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Help with C is Euler's constant and $\Gamma(0)=\infty$ in paper I am referring to a paper by S. Nadarajah & S. Kotz.
The notation is simple enough to understand, however i having trouble with $C$ is Euler’s constant and $\Gamma(0)=\infty$
by equation (2.3) I have
$$F(z)= \displaystyle\lambda\int_{0}^{\infty}y^{-1-1}exp\left(-\frac{\lambda}{y}\right)erfc\left(-\frac{z}{\sqrt{2}\sigma} y\right)~dy - 1$$
by lemma 1(defined $\displaystyle p>0 , \alpha < 0 , \arrowvert \text{arg}(c) \arrowvert < \frac{\pi}{4}$) , then
\begin{align}
\displaystyle F(z) &= \lambda\left[\frac{1}{\lambda} + \frac{2z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) + \frac{z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},\frac{1}{2},1;-\frac{\lambda^2z^2}{8\sigma^2}\right) \right. \\
& \hspace{10mm} \left. + \frac{z^2 \lambda}{4\sqrt{\pi}\sigma^2}\Gamma \left(-\frac{1}{2}\right) G\left(1;2,\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) \right]-1
\end{align}
note that $\Gamma(-\frac{1}{2})=-2\sqrt{\pi}$ and $\Gamma(0)=\infty$
so $~~~~~~~~\displaystyle F(z) = \frac{\lambda z}{\sqrt{2}\sigma}\left[\frac{3\Gamma(0)}{\sqrt{\pi}}G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) - \frac{\lambda z}{\sqrt{2}\sigma}G\left(1;2,
\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right)\right] $
but the equation is not equal to equation(2.1)
I would be really happy if someone could help me.help please
|
Notes:
*
*The paper referenced is unclear about the parameters $p$ and how $z$ is defined. It would seem that somehow $p \sim \Gamma(0)$. This would work if another parameter is defined, say $\alpha$, such that $p = \alpha \, \Gamma(0) \to \beta$ where $\beta$ is a finite quantity.
*It is of interest to note that
\begin{align}
I &= \int_{0}^{\infty} e^{-\frac{p}{x}} \, erfc(c x) \, \frac{dx}{x^{2}} \\
&= \frac{1}{p} - \frac{3 \, C \, \Gamma(0)}{\sqrt{\pi}} \, G\left(\frac{1}{2}; \frac{3}{2}, 1, \frac{1}{2}; - \frac{c^{2} p^{2}}{4} \right) - p \, C^{2} \, G\left(1; 2, \frac{3}{2}, \frac{3}{2}; - \frac{c^{2} p^{2}}{4} \right)
\end{align}
and upon making the change of variable $u = p/x$ then performing integration by parts, then making the change of variable $t = (cp)/u$ the integral becomes
\begin{align}
I = \frac{2 \, c^{2} \, p^{2}}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- t^{2} - \frac{c p}{t}} \, \frac{dt}{t}
\end{align}
Let
\begin{align}
I &= \int_{0}^{\infty} e^{-\frac{p}{x}} \, erfc(c x) \, \frac{dx}{x^{2}} \\
\end{align}
and make the change $u = \frac{p}{x}$ to obtain
$$I = \int_{0}^{\infty} e^{-u} \, erfc\left(\frac{c p}{u}\right) \, du.$$
Now, by integration by parts, where
$$\partial_{u} \, ercf\left(\frac{c p}{u}\right) = \frac{2 c p}{\sqrt{\pi} \, u} \, e^{- \frac{c^{2} p^{2}}{u^{2}}},$$
the integral becomes
$$I = \left[ - e^{-u} \, erfc\left(\frac{c p}{u}\right) \right]_{0}^{\infty} + \frac{2 c p}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- \frac{c^{2}p^{2}}{u^{2}} - u} \, \frac{du}{u}.$$
Since $erfc(\infty) = 0$ then the non-integral term vanishes and upon making the change $x = \frac{c p}{u}$ the resulting integral becomes
$$I = \frac{2 c^{2} p^{2}}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- x^{2} - \frac{c p}{x}} \, \frac{dx}{x}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating limit-sequences So i have this limit to calculate:
$\lim_{n\to\infty}\frac{[x] + [2x]+ ... +[nx]}{n^2}$
And i tried to make some boundaries and got this two limits:
$\lim_{n\to\infty}\frac{[x] + [x]+ ... +[x]}{n^2}$
$\lim_{n\to\infty}\frac{[nx] + [nx]+ ... +[nx]}{n^2}$
But i am not sure if it's correct,while the first limit is 0,and the second one is not.
And also those [ ] are floor functions.
Any help would be appreciated.
|
$$x-1\leqslant \left \lfloor x \right \rfloor< x\\2x-1\leqslant \left \lfloor 2x \right \rfloor< 2x\\3x-1\leqslant \left \lfloor 3x \right \rfloor< 3x\\...\\\\nx-1\leqslant \left \lfloor nx \right \rfloor< nx$$
now sum of them
$$(x+2x+3x+..+nx)-(1+1+1..1)\leq \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor< (x+2x+3x+..+nx)\\ \frac{n(n+1)}{2}x -n\leq \left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor< \frac{n(n+1)}{2}x $$
now divide by $n^2$
$$\frac{n(n+1)}{2n^2}x -\frac{n}{n^2}\leq \frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2}< \frac{n(n+1)}{2n^2}x$$
now apply limit ,by squeeze theorem
$$\lim_{n \to \infty} \frac{n(n+1)}{2n^2}x -\frac{n}{n^2}\leq \lim_{n \to \infty}\frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2}< \lim_{n \to \infty}\frac{n(n+1)}{2n^2}x\\ \frac{1}{2}x \leq \lim_{n \to \infty}\frac{(\left \lfloor x \right \rfloor+\left \lfloor 2x \right \rfloor+...+\left \lfloor nx \right \rfloor)}{n^2} <\frac{1}{2} x$$ so the limit =$\frac{1}{2}x$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $a_{i}$ and subsequently substitute them into the integral.
It can be observed that for finding this constants we may use following trick.
Suppose we have to find $a_1$ then for time being remove the term beneath $a_1$ from the relation $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$ now put the value of $x$ that would appear if the removed term is equated to $0$ in $\frac{x}{(x-1)(2x+1)(x+3)}$. This the value of constant $a_1$.
How does this works ?
|
Notice, the partial fractions $$\frac{x}{(x-1)(2x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{2x+1}+\frac{C}{x+3}$$ On solving for $A, B, C$, we get $A=\frac{1}{12}$, $B=\frac{2}{15}$ & $C=-\frac{3}{20}$
$$=\frac{1}{12(x-1)}+\frac{2}{15(2x+1)}-\frac{3}{20(x+3)}$$
Hence, we have $$\int \frac{x}{(x-1)(2x+1)(x+3)}=\int\frac{1}{12(x-1)}+\frac{2}{15(2x+1)}-\frac{3}{20(x+3)}dx=$$ $$=\frac{1}{12}\ln(x-1)+\frac{1}{15}\ln(2x+1)-\frac{3}{20}\ln(x+3)+c$$
|
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|
Finding $\iint_S {z \:ds}$ for some $S$ $$\iint_S {z \:ds}$$
In this double integral above, $S$ is the part of a sphere, $x^2+y^2+z^2=1$, which lies above the cone, $z=\sqrt{x^2+y^2}$. How can I calculate the above double integral.
Can someone help me to solve this? I got $\frac{\pi}{3\sqrt{2}}$ as my answer. Can someone verify this
|
This problem is actually best done in cylindrical coordinates. Here when we let $x = r\cos\theta$ and $y = r\sin\theta$ we can rewrite our $z$-limits as $z = r$ and $r^2 +z^2 = 1$. Note that when $z =1$ we have $r = 0$, and similarly when $z = r$ we get that $r^2 + r^2 = 1$ implying that $r = \frac{1}{\sqrt{2}}$. Therefore our integral can be written
\begin{eqnarray*}
\int_0^{2\pi} d\theta \int_0^{1/\sqrt{2}} \int_r^{\sqrt{1-r^2}}rzdzdr & = & 2\pi\int_0^{1/\sqrt{2}}\frac{r}{2}(1-r^2 - r^2)dr \\
& = & \pi\int_0^{1/\sqrt{2}}r - 2r^3 dr \\
& = & \pi\left (\frac{1}{2}\cdot \frac{1}{2} - \frac{1}{2}\cdot \frac{1}{4} \right )\\
& = & \pi\left (\frac{2}{8} - \frac{1}{8} \right ) \\
& = & \frac{\pi}{8}.
