}\tan^{-1}\left(\frac{\tan\left(\frac{\pi}{4}\right)}{\sqrt{\sqrt{5}-2}}\right)+\frac{3+\sqrt{5}}{\sqrt{5}\sqrt{\sqrt{5}+2}}\tanh^{-1}\left(\frac{\tan\left(\frac{\pi}{4}\right)}{\sqrt{\sqrt{5}+2}}\right)\right)\approx -8.734.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Combinatorial summation problem: Find $ \sum_{k=1}^{n} k {n+k-1 \choose 3} $ Can we get a neat form for this summation:
$$ \sum_{k=1}^{n} k {n+k-1 \choose 3}? $$
General combinatorial tricks didn't work.
|
Let's try:
$$\begin{aligned}
\sum_{k=1}^n k\binom{n+k-1}{3} &= \sum_{k=1}^n \sum_{i=1}^k \binom{n+k-1}{3} \\
&= \sum_{i=1}^n \sum_{k=i}^n \binom{n+k-1}{3}.
\end{aligned}$$
Using the known relation
$$
\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}
$$
you can obtain
$$\begin{aligned}
\binom{n+i-1}{4} + \sum_{k=i}^n \binom{n+k-1}{3}
&= \binom{n+i-1}{4} + \binom{n+i-1}{3} + \binom{n+(i+1)-1}{3}+\dots + \binom{n+n-1}{3} \\
&= \binom{n+(i+1)-1}{4} + \binom{n+(i+1)-1}{3}+\binom{n+(i+2)-1}{3}+\dots + \binom{n+n-1}{3} \\
&=\binom{n+(i+2)-1}{4} + \binom{n+(i+2)-1}{3}+\binom{n+(i+3)-1}{3}+\dots + \binom{n+n-1}{3} \\
&= \dots \\
&=\binom{n+n-1}{4}+\binom{n+n-1}{3} \\
&= \binom{2n}{4} \\
\end{aligned}$$
so
$$
\sum_{k=1}^n k\binom{n-k+1}{3} = \sum_{i=1}^n \binom{2n}{4}-\binom{n+i-1}{4} $$
and applying the same rule as before,
$$
= n\binom{2n}{4} - \binom{2n}{5} + \binom{n}{5}
$$
|
{
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"url": "https://math.stackexchange.com/questions/2262335",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
|
Set $\sin^2(x) = 1 - \cos^2(x) = y^2$ Then you have
$$
v=\sqrt{a^2(1-y^2)+b^2 y^2}+\sqrt{b^2 (1-y^2)+a^2 y^2}
$$
One extremum is obtained (differentiate w.r.t. y) at
$$
\sqrt{a^2(1-y^2)+b^2 y^2}= \sqrt{b^2 (1-y^2)+a^2 y^2}
$$
or
$$
y^2 = 1/2
$$
The other extremum is obtained at $y=0$.
So the difference of $v^2$ between maximum and minimum is
$$
2(a^2 + b^2) - (a+b)^2 = (a-b)^2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2263431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$
Let $h$ be $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
{]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.
I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(\pi)-\cos(\beta)\sin(\pi) = -\sin(\beta)$$
$$\\$$
$$-\sin(\beta) = \frac{1}{3} \Leftrightarrow \\
\sin(\beta) = - \frac{1}{3}$$
$$\\$$
$$\cos \beta = \sqrt{1-\sin^2(\beta)} = \sqrt{1-(-\frac{1}{3})^2} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$
And so $h(\beta)$ is :
$$\frac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} = \\
\frac{\sqrt{2}}{2\sqrt{2}-3}$$
But my book says the solution is $6\sqrt{2}-8$. What went wrong?
|
You answer is very close.
Solve for$\beta$
Simplify
$$
\sin \left( \beta - \pi \right) = - \sin \beta = \frac{1}{3}
$$
The solutions are
$$
\beta = - \arcsin \frac{1}{3} + 2\pi k, \quad \beta = \pi + \arcsin \frac{1}{3} + 2 \pi k
$$
where $k$ is an integer.
The constraint given in the problem is satisfied when
$$
\boxed{\beta = \pi +\sin ^{-1}\left(\frac{1}{3}\right)}
$$
Compute the cosine
$$
\cos \beta =
- \sqrt{1-\sin^{2}\beta} =
- \sqrt{1 -\left(\frac{1}{3}\right)^{2}}=
-\frac{2 \sqrt{2}}{3}
$$
The external minus sign is required because of the constraint on $\beta$.
Write solution.
Substitute
$$
\frac{\cos \beta}{\cos \beta - 1} =
-\frac{2 \sqrt{2}}{3 \left(-\frac{2 \sqrt{2}}{3}-1\right)} =
-\frac{2 \sqrt{2}}{-3-2 \sqrt{2}} =
\boxed{6 \sqrt{2}-8}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\cos{y}=x+\frac{1}{x}$, possible for any values of $y$? If for real values of x, $\cos{y}=x+\frac{1}{x}$,
Then how do we prove that no value of $y$ is possible ?
|
The value of $\cos(y)$ is between $-1$ and $1$. Therefore, for this equality to hold, $x+\frac{1}{x}$ must be between $-1$ and $1$.
*
*If $x>1$, then $x+\frac{1}{x}>x>1$.
*If $x=1$, then $x+\frac{1}{x}=2>1$.
*If $0<x<1$, then $\frac{1}{x}>1$, so $x+\frac{1}{x}>\frac{1}{x}>1$.
*If $x=0$, then $x+\frac{1}{x}$ is not defined.
*If $-1<x<0$, then $\frac{1}{x}<-1$ so $x+\frac{1}{x}<\frac{1}{x}<-1$.
*If $x=-1$, then $x+\frac{1}{x}=-2<-1$.
*If $x<-1$, then $x+\frac{1}{x}<x<-1$.
Since $x+\frac{1}{x}$ is either greater than $1$ or less than $-1$, the equality can never hold (at least in the reals).
Alternately, one could use calculus and compute minima and limits as $x\rightarrow\pm\infty$ and $x\rightarrow 0^{\pm}$. However, since the necessary inequality can be proved with only elementary facts, I chose this method.
|
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"url": "https://math.stackexchange.com/questions/2266973",
"timestamp": "2023-03-29T00:00:00",
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|
Find the signal and zeros of $\cos x - \sin x$
Let $f$ be a function of domain $\mathbb{R}$ defined by:
$$f(x) = \cos x - \sin x$$
Find the signal and zeros of the function.
First I tried to find the zeros of $f(x)$:
$$0 = \cos x - \sin x \Leftrightarrow \\
0 = \sqrt{1-\sin^2x}-\sin x \Leftrightarrow \\
\sin x = \sqrt{1-\sin^2x} \Leftrightarrow \\
\sin^2x=1-\sin^2x\Leftrightarrow \\
2\sin^2x=1 \Leftrightarrow \\
\sin x = \frac{1}{\sqrt{2}} \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \pi - \frac{\pi}{4}+2k\pi \Leftrightarrow \\
x = \frac{\pi}{4}+2k\pi \lor \frac{3\pi}{4}+2k\pi $$
But then when I check the value of $f(x)$ for the answers above, $\frac{\pi}{4}+2k\pi$ gives me a zero while $\frac{3\pi}{4}+2k\pi$ gives me a minimum. Why is that?
As for the signal, $\frac{\pi}{4}+2k\pi$ gives me the zero on the left side when the function becomes negative.
My book states the solution is
Zeros:
$x = \frac{\pi}{4}+k\pi, k\in \mathbb{R}$
$f$ is negative in the intervals
$$]\frac{\pi}{4}+k2\pi,\frac{5\pi}{4}+k2\pi[,k\in\mathbb{Z};$$
$f$ is positive in the intervals
$$]-\frac{3\pi}{4}+k2\pi,\frac{\pi}{4}+k2\pi[,k\in\mathbb{Z}$$
How do I solve this?
|
$f(x)=\cos x−\sin x = \sqrt {2} \cos (x+\frac \pi 4)$
more generally
$a\cos x+b\sin x = \sqrt {a^2 + b^2} \cos (x - \tan^{-1}\frac ba)$
|
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|
Substitution Integral: $\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx$ $$\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx $$ I tried with substitution $t=3-x$ but I don't know what to do next.
Also how do I solve this limit: $$\lim_{x\to\infty}\frac{1}{x}\int_0^x\frac{dt}{2+\cos t} $$ I saw that most problems like this are solved by putting the period of the function in front,would it work here?
|
Hint:
(1) $\displaystyle \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}}=\frac{\sqrt{x}(\sqrt{x}+\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})(\sqrt{x}+\sqrt{3-x})}$
(2) As the integrand is even and has a period of $2\pi$, $\displaystyle\int_{k\pi}^{(k+1)\pi}\frac{dt}{2+\cos t} =\int_{0}^{\pi}\frac{dt}{2+\cos t} $ for all integers $k$.
Let $\displaystyle\int_{0}^{\pi}\frac{dt}{2+\cos t} =l$. As the integrand is strictly positive, $\displaystyle\int_{0}^{x}\frac{dt}{2+\cos t} $ is strictly increasing.
If $n\pi\le x<(n+1)\pi$,
$$nl\le \int_{0}^{x}\frac{dt}{2+\cos t}<(n+1)l $$
By letting $u=\tan\frac{t} {2}$,
$$\int_{0}^{\pi}\frac{dt}{2+\cos t} =2\int_0^\infty\frac{du}{3+u^2}=\frac{\pi}{\sqrt{3}}$$
So,
$$\frac{n\pi}{\sqrt{3}}\le \int_{0}^{x}\frac{dt}{2+\cos t}<\frac{(n+1)\pi}{\sqrt{3}} $$
$$\frac{x-\pi}{\sqrt{3}}< \int_{0}^{x}\frac{dt}{2+\cos t}<\frac{x+\pi}{\sqrt{3}} $$
$$\frac{1-\frac{\pi}{x}}{\sqrt{3}}< \frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t}<\frac{1+\frac{\pi}{x}}{\sqrt{3}} $$
So, $\displaystyle \lim_{x\to\infty}\left[\frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t}\right]=\frac{1}{\sqrt{3}}$.
|
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|
maximum of combinatorial function Find the maximum of $\binom{30-k}{x} {(\frac{4}{5})}^{30-k}$ w.r.t. $k$, and find $k$ as a function of $x$ when the function maximises.
Note that as $k$ increases $\binom{30-k}{x}$ decreases and ${(\frac{4}{5})}^{30-k}$ increases.
|
When $k=30-x$, the binomial coefficient equals $1$, and the power of $\frac45$ is simply $\left(\frac45\right)^x$. Each time we decrease $k$ by $1$, the binomial coefficient increases by the ratio $\frac{\binom{n+1}{x}}{\binom{n}{x}}$ for some $n$, while the power of $\frac45$ decreases by a ratio of $\frac45$. Thus, we need to find the value of $n$ where $\frac{\binom{n+1}{x}}{\binom{n}{x}}<\frac54$ for the first time. That ratio simplifies to $\frac{n+1}{n+1-x}$. So, we calculate:
$$\begin{align}\frac{n+1}{n+1-x} &= \frac{5}{4}\\
4(n+1) &= 5(n+1-x)\\
4n + 4 &= 5n + 5 - 5x \\
5x - 1 &= n
\end{align}$$
It appears that the expression increases until $30-k=n=5x-1$, takes on the same value at $30-k=n+1=5x$, and then decreases from there. These correspond to the $k$ values $31-5x$ and $30-5x$. If $x=6$, the function is maximized at $k=0$ and $k=1$, for $x>6$, the maximum is only at $k=0$.
|
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|
Perhaps a Pell type equation Find all pairs of positive integers $(a,b)$ that satisfy $13^a+3=b^2$.
If $a$ is even then $3=(b-13^{a/2})(b+13^{a/2})$ which have no solutions. Now if the case $a=2k+1$ is odd then $b^2-13.(13^k)^{2}=3$ I cant proceed from here, please help. Any other methods for the latter case?
|
A solution using elliptic curves, or Mordell equations, as Will Jagy suggests in the comments.