\end{eqnarray*}
|
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|
Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-8r^2+8=0.\tag{$\star$}$$
The rational zeros theorem tells us that the only possible rational solutions to ($\star$) are $±1$, $±2$, $±4$, $±8$.
I do not know where to go from here. Please help me complete this proof.
|
\begin{align*}r=\sqrt{6+4\sqrt 2}-\sqrt 2&\Rightarrow r^2+2r\sqrt 2+2=6+4\sqrt2\iff r^2-4=2\sqrt 2(2-r)\\
&\Rightarrow(r^2-4)^2=8(r-2)^2\iff(r-2)^2((r+2)^2-8)=0\\
\end{align*}
The only rational root of this polynomial is $r=2$. We have to explain why $r$ can't be one of the irrational roots $\;-2(1\pm\sqrt2)$.
Anyway it can't be the negative one: $r>0$ since $6+4\sqrt 2>2$. The other root is ${}<1$ since $(1+2)^2-8>0$. However, it's easy to see that $r>1$ since $\sqrt{6+4\sqrt 2}-\sqrt 2>\sqrt 9-\sqrt 2>3-2$.
Consequently $r$ has to be equal to $2$.
Note: It's a standard exercise in high school to observe that
$$\sqrt{6+4\sqrt 2}=\sqrt{(2+\sqrt2)^2}=2+\sqrt 2,\enspace\text{whence}\quad r=2+\sqrt 2-\sqrt 2.$$
|
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|
Solution to equation $4n+1 = (2k+1)^2-(2u)^2$ over the natural numbers(!) Let $n \in \mathbb{N}$ be given, then I want to know if there are $k ,u \in \mathbb{N}$ (with $k \neq n, u \neq n$) such that $4n+1 = (2k+1)^2-(2u)^2$ holds.
What makes it difficult for me is the fact that I am looking for solutions over the natural numbers. Otherwise one would see that there are infinitely many solutions (for example by completing the square).
|
Given
$$
4 n + 1 = \Big( 2 k + 1 \Big)^2 - \Big( 2 u \Big)^2.
$$
Set
$$
k = u + a,
$$
then we get
$$
n = a^2 + a + \big( 2 a + 1 ) u.
$$
The case $a=0$ is the solution
$$
n = u = k,
$$
which is to be excluded.
For every $a > 0$ we have a linear relation between $n$ and $u$.
$$
\begin{array}{l|ll}
a & n\\
\hline
1 & 2 + 3u\\
2 & 6 + 5u\\
3 & 12 + 7u\\
4 & 20 + 9u & = 2 + 9(u + 2)\\
5 & 30 + 11u\\
6 & 42 + 13u\\
7 & 56 + 15u & = 2 + 3 ( 5 u + 18 ) = 6 + 5(3 u + 10)\\
\vdots & \vdots
\end{array}
$$
The equation is not general solvable for a given $n$.
In case $n$ is even, we can find solutions for:
$$
n = 8, 14, 16, 20, 26 (3\times),32, 36, 38 (2\times),40, 44, 46, 50, 52, 54, 56 (3\times),62,\\
66, 68 (2\times),74 (2\times),76, 82, 86 (2\times),92, 94, 96 (3\times), \cdots
$$
For example
$$
\begin{eqnarray}
4 \times 26 + 1 &=& \big( 2 \times 9 + 1 \big)^2 - \big( 2 \times 8 \big)^2\\
&=& \big( 2 \times 6 + 1 \big)^2 - \big( 2 \times 4 \big)^2\\
&=& \big( 2 \times 5 + 1 \big)^2 - \big( 2 \times 2 \big)^2
\end{eqnarray}
$$
|
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|
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
No induction is necessary: it's a matter of a telescoping sum, if you write:
$$\frac1{k(k+2)}=\frac12\Bigl(\frac 1k-\frac1{k+2}\Bigr)$$
Apply this decomposition to the above sum:
\begin{align*}
\sum_{k=1}^n\frac1{k(k+2)}&=\frac12\sum_{k=1}^n\Bigl(\frac 1k-\frac1{k+2}\Bigr)\\
&=\frac12\Bigl(1\color{red}{-\frac13}+\frac12\color{red}{-\frac14+\frac13-\frac15+\dots+\frac1{n-1}}-\frac1{n+1}\color{red}{+\frac 1n}-\frac1{n+2}\Bigr)\\
&=\frac12\Bigl(1+\frac12-\frac1{n+1}-\frac1{n+2}\Bigr)=\frac34-\frac{n+2+n+1}{2(n+1)(n+2)}.
\end{align*}
|
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|
Rewriting approximated terms The following data are inferred from a presentation slide, so I do not much info.
Using linear approximation and log rules $\sqrt x $ can be rewritten as $\frac{x+1}{2}$, where $(1 \leq x \lt 2) $ ... (1)
I understood the Eq. (1).
Now the slide says, in the range $(2 \leq x \lt 4)$,
$ \sqrt x = 2^{\frac{log_2(x)}{2}} \approx 2^{\frac{(1+x-1)}{2}}$ ... (2)
Could some one tell me, how and why do we get the term `$x$' approximated to $ (1+x-1) $ in Eq. (2) ?
Note! I am aware that using linear approximation $log_2(x) \approx x-1 $
Thank You.
|
Probably there is a typo in
$ \sqrt x \approx 2^{\frac{(1+x-1)}{2}}$
because it is not a correct approximate in the range $2 \leq x < 4$
Case of $x=2 $ then $ \sqrt 2 \approx 2^{\frac{(1+2-1)}{2}}=2$ is false.
Case of $x=4 $ then $ \sqrt 4 \approx 2^{\frac{(1+4-1)}{2}}=4$ is false.
If we use the Taylor's series :
$$f(x)=f(x_1)+f'(x_1)(x-x_1)+...$$
with $x_1=2$ and $f(x)=\sqrt{x}$ hense $f'(x)=\frac{1}{2\sqrt{x}}$
we obtain :
$$f(x)\approx \sqrt{2}+\frac{1}{2\sqrt{2}}(x-2)$$
Case of $x=2 $ then $ \sqrt 2 \approx \sqrt{2}+\frac{1}{2\sqrt{2}}(2-2)=\sqrt{2}$ is OK.
Case of $x=4 $ then $ \sqrt 4 \approx \sqrt{2}+\frac{1}{2\sqrt{2}}(4-2)=\frac{3\sqrt{2}}{2} \simeq 2.12$ is OK.
Another kind of approximate on the form $\sqrt{x} \approx 2^{ax+b}$ is identifed so that $2^{2a+b}=\sqrt{2}$ and $ 2^{4a+b}=2$ leading to $a=\frac{1}{4}$ and $b=0$ :
$$\sqrt{x}\approx 2^{x/4}$$
on the range$2 \leq x < 4$
but the preceeding approximate $f(x)\approx \sqrt{2}+\frac{1}{2\sqrt{2}}(x-2)$ is better.
Of course, even better approximates can be obtained in considering more terms of the Taylor's series.
|
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|
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also $a^3+b^3=(a+b)^3-3ab(a+b)$. But both ways aren't working.
Please explain how do I solve these types of questions analytically.
|
Given $$x^3+y^3=72$$
$$xy=8$$
Notice,
$$(x+y)^3=x^3+y^3+3xy(x+y)$$ $$(x+y)^3=72+3(8)(x+y)$$
$$(x+y)^3-24(x+y)-72=0$$
Above is the cubic equation in terms of $(x+y)$ which has one real root $6$ Hence, we get $$x+y=6$$
Now, $$(x-y)^2=(x+y)^2-4xy$$
$$(x-y)=\pm\sqrt{(x+y)^2-4xy}$$
$$x-y=\pm\sqrt{(6)^2-4\times8 }$$
$$=\pm\sqrt{4}=\pm 2$$
Hence, we have
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x-y=\pm 2}}$$
|
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|
Solutions for differential equation The motion of a particle moving along the x-axis obeys the differential equation:
$ \ddot x - 4 \dot x + 4x =-t^2 $
Find the solution for $ x(t)$, given $ x(0)=0 $ and $ \dot x (0) = 0 $.
Can this be solved by treating it as a second-order non-homogenous differential equation? If so, re-formatting the questions equation, would it be:
$ \frac{d^2x}{dt^2} - 4 \frac{dx}{dt}+4x=-t^2$ ?
|
Using constant coefficient method we solve for two solutions $x_h$ and $x_p$ where the general solution is $x_g=x_h+x_p$
Solving for $x_h$ implies that we solve the homogeneous system $x'' -4x'+ 4x=0$ which has the characteristic equation
$$\lambda^2 -4\lambda +4 = 0 \implies ( \lambda-2)^2=0 \implies \lambda_{1,2}=2 $$
So, $$x_h(t)=c_1e^{2t}+c_2te^{2t}$$
Solving for $x_p$ we use the method of undetermined coefficients and we let $$x_p(t)= \alpha t^2+\beta t +\omega $$ and we solve the original DE.
We have:
$$x_p'=2\alpha t+\beta$$ $$x''_p=2\alpha$$
PLugging that into the non homogeneous DE:
$$ 2\alpha - 4( 2\alpha t+\beta ) + 4(\alpha t^2+\beta t +\omega) = -t^2$$
$$ (4\alpha)t^2 + (-8\alpha+4\beta) t + (2\alpha - 4\beta+4\omega) = -t^2$$
$$ \implies 4\alpha = -1 \implies \alpha= \frac{-1}{4}$$$$ \implies-8\alpha+4\beta=0 \implies \beta = \frac{-1}{2} $$$$ \implies 2\alpha - 4\beta+4\omega=0 \implies \omega= \frac{-3}{8}$$
So in we have $$ x_p(t)= \frac{-1}{4}t^2 - \frac{1}{2}t -\frac{3}{8} $$
So the general solution is:
$$x_g(t)= x_h + x_p = c_1e^{2t}+c_2te^{2t} + \frac{-1}{4}t^2 - \frac{1}{2}t -\frac{3}{8} $$
Finally, given the initial conditions $x(0)=0$ and $x'(0)= 0 $, we find that the constants are $c_1=3$ and$ c_2=-2$ to get the unique solution:
$$ x(t)= 3e^{2t}-2te^{2t} + \frac{-1}{4}t^2 - \frac{1}{2}t -\frac{3}{8} $$
|
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|
Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$ factors and is
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\dots \left(1-\frac{i-1}{n}\right)\left(1-\frac{i^2}{n}\right).$$
It seems, amazingly, that the answer is 0. How can one show this?
|
This might get you started:
$$\begin{align}
&\left(1-{1^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2^2\over n}\right)+\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots\\
&\quad=\left(1-{1\over n}\right)\left(1+\left(1-{2^2\over n}\right)+\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(\left(2-{2^2\over n}\right)+\left(1-{2\over n}\right)\left(1-{3^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(2+\left(1-{3^2\over n}\right)+\left(1-{3\over n}\right)\left(1-{4^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(\left(3-{3^2\over n}\right)+\left(1-{3\over n}\right)\left(1-{4^2\over n}\right)+\cdots \right)\\
&\quad=\left(1-{1\over n}\right)\left(1-{2\over n}\right)\left(1-{3\over n}\right)\left(3+\left(1-{4^2\over n}\right)+\left(1-{4\over n}\right)\left(1-{5^2\over n}\right)+\cdots \right)\\
\end{align}$$
|
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"url": "https://math.stackexchange.com/questions/1394580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Proof of the identity: $c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}$ Trigs is not my strongest apparently...
I need to prove $c\sin \frac{A-B}{2} = (a-b) \cos \frac{C}{2}$ for a general triangle $ABC$.
Here is what I do, or rather, here is how I fail at proving it:
$\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}$, so $\displaystyle{\frac{\sin \frac{A-B}{2}}{\sin \frac{A+B}{2}} \equiv \displaystyle{\frac{a-b}{c}}}$.
This implies: $\displaystyle{\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \equiv \frac{a-b}{c}}$.
Now, imagine graphing an angle bisector from angle $A$ and then from angle $B$, the point where they intersect, let's call it $K$. From that point drop a perpendicular on $AB$, let's call that point $L$. Hence, $\tan \frac{A}{2} = \frac{KL}{AL}$ and $\tan \frac{B}{2} = \frac{KL}{LB}$. Plugging those in, gives us: $$\frac{LB-AL}{c} \equiv \frac{a-b}{c}$$ And now I have no clue how to show that $LB-AL = a-b$.
If you could let me know how to show that and/or you know a better way of proving the identity, please share.
|
Using sine law, $$\dfrac{a-b}c=\dfrac{\sin A-\sin B}{\sin C}$$
Using Prosthaphaeresis & Double Angle Formula,
$$\dfrac{\sin A-\sin B}{\sin C}=\dfrac{2\sin\dfrac{A-B}2\cos\dfrac{A+B}2}{2\sin\dfrac C2\cos\dfrac C2}$$
Now $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2\implies\cos\dfrac{A+B}2=?$
Finally, if $\sin\dfrac C2=0,\dfrac C2=n\pi\iff C=2n\pi$ where $n$ is any integer
But $0<C<\pi\implies\sin\dfrac C2\ne0$
|
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|
Calculating $\sum_{n=1}^{\infty}\ln^2 \!\left(1+\frac1{2n}\right) \!\ln^2\!\left(1+\frac1{2n+1}\right)$ Based upon Oloa's question here Evaluating $\sum_{n \geq 1}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$ I was thinking if we possibly can get a nice way to evaluate the series
$$\sum_{n=1}^{\infty}\ln^2 \!\left(1+\frac1{2n}\right) \!\ln^2\!\left(1+\frac1{2n+1}\right).$$
Maybe using the same telescoping idea or it simply doesn't work? What then?
|
It can be shown that:
\begin{align}
\sum_{n=1}^{\infty} \ln^{2}\left(1+\frac{1}{2n}\right) \, \ln^{2}\left(1 + \frac{1}{2n+1}\right) = - \frac{\ln^{4} 2}{2} + 2 \, \sum_{n=1}^{\infty} \ln^{2}\left(1 + \frac{1}{n}\right) \, \ln\left(1+\frac{1}{2n}\right) \, \ln\left(1 + \frac{1}{2n+1}\right).
\end{align}
The remaining summation may be presentable in a form not based on numerical estimation. In the event a few decimal place accuracy is sought the following identity is presented. Accurate to 9 decimal places the series is given by
\begin{align}
\sum_{n=1}^{\infty} \ln^{2}\left(1+\frac{1}{2n}\right) \, \ln^{2}\left(1 + \frac{1}{2n+1}\right) = \left( \frac{1}{2} - \frac{1}{5^{\frac{32}{37}} \, \gamma} \right) \, \ln^{4}2 \approx 0.01600318562\cdots
\end{align}
where $\gamma$ is the Euler-Mascheroni constant.
Proof:
Let
$$f(x) = \ln\left(1 + \frac{1}{x}\right) \tag{1}$$
then
$$f^{2}(n) = ( f(2n) + f(2n+1) )^{2} = f^{2}(2n) + f^{2}(2n+1) + 2 \, f(2n) \, f(2n+1) \tag{2}$$
which yields
$$f(2n) \, f(2n+1) = \frac{1}{2} \, [ f^{2}(n) - f^{2}(2n) - f^{2}(2n+1) ]. \tag{3}$$
Squaring both sides provides
\begin{align}
& f^{2}(2n) \, f^{2}(2n+1) \\
& \hspace{5mm} = \frac{1}{4} \, [ f^{4}(n) + f^{4}(2n) + f^{4}(2n+1) - 2 \, f^{2}(n) \, ( f^{2}(2n) + f^{2}(2n+1) ) + 2 \, f^{2}(2n) \, f^{2}(2n+1) ]
\end{align}
or
$$f^{2}(2n) \, f^{2}(2n+1) = \frac{1}{2} \, [ f^{4}(n) + f^{4}(2n) + f^{4}(2n+1) - 2 \, f^{2}(n) \, ( f^{2}(2n) + f^{2}(2n+1) ) ] \tag{4}$$
By making use of (1) this reduces to
$$f^{2}(2n) \, f^{2}(2n+1) = - \frac{1}{2} \, [f^{4}(n) - f^{4}(2n) - f^{4}(2n+1) ] + 2 \, f^{2}(n) \, f(2n) \, f(2n+1). \tag{5}$$
Now, summing over the index then it is seen that:
\begin{align}
S &= \sum_{n=1}^{\infty} f^{2}(2n) \, f^{2}(2n+1) \\
&= - \frac{1}{2} \, \sum_{n=1}^{\infty} \left[ f^{4}(n) - f^{4}(2n) - f^{4}(2n+1) \right] + 2 \, \sum_{n=1}^{\infty} f^{2}(n) \, f(2n) \, f(2n+1) \\
&= - \frac{1}{2} \, f^{4}(1) + + 2 \, \sum_{n=1}^{\infty} f^{2}(n) \, f(2n) \, f(2n+1) \tag{6}
\end{align}
From (6) the first statement presented is obtained. As to the remaining summation it is conjectured that it is also of the form $A_{0} \, \ln^{4}2$.