There are three cases:
$1)$ $\ $ $a=3k$, $k\in\mathbb Z^+$. Then $\left(13^k\right)^3+3=b^2$ is a Mordell equation.
See http://oeis.org/search?q=mordell&language=english&go=Search,
in particular (this link is inside the link), e.g., http://oeis.org/A001014/a001014.txt
$b=\pm 2$, $13^k=1$, $k=0$, $a=0$. No solutions, because $k\in\mathbb Z^+$.
$2)$ $\ $ $a=3k+1$, $k\in\mathbb Z$, $k\ge 0$.
Then $\left(13^{k+1}\right)^3+507=(13b)^2$.
$13^{k+1}=13$, $13b=\pm 52$, $k=0$, $a=1$, $b=\pm 4$.
$3)$ $\ $ $a=3k+2$, $k\in\mathbb Z$, $k\ge 0$.
Then $3\cdot 13^4>10000$, so a different method is needed.
Then $13^3\equiv 1\pmod{61}$, $13^2\equiv 47\pmod{61}$
$\left(13^{3}\right)^k\cdot 13^2+3\equiv b^2\pmod{61}$
$50\equiv b^2\pmod{61}$
$2\equiv \left(b\cdot 5^{-1}\right)^2\pmod{61}$
because $\gcd(5,61)=1$.
A contradiction, because $2$ is not a quadratic residue mod $61$, because $61$ is a prime of the form $8k+5$. See Quadratic Reciprocity.
|
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|
Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$ Evaluate: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$.
......
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
It takes $\infty - \infty $ form when $x=\infty $.
Now,
$$=\lim_{x\to \infty } (\sqrt {x-a} - \sqrt {bx})\times \dfrac {\sqrt {x-a}+\sqrt {bx}}{\sqrt {x-a}+\sqrt {bx}}$$
$$=\lim_{x\to \infty } \dfrac {x-a-bx}{\sqrt {x-a}+\sqrt {bx}}$$
|
Take $\sqrt{x}$ out to get:
$$L=\lim_\limits{x\to\infty} \sqrt{x}\left(\sqrt{1-\frac{a}{x}}-\sqrt{b}\right)=\lim_\limits{x\to\infty} \sqrt{x}\left(1-\sqrt{b}\right).$$
If $b=1$, $L=0$.
If $b>1$, $L=-\infty$.
If $0\le b<1$, $L=+\infty$.
|
{
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|
Cauchy random variable question If X is a Cauchy random variable it has the density function as follows:
$f_X(x):= \frac{1}{\pi}\cdot\frac{1}{1+(x-\Theta)^2 }$ con $x \in R$ e $- \Theta<x<\Theta$
I have to demostrate that the random variable $Y=\frac{1}{X}$ is a Cauchy random variable too.
$$F_Y (a)=P(Y<a)=P(\frac{1}{X}<a)=P(X>\frac{1}{a})=1-F_x(\frac{1}{a})=1- \int_{- \infty}^{\frac{1}{a}} \frac{1}{\pi}\cdot\frac{1}{1+(x-\Theta)^2 }\, dx=1-\frac{1}{\pi} *[arctg(x- \Theta)]_{- \infty}^{\frac{1}{a}} =1-\frac{1}{\pi} *[arctg(\frac{1}{a} - \Theta)-(-\frac{\pi}{2}) ]=1-\frac{1}{\pi} *(\frac{\pi}{2}+arctg(\frac{1}{a} - \Theta) )=1-\frac{1}{2} -\frac{1}{\pi}*arctg(\frac{1}{a} - \Theta )=\frac{1}{2} -\frac{1}{\pi}*arctg(\frac{1}{a} - \Theta )$$
Throught the identity $\frac{\pi}{2}=\arctan (x)+\arctan (\frac{1}{x})$.
I can write the previous identity in
$$F_Y (a)=
\frac{1}{\pi}\cdot\arctan(\frac{x}{1-x*\theta})$$
Then I can derivate $F_Y (a)$ to find $f_Y (a)$
$$f_Y (a)=\frac{1}{\pi}\cdot\frac{1}{1+\frac{x}{(1- \Theta*x)^2}}$$p
Now I don't know how to continue
can someone help me if I have done some mistakes?
|
Let $h(x)=x^{-1}$. Then
\begin{align}
f_Y(y)&=f_X(x)(h^{-1}(y))\times\left\lvert \frac{dh^{-1}(y)}{dy}\right\rvert \\
&=\frac{1}{\pi}\times\frac{1}{1+(1/y-\theta)^2}\times\frac{1}{y^2} \\
&=\frac{1}{\pi}\times\frac{1}{\gamma\left(1+[\gamma^{-1}(y-\tau)]^2\right)},
\end{align}
where $\gamma\equiv (1+\theta^2)^{-1}$ and $\tau\equiv\gamma\cdot\theta$.
Alternatively (assume for simplicity that $\theta=0$), for $y>0$,
$$
\mathsf{P}(Y\le y)=\mathsf{P}(X\le 0)+\mathsf{P}(X\ge y^{-1})=1-\frac{\arctan(y^{-1})}{\pi}
$$
and for $y\le 0$,
$$
\mathsf{P}(Y\le y)=\mathsf{P}(y^{-1}\le X\le 0)=-\frac{\arctan(y^{-1})}{\pi}.
$$
Thus,
$$
f_Y(y)=F_Y'(y)=\frac{1}{\pi}\times \frac{1}{1+y^2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are positive real numbers $x,y$ allowed to be taken out during this proof?
Prove $$\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.
Can I simplify $\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$
and then $(x^2 - y^2)(x-y) \ge 0$ cancelling out the $xy$?
Is this valid because then
\begin{align}
(x+y)(x-y)(x-y)&\ge 0\cdot(x-y)^2\ge 0
\end{align}
which is always true and then prove backwards from here?
|
If $x,y$ are positive, then $xy>0$; so, when you have
$$\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$$
you're allowed to multiply by $xy$ (and the $\ge $ won't flip), to get
$$(x^2 - y^2)(x-y) \ge 0$$
And from that indeed you can get
$$(x^2 - y^2)(x-y)=(x+y)(x-y)(x-y)=(x+y)(x-y)^2 \ge 0$$
which is true, since $x+y>0$ and $(x-y)^2\ge 0$.
|
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|
Find the SUM of all the numbers of S. Let S be the set of all three digits numbers. Such that
*
*The digits in each number are all from the set
$\left\{1,2,3,\ldots,9\right\}$.
*Exactly one digit in each number is even.
Then, find the SUM of all the numbers of S.
I have tried that,
The only even digit can have any of the three positions,
And the digit itself has $4$ choices $\left(2, 4, 6\ \mbox{or}\ 8\right)$. The other two digits can be filled in $5 \times 5 = 25$ ways.
Then the number of elements in S is $300$.
But I failed to find the total sum of the elements. Please help me to find this. Thanks in advance.
|
Note that if you permute $3$ distinct digits $x,y,z$, then the sum of the six 3-digit number is $222\cdot(x+y+z)$. If you permute $3$ digits $x,x,z$, with $x\not=z$ then the sum of the three 3-digit number is $111\cdot(2x+z)$. Therefore your sum is
$$222\sum_{x<y\ x,y\in\{1,3,5,7,9\}}\sum_{z\in\{2,4,6,8\}}(x+y+z)+111\sum_{x\in\{1,3,6,7,9\}}\sum_{z\in\{2,4,6,8\}}\cdot(2x+z)$$
that is
$$222\cdot 4\sum_{x<y\ x,y\in\{1,3,5,7,9\}}(x+y)+111\cdot 4\cdot 2\sum_{x\in\{1,3,6,7,9\}}x+(222\cdot 10+111\cdot5)\sum_{z\in\{2,4,6,8\}}z$$
and we finally find
$$888\cdot 100+888\cdot 25
+2775\cdot 20=166500.$$
|
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|
How to prove the following in general: $\lim\limits_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$? To prove
$$\lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
The proof is easy when we take $n$ as positive integer and $a$ any positive real number.
In my book it is given that the result is true even when $n$ is any rational number and $a$ any positive real number. But the proof is not given.
Please provide some hint to construct the proof for general case.
|
$$\begin{array}{rcl}
\displaystyle \lim_{x\rightarrow a} \frac{x^n-a^n}{x-a}
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p/q}-a^{p/q}}{x-a} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x^{p/q}-a^{p/q}\right)\left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^p - a^p}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{\left(x-a\right) \left(x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}\right)}{\left(x-a\right) \left(x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}\right)} \\
&=& \displaystyle \lim_{x\rightarrow a} \frac{x^{p-1} + x^{p-2}a + \cdots + xa^{p-2} + a^{p-1}}{x^{(q-1)p/q} + x^{(q-2)p/q}a^{p/q} + \cdots + x^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\
&=& \displaystyle \frac{a^{p-1} + a^{p-2}a + \cdots + aa^{p-2} + a^{p-1}}{a^{(q-1)p/q} + a^{(q-2)p/q}a^{p/q} + \cdots + a^{p/q}a^{(q-2)p/q} + a^{(q-1)p/q}} \\
&=& \displaystyle \frac{pa^{p-1}}{qa^{(q-1)p/q}} \\
&=& \displaystyle \frac p q a^{(p-1)-(q-1)p/q} \\
&=& \displaystyle \frac p q a^{p-1-p+p/q} \\
&=& \displaystyle \frac p q a^{p/q-1} \\
&=& \displaystyle n a^{n-1} \\
\end{array}$$
|
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|
Find integer part of the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$ for $x+y=3$
If $M$ is the maximum value of $x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$. Subject to $x+y=3$. Find the value of $\lfloor M \rfloor$.
I think the maximum of $M$ occurred when $x=y=3/2$. And I just put the value of that in M and found 34. What is the original answer with good solution ?? Somebody please help me.
|
$x^{4}y+x^{3}y+x^{2}y+xy+xy^{2}+xy^{3}+xy^{4}=xy+xy(x+y)+xy(x^{2}+y^{2})+xy(x^{3}+y^{3})=xy+3xy+xy((x+y)^{2}-2xy)+xy((x+y)^{3}-3x^{2}y-3xy^{2})=4xy+xy(9-2xy)+xy(27-3xy(x+y))=13xy-2(xy)^{2}+xy(27-9xy)=40xy-11(xy)^{2}.$
Now take $xy=k$ and you have $k\leq \frac{9}{4}$. Now our expression is $40k-11k^{2}$. Using derivative calculate $k$ such that the expression is maximum and check that the value of $k$ is less than $\frac{9}{4}$
|
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|
Easy way to find out limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$? What's an easy way to find out the limit of $a_n = \left (1+\frac{1}{n^2} \right )^n$ for $n \rightarrow \infty$?
I don't think binomial expansion like with $\left (1-\frac{1}{n^2} \right )^n = \left (1+\frac{1}{n} \right )^n \cdot \left (1-\frac{1}{n} \right )^n$ is possible.
And Bernoulli's inequality only shows $\left (1+\frac{1}{n^2} \right )^n \geq 1 + n \cdot \frac{1}{n^2} = 1 + \frac{1}{n} \geq 1$ which doesn't seem to help as well.
|
Using the binomial theorem and the fact that $\binom{n}{k}\leq\frac{n^k}{k!}$, we have: $$1\leq\left(1+\frac1{n^2}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac1{n^{2k}}\leq\sum_{k=0}^n\frac{n^k}{k!n^{2k}}=\sum_{k=0}^n\frac{1}{k!n^k}\leq1+\frac1n+\sum_{k=2}^n\frac1{n^2}\leq1+\frac2n.$$ Therefore, by the squeeze theorem, the limit is $1$.
|
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|
What is the limit of this sequence? How does it relate to the exponential function?
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$
I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.