|
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|
Derivation of Series Expansions
I have the series expansion for
$$I=\frac{1}{\sqrt{1-x^2}}$$
$$|x|\lt1$$
which is
$$1+\sum_{k=1}^\infty\frac{1.3.5.7......(2k-1)x^{2k}}{2^kk!}$$
or so I am assured. Given the above what is the series expansion for
$$\int_0^x\frac{dt}{\sqrt{1-t^2}}$$
can I simply substitute $k+1$ for $k$ and integrate the $x$ terms in the expansion above? I hope I am not completely off-course. I have made a stab in the dark which is
$$\sin^{-1}{x}=\sum_{k=0}^\infty{\frac{x^{1+2k}\frac{1}{2}k}{k!+2kk!}}$$
|
You need to use the following theorem:
A power series can be integrated term by term in the interior of region of convergence.
If we restrict to real variables then a power series of the from $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \cdots = \sum_{n = 0}^{\infty}a_{n}x^{n}\tag{1}$$ has region of convergence as an interval of type $(-R, R)$ so that the series converges for $|x| < R$. It may also converge for $x = \pm R$ but that is not important here.
The above theorem says that we can integrate the equation $(1)$ and get $$\int_{0}^{x}f(t)\,dt = a_{0}x + a_{1}\frac{x^{2}}{2} + a_{2}\frac{x^{3}}{3} + \cdots = \sum_{n = 1}^{\infty}\frac{a_{n - 1}}{n}x^{n}\tag{2}$$ and this is valid for all $x \in (-R, R)$ (or $|x| < R$).
In your case we have $$\frac{1}{\sqrt{1 - x^{2}}} = 1 + \sum_{n = 1}^{\infty} \frac{1\cdot 3\cdot 5\cdots (2n - 1) x^{2n}}{2^{n}n!}$$ which is valid for $|x| < 1$ and we can integrate it to get the series for $\sin^{1}x$ which will also be valid for $|x| < 1$.
However there is catch. Unless you are aware of the theorem (by "awareness" I mean "a complete understanding of its proof") mentioned in beginning of this answer, you should not use the this technique. Rather there are special mechanisms for each specific series.
In your case of $\sin^{-1}x$ Hardy shows us the way in his book "A Course of Pure Mathematics" on page $339$ where he gives a reduction formula for integral $$\int_{0}^{x}\sin^{2n - 1}t\,dt$$ in the form $$1 = \cos x + \frac{1}{2}\cos x\sin^{2}x + \cdots + \frac{1\cdot 3\cdots (2n - 3)}{2\cdot 4\cdots (2n - 2)}\cos x\sin^{2n - 2}x + r_{n}\tag{3}$$ where $$r_{n} = \frac{3\cdot 5\cdots (2n - 1)}{2\cdot 4\cdots (2n - 2)}\int_{0}^{x}\sin^{2n - 1}t\,dt$$ and integrating equation $(3)$ we get $$\alpha = \sin \alpha + \frac{1}{2}\cdot\frac{\sin^{3}\alpha}{3} + \cdots + \frac{1\cdot 3\cdots (2n - 3)}{2\cdot 4\cdots (2n - 2)}\cdot\frac{\sin^{2n - 1}\alpha}{2n - 1} + R_{n}\tag{4}$$ where $$R_{n} = \int_{0}^{\alpha}r_{n}\,dx = \frac{3\cdot 5\cdots (2n - 1)}{2\cdot 4\cdots (2n - 2)}\int_{0}^{\alpha}(\alpha - x)\sin^{2n - 1}x\,dx\tag{5}$$ from which we get $$0 \leq R_{n} \leq \frac{1\cdot 3\cdots (2n - 1)}{2\cdot 4\cdots (2n)}\alpha\sin^{2n}\alpha\tag{6}$$ for $0 \leq \alpha \leq \dfrac{\pi}{2}$. Letting $n \to \infty$ in $(4)$ and noting than $R_{n} \to 0$ as $n \to \infty$ it follows that (on putting $\sin \alpha = u$) $$\sin^{-1}u = u + \frac{1}{2}\frac{u^{3}}{3} + \frac{1\cdot 3}{2\cdot 4}\frac{u^{5}}{5} + \cdots\tag{7}$$ where $0 \leq u \leq 1$. Since both sides are odd functions of $u$ the formula also holds for $-1 \leq u \leq 0$ or thus for all $|u| \leq 1$.
|
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|
How to prove the trigonometric identity $\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$ I am doing some practice questions for a Math class and I was told that similar questions would be in the exam. So I need to learn this but I have no idea where to even start with this question:
$$\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$$
Hint: use standard factorization for the difference of 2 cubes, e.g. $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
Help please I find that looking at the working when someone does these questions allows me to learn the method. Thanks in advance.
|
Start from the left-hand side:
\begin{align}
\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1
&=\frac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}
+\frac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}-1\\
&=\frac{\cos^2x}{\sin x(\cos x-\sin x)}
-\frac{\sin^2x}{\cos x(\cos x-\sin x)}-1\\
&=\frac{\cos^3x-\sin^3x}{\sin x\cos x(\cos x-\sin x)}-1\\
&=\frac{\cos^2x+\sin x\cos x+\sin^2x}{\sin x\cos x}-1\\
&=\frac{1}{\sin x\cos x}
\end{align}
|
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|
Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ to be possible to find an inverse of $15 \pmod{88}$.
\begin{align*}
88 & = 5 \times 15 + 13\\
15 & = 1 \times 13 + 2\\
13 & = 6 \times 2 + 1\\
2 & = 2 \times 1 + 0
\end{align*}
So,
$$\gcd(88, 15) = 1$$
Now, we need to write this into the form:
$$\gcd(88, 15) = 88x + 15y.$$
And find $x$ and $y$.
\begin{align*}
1 & = 13(1) + 2(-6)\\
& = 13(7) + 15(-6)\\
& = 88(7) + 15(-41)
\end{align*}
So, $x = 7$ and $y = -41$.
So, an inverse of $15 \pmod{88} = -41$. Now, I need to find an inverse that is between $0$ and $87$. What is a good easy approach to find other inverses? Any ideas please?
|
$x$ is an inverse of $15 \pmod{88}$ if and only if $x=88n-41$ for some (any) integer $n$, because it is necessary and sufficient that $15(-41+x)$ is divisible by $88$, but this is equivalent to $(-41+x)$ being divisible by $88$.
|
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|
Deducing $\sum_{r=1}^{n}r$ from sine summation formula We know the famous formula
$$\sum_{r=1}^{n}\sin r\theta=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}\ .$$
I have come across a question that use the above result to find $\sum_{r=1}^{n}r$. I thought but could not deduce how this can be done.Is there a way to find $\sum_{r=1}^{n}r$ from the given sine summation series. Thanks.
|
$\sum_{r=1}^{n}\sin r\theta
=\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}
$
My first thought on seeing this
is to somehow use
$\lim_{x \to 0} \frac{\sin x}{x}
= 1
$.
Let's divide both sides by
$\theta$:
$\sum_{r=1}^{n}\frac{\sin r\theta}{\theta}
=\frac{\sin \frac{n\theta}{2}\csc\frac{\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta}
=\frac{\sin \frac{n\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta\sin\frac{\theta}{2}}
$.
The left side,
as $\theta \to 0$,
becomes
$\sum_{r=1}^{n} r
$,
since
$\frac{\sin r\theta}{\theta}
=r\frac{\sin r\theta}{r\theta}
$.
This is a good sign.
The right side is
$\begin{array}\\
\frac{\sin \frac{n\theta}{2}\sin\frac{(n+1)\theta}{2}}{\theta\sin\frac{\theta}{2}}
&=\frac{n\theta}{2}\frac{\sin \frac{n\theta}{2}}{\frac{n}{2}\theta}
\frac{(n+1)\theta}{2}\frac{\sin \frac{(n+1)\theta}{2}}{\frac{n+1}{2}\theta}
\frac1{\theta\sin\frac{\theta}{2}}\\
&=\frac{n}{2}\frac{\sin \frac{n\theta}{2}}{\frac{n}{2}\theta}
\frac{(n+1)\theta}{2\sin\frac{\theta}{2}}\frac{\sin \frac{(n+1)\theta}{2}}{\frac{n+1}{2}\theta}\\
&\to \frac{n(n+1)}{2}
\quad\text{ as }\theta \to 0 \text{ (repeatedly using } \lim_{x \to 0} \frac{\sin x}{x}
= 1)\\
\end{array}
$
|
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|
Evaluation of trignometric limit I want to find the following limit without L'Hospital.
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.
|
As the OP suggested openness to asymptotic analysis, we proceed accordingly.
First, we note the expansion of the numerator is given by
$$\begin{align}
1+\tan^{1/3} x&=1+\left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\
&=\frac23 (x-3\pi/4)+O(x-3\pi/4)^2 \tag 1
\end{align}$$
whereas the expansion of the denominator is given by
and
$$\begin{align}
1-2\cos^2x&=-\cos 2x\\\\
&=-2(x-3\pi/4)+O(x-3\pi/4)^3 \tag 2
\end{align}$$
Thus, putting $(1)$ and $(2)$ together yields
$$\begin{align}
\frac{1+\tan^{1/3} x}{1-2\cos^2x}&=\frac{\frac23 (x-3\pi/4)+O(x-3\pi/4)^2}{-2(x-3\pi/4)+O(x-3\pi/4)^3}\\\\
&=-\frac13+O(x-3\pi/4)\\\\
&\to -\frac13\,\,\text{as}\,\,x\to 3\pi/4
\end{align}$$
Therefore, we have that the limit of interest is
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 3\pi/4}\frac{1+\tan^{1/3} x}{1-2\cos^2x}=-\frac13}$$
NOTES:
To arrive at $(1)$, we first note that for $f(x)=\tan x$, $f'(x)=\sec^2 x$ and $f''(x)=2\sec^2 x\tan x$.
Therefore, $f(3\pi/4)=-1$ and $f'(3\pi/4)=\frac12$ so that
$$\tan x=-1+2(x-3\pi/4)+O(x-3\pi/4)^2$$
Then, using the expansion for $(-1+x)^{1/3}=-1+\frac13x+O(x^2)$, we see that
$$\begin{align}
\tan^{1/3} x&= \left(-1+2(x-3\pi/4)+O(x-3\pi/4)^2\right)^{1/3}\\\\
&=-1+\frac23(x-3\pi/4)+O(x-3\pi/4)^2
\end{align}$$
To arrive at $(2)$, we first note the trigonometric identity $\cos 2x=2\cos^2-1$.
Then, for $g(x)=1-2\cos^2 x=-\cos 2x$, we have $g'(x)=2\sin 2x$, $g''(x)=4\cos 2x$, and $g'''(x)=-8 \sin 2x$.
Therefore, $g(3\pi/4)=0$ and $g'(3\pi/4)=-2$, and $g''(3\pi/4)=0$ so that
$$-\cos 2x=-2(x-3\pi/4)+O(x-3\pi/4)^3$$
|
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|
If $\frac{dy}{dx}\frac{dx}{dy} = 1$, does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? I know $\frac{dy}{dx}\frac{dx}{dy} = 1$ because the chain rule says $1 = \frac{dy}{dy} = \frac{dy}{dx}\frac{dx}{dy}$. But does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? Or would that be too good to be true?
|
What is true is that (assuming always that $y$ is a one-to-one function of $x$ on some interval)
$$ \eqalign{\dfrac{d^2 x}{dy^2} &= \dfrac{d}{dy} \dfrac{dx}{dy} = \dfrac{dx}{dy} \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right)^{-1}\cr
&= - \left(\dfrac{dy}{dx}\right)^{-3} \dfrac{d^2 y}{dx^2} }$$
In order to have $\dfrac{d^2 x}{dy^2} \dfrac{d^2 y}{dx^2} = 1$, you'd need
$$ \left(\dfrac {d^2 y}{dx^2}\right)^2 = - \left( \dfrac{dy}{dx}\right)^3$$
The solutions of this differential equation include
$$ y = \dfrac{4}{a+x} + b$$
for constants $a,b$.
|
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|
Solve $\cos \frac{4x}{3}=\cos x+1$
Solve the equation \begin{equation} \cos \frac{4x}{3}=\cos x+1\tag 1\end{equation}
I had tried by taking $\cos\dfrac x3=t$ and from this we have $\displaystyle\cos\frac{4x}3=2\left(2t^2-1\right)^2-1; \cos x=4t^3-3t$
$(1) \iff t\left(8t^3-4t^2-8t+3\right)=0$
But I can't solve $8t^3-4t^2-8t+3=0$ because it gives me the approximate roots when I need exact roots
|
HINT:
Notice, we have $$\cos \frac{4x}{3}=\cos x+1$$ Let, $x=3t$ , we get
$$\cos 4t=\cos 3t+1$$ $$2\cos^2 2t-1=4\cos^3 t-3\cos t+1$$
$$2(2\cos^2 t-1)^2-4\cos^3 t+3\cos t-2=0$$
$$8\cos^4 t-8\cos^2t+2-4\cos^3 t+3\cos t-2=0$$
$$\cos t(8\cos^3 t-4\cos^2t+2-8\cos t+3)=0$$
$$\iff \cos t=0\iff t=\frac{\pi}{2}\iff x=\frac{3\pi}{2}$$
$$\iff 8\cos^3 t-4\cos^2t+2-8\cos t+3=0$$
Can you solve this cubic equation for $\cos t$?
|
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|
Find the number of natural number solutions of $a+2b+c=100$
Find the number of natural number solutions of $a+2b+c=100$
I remember something like stars and bars if the equation I change to $a+b_{1}+b_{2}+c=100$
then i get $\dbinom{99}{3}$ ways.
If the equation is like $a_{1}+a_{2}+a_{3}+a_{4}=m$ I can use $\dbinom{m-1}{3}$ ways
how ever I am not familier with variation in the problem.
I look for a short and simple way.
I have studied maths upto $12$th grade. thanks.
|
I will assume that by the natural numbers that you mean the positive integers (as opposed to the nonnegative integers).
Observe that $a + c = 100 - 2b$ is an even number. Thus, $a$ and $c$ must have the same parity.
Case 1: The numbers $a$ and $c$ are both even. Let $a = 2u$; let $c = 2v$. Then $u, v \in \mathbb{N}$. Moreover,
\begin{align*}
a + 2b + c & = 100\\
2u + 2b + 2v & = 100\\
u + b + v & = 50 \tag{1}
\end{align*}
which is an equation in positive integers. The number of solutions of equation 1 is the number of ways we can place two addition signs in the $49$ spaces between consecutive ones in a row of $50$ ones, which is $\binom{49}{2}$.
Case 2: The numbers $a$ and $c$ are both odd. Let $a = 2s - 1$; let $b = 2t - 1$. Then $s, t \in \mathbb{N}$. Moreover,
\begin{align*}
a + 2b + c & = 100\\
2s - 1 + 2b + 2t - 1 & = 100\\
2s + 2b + 2t & = 102\\
s + b + t & = 51 \tag{2}
\end{align*}
which is an equation in positive integers. The number of solutions of equation 2 is the number of ways two addition signs can be placed in the $50$ spaces that appear between consecutive ones in a row of $51$ ones, which is $\binom{50}{2}$.