Edit:
I'm trying to prove a result that would be true if
$(1+x^2/n^2)^n -1 = x^kO(x/n).$
|
$$A_n=\left(1+\frac{x^2}{n^2}\right)^n\implies \log(A_n)=n \log\left(1+\frac{x^2}{n^2}\right)$$ Now, using Taylor series for large values of $n$ $$\log\left(1+\frac{x^2}{n^2}\right)=\frac{x^2}{n^2}-\frac{x^4}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ $$ \log(A_n)=\frac{x^2}{n}-\frac{x^4}{2 n^3}+O\left(\frac{1}{n^6}\right)$$ Taylor again using $$A_n=e^{\log(A_n)}=1+\frac{x^2}{n}+\frac{x^4}{2 n^2}+\frac{x^4 \left(x^2-3\right)}{6 n^3}+\frac{x^6
\left(x^2-12\right)}{24 n^4}+O\left(\frac{1}{n^5}\right)$$ For illustration purposes, let us use $x=10$ and $n=100$ $$A_{100}=\left(\frac{101}{100}\right)^{100}\approx 2.70481$$ while the above expansion would give $$\frac{1619}{600}\approx 2.69833$$
|
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|
How to solve integral arctan I solved this integral: $$ \int_0^{\pi/2}\frac{dx}{5+4\cos x} $$ and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct? how do I find the answer in $\pi$?
|
and I obtained $\frac{2}{3} \arctan \frac{1}{3}$ is it correct?
Yes, that is correct!
How do I find the answer in $\pi$?
What do you mean, "in $\pi$"?
Perhaps the model answer looks different (containing $\pi$)? You can use the property:
$$\boxed{\arctan a + \arctan \frac{1}{a} = \frac{\pi}{2}} \implies \color{blue}{\arctan\frac{1}{3} = \frac{\pi}{2}-\arctan 3 }$$
to rewrite as follows:
$$\frac{2}{3}\color{blue}{\arctan\frac{1}{3}} = \frac{2}{3}\left(\color{blue}{ \frac{\pi}{2}-\arctan 3} \right) = \frac{\pi}{3}-\frac{2}{3}\arctan 3$$
|
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|
Put fraction in "arctan-friendly" form I would like to put $\int\frac{1}{(2x^2+x+1)}dx$ into something like $\int\frac{1}{(u^2+1)}dx$. What is the quickest way to proceed? I know that previous fraction can be rewritten as $2t^2+t+1 = \frac{7}{8}\left( \left( \frac{4t+1}{\sqrt{7}} \right)^2 +1 \right)$, but I don't have any explaination from where this comes from.
Finally, the integral yields
$$\int_b^a \frac{7}{8} \left( \left(\frac{4t+1}{\sqrt{7}} \right)^2+1 \right)dt = \frac{2}{\sqrt{7}}\left[\arctan \left(\frac{4t+1}{\sqrt{7}}\right)\right]^a_b $$
|
Note that
$$(ax+b)^2=a^2x^2+2abx+b^2$$
Now we want to complete the square on $2x^2+x+1$. We then have $a^2=2, 2ab=1 $ $\implies 4a^2b^2=1 \implies b^2=\frac 1 8$. Thus, we write
$$2x^2+x+1 = \left(2x^2+x+\frac 1 8\right)+\frac 7 8 = \left(\sqrt 2 x+\frac{1}{2\sqrt 2}\right)^2+\frac 7 8$$
Thus, letting $\sqrt2 x+\frac{1}{2\sqrt 2}=\sqrt{\frac 7 8}\tan\theta$, our integral becomes
$$\int \frac{1}{\frac 7 8 \tan^2 \theta+\frac 7 8}\cdot \frac {\sqrt{7}} 4\sec^2\theta d\theta$$
|
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|
Find the probability that "A" wins the game. In a game, A and B play against each other. They throw a die alternatively until they get the face of 6. Once one gets 6, he wins. i.e., there is only one winner and they keep playing until one of them wins.
Consider that B plays first.
What is the probability of A winning?
I don't even have a clue how to start, but I'm quite sure that it is concerned with the geometric progression.
Pls do help me, any answers will be greatly appreciated. ( but if possible, pls do explain in detail )
|
For $A$ to win on their first turn we need $B$ to not get a 6 and $A$ to get a 6, for probability $\frac{5}{6}\cdot\frac{1}{6}$.
For $A$ to win on their second turn, we need both to miss on their first try, $B$ to miss on their second try, and $A$ to get 6 on their second try, for probability $\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}$.
For $A$ to win on their third turn, we need $B$, $A$, $B$, $A$, $B$ to miss in that order and then $A$ to get a 6, for probability $\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}$.
This pattern continues infinitely so the probability $A$ wins the whole thing is $$\frac{5}{6}\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}+\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}+\cdots = \dfrac{\frac{5}{6}\cdot\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2} = \dfrac{5}{11}.$$
|
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Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using 3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$
Is everything right here?
But how to check it for convergence?
|
To assume that a series is convergent to compute the value is not valid Ixion. i.e. $S=1+2+4+6+8+...$ is obviously not convergent although $S=1+2(1+2+3+4+...)=1+2S$ would imply $S=-1$.
|
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|
Calculate the sum of $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ I would like to calculate the sum for $x\in(-1,1)$ of this: $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$
So far I managed that
$$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$
and
$$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$
and
$$ \int -\log(1-x) dx \overset{c}{=} x+(1-x)\log(1-x). $$
But I really don't know if I can put it somehow together or not.
|
Using partial fraction decomposition $$\frac 1{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ So
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty} \frac{x^{n+1}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x\sum_{n=1}^{\infty} \frac{x^{n}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x \log(1-x)-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$$ Now, $$\frac d{dx} \left(\frac{x^{n+1}}{n+1}\right)=x^n\implies \frac d{dx} \left(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} \right) =\frac{x}{1-x}=\frac{x-1+1}{1-x}=\frac{1}{1-x}-1$$ Now, integrate to get $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=-\log(1-x)-x$$
|
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|
Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$$
Note that:
$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$$
Now we have $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$, i.e.$\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ and $\tan h \sim_{0}h.$
I'm stuck.
Update: here is another way :
\begin{aligned}
\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\
&=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
\end{aligned}
Thus
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left(\dfrac{\pi x}{2a}\right)} =e^{\dfrac{2}{\pi}} $$
*
*Am i right beside i'm intersted in way which use equivalents
|
$$\lim _{ x\to a } \left( 2-\frac { x }{ a } \right) ^{ \tan \left( \frac { \pi x }{ 2a } \right) }=\lim _{ x\to a }{ \left[ { \left( 1+\left( 1-\frac { x }{ a } \right) \right) }^{ \frac { a }{ a-x } } \right] } ^{ \frac { a-x }{ a } \tan \left( \frac { \pi x }{ 2a } \right) }=\\ ={ e }^{ \lim _{ x\rightarrow a }{ \frac { a-x }{ a } \tan \left( \frac { \pi x }{ 2a } \right) } }={ e }^{ \lim _{ x\rightarrow a }{ \frac { a-x }{ a\cot { \left( \frac { \pi x }{ 2a } \right) } } } }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow a }{ \frac { -1 }{ -a\frac { 1 }{ \sin ^{ 2 }{ \left( \frac { \pi x }{ 2a } \right) } } \left( \frac { \pi }{ 2a } \right) } } }=\color{blue}{{ e }^{ \frac { 2 }{ \pi } }}$$
|
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|
From $a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$ to $a_n$ to be proven by induction
Find and Prove by induction an explicit formula for $a_n$ if $a_1=1$ and for $n \geq 1$
$$a_{n+1}=(1+ \frac{1}{n})^n \cdot a_n$$
Checking the pattern:
$$a_1=1$$
$$a_2= 2 \cdot 1$$
$$a_3= (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_4= (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
$$a_5= (\frac{5}{4})^4 \cdot (\frac{4}{3})^3 \cdot (\frac{3}{2})^2 \cdot 2 \cdot 1$$
I can see the general pattern in the index as $(...)^{(n-1)!}$
and I am tempted to state that it is $( \frac{n!}{(n-1)!})^{(n-1)!}
$
Should this be correct, I do not see the factorial manipulation that allow me to write this.
Otherwise, what would the formula be? What would the approach be?
Much appreciated
|
Rewrite:
$$\frac{a_{n+1}}{(n+1)^{n+1}}=\frac{a_n}{n^n} . \frac{1}{n+1}$$
Thus:
$$\frac{a_{n+1}}{(n+1)^{n+1}}=\frac{a_n}{n^n}.\frac{1}{n+1}
=\frac{a_{n-1}}{n^{n-1}}.\frac{1}{n}.\frac{1}{n+1}
=\frac{a_{n-2}}{n^{n-2}}.\frac{1}{n-1}.\frac{1}{n}.\frac{1}{n+1}$$
$$=...$$
$$=\frac{a_1}{1}.\frac{1}{(n+1)!}$$
Therefore:
$$a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}$$
Of course once you know the result you can use induction to prove it.
|
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|
Prove that the following are real numbers
*
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z)}{|z|}$$ which is real number as a division of two real numbers
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
$$\overline{z^3\cdot\overline{z}+z\cdot\overline{z}^3}$$
Due to: $\overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$ and $\overline{z_{1}\cdot z_{2}}=\overline{z_{1}}\cdot\overline{z_{2}}$
we get:
$$\overline{z}^3\cdot{z}+\overline{z}\cdot{z}^3$$
Now we can say that we got the same elements and $(z=\overline{z})$ therefore it is real number?
Is the reasoning in both of these correct?
|
we have $$\frac{z+\overline{z}}{z\overline{z}}=\frac{2x}{x^2+y^2}$$ if $$z=x+iy$$
|
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|
Compute $\sum\limits_{n=1}^\infty\frac{b_{n-1}}{n 2^n}$ where $b_n=\sum\limits_{k=0}^n\frac{2^k}{k+1}$ Define the sequence $a_n := \frac{2^n}{n+1}$. Then, define the sequence $b_n$, the partial sums of $a_n$; i.e.:
$$b_n=\sum_{k=0}^{n} {a_k}$$
The problem is to compute:
$$\sum_{n=1}^{\infty} {\left(\frac{b_{n-1}}{n 2^n}\right)}$$
This is very difficult to estimate using a computer algebra system. On Maple, I cannot complete an estimate of $100000$ terms under a minute. (However, the sum of the first $1000$ terms is approximately $1.23270055$.)
I am wondering whether the double sum can be converted into a series convolution of some sort; the possibility of getting a $2^{n-m}$ term seems to help that prospect.
Thanks!
|
Consider the following.
Given:
\begin{align}
b_{n} &= \sum_{k=0}^{n} \frac{2^{k}}{k+1} \\
S &= \sum_{n=1}^{\infty} \frac{b_{n-1}}{2^{n} \, n}.
\end{align}
Now:
$$b_{n} = \frac{1}{2} \, \sum_{k=1}^{n+1} \frac{2^{k}}{k} $$
which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{2^{k}}{2^{n} \, n \, k} = \frac{1}{2} \, \sum_{n,k=1}^{\infty} \frac{1}{2^{n} \, k \, (n+k)} = \frac{\pi^{2}}{8}.
\end{align}
An alternate view is:
\begin{align}
b_{n} &= \sum_{k=0}^{n} \frac{2^{k}}{k+1} = \sum_{k=0}^{n} 2^{k} \, \int_{0}^{1} t^{k-1} \, dt \\
&= \int_{0}^{1} \left( \sum_{k=0}^{n} (2 t)^{k} \right) \, dt \\
&= \int_{0}^{1} \frac{1 - (2 t)^{n+1}}{1-2 t} \, dt.
\end{align}
Now,
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{2^{n} \, n} \, \int_{0}^{1} \frac{1 - (2 t)^{n}}{1-2 t} \, dt \\
&= \int_{0}^{1} \frac{\ln(2) + \ln(1-t)}{1-2 t} \, dt \\
&= \left[ \frac{1}{2} \, Li_{2}(2 t -1) \right]_{0}^{1} = \frac{1}{2} \, (Li_{2}(1) - Li_{2}(-1) ) \\
&= \frac{1}{2} \, \left(\frac{\pi^2}{6} + \frac{\pi^2}{12} \right) = \frac{\pi^{2}}{8}.
\end{align}
Note: $Li_{2}(x)$ is the dilogarithm function.
|
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|
Compute $\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$
Compute$$\int_0^\frac {\pi} {2} \frac {\cos x}{(1+\sqrt{\sin (2x)})^2}\,dx$$
I think there's no closed form antiderivative of it. I tried WolframAlpha but it didn't help much.
|
By setting $x=\frac{t}{2}$, then exploiting symmetry:
$$ I=\int_{0}^{\pi/2}\frac{\cos(x)}{\left(1+\sqrt{\sin(2x)}\right)^2} = \frac{1}{2\sqrt{2}}\int_{0}^{\pi/2}\frac{\sqrt{1+\cos(x)}+\sqrt{1-\cos(x)}}{\left(1+\sqrt{\sin(x)}\right)^2}\,dx $$
and the last integral can be written in the following form:
$$ I = \frac{1}{2}\int_{0}^{\pi/2}\frac{\sqrt{1+\sin x}}{1+\sin x+2\sqrt{\sin x}}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\sqrt{1+x}}{\sqrt{1-x^2}}\cdot\frac{dx}{(1+\sqrt{x})^2} $$
or the following one:
$$ I = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{1-x}(1+\sqrt{x})^2} = \frac{1}{2}\int_{0}^{1}\frac{dx}{\sqrt{x}(1+\sqrt{1-x})^2}=\int_{0}^{1}\frac{dx}{\left(1+\sqrt{1-x^2}\right)^2}$$
that also equals
$$ I = \int_{0}^{\pi/2}\frac{\cos t}{(1+\cos t)^2}\,dt = \int_{0}^{2}\frac{\frac{1}{4}-\frac{u^4}{64}}{1+\frac{u^2}{4}}\,du = \frac{1}{16}\int_{0}^{2}\left(4-u^2\right)\,du = \color{red}{\frac{1}{3}}$$
by Weierstrass substitution $t=2\arctan\frac{u}{2}$.
|
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|
Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
|
$$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$
$$=\sum_{k=1}^n 4!\binom{k+3}{4}$$
$$=4!\sum_{k=1}^n \binom{k+3}{4}$$
Using the hockey-stick identity:
$$=4!\binom{k+4}{5}$$
$$=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
|
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|
Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\ c&d \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmatrix} a^2+bc &b(a+d) \\ c(a+d)&d^2+bc \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
So
$$a^2+bc=1 \tag{1}$$
$$b(a+d)=2 \tag{2}$$
$$c(a+d)=3 \tag{3}$$
$$d^2+bc=4 \tag{4}$$ From $(2)$ and $(3)$ we get
$$\frac{b}{c}=\frac{2}{3}$$
Let $b=2k $ and $c=3k$ Then
$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$ So
$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\ 3k&\sqrt{4-6k^2} \end{bmatrix}$$
So
$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\ 3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4 \end{bmatrix}=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$$
Now we have to solve for $k$ using equation
$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also
$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$
$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$
Subtracting $(6)$ from $(5)$ we get
$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$ Squaring Both sides we get
$$33k^4-10k^2+1=0$$ we get
$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?
|
Just variation of the loup blanc's answer...
Interesting it would be also to check whether it is possible to use
$B^2-\text{trace}(B)B+\det(B)I=0$ ?
Indeed
$B^2+\det(B)I= A\pm\sqrt{\det(A)}I=\text{trace}(B)B=tB $
If we name $ C=A\pm\sqrt{\det(A)}I$ ( $C$ can be directly calculated) then $\text{trace}(C/t)=t$, hence
$t^2=\text{trace}(C )$.
Also $4$ solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int \sqrt{2x + 3} dx$ Integrate: $\int \sqrt{2x + 3} \ dx$
Doing some guesswork gives me: $\frac{1}{2}(2x + 3)^{3/2}$ to try. Differentiating this gives: $2\frac{3}{2}\frac{1}{2}(2x + 3)^{1/2}$ = $\frac{6}{4}(2x + 3)^{1/2}$
Too bad the answer should be: $\frac{1}{3}(2x + 3)^{1/2}$, so clearly I'm missing some factor of 2. Where am I going wrong?
|
Here is how you could approach this:
Your answer is within a constant factor of the real answer, so you have
$$\int \sqrt{2x+3}dx=c*(2x+3)^{\frac{3}{2}}$$
When we differentiate $c*(2x+3)^{\frac{3}{2}}$ we want to get $\sqrt{2x+3}$, so differentiate and solve for $c$:
$$\sqrt{2x+3}=\frac{d}{dx}\big(c*(2x+3)^{\frac{3}{2}}\big)$$
Use the chain rule:
$$\sqrt{2x+3}=\frac{3}{2}*2*c*\sqrt{2x+3}$$
then solve for $c$:
$$1=\frac{3}{2}*2*c$$
$$1=3c$$
$$c=\frac{1}{3}$$
That's one way to think about it.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate product $\prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right)$? How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right) $$....Can I take a geometric series and compare it with that?
|
Multiply through by $\frac12=1-\frac1{2^{2^0}}$ to get
$$\begin{align}\left(1-\frac1{2^{2^0}}\right)\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)&=\left(1-\frac1{2^{2^0}}\right)\left(1+\frac1{2^{2^0}}\right)\prod_{k=1}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^1}}\right)\left(1+\frac1{2^{2^1}}\right)\prod_{k=2}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^2}}\right)\left(1+\frac1{2^{2^2}}\right)\prod_{k=3}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^3}}\right)\left(1+\frac1{2^{2^3}}\right)\prod_{k=4}^n\left(1+\frac1{2^{2^k}}\right)\\&=\quad\vdots\\&=\left(1-\frac1{2^{2^{n+1}}}\right)\end{align}$$
$$\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)=2-\frac1{2^{2^{n+1}-1}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find an upperbound for the rational function I know
$$\lim_{(x, y)\to (0,0)} \frac{x^3 + y^4}{x^2 + y^2} = 0$$ so
$$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert \le f(x, y)$$ for some simpler $f(x, y)$ whose limit is also $0$.
How do I find the function $f(x, y)$? In other words, how do I get the upper bound?
|
Let $x=r\cos{t}, y=r\sin{t}$, then $$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert=|r(\cos^3{t}+r\sin^4{t})|\leq |r||1+r| = \sqrt{x^2+y^2}+x^2+y^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration $\int \frac{\sec(x)}{\tan^2(x) + \tan(x)}$ How to evaluate the following integral $$\int \frac{\sec(x)}{\tan^2(x) + \tan(x)} \ dx$$
Thank you.
|
$$\int\frac{\sec{x}}{\tan^2x+\tan{x}}dx=\int\frac{\cos{x}}{\sin{x}(\sin{x}+\cos{x})}dx=$$
$$\int\left(\frac{1}{\sin{x}}-\frac{1}{\sin{x}+\cos{x}}\right)dx=-\int\left(\frac{d(\cos{x})}{\sin^2{x}}-\frac{d\left(\cos\left(\frac{\pi}{4}+x\right)\right)}{\sqrt2\sin^2\left(\frac{\pi}{4}+x\right)}\right)=$$
$$=-\int\left(\frac{d(\cos{x})}{(1-\cos{x})(1+\cos{x})}-\frac{d\left(\cos\left(\frac{\pi}{4}+x\right)\right)}{\sqrt2\left(1-\cos\left(\frac{\pi}{4}+x\right)\right)\left(1+\cos\left(\frac{\pi}{4}+x\right)\right)}\right)=$$
$$=-\ln\sqrt{\frac{1+\cos{x}}{1-\cos{x}}}+\frac{1}{\sqrt2}\ln\sqrt{\frac{1+\cos\left(\frac{\pi}{4}+x\right)}{1-\cos\left(\frac{\pi}{4}+x\right)}}+C=$$
$$=-\ln|\cot\frac{x}{2}|+\frac{1}{\sqrt{2}}\ln|\cot\left(\frac{\pi}{8}+\frac{x}{2}\right)|+C.$$
|
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|
Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ $$\sum _{n=1}^{\infty}\frac{n}{5^n}$$
I tried to plug in $n=1,2,3,4,...$ but I can't use common ratio to solve problem.
I think there is another way like using differentiation or integral but I don't no exactly what to do.
|
I am not using calculus let the sum be $S$ thus $S=\frac{1}{5}+\frac{2}{5^2}$...now $\frac{S}{5}=\frac{1}{5^2}+\frac{2}{5^3}+...$ subtracting two we have $\frac{4S}{5}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...$ which is GP with common ratio less than 1 using formula that sum of such gp is $\frac{a}{1-r}$ we have $\frac{4S}{5}=\frac{1}{4}$ thus $S=\frac{5}{16}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers. Find all positive integers $n \le 6$, such that the equation $a^n+b^n=c^n+n$ has solutions over the integers.
My attempt:
For $n=1$, trivially there are integer solutions.
For $n=2$, we have the solutions
$$a=b=\pm 1, c=0.$$
For $n=3$, we have the solution
$$a=b=1, c = -1.$$
Anyone can complete the proof for $n=4,5,6$?
|
If $n$ is such that $2n+1$ is prime, then the only $n$th powers mod $2n+1$ are $0$ and $\pm 1$ (since the group of units mod $2n+1$ is cyclic of order $2n$). It follows easily that $a^n+b^n=c^n+n$ has no solutions mod $2n+1$ if $n>3$. In particular, this handles the cases $n=5$ and $n=6$ (as well as infinitely many other larger values of $n$).
For $n=4$, you need a more ad hoc argument, since $2n+1$ is not prime. Working mod $8$ does the trick, since the only $4$th powers (in fact, the only squares) mod $8$ are $0$ and $1$.
|
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|
If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The value of k is given to be 4, but I am unable to derive that result.
Kindly provide me with some insight, or with a step-by-step solution.
Thanks in advance,
Abhigyan
|
\begin{align}
k&=\frac{4\cos^2\frac\pi7-1}{\cos\frac{\pi}{7}(2\cos^2\frac\pi7-1)}\\
&=\frac{2\cos\frac{2\pi}{7}+1}{\cos\frac{2\pi}{7}\cos\frac{\pi}{7}}\\
&=\frac{2\cos\frac{2\pi}{7}+1}{\frac{1}{2}(\cos\frac{3\pi}{7}+\cos\frac{\pi}{7})}\\
&=\frac{4\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{3\pi}{7}+\sin\frac{\pi}{7}\cos\frac{\pi}{7}}\\
&=\frac{2\sin\frac{3\pi}{7}-2\sin\frac{\pi}{7}+2\sin\frac{\pi}{7}}{\frac{1}{2}(\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{2\pi}{7})}\\
&=\frac{4\sin\frac{3\pi}{7}}{\sin\frac{4\pi}{7}}\\
&=4
\end{align}
|
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|
How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point?
used partial fraction:
$$\frac{n^2}{n^3+1}=\frac{1}{3}\cdot\frac{2n-1}{n^2-n+1}+\frac{1}{3}\cdot \frac{1}{n+1}$$
I'm stuck with the first term in the partial fraction , the second term simply yields $1-\log(2)$
|
$$
\begin{align}
&\sum_{n=1}^\infty(-1)^{n-1}\frac{n^2}{n^3+1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n+e^{-2\pi i/3}}\right)\tag{1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n-1-e^{2\pi i/3}}\right)\tag{2}\\
&=\frac13\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+1}
-\frac13\sum_{n=-\infty}^\infty\frac{(-1)^n}{n+e^{2\pi i/3}}\tag{3}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(\pi e^{2\pi i/3}\right)\tag{4}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(-\frac\pi2+i\frac{\pi\sqrt3}2\right)\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{1-\log(2)}3+\frac\pi3\operatorname{sech}\left(\frac{\pi\sqrt3}2\right)}\tag{6}
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $1+e^{2\pi i/3}+e^{-2\pi i/3}=0$
$(3)$: $n\mapsto1-n$ sends $\frac{(-1)^n}{n-1-e^{2\pi i/3}}\mapsto\frac{(-1)^n}{n+e^{2\pi i/3}}$ and $\{1,2,3,\dots\}\mapsto\{0,-1,-2,\dots\}$
$(4)$: Equation $(3)$ from this answer
$(5)$: expand
$(6)$: evaluate
|
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|
Tom Apostol - Calculus Vol. 1: Method of Exhaustion in Introduction I'm new to proof based math, and I really want to get better from pursuing the discipline by myself. I actually had a fear, last year, of proof-based math after my first semester in college. It was pretty brutal - I had never seen or experienced proofs before, and there was no attempt to help the students familiarize themselves. It seemed like either we knew the methods of proof, and how to write a good proof from the beginning, or spend all of your time being utterly confused throughout the material of the class. It was during this time I came to appreciate that I really need math literacy, and to get more experience with the techniques and underlying assumptions that are implicit in reading certain proofs. So, I'm now trying to get over this fear, by tackling the same content with the same textbook - Tom M. Apostol's Calculus Vol. 1.