Thus, the number of solutions of the equation $a + 2b + c = 100$ in the positive integers is
$$\binom{49}{2} + \binom{50}{2} = 1176 + 1225 = 2401$$
|
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|
Show using logarithms that the first equation can be transformed into the second. Show using logarithms that if $y^k = (1-k)zx^k(a)^{-1}$ then $y = (1-k)^{(1/k)}z^{(1/k)}x(a)^{(-1/k)}$.
|
Take logarithm of both sides of the equation $\,y^k = \left(1-k\right)\,z\,x^k\,\left(a\right)^{-1},\,$ get
$$
\begin{aligned}
y^k = \left(1-k\right)\,z\,x^k\,a^{-1}
&\implies
\ln\left( y^k \right) = \ln\left(\left(1-k\right)\,z\,x^k\,\left(a\right)^{-1} \right)
\\
&\implies
k\ln y = \ln\left(1-k \right) + \ln z + \ln\left(x^k \right) -\ln a
\\
&\implies
k\ln y = \ln\left(1-k \right) + \ln z + k\ln x -\ln a
\\
&\implies
\ln y = \frac{1}{k}\ln\left(1-k \right) + \frac{1}{k}\ln z + \frac{k}{k}\ln x-\frac{1}{k}\ln a
\\
&\implies
\ln y = \ln\left(\left(1-k \right)^\frac{1}{k}\right) + \ln \left(z^\frac{1}{k}\right) + \ln x -\ln \left(a^\frac{1}{k}\right)
\\
&\implies
\ln y = \ln\left(\left(1-k \right)^\frac{1}{k} \,z^\frac{1}{k}\, x \,a^\frac{-1}{k}\right)
\\
&\implies
y = \left(1-k \right)^\frac{1}{k} \,z^\frac{1}{k}\, x \,\left(a\right)^\frac{-1}{k}
\end{aligned}
$$
Q.E.D.
|
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|
Solve $\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$ for $a/b$ The question:
$$\log_9 (a) + \log_{12} (b) = \log_{16} (a+b)$$
solve for $a/b$.
It gives hints:
put it all in terms of x.
$$9^x=a$$
$$12^x=b$$
$$16^x=a+b$$
Now prove that:
$b^2=a(a+b)$ I did and it does equal.
Then it tells me to divide both sides of $b^2=a(a+b)$ by $b^2$. Here is what I got:
$$\frac{b^2}{b^2} = \frac{a(a+b)}{b^2}$$
$$1 = \frac{a}{b}*\frac{a+b}{b}$$
$$\frac{1}{a+b}*\frac{1}{b}=\frac{a}{b}$$
$$\frac{1}{b(a+b)}=\frac{a}{b}$$
$$\frac{1}{12^x16^x}=\frac{9^x}{12^x}$$
$$\frac{1}{192^x}=\frac{3^x}{4^x}$$
$$1=\frac{576^x}{4^x}$$
$$1=144^x$$
$$x=\log_{144}(1)=\frac{\log(1)}{\log(144)}=0$$
Therefore:
$$\frac{a}{b}=\frac{9^x}{12^x}=\frac{9^0}{12^0}=\frac{1}{1}=1$$
so $a/b = 1$ is what I got. But I looked this question up on the internet and apparently it has to be $a/b=$the golden ratio (there was a hint at the end of the revision sheet saying $a/b$ and sunflower). But I can't figure it out. I always end up with $a/b = 1$ whatever I do. Help please.
|
Hint : You have the following equality
$$1 = \frac{a}{b} \times \frac{a+b}{b}, $$
and observe that $\frac{a+b}{b} = \frac{a}{b}+\frac{b}{b} = \frac{a}{b} + 1$, now we define $Y=\frac{a}{b}$, to get that $Y$ is a solution of
$$1 = Y\times(Y+1)\qquad \Leftrightarrow\qquad Y^2+Y-1=0. $$
|
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|
Proving that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator
Prove that $e^{\pi}-{\pi}^e\lt 1$ without using a calculator.
I did in the following way. Are there other ways?
Proof : Let $f(x)=e\pi\frac{\ln x}{x}$. Then,
$$e^{\pi}-{\pi}^e=e^{f(e)}-{e}^{f(\pi)}\tag1$$
Now,
$$f'(x)=\frac{e\pi(1-\ln x)}{x^2},\quad f''(x)=\frac{e\pi (2\ln x-3)}{x^3},\quad f'''(x)=\frac{e\pi (11-6\ln x)}{x^4}.$$
Since $f'(x)\lt 0$ for $e\lt x\lt\pi$, one has $f(e)\gt f(\pi)$. By Taylor's theorem, there exists a point $c$ in $(e,\pi)$ such that
$$\begin{align}f(\pi)&=f(e)+(\pi-e)f'(e)+\frac{(\pi-e)^2}{2}f''(c)\\&\gt f(e)+(\pi-e)\cdot 0+\frac{(\pi-e)^2}{2}\cdot \frac{e\pi(2\ln e-3)}{e^3}\\&=f(e)-\frac{\pi(\pi-e)^2}{2e^2}\tag2\end{align}$$
because $f'''(x)\gt 0\ (e\lt x\lt \pi)$ implies $f''(c)\gt f''(e)$.
By the mean value theorem and $(2)$,
$$e^{f(e)}-e^{f(\pi)}\lt (f(e)-f(\pi))e^{f(e)}\lt \frac{\pi (\pi-e)^2}{2e^2}\cdot e^{\pi}=\frac{e\pi(\pi-e)^2}{2}\cdot e^{\pi-3}\tag3$$
Since $e^x\lt \frac{1}{1-x}\ (0\lt x\lt 1)$ and $0\lt \pi-3\lt 1$,
$$e^{\pi-3}\lt\frac{1}{4-\pi}\tag4$$
From $(1)(3)(4)$,
$$e^{\pi}-{\pi}^e\lt \frac{e\pi(\pi-e)^2}{2}\cdot\frac{1}{4-\pi}\lt\frac{3\times\frac{22}{7}\left(\frac{22}{7}-2.718\right)^2}{2}\cdot\frac{1}{4-\frac{22}{7}}\lt 0.993\lt 1$$
|
Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx
f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
$$
where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.
Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$. Then we have taken off too much slack:
$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$
Using the continued fractions of $e$ and $\pi $ does give less than one and check with a calculator:
$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$
Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.
How good is this approximation? Somehow we should compute how quickly $f$ changes with $x$ and $y$:
$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$
The $y$ partial derivative is similar, except it is positive instead of negative.
It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:
$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and } \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$
These would be our estimates of $\epsilon_1$ and $\epsilon_2$. Then using a tiny bit of multivariable calculus:
$$ \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
\approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$
This should be enough to establish $|f(\pi,e)| < 1$.
This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.
|
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|
What is the probability that a natural number is a sum of two squares? Some natural numbers can be expressed as a sum of two squares:
$$2=1^2+1^2$$
$$25=3^2+4^2$$
$$50=7^2+1^2$$
If one chooses a random natural number, what would be the probability that that number is a sum of two squares? Is it zero?
I read about Lagrange´s theorem on squares, but it looks it can´t be useful here.
NOTE 1: "Square" means "square of a natural number".
NOTE 2: I am aware that the expression "random natural number" is not a strict math notion. However, as I said in a comment, one can adopt a reasonable strict definition, which is not difficult to devise at all. It is mentioned also in an answer below.
NOTE 3: A related question on SE: How to determine whether a number can be written as a sum of two squares?
|
Consider the function $r: \mathbb{N} \to \mathbb{C}$:
$$
r(n) = \begin{cases}
1 &n \text{ is the sum of two squares (0 is a square)} \\
0 &\text{otherwise}.
\end{cases}
$$
One way to define the "probability that a random integer is the sum of two squares" would be to consider the distribution on the integers where $n$ selected with probability proportional to $n^x$, for $x > 1$, and then to take the limit as $x \to 1$. That is, we can try to compute:
$$
\lim_{x \to 1} \frac{\sum_{i=1}^\infty \frac{r(n)}{n^x}}{\sum_{i=1}^\infty \frac{1}{n^x}} \tag{1}.
$$
As mentioned here, $r(n)$ is $1$ iff every prime of the form $4k-1$ occurs an even number of times in $n$.