In the book, in the introduction portion I $1.3$, the method of exhaustion is used to calculate the area under a parabolic segment of function $f(x)=x^2$ and base length $b$.
Near equation I$.10$, we proceed to prove that $\frac{b^3}{3}$ is the only value between $s_n$ and $S_n$, the areas below and above the segment respectively, and hence must be the value of the area under the parabolic segment from $0$ to $b$.
My confusion of this part of the proof comes in when he assumes that $A>b^3/3$ to find the following contradiction: $$n < \frac{b^3}{A - b^3 / 3}$$
for $\{n\quad|\quad n\in \mathbb Z \qquad and \ \qquad n\gt0\}$, where $A$ is the guessed value for the area under the segment. The contradiction lies in the fact that obviously $$n\ge \frac{b^3}{A-b^3/3}$$
but I'm not sure why -- it seems like it's reasonable for $\frac{b^3}{A-b^3/3}$ to be larger than 2 if we assume $b$ is some integer greater or equal to 1, and $A-\frac{b^3}{3}$ is at least less than half of 1, and n is allowed to be 1, so how does this contradict what has been said before?
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Remember, he states before it that, for all $n\ge1$, $A < \frac{b^3}{3} + \frac{b^3}{n}$. This is actually the same with the statement:
$$∀n\ge1,\Biggl(A < \frac{b^3}{3} + \frac{b^3}{n}\Biggr)$$
Thus since we assumed that $A > \frac{b^3}{3}$, this yields to the fact that $A - \frac{b^3}{3} > 0$, thus we can change something inside the universal quantifier statement (like multiplying with $\frac{x}{x}$ or any other algebra operation).
So now, $∀n \ge 1, \bigl(A < \frac{b^3}{3} + \frac{b^3}{n}\bigr)$ means $∀n\ge1, \bigl(A - \frac{b^3}{3} < \frac{b^3}{n}\bigl)$, since the LHS is positive, we divide both sides and get $∀n \ge 1, \biggl(1 < \frac{\frac{b^3}{n}}{A - \frac{b^3}{3}}\biggr)$, we proceed to multiply both sides by $n$, which is okay, since $n \ge 1$ we finally arrives at a contradiction, namely:
$$∀n \ge 1 \Biggl(n < \frac{b^3}{A - \frac{b^3}{3}}\Biggr)$$.
How to see this contradiction more clearly? First, note that the RHS is positive, so $\frac{b^3}{A - \frac{b^3}{3}} > 0$, we add $1$ to both sides and get $\frac{b^3}{A - \frac{b^3}{3}} + 1 > 1$, let $n = \frac{b^3}{A - \frac{b^3}{3}} + 1$, which is larger than $1$ (so this is one of the quantified value of the universal quantifier), but this $n$ is not smaller than $\frac{b^3}{A - \frac{b^3}{3}}$, for if it does, then $1 < 0$ which is not possible, therefore, a contradiction, as required. You can do the same with the other direction, showing that the only possibility is for $A = \frac{b^3}{3}$.
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Need to prove formula for $\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$ I need help showing that for $|z|<1$ we have
$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}}$ = $\frac{z}{1-z}$
I tried using $\sum_{n=0}^{\infty} (-z^{2^k})^{n} = \frac{1}{1+z^{2^k}}$ but that hasn't gotten me very far. Any help is appreciated!
|
Let
$$\sum_{k=0}^{\infty} \frac{2^{k} z^{2^k}}{1 + z^{2^k}} = \sum_{k=0}^{\infty} a_n z^n$$
and examine contributions to the $a_n$ from the various terms in the $k$ sum.
For $n=0$ none of the terms in the sum contributes (the $k=0$ contribution is
$z-z^2+z^3-z^4+\ldots$).
For $n$ odd only the $k=0$ term contributes, and thus for $n$ odd, $a_n = 1$.
The $k=1$ term of the sum contributes to $(-1)^{n/2+1}2^1$ to each $a_n$ for $n$ even. Higher $k$ terms give no contribution to $a_2$, or indeed to any
$a_{4p+2}$, thus for $p\in\Bbb Z$
$$
a_{4p+2} = (-1) + 2 = 1 $$
The contributions of the $k=1$ term of the sum to each $a_{4p}$ is $-2$.
The $k=2$ term of the sum contributes to $(-1)^{n/4+1}2^2$ to each $a_n$ for $n$ divisible by $4$. Higher $k$ terms give no contribution to any
$a_{8p+4}$, thus for $p\in\Bbb Z$
$$
a_{8p+4} = (-1) + (-2) + 4 = 1 $$
The contributions of the $k=2$ term of the sum to each $a_{8p}$ is $-4$.
The $k=3$ term of the sum contributes to $(-1)^{n/8+1}2^3$ to each $a_n$ for $n$ divisible by $8$. Higher $k$ terms give no contribution to any
$a_{16p+8}$, thus for $p\in\Bbb Z$
$$
a_{16p+8} = (-1) + (-2) + (-4) + 8 = 1 $$
The contributions of the $k=2$ term of the sum to each $a_{8p}$ is $-8$.
Without going to the effort to formally introduce induction here, we can see the pattern: For any even $n>0$, the net contributions always add to $1$, and for odd $n$ we already saw a value of $a_n = 1$. Thus the result
is
$$
\sum_{k=1}^\infty z^n = \frac{z}{1-z}
$$
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Summation of product of three terms in AP Here is an interesting relationship:
$$\sum_{r=1}^n r(r+m)(r+2m)=\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m} r\right)\tag{1}$$
$$\text{i.e.} \;\; 1\cdot(1+m)(1+2m)+2\cdot (2+m)(2+2m)+3\cdot (3+m)(3+2m)+\cdots +n(n+m)(n+2m)=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+2m))$$
which gives the following for the first few values of $m$:
$$\scriptsize\begin{align}
&1\cdot 1\cdot 1+2\cdot 2\cdot2+3\cdot3\cdot3+\cdots+n\cdot n\cdot n&&=(1+2+3+\cdots+n)(1+2+3+\cdots+n)\\
&1\cdot2\cdot3+2\cdot3\cdot4+3\cdot5\cdot7+\cdots+n(n+1)(n+2)&&=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+2))\\
&1\cdot3\cdot5+2\cdot4\cdot6+3\cdot5\cdot7+\cdots+n(n+2)(n+4)&&=(1+2+3+\cdots+n)(1+2+3+\cdots+(n+4))\\ \end{align}$$
This can be proven by expanding both sides and finding that the result is
$$\frac {n(n+1)}2\cdot \frac {(n+2m)(n+2m+1)}2\tag{2}$$
Question
Can the result can be proven without expansion to the closed form but only by manipulating summands and limits, i.e. prove $(1)$ without first expanding to $(2)$?
Some interesting but commonly-known results follow from this.
Setting $m=0$ gives the "sum of cubes as square of sum of integers" result:
$$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2$$
Setting $m=1$ gives the "discrete integral"
$$\sum_{r=1}^n r^\overline{3}=\frac{\;n^\overline{4}}4$$
where $r^\overline{a}$ is the symbol for the rising factorial.
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I will use the identity
$$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2$$
It can be proved without direct expansion. (https://math.stackexchange.com/a/1215805/268334)
\begin{align}
&\;\sum_{r=1}^n r(r+m)(r+2m)\\
=&\;\sum_{r=1}^n [(r+m)^3-m^2(r+m)]\\
=&\;\sum_{r=1}^{n+m}r^3-\sum_{r=1}^mr^3-m^2\sum_{r=1}^n (r+m)\\
=&\;\left(\sum_{r=1}^{n+m}r\right)^2-\left(\sum_{r=1}^mr\right)^2-m^2\sum_{r=1}^n (r+m)\\
=&\;\left(\sum_{r=1}^{n+m}r-\sum_{r=1}^mr\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr\right)-m^2\sum_{r=1}^n (r+m)\\
=&\;\left(\sum_{r=1}^n (r+m)\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr\right)-m^2\sum_{r=1}^n (r+m)\\
=&\;\left(\sum_{r=1}^n (r+m)\right)\left(\sum_{r=1}^{n+m}r+\sum_{r=1}^mr-m^2\right)\\
=&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-\sum_{r=1}^m(r+m+n)+\sum_{r=1}^mr-m^2\right)\\
=&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-m(m+n)-m^2\right)\\
=&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-\sum_{r=1}^m(r+m+n)+\sum_{r=1}^mr-m^2\right)\\
=&\;\left(\sum_{r=1}^n r+mn\right)\left(\sum_{r=1}^{n+2m}r-m(2m+n)\right)\\
=&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+mn\sum_{r=1}^{n+2m}r-m(2m+n)\sum_{r=1}^nr-m^2n(2m+n)\\
=&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+\frac{1}{2}mn(n+2m)(n+2m+1)\\
&\qquad-\frac{1}{2}m(2m+n)n(n+1)-m^2n(2m+n)\\
=&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)+\frac{1}{2}mn(n+2m)(n+2m+1-n-1-2m)\\
=&\;\left(\sum_{r=1}^n r\right)\left(\sum_{r=1}^{n+2m}r\right)
\end{align}
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Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
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Suppose
$a^b-b^a = n$
with all variables
positive integers.
If $n=1$,
this is Catalan's problem
for which the only solution is
$a=3, b=2$.
Assume
$n \ge 2$.
I will show that there
at most a finite number of solutions
and give bounds for
$a$ and $b$
in terms of $n$.
For $n=1008$,
the only solution is
$a=n+1, b=1$.
If $b=1$,
then
$a-1 = n$,
so $a = n+1$.
Then
$a-b = n
\equiv 0 \bmod n$.
If $b=2$,
then
$a^2-2^a = n$.
Since
$a^2 \le 2^a$
for
$a\le 2$
and
$a \ge 4$,
we must have
$a=3$
so the only $n$ that works
is
$n =3^2-2^3 = 1$.
If
$b \ge 3$,
if
$a > b$
then
$b^a > a^b$
(since
$b^{1/b} > a^{1/a}$
)
so no solution.
Therefore
if $b \ge 3$
then $a < b$.
We can't have $a=1$.
If $a=2$
then
$2^b-b^2 = n$.
Since the left side
is increasing for
$b \ge 4$,
there is at most one solution.
For $n=1008$,
there is no solution.
If $a=3$
then
$3^b-b^3 = n$.
Since the left side
is increasing for
$b \ge 3$,
there is at most one solution.
For $n=1008$,
there is no solution.
Suppose
$a \ge 4$
so $b \ge 5$.
Let $b = a+c$
where $c \ge 1$.
Then,
since
$(1+1/x)^x < e$
and
$u^r - v^r
\ge (u-v)v^{r-1}
$
if
$u > v > 1$
and
$r > 1$,
$\begin{array}\\
n
&=a^{a+c}-(a+c)^a\\
&=a^a(a^{c}-(1+c/a)^a)\\
&\gt a^a(a^{c}-e^c)\\
&\gt a^a(a^{c}-(a-2)^c)\\
&\gt 2a^ac(a-2)^{c-1}\\
&\ge 2a^a\\
\end{array}
$
This gives a
finite number of
possible values for $a$.
If
$n=1008$,
the only possibility is
$a=4$
(since
$4^4 = 256$
and
$5^5 = 3125$).
If $a=4$,
this is
$n
\gt 2\cdot 256c2^{c-1}
\ge 512c
$
so the only possibility is
$c=1$
so $b= 5$.
This does not work
(to see without computation:
$4^5-5^4$ is odd).
|
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|
A long trigonometric/hyperbolic integral Evaluate:
$$ \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}=\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx=-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
In the first integral let $x=\sin\theta$, $dx=\cos\theta d\theta$
In the second integral let $x=\sinh \varphi$, $dx=\cosh\varphi d\varphi$
$$-\frac{1}{2}\left(\int \frac{\sqrt{1-\sin^2\theta}}{\sin^2\theta}\cos\theta d\theta-\int\frac{\sqrt{\sinh^2\varphi+1}}{\sinh^2\varphi}\cosh\varphi d\varphi\right)=-\frac{1}{2}\left(\int \frac{\cos^2\theta}{\sin^2\theta}d\theta-\int\frac{{\cosh^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(\int \frac{1-\sin^2\theta}{\sin^2\theta}d\theta-\int\frac{{1+\sin^2\varphi}}{\sinh^2\varphi}d\varphi\right)=-\frac{1}{2}\left(-\cot\theta -\theta+\coth\varphi-\varphi\right)+c=\frac{1}{2}\left(\cot(\arcsin x)+\arcsin x-\coth(arcsinh x)+arcsinh x\right)+c$$
I'm not sure if it is right. Thanks in advance.
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$$I = \int \frac {dx}{\sqrt{1-x^2}+\sqrt{x^2+1}}$$
$$\int \frac{\sqrt{1-x^2}-\sqrt{x^2+1}}{-2x^2}dx$$
$$-\frac{1}{2}\left(\int \frac{\sqrt{1-x^2}}{x^2}dx-\int\frac{\sqrt{x^2+1}}{x^2}dx\right) $$
Apply integration By Parts:
$$I_1 = \int \frac{\sqrt{1-x^2}}{x^2}dx$$
$$\sqrt{1 - x^2}\left( -\frac{1}{x}\right) - \int -\frac{x}{\sqrt{1-x^2}}\left( -\frac{1}{x}\right)dx$$
$$-\frac{\sqrt{1 - x^2}}{x} - \int \frac{1}{\sqrt{1-x^2}}dx$$
$$I_1 = -\frac{\sqrt{1 - x^2}}{x} - \arcsin(x) +C$$
Apply integration By Parts:
$$I_2 = \int\frac{\sqrt{x^2+1}}{x^2}dx$$
$$\int\frac{\sqrt{x^2+1}}{x^2}dx$$
$$\sqrt{x^2+1}\left( -\frac{1}{x}\right) - \int \frac{x}{\sqrt{x^2+1}}\left( -\frac{1}{x}\right)dx$$
$$-\frac{\sqrt{x^2+1}}{x} + \int \frac{1}{\sqrt{x^2+1}}dx$$
$$I_2-\frac{\sqrt{x^2+1}}{x} + \operatorname{arcsch}(x) + C$$
Also
$$I = -\frac{1}{2}\left( \left(-\frac{\sqrt{1 - x^2}}{x} - \arcsin(x)\right) - \left( -\frac{\sqrt{x^2+1}}{x} + \operatorname{arcsch}(x) \right)\right) + C$$
$$I = -\frac{1}{2}\left( -\frac{\sqrt{1 - x^2}}{x} - \arcsin(x) + \frac{\sqrt{x^2+1}}{x} - \operatorname{arcsch}(x) \right) + C$$
And the solution coincides with WolframAlpha :D
$$\int \frac{dx}{\sqrt{1-x^2}+\sqrt{x^2+1}} dx = \frac{\sqrt{1 - x^2} - \sqrt{1 + x^2} + x \sin^{-1}(x) + x \sinh^{-1}(x))}{2 x} + C$$
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How to prove by induction that for $n$ $ \in \mathbb N $ , $ 2 n - 18 < n^2-8n +8 $? Question:
Prove for $n$ $ \in \mathbb N $ , $ 2 n\ -\ 18\ <\ n^2-8n\ +8 $
My attempt:
$ Base\ Case:\ n\ =\ 1,\ it\ holds. $
$I.H:\ Suppose\ 2k-18\ <\ k^2-8k+8,\ where\ k\ is\ a\ natural\ number.$
$ Then,\ \left(k+1\right)^2-8\left(k+1\right)+8\ =\ k^2+2k+1-8k-8+8\ >2k-18+2k+1-8$
I am stuck here. Any help would be appreciated. Thanks.
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Assuming $\quad2n-18 < n^2-8n+8\quad $ then
$\begin{array}{ll}
2(n+1)-18 = (2n-18)+2 < & (n^2-8n+8)+2\\
&=[n^2-6n+1]-2n+9\\
&=[(n+1)^2-8(n+1)+8]-(2n-9)\\
\end{array}$
So the induction works when $2n-9\ge 0$ that is when $n\ge 5$
You'll get to verify the initial step for $n=5\ :\ \begin{cases}2n-18=-8\\
n^2-8n+8=25-40+8=-7\end{cases}$
For the cases $n\le 4$ you have to verify them manually.
Or go for the algebraic solution $n^2-10n+26=(n-5)^2+1>0$.
|
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|
Determine if $\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2}$ exist What I tried:
Let $$\ f(x,y) = \frac{x^3y^4}{(x^4 + y^2)^2}$$
For points of the form$\ (x,0)$ then $\ f(x,0)=0$, similarly, for$\ (0,y)$ then $\ f(0,y)=0$, so lets suppose that:
$$\lim_{(x,y)\to(0,0)} \frac{x^3y^4}{(x^4 + y^2)^2} =0$$
So, for$\ ε>0$ if$\ δ=ε$ we have to prove that:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|<ε$$
But I'm having a hard time trying to prove the last part, I tried:
$$ |\frac{x^3y^4}{(x^4 + y^2)^2} -0|=|\frac{x^3y^4}{(x^4 + y^2)^2}| ≤ |\frac{(x^3y^4)(x^4 + y^2)^2}{(x^4 + y^2)^2}|=|(x^3y^4)|$$
|
Let's rewrite $y=ux^2$ in order to render the denominator homogeneous.
Note: $u$ is variable, write it $u=\frac y{x^2}$ if you prefer.
$\displaystyle f(x,y)=\frac{x^3u^4x^8}{(x^4+u^2x^4)^2}=\underbrace{\frac{u^4}{(1+u^2)^2}}_\text{bounded}\ \underbrace{x^3}_{\to 0}\to 0$
|
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|
Does this function have a closed form? $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$
$$k_b(x)=\sum^\infty_{n=0}{\frac{(n+x)^2}{n!}}$$
I noticed these functions closely relate to $e$. By looking at them I was able to determain a closed from for $k_b$.
$$k_b(x)=e(x^2+2x+2)$$
Though I still have not found a short form for $k_a$. I do think it exists. Some integer inputs are as follows.
$k_a(0)=2e$
$k_a(1)=e-1$
$k_a(2)=2e-5$
$k_a(3)=5e-\frac{27}{2}$
Can any of you take a stab at it?
|
This is too long for a comment.
Considering $$k_a(x)=\sum^\infty_{n=0}{\frac{n^2}{(n+x)!}}$$ and using the result from Wolfram Alpha, it seems that the expression could simplify to
$$k_a(x)=\frac{(2+x-x^2)+e\,(2-2x+x^2)\left(\Gamma (x+2)-(x+1)\, \Gamma (x+1,1) \right) }{\Gamma (x+2)}$$ which, as user404188 noticed, for integer values of $x$, reduces to $$k_a(x)=(x^2-2x+2)\,e + \mathrm{rational}$$ as shown in the table.
$$\left(
\begin{array}{cc}
x & k_a(x) \\
0 & 2 e \\
1 & -1+e \\
2 & -5+2 e \\
3 & -\frac{27}{2}+5 e \\
4 & -\frac{163}{6}+10 e \\
5 & -\frac{1109}{24}+17 e \\
6 & -\frac{2827}{40}+26 e \\
7 & -\frac{14483}{144}+37 e \\
8 & -\frac{685007}{5040}+50 e \\
9 & -\frac{2374691}{13440}+65 e \\
10 & -\frac{80885629}{362880}+82 e
\end{array}
\right)$$
The expression can rewrite $$k_a(x)=e \left(x^2-2 x+2\right)-\frac{x^2-x-2}{\Gamma (x+2)}-e \left(x^3-x^2+2\right)\frac{ \Gamma (x+1,1)}{\Gamma (x+2)}$$
|
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|
Prove: If n is a perfect square, then n+3 is not a perfect square Use either direct proof, proof by contrapositive, or proof by contradiction.
Using proof by contradiction method
Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)
There exists integers x and y such that $n = x^2$ and $n+3 = y^2$
Then $x^2 + 3 = y^2$
Then $3 = y^2 - x^2$
Then $3 = (y-x)(y+x)$
Then $y+x = 3$ and $y-x=1$
Then $x = 1, y = 2$
Since $x = 1$, that implies $n = 1$
this is how far I got
Anyone know what I should do now?
|
$$
3=y^2-x^2 \ge (x+1)^2-x^2= 2x+1
$$
which is false if $x>1$.
|
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Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+11$ . Where did I go wrong ? Thanks !!
|
Using production rule and rule for derivative of complicated function:
$$(f\cdot g)' = f'\cdot g + f \cdot g' \\ f(\varphi(x))' = f' \cdot (\varphi(x))'$$
We have:
$$(2x-1)'\cdot(x+3)^{\frac{1}{2}}+(2x-1)\cdot((x+3)^{\frac{1}{2}})'\cdot(x+3)'$$
$$(2x-1)' = 2 \\ ((x+3)^{\frac{1}{2}})' = \frac{1}{2}\cdot(x+3)^{\frac{-1}{2}} \\ (x+3)' = 1$$
You can do the rest I think :)
|
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|
Find General Solution of First Order DE.
Find general solution of
$$y'+\frac{2y}{x}=\frac{3}{x^2}$$
and find solution for which
$$y(2)=-1$$
I know
$$xy'+\frac{r(x)y}{x}=3r(x)x$$
$$(ry')=ry'+r'y$$$$r'=\frac{r}{x},\frac{r'}{r}=\frac{1}{x},ln(r)=ln(x),r=x$$$$\frac{xy'}{2}+y=\frac{3}{2x}$$
$$\frac{(xy)'}{2}=\frac{3}{2x}$$$$\frac{xy}{2}=\int \frac{3}{x}dx=3ln|x|+c$$
but where to from here?
|
It is a non-homogeneous linear ODE and IVP. Find the general solution in two steps:
1) Solve the homogeneous equation:
$$y'+\frac{2y}{x}=0 \Rightarrow\frac{1}{y}dy=-\frac{2}{x}dx \Rightarrow y=\frac{C}{x^2}.$$
2) Assume $y=\frac{C(x)}{x^2}:$
$$y'+\frac{2y}{x}=\frac{3}{x^2} \Rightarrow \frac{C'(x)x^2-2xC(x)}{x^4}+\frac{2C(x)}{x^3}=\frac{3}{x^2} \Rightarrow$$
$$C'(x)=3 \Rightarrow C(x)=3x+A$$
Thus: $y=\frac{C(x)}{x^2}=\frac{3x+A}{x^2}$ is a general solution.
To find a particular solution (that is, the value of the constant $A$), use the initial value $y(2)=-1:$
$$\frac{3\cdot 2+A}{2^2}=-1 \Rightarrow A=-10.$$
Thus: $y=\frac{3x-10}{x^2}$ is a particular solution.
|
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|
Solve this Differential Equation in y and x.
Solve the following DE:
$$y'=\frac{y+y^2}{x+x^2}$$
in particular
$$y(2)=1$$$$\frac{y'}{y+y^2}=\frac{1}{x+x^2}$$$$\int \frac{dy}{y+y^2}=\int \frac{dx}{x+x^2}$$$$y'dx=dy$$$$ln(y)-ln(y+1)=ln(x)-ln(x+1)+C$$
but where to from here?
|
This gives
\begin{align*}
\frac{y}{y+1} &=e^{C}\frac{x}{x+1}\\
2y&=\frac{e^Cx+x+1}{e^Cx-x-1}.