It follows that $r(n)$ is multiplicative, and
\begin{align*}
\frac{\sum_{i=1}^\infty \frac{r(n)}{n^x}}{\sum_{i=1}^\infty \frac{1}{n^x}}
&= \frac{\prod_{p \text{ prime}} \sum_{i=0}^\infty \frac{r(p^i)}{p^{ix}}}{\prod_{p \text{ prime}} \sum_{i=0}^\infty \frac{1}{p^{ix}}} \\
&= \prod_{p \text{ prime}} \frac{1 + \frac{r(p)}{p^x} + \frac{r(p^2)}{p^{2x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \\
&= \prod_{p \equiv 3 \pmod{4}} \frac{1 + 0 + \frac{1}{p^{2x}} + 0 + \frac{1}{p^{4x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots}
\prod_{p \equiv 1,2 \pmod{4}} \frac{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \\
&= \prod_{p \equiv 3 \pmod{4}} \frac{1 + \frac{1}{p^{2x}} + \frac{1}{p^{4x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \\
&= \prod_{p \equiv 3 \pmod{4}} \frac{1 \big/ \left(1 - \tfrac{1}{p^{2x}}\right)}{1 \big/ \left(1 - \tfrac{1}{p^{x}}\right)} \\
&= \prod_{p \equiv 3 \pmod{4}} \frac{1}{1 + p^{-x}}. \\
\end{align*}
Now if you plug in $x = 1$, you get the product
$$
\prod_{p \equiv 3 \pmod{4}} \frac{p}{p + 1}
= \prod_{p \equiv 3 \pmod{4}} \left( 1 - \frac{1}{p + 1} \right)
= 0,
$$
by the reasoning of Hagen von Eitzen.
So by this definition of probability, the probability that a random integer is a sum of two squares is zero.
|
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|
If $a,b,c,d,e,f$ are non negative real numbers such that $a+b+c+d+e+f=1$, then find maximum value of $ab+bc+cd+de+ef$ $(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$
$ab+bc+cd+de+ef \le Y$ since all are positive.
Therefore $1\ge X+(ab+bc+cd+de+ef)$
Edit: From AM GM inequality, $$X\ge 6(abcde)^{1/3}$$and for am-gm inequality for a,b,c,d,e,f: $$1\ge 6(abcdef)^{1/6}$$
Hence $X\ge 6.\frac{1}{36}$ and $X\ge \frac{1}{6}$.
Is this correct?
|
I will give you an example with 4 variables to show how to solve such problems:
$a+b+c+d=1,b=1-(a+c)-d, ab+bc+cd=b(a+c)+cd=(1-(a+c))(a+c)-d(a+c)+cd=(1-(a+c))(a+c)-ad \le \left(\dfrac{(1-(a+c))+(a+c)}{2}\right)^2-ad \le \dfrac{1}{4}$
when $d=0,a+c=\dfrac{1}{2},b=\dfrac{1}{2}$ you will get maximum value.
|
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|
Why doesn't quadratic formula lead to a the correct factored form of the original equation? Applying the quadratic formula to $2x^2-3x+1$ we have
\begin{eqnarray*}
a&=&2 \\
b&=&-3 \\
c&=&1
\end{eqnarray*}
which gives me two roots:
\begin{eqnarray*}
x_1&=&1 \\
x_2&=&\tfrac{1}{2}
\end{eqnarray*}
Therefore you can re-write the original quadratic as:
$$(x-1)\left(x-\tfrac{1}{2}\right)$$
However, if you actually multiply this out then you get:
$x^2-\frac{3}{2}x + \frac{1}{2}$ which is not the same as $2x^2-3x+1$.
So why the discrepancy? Aren't you supposed to get the original equation?
|
How can $x=\frac{1}{2}$ and $x=1$ both be solutions to $x^2 - \tfrac{3}{2}x+\tfrac{1}{2} = 0$ and $2x^2-3x+1=0$?
The simple fact is that $2x^2-3x+1 \equiv 2\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so
$2x^2-3x+1 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.
By a similar argument: $4x^2-6x+2 \equiv 4\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so
$4x^2-6x+2 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.
In general, for any non-zero number $k$, we have
$ax^2+bx+c=0$ if, and only if, $(ka)x^2+(kb)x+(kc)=0$. Why not try to prove this?!
Even more generally, for any non-zero number $k$: $\mathrm{f}(x)=0$ if, and only if, $k\cdot \mathrm{f}(x)=0$.
|
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|
How do I solve $x^4-3x^2+2=0$? How do I solve $x^4-3x^2+2=0$ ?
I would appreciate some kind of hint here. I have no clue how to start this problem.
|
We can complete squares as follows:
\begin{align*}
x^4-3x^2+2&=0\\
x^4-3x^2&=-2\\
x^4-3x^2+\color{green}{\frac{9}{4}}&=-2+\color{green}{\frac{9}{4}}\\
\left(x^2-\frac{3}{2}\right)^2&=\frac{1}{4}\\
\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}&=0\\
\end{align*}
Now we can factor this difference of squares:
\begin{align*}
\left(x^2-\frac{3}{2}\right)^2-\frac{1}{4}&=0\\
\left(x^2-\frac{3}{2}+\frac{1}{2}\right)\left(x^2-\frac{3}{2}-\frac{1}{2}\right)&=0\\
\left(x^2-1\right)\left(x^2-2\right)&=0\\
(x+1)(x-1)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)&=0\\
\end{align*}
Hence, the solutions are $\pm 1$ and $\pm\sqrt{2}$.
|
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|
Integrate $\tan^2(\frac{\pi}{12} \cdot y)$ Integrate $\tan^2(\frac{\pi}{12} \cdot y)$
Wolfram gives the answer:
$$\frac{12 \tan(\frac{\pi y}{12})}{\pi} -y + \text{constant}$$
But why is the value of $\tan^2$ not getting differentiated?
According to this rule, the answer should be:
$$\frac{\tan(\frac{\pi}{12} \cdot y)}{\frac{\pi}{12} \cdot (2-1)}$$
($\dfrac{\pi}{12}$ is a constant so it's not differentiated)
Also, what is meant by aX? In my case, what's my "aX"? $\frac{\pi}{12} \cdot y$? So does the "a" simply means what's before the variable y? $\frac{\pi}{12}$?
|
Let $\frac{\pi}{12}\cdot y=t\implies \frac{\pi}{12}dy=dt\iff dy=\frac{12}{\pi}dt$
We have $$\int \tan^2\left(\frac{\pi}{12}\cdot y\right)dy=\int \tan^2 (t) \frac{12}{\pi}dt$$ $$=\frac{12}{\pi}\int \tan^2(t)dt$$
$$=\frac{12}{\pi}\int (\sec^2(t)-1)dt$$
$$=\frac{12}{\pi}\int\sec^2(t)dt-\frac{12}{\pi}\int dt$$
$$=\frac{12}{\pi}\tan(t)-\frac{12}{\pi}\cdot t+C$$
Setting $t=\frac{\pi}{12}\cdot y$
$$=\frac{12}{\pi}\tan\left(\frac{\pi}{12}\cdot y\right)-\frac{12}{\pi}\cdot \frac{\pi}{12}\cdot y+C$$
$$=\color{red}{\frac{12}{\pi}\tan\left(\frac{\pi}{12}\cdot y\right)-y+C}$$
|
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|
Finding a matrix that satisfy a givven inner product
Let $V=M_2(\mathbb{R})$
with $\left\langle A,B\right\rangle:=\text{tr}\left(AB^*\right)$ and let $A=\left(\begin{array}{cc} 1 & -3 \\ -2 & 2 \end{array}\right)$ find $B\in M_2(\mathbb{R})$ such that $B$ is orthonormal and orthogonal to $A$
So we need the both $||B||=1$ and $\text{tr}\left(\left(\begin{array}{cc} 1 & -3 \\ -2 & 2 \end{array}\right)\left(\begin{array}{cc} a & c \\ b & d \end{array}\right)\right)=0$
So we have $\sqrt{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{cc} a & c \\ b & d \end{array}\right)}=1$
$a^2+b^2+c^2+d^2=1$ and $a-3b-2c+2d=0$ how should I continue?
|
$||B||=1\implies a^2+b^2+c^2+d^2=1$. $A\perp B=0\implies a-3b-2c+2d=0$. Pick $a,b,c,d$ in such a way that $a=\sqrt{0.5}\cos\phi,b=\sqrt{0.5}\sin\phi,c=\sqrt{0.5}\cos\theta,d=\sqrt{0.5}\sin\theta$ so that the $1$-st equality is taken care of. If we pick $\theta,\phi$ in such a way that $a=3b$ and $c=d$ we're done. $\phi=\arctan\left({3^{-1}}\right)$ and $\theta={\pi\over4}$ do the job.
|
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|
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$.