\end{align*}
Now use $y(2)=1$, to get $e^C$.
|
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|
How do I find the term of a recursive sequence? I have $\{a_n\}$ the following sequence:
$a_1 =-1$
$a_k a_{k+1} = - a_k - \frac{1}{4}$
How can I find $a_1 a_2 \cdots a_n$?
|
$a_{k+1} = -\frac{4a_k+1}{4a_{k}}$
$a_2 = -3/4$
$a_3 = -2/3=-4/6$
$a_4 = -5/8$
$a_5 = -3/5 = -6/10$
$a_6 = -7/12$
So, we prove that $a_n = -(n+1)/(2n)$
Base case: $a_1 = -1$
IH: $a_{k} = -(k+1)/(2k)$
Then, $a_{k+1} = -\frac{4a_k+1}{4a_{k}} = -\frac{-2(k+1)/k + 1}{-2(k+1)/k} = -\frac{-2k-2+k}{-2(k+1)} = -((k+1)+1)/(2(k+1))$
Take it from here.
|
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|
For how many integers $(3n)^4-(n-10)^4$ is a perfect square?
For how many integers $n$ the expression
$$(3n)^4-(n-10)^4$$
becomes a perfect square?
$$$$One way is: Let $x=3n$ and $y=n-10$ to get $$x^4-y^4=z^2.$$ This equation does not have non-trivial solutions in integers (I do not want to discuss about this).
I was trying to solve this problem with simpler elementary methods. For example we can write the equation as
$$40(n+5)(2n-5)(n^2-2n+10)=m^2.$$
Observe that $40 \mid m^2$. Thus $20 \mid m$. Write $m=20k$ to get
$$(n+5)(2n-5)(n^2-2n+10)=10k^2,$$
Where $k$ is a non-negative integer. Look mod $5$. We get $n^3(n-2) \equiv 0 \pmod{5}$. So we have either $n \equiv 0 \pmod{5}$ or $n \equiv 2 \pmod{5}$.
If $n \equiv 0 \pmod{5}$ then let $n=5N$ to get
$$125(N+1)(2N-1)(5N^2-2N+2)=10k^2$$
It follows that $25 \mid k^2$ and $k=5K$. Now we have the following equation $$(N+1)(2N-1)(5N^2-2N+2)=2K^2$$
Is this getting simpler or harder this way? Do you have a suggestion?
|
$$(3n)^4-(n-10)^4=m^2\Rightarrow\begin{cases} 9n^2=x^2+y^2\\n^2-20n+100=x^2-y^2\text{ or}\space 2xy\end{cases}$$ WLG we can consider $n,x,y$ without common factors then $n$ cannot be even because the well known parametrization of Pythagorean triples we used above.
$$n\text{ odd}\Rightarrow 10n^2-20n+100=2x^2\Rightarrow 5n^2-10n+50=x^2$$
Then $$5n^2-10n+50=25x_1^2\Rightarrow n^2-2n+10=5x_1^2\iff(n-1)^2+3^2=5x_1^2$$ This implies $5x_1^2=25$ which is absurde.
|
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|
If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to proceed anymore. Thanks in advance.
|
This solution builds on the idea mentioned by @Bernard in one of the comments above.
Let $u=x+\dfrac 1x$.
$$y=\frac {x^2+1}{x^2+x+1}=\frac {x+\frac 1x}{x+1+\frac 1x}=\frac u{u+1}=\frac 1{\frac 1u+1}$$
For $x>0$, we have $u>0$, and $u_{\text{min}}=2$*.
*
*Hence $y_\text{min}=\dfrac 1{\frac 1{u_{\text{min}}}+1}=\dfrac 23$
For $x<0$, we have $u<0$, and $u_{\text{max}}=-2$*.
*
*Hence $y_\text{max}=\dfrac 1{\frac 1{u_\text{max}}+1}=2$
In summary,
$$\begin{align}
\frac 23&\le &y&\le 2\\
-\frac 23&\le &y-\frac 43&\le \frac 23\\
& &\bigg| y-\frac 43\bigg|&\le \color{red}{\frac 23}
\end{align}$$
Graph below shows $y=\frac {x^2+1}{x^2+x+1}$.
*can be shown easily using simple calculus.
|
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|
how to enumerate and index partial permutations with repeats I know how to count permutations of, say, p red balls and q white balls and r blue balls. I also know how to count permutations of part of a set of distinct objects. But I don't know an efficient way to count partial permutations with repetition.
To illustrate what I mean: suppose I have a supply of eight As, seven Bs and six Cs, and want to know how many sequences of length 12 can be made from these. I can do it by adding up the permutations of 8 As and 4 Bs, 8 As and 3 Bs and one C, ... but that's tedious (this example has 45 cases). Is there a better way?
Finally, is there a procedure to generate the nth such sequence in lexicographic order, without cranking out all of them?
|
We can count the number of different sequences with the help of exponential generating functions.
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series we can calculate the wanted number of sequences as
\begin{align*}
12!&[z^{12}](1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^8}{8!})(1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\
\frac{z^7}{7!})\\
&\qquad\cdot(1+\frac{z}{1!}+\frac{z^2}{2!}+\cdots+\frac{z^6}{6!})\\
&=484110
\end{align*}
The calculation was done with the help of Wolfram Alpha.
Note: The result coincides with the calculation of @N.F.Taussig. We obtain
\begin{align*}
\binom{12}{8}2^4&+\binom{12}{7}2^5+\binom{12}{6}2^6+\binom{12}{5}\left(2^7-1\right)\\
&+\binom{12}{4}\left[2^8-2\binom{8}{8}-\binom{8}{7}\right]\\
&+\binom{12}{3}\left[2^9-2\binom{9}{9}-2\binom{9}{8}-\binom{9}{7}\right]\\
&+\binom{12}{2}\left[2^{10}-2\binom{10}{10}-2\binom{10}{9}-2\binom{10}{8}-\binom{10}{7}\right]\\
&+\binom{12}{1}\left[\binom{11}{5}+\binom{11}{6}+\binom{11}{7}\right]\\
&+\binom{12}{0}\left[\binom{12}{6}+\binom{12}{7}\right]\\
&=7920+25344+59136+100584+121770\\
&\qquad+100320+52272+15048+1716\\
&=484110
\end{align*}
|
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|
Weird differential equation How to solve the given differential equation(given $ f(1)=e$ and $f(0)=0$, $f'(0)=1$,$f''(0)=0.$)
$$f''(x)=2xf'(x)+4f(x).$$
Where $f'(x)$ is first derivative and $f''(x)$ is second derivative.
I was trying to guess that the function is exponential(of form of $e^{x^2}$) but couldn't get the final function.
|
Frobenius Method:
To solve $y''-2xy'-4y=0$ consider Frobenius solution $y = \sum_{n=0}^{\infty} a_n x^{n+r}$ hence $y' = \sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}$ and $y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}$ hence:
$$ \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}-2x\sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}-4\sum_{n=0}^{\infty} a_n x^{n+r}=0 $$
Let $k+r=n+r-2$ so $n=k+2$ and we find:
$$ \sum_{k=-2}^{\infty} (k+2+r)(k+1+r)a_{k+2} x^{k+r}-\sum_{n=0}^{\infty} 2(n+r)a_n x^{n+r}-4\sum_{n=0}^{\infty} a_n x^{n+r}=0 $$
Splitting off the $k=-2,-1$ terms we form a common sum for $n=0,1,...$
$$ r(r-1)a_0x^{r-2} + (1+r)ra_1x^{r-1} +
\sum_{n=0}^{\infty} \left( (n+2+r)(n+1+r)a_{n+2} - 2(n+r)a_n -4a_n \right)x^{n+r}=0 $$
cleaning it up a bit,
$$ r(r-1)a_0x^{r-2}+r(1+r)a_1x^{r-1}+
\sum_{n=2}^{\infty} \left( (n+2+r)(n+1+r)a_{n+2} - 2(n+r+2)a_n \right)x^{n+r}=0 $$
So, apparently, $r=0$ or $r=1$. We want $y(0)=0$ and $y'(0)=1$ and $y''(0)=0$. I'll investigate the $r=1$ case which implies $a_1=0$. Moreover, for $n=0,1,\dots $
$$ (n+3)(n+2)a_{n+2} - 2(n+3)a_n = 0 \ \ \Rightarrow \ \ a_{n+2} = \frac{2a_n}{n+2} $$
Thus, $a_1=0$ implies $a_{2k+1}=0$ for $k=1,2,\dots$. On the other hand,
$$ a_2 = \frac{2a_0}{2}=a_0, \ \ a_4 = \frac{2a_2}{4} = \frac{a_0}{2}, \ \ a_6 = \frac{2a_4}{6} = \frac{a_0}{6}, \ \ a_8 = \frac{2a_6}{8} = \frac{a_0}{24} $$
Thus,
\begin{align}
y &= a_0x+a_2x^3+a_4x^5+a_6x^7+a_8x^9+ \cdots \\ \notag
&= a_0x\left( 1+x^2+\frac{1}{2}x^4+\frac{1}{6}x^6+\frac{1}{24}x^8+ \cdots \right) \\ \notag
\end{align}
Note $y''(0)=0$ is true for this solution and you can see the term in parenthesis above is simply the power series of $e^{x^2}$ hence
$$ y = a_0xe^{x^2} $$
finally, as $y(1)=a_0(1)e = e$ we deduce $a_0=1$ and conclude $y=xe^{x^2}$.
|
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|
Find coefficient of $x^2$ in a complicated expansion
Find the coefficient of $x^2$ in the expansion of $(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$
I noticed that since we only care about the coefficient of $x^2$, only the coefficients of $x^2$ inside the square brackets, as well as the constant terms, will matter. From there I am stuck when it comes to expanding $(1+2x+3x^2)^6$ as I can't see an easy way to use the binomial theorem (for example by factorising $1+2x+3x^2$, which I can't do as there are no real roots). Is there a way to find the coefficient of $x^2$ in this expansion (specifically $(3x^2+2x+6)^6$ as I can get the answer from there)?
|
Note that$(4-x^2)[(1+2x+3x^2)^6+(1+4x^3)^5]$ is equal to
$$4[(1+2x)^6+6\cdot 3x^2+1+o(x^2)]-x^2\cdot (1+1+o(x))\\
=4[(1+6\cdot 2x+\binom{6}{2}(2x)^2+18x^2+1+o(x^2)]-2x^2+o(x^2)\\
=8+48x+310x^2+o(x^2)$$
where we used the Binomial Theorem and the little-o notation.
|
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|
Baby Rudin: Theorem 3.20 (c) and (d) In theorem 3.20, Rudin offers a proof to the limits of the following sequences:
$$u_n = n^{\frac{1}{n}}$$
$$v_n = \frac{n^{\alpha }}{(1+p)^n}$$ with $\alpha$ a real number and $p > 0$.
In both the proofs, Rudin uses inequalities which I am not familiar with. For $u_n$, the following inequality is used:
$$(1+x)^n \ge \frac{n(n-1)}{2}x^2$$
For $v_n$:
$${n \choose k}p^k > \frac{n^kp^k}{2^kk!}\quad \quad(n>2k) $$
I have painstakingly come up with proofs for both of those inequalities, however, and this is especially true for the second inequality, my proofs are not very elegant nor do they give me any idea of the inequality came to be.
Surely Rudin doesn't pull these inequalities out of thin air. Where can I find a more comprehensive derivation of these inequalities? Where do they come from exactly (which field)?
Update:
Here are my proofs. They both rely on induction.
For the first one:
Suppose $x\ge0$. Then, for $n = 2$:
$$ (1+x)^2 = 1+2x+x^2 $$
$$\frac{2(2-1)}{2} x^2 = x^2$$
Since $x\ge0$, we have $(1+x)^n \ge \frac{n(n-1)}{2}x^2 $ for n = 2.
Suppose our inequality is true for a certain n. Then:
$$ (1+x)^n + nx^2 \ge \frac{n(n+1)}{2}x^2$$
Let $D = (1+x)^{n+1} - (1+x)^n - nx^2 $. Since $\forall n \in \mathbb{N}, \forall m = 0,1, ... n, {n+1 \choose m} \ge {n \choose m} $, and ${n+1 \choose 2} - {n \choose 2} - n = 0$, we have:
$$D \ge 0 \Rightarrow (1+x)^{n+1} \ge (1+x)^n + nx^2$$
So:
$$(1+x)^{n+1} \ge \frac{n(n+1)}{2}x^2$$ The theorem follows by induction.
For the second inequality:
We want to prove that $n(n-1)...(n-k+1) > \frac{n^k}{2^k}$, with $n > 2k$
Let $n\in\mathbb{N}, k$ and integer greater than 0 such that $2n > k$. Then for $k=1$, we have: $$n > \frac{n}{2} $$
Now suppose our inequality holds for a certain $k$. Then:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k)$$
Since $n > 2k$, $n-k > \frac{n}{2} > 0$ such that:
$$n(n-1)...(n-k+1)(n-k) > \frac{n^k}{2^k}(n-k) > \frac{n^{k+1}}{2^{k+1}} $$
By induction, our inequality holds for all $k$ such that $n > 2k$
|
For the first one, assuming $x\geq0$ :
$$
\begin{align}
(1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\
&=1+nx+\frac{n(n-1)}{2}x^2+\cdots+x^n \\
&\geq \frac{n(n-1)}{2}x^2
\end{align}
$$
For the second one :
$$
\begin{align}
\binom{n}{k}&= \frac{n(n-1)\cdots(n-k+1)}{k!} \\
&=\frac{1\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)}{k!}n^k \\
&>\frac{(1/2)(1/2)\cdots(1/2)}{k!}n^k \\
&=\frac{n^k}{2^kk!}
\end{align}
$$
|
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|
$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$ I was trying to calculate the following limit:
$$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$$
and, feeding it into WolframAlpha, I found the limit is $0$.
But; the function isn't defined from the left of $3$
My knowledge is that the function must be defined in an open interval around $\{ x_0\}$ ; "neighbourhood"
So, is the answer of WolframAlpha correct, or does the limit not exist?
|
First, note $x$ must be not less than $3$ for a root to be defined, and greater than $3$ for a denominator to be nonzero (hence a function defined). So any 'open interval' you may consider must be a subset of the domain, which is, at most, $(3,\infty)$.
Then:
$$\begin{align}
\frac{x^3-27}{\sqrt{x-3}} & = \frac{(x-3)(x^2+3x+9)}{\sqrt{x-3}} \\
& = \frac{\sqrt{x-3}\,\sqrt{x-3}\,(x^2+3x+9)}{\sqrt{x-3}} \\
& = \frac{\sqrt{x-3}}{\sqrt{x-3}}\cdot\sqrt{x-3}\,(x^2+3x+9) \\
& = \sqrt{x-3}\,(x^2+3x+9)
\end{align}$$
Now the existence of a limit is quite obvious, isn't it?
|
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|
Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors. Find a basis for the subspace of $\mathbb R^4$ spanned by the 4 vectors.
$\alpha_1=(1,1,2,4),\alpha_2=(2,-1,-5,2),\alpha_3=(1,-1,-4,0),\alpha_4=(2,1,1,6)$
Check for linear dependence : $ c_1\alpha_1+c_2\alpha_2+c_3\alpha_3+c_4\alpha_4=0$
$\left[\begin{matrix}
1 & 2 & 1 & 2 & 0 \\ 1 & -1 & -1 & 1 & 0 \\ 2 & -5 & -4 & 1 & 0 \\ 4 & 2 & 0 & 6 & 0 \\
\end{matrix}\right]$ after row reduction became $\left[\begin{matrix}
1 & 0 & -1/3 & 4/3 & 0 \\ 0 & 1 & 2/3 & 1/3 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\
\end{matrix}\right]$
$\Rightarrow c_3,c_4$ can be any value, and $c_1=(1/3)c_3+(4/3)c_4, c_2=-(2/3)c_3-(1/3)c_4$, the given vectors are dependent.
But how to find basis?
$\Box\left[\begin{matrix}1/3 \\ -2/3 \\ 1 \\ 0\end{matrix}\right]+\Box \left[\begin{matrix}-4/3\\-1/3\\0\\1\end{matrix}\right]=\alpha_k, k=1,2,3,4$ is not helping
|
Operate row reduction on the transposed matrix, i.e. write the vectors as row vectors:
$$\begin{bmatrix}
1&1&2&4\\
2&-1&-5&2\\
1&-1&-4&0\\
2&1&1&6
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0&-3&-9&-6\\
0&-2&-6&-4\\
0&-1&-3&-2
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}
1&1&2&4\\
0& 1&3&2\\
0&-2&-6&-4\\
0&-1&-3&-2
\end{bmatrix}
\rightsquigarrow\begin{bmatrix}
1&1&2&4\\
0& 1&3&2\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}$$
Thus the vectors $\;\alpha_1=(1,1,2,4)$ and $\;\beta_2=(0,1,3,2)$ constitute a basis of the subspace spanned by $\;\alpha_1, \alpha_2, \alpha_3, \alpha_4$.
As $\beta_2=\alpha_1-2\alpha_2$, we might as well take $(\alpha_1, \alpha_2)$ as a basis. Or, proceeding a little further in row reduction, set $\;\beta_1=\alpha_1-\beta_2=(1,0,-1,2)$ and choose $\;(\beta_1,\beta_2)$ as a basis.
|
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|
Two unit vectors $x,y \in \mathbb{R}^n$ satisfying $x.y = \frac{\sqrt{3}}{2}$ Two unit vectors $x$ and $y$ in $\mathbb{R}^n$ satisfy x.y= $\frac{\sqrt{3}}{2}$
Evaluate
A) angle between $x$ and $y$ in Radian.
I got $\frac{\pi}{6}$
B) $y.(3x-5y)$
Looked at solving for $x$ and $y$, but think am going down completely the wrong way. In solving for $x$ and $y$ i isolated for $x$ to get $x =\frac{5y}2$
C)the length of $7x+2y$
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Well, for the first question, we have - where $\phi$ is the angle between $x$ and $y$:
$$\cos\phi=\frac{x\cdot y}{|x||y|}=\frac{\sqrt{3}}{2}\Rightarrow\phi=\frac{\pi}{6}$$
since $\phi\in[0,\pi)$
For the second question, we have:
$$y\cdot(3x-5y)=y\cdot(3x)-5y\cdot y=3x\cdot y-5y^2=2\frac{\sqrt{3}}{2}-5|y|^2=3\frac{\sqrt{3}}{2}-5$$
For the third question, we have:
$$\begin{align*}|7x+2y|^2=&(7x+2y)^2=49x^2+28x\cdot y+4y^2=49|x|^2+28\frac{\sqrt{3}}{2}+4|y|^2=\\=&49+14\sqrt{3}+4=53+14\sqrt{3}\end{align*}$$
So
$$|7x+2y|=\sqrt{53+14\sqrt{3}}$$
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"url": "https://math.stackexchange.com/questions/2348687",
"timestamp": "2023-03-29T00:00:00",
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Squares and Rationalization I was solving this question, and I'm hitting a wall.
If ${1\over{\sqrt{2011+\sqrt{2011^2-1}}}}=\sqrt{m}-\sqrt{n}$, where $m$ and $n$ are positive integers, what is the value of $m+n$?
I tried to solve this question with two approaches:
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
$={\sqrt{2011+\sqrt{2011^2-1}}\over{2011+\sqrt{2011^2-1}}}$
$={({\sqrt{2011+\sqrt{2011^2-1}})(\sqrt{2011^2-1}-2011)}\over{2011^2-2012}}$
Squaring,
$m+n-2\sqrt{mn}$
$={{{{(2011+\sqrt{2011^2-1}})\{4022(2011-\sqrt{2011^2-1})-1\}}}\over{2011^4+2012^2-2\cdot{2011^2}\cdot{2012}}}$
This goes nowhere
And
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
Squaring
${m+n-2\sqrt{mn}}$
$={{\sqrt{2011^2-1}-2011}\over{(2011)(2010)-1}}$
Which again, probably goes nowhere.
Can anyone help?
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Hint
$\dfrac1{\sqrt{2011+\sqrt{2011^2-1}}}$
$=\dfrac1{\sqrt{2011+2\sqrt{1005*1006}}}$
$=\dfrac1{\sqrt{\sqrt{1005}+\sqrt{1006}}^2}$
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"url": "https://math.stackexchange.com/questions/2352563",
"timestamp": "2023-03-29T00:00:00",
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Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not . Conclude whether the limit $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})+\sin(2x)] $ exists or not .
Answer:
Since $ \lim_{x \rightarrow \infty} [\ln(1+\frac{1}{x})]=0 , \ \ and \ \ - 1\leq \sin(2x) \leq 1 $, the given limit oscillates between $ -1 \ \ to \ \ 1 $.
So the limit does not exists .
I need confirmation about my work. Any help is there ?
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Let us consider the following two sequences:
$$a_n=n\pi+\frac{\pi}{4}$$
$$b_n=n\pi+\frac{3\pi}{4}$$
and let $f(x)=\ln\left(1+\frac{1}{x}\right)+\sin(2x)$
So, we see that:
$$\begin{align*}f(a_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+1\overset{n\to\infty}{\longrightarrow}1
\end{align*}$$
and
$$\begin{align*}f(b_n)=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(2n\pi+\frac{3\pi}{2}\right)=\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)+\sin\left(\frac{3\pi}{2}\right)=\\=&\ln\left(1+\frac{1}{n\pi+\frac{\pi}{2}}\right)-1\overset{n\to\infty}{\longrightarrow}-1
\end{align*}$$
So, the limit $\lim\limits_{x\to+\infty}f(x)$ does not exist since we have shown that $f$ has at least two accumulation points as $x\to+\infty$.
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"url": "https://math.stackexchange.com/questions/2353103",
"timestamp": "2023-03-29T00:00:00",
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Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?
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Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.
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"url": "https://math.stackexchange.com/questions/2354401",
"timestamp": "2023-03-29T00:00:00",
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Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but clearly this will not work because for $n=3$ we have $3\cdot2^3+1=3\cdot8+1=24+1=25$ which is clearly not divisible by $3$.
Any ideas how to evaluate a formula that generates all the $n$'s for which $n2^n+1$ is divisible by $3$?
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For any natural number $ n $ we have that $ n \equiv i $ mod $ 3 $ where $ i \in \{0,1,2\} $. So let us consider each case.
If $ n \equiv 0 $ mod $ 3 $ then clearly $ n2^{n}+1 \equiv 1 $ mod $ 3 $ so this is not a valid case.
Suppose now that $ n \equiv 1 $ mod $ 3 $ and write $ n=3k+1 $. Then $$ n2^{n}+1=(3k+1)2^{3k+1}+1 \equiv 2^{3k+1}+1 (mod 3) \equiv 2 \cdot 8^{k}+1 (mod 3 ) \equiv 2 \cdot (-1)^{k}+1 (mod 3) $$
Therefore $ k $ must be even in this case so $ k=2l $ for some integer $ l $ thus $ n=6l+1 $.
Finally, if $ n \equiv 2 (mod 3) $, write $ n=3k+2 $. Then we have $$ n2^{n}+1=(3k+2)\cdot 2^{3k+2}+1 \equiv 2 \cdot 2^{3k+2}+1 (mod 3)\equiv 8^{k+1}+1 (mod 3) \equiv (-1)^{k+1}+1 (mod 3) $$
Hence in this case, we must again have that $ k $ is even i.e. $ k=2l $ for some integer $ l $ so $ n=6l+2 $.
Therefore we conclude that the only positive integers $ n $ satisfying the condition in the hypothesis are those $ n's $ congruent to $ 1 $ or to $ 2 $ modulo $ 6 $.
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