I tried to solve this question. The lines are $x-2y+3=0$ and $x+2y+3=0$ which intersect at $(-3,0)$ and the circle has its center at $(0,0)$ and radius $\sqrt5$. The figure was asymmetric and calculations were difficult. So I shifted the origin to $(-3,0)$. In the new system the equations become $X+2Y=0$ and $X-2Y=0$ and the circle equation becomes $(X-3)^2+Y^2=5$. Now I found the point of intersection of lines and the circle which comes out to be $(\frac{4}{5},\frac{2}{5})$
Now when I integrate and required area is 4 times the integral $\int_{0}^{4/5}\frac{x}{2}dx+\int_{4/5}^{\sqrt5}\sqrt{5-(x-3)^2}dx$
but the answer $5(\pi-\arcsin\frac{4}{5})+\frac{24}{5}$ is still elusive and the calculations have not simplified even after origin shifting. Is the origin shifting futile. What is the correct way to solve it? Please help me.
|
The line of equation $x-2y+3=0$ intersects the circle of equation $x^2+y^2=5$ at
$A(-\frac{11}{5},-\frac{2}{5})$ and $B(1,2)$. It follows that
$AB=\frac{8}{\sqrt{5}}$.
$\qquad\qquad\qquad$
Using the notation of the figure we have $\sin\theta=\frac{AB}{2OA}=\frac{4}{5}$. That is $\theta=\arcsin\frac{4}{5}$. Moreover, $OH^2=5-\frac{16}{5}=\frac{9}{5}$, or $OH=\frac{3}{\sqrt5}$.
Now, $${\cal S}=\frac{\theta}{2}OA^2-\frac{1}{2}AB\cdot OH
=\frac{5}{2}\arcsin\frac{4}{5}-\frac{4}{\sqrt5}\frac{3}{\sqrt5}=
\frac{5}{2}\arcsin\frac{4}{5}-\frac{12}{5}$$
Finally the desired area is
$${\frak A}=\pi(\sqrt5)^2-2{\cal S}=5\left(\pi-
\arcsin\frac{4}{5}\right)+\frac{24}{5}.$$
Which is the right answer.
|
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|
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
One root is $x=1$ (checking functions $\arccos$ and $\mathrm{sgn}$).
Second root is $x=0.442$.
How to find the second root?
How to check if this equation only has two roots?
|
Taking $x=\tan \theta$ and noting that $2x/(1+x^2)=\sin 2\theta$ will simplify the equation to $$\pi/2-2\theta+\sin 2\theta\cdot\mathrm{sgn}(\tan^2\theta-1)=0\\\implies f(\theta)+g(\theta)=0$$ where $$f(\theta)=\pi/2-2\theta,\ g(\theta)=\sin 2\theta\cdot \mathrm{sgn}(\tan^2\theta-1)=\left\{\begin{array}
&
\sin 2\theta & \mathrm{if}\ \theta\ge \pi/4\\
-\sin 2\theta & \mathrm{if}\ \theta< \pi/4
\end{array}
\right.$$ And then we can graphically find the solution by plotting the two functions.
|
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|
Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible? Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?
What is the best/simplest/elementary way to approach this? Any solutions or hints are greatly appreciated.
|
Because the modulus is so small, we can do this the hard way.
All arithmetic will be done in $\mathbb Z_7.$
$p(x)=6x^4+3x^3+6x^2+2x+5$
$-p(x) = x^4 + 4x^3 + x^2 + 5x + 2$
$-p(x-1) = x^4 + 2x^2 + 4x + 2$
Now we suppose that
\begin{align}
-p(x-1)
&= (x^2 + ax + b)(x^2 + cx + d)\\
&= x^4 + (a+c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd
\end{align}
We end up with the equations
\begin{align}
a+c &= 0\\
ac + b + d &= 2\\
ad+bc &= 4\\
bd &= 2
\end{align}
Iterating through $b$, we finally get $(a,b,c,d) = (4,1,3,2)$.
So
\begin{align}
-p(x-1) &= (x^2 + 4x + 1)(x^2 + 3x + 2)\\
p(x-1) &= -(x^2 + 4x + 1)(x^2 + 3x + 2)\\
p(x) &= -(x^2 + 5x + 3)(x^2 + 6x + 3)\\
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving an integral with substitution method I want a hint for the following integral:
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$$
where $a$ is a real constant.
In fact,
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}} = -\frac{\sqrt{1- a^{2}+x^{2}}}{x(1- a^{2})}$$
why?
|
Here's one approach:
$$ \frac{1}{x^2 \sqrt{1-a^2 + x^2}} = \frac{1}{x^3\sqrt{\frac{x^2-a^2+1}{x^2}}}$$
$$ =\frac{1}{x^3\sqrt{\frac{1-a^2}{x^2} + 1}} $$
Let $\theta = \frac{1-a^2}{x^2} + 1$
$$ \frac{d\theta}{dx} = \frac{-2(1-a^2)}{x^3} \ \implies \ dx = \frac{-x^3}{2(1-a^2)} \ d\theta $$
The integral then reduces to:
$$ \frac{-1}{2(1-a^2)}\int \frac{1}{\sqrt{\theta}}\ \ d\theta = \frac{-1}{2(1-a^2)} 2 \sqrt{\theta} + c= -\frac{\sqrt{1-a^2+x^2}}{x(1-a^2)} + c $$
$$ = \frac{\sqrt{1-a^2+x^2}}{x(a^2-1)} + c$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of ways to select $2$ similar and $3$ different letter = $\dbinom{4}{1}\times \dbinom{4}{3}=16$.
Number of ways of selecting $2$ similar + $2$ more similar letter and $1$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and $2$ different letter = $\dbinom{4}{2}\times \dbinom{3}{1}=18$.
Number of ways to select $3$ similar and another $2$ other similar = $\dbinom{3}{1}\times \dbinom{3}{1}=9$
Number of ways to select $4$ similar and $1$ different letter = $\dbinom{2}{1}\times \dbinom{4}{1}=8$
Ways of selecting
$5$ similar letters = $1$
Total ways = $1+16+18+18+9+8+1= 71$
Well I have the solution But I am not able to fully understand it.
Or if their could be an $\color{red}{\text{alternate way}}$ than it would be great.
I have studied maths up to $12$th grade.
|
If Order is Unimportant
The number of ways to choose $5$ letters (if their order is unimportant) is the coefficient of $x^5$ in
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{x^2}}^{1\text{ E}}
\overbrace{\left(1+x+x^2\right)}^{2\text{ D's}}
\overbrace{\left(1+x+x^2+x^3\right)}^{3\text{ C's}}
\overbrace{\left(1+x+x^2+x^3+x^4\right)}^{4\text{ B's}}
\overbrace{\left(1+x+x^2+x^3+x^4+x^5\right)}^{5\text{ A's}}\\
&=\frac{1-x^2}{1-x}\frac{1-x^3}{1-x}\frac{1-x^4}{1-x}\frac{1-x^5}{1-x}\frac{1-x^6}{1-x}\\
&=\frac{1-x^2-x^3-x^4+O\left(x^7\right)}{(1-x)^5}\\[3pt]
&=\small\left[1-x^2-x^3-x^4+O\!\left(x^7\right)\right]\!\left[1+5x+15x^2+35x^3+70x^4+126x^5+210x^6+O\!\left(x^7\right)\right]\\[9pt]
&=1+5x+14x^2+29x^3+49x^4+71x^5+90x^6+O\!\left(x^7\right)\tag{1}
\end{align}
$$
where we used the Binomial Theorem for $(1-x)^{-5}$ above.
The coefficient of $x^5$ in $(1)$ is $71$.
If Order is Important
If the order of the letters is important, we can compute the exponential generating function with
$$
\begin{align}
&\small\overbrace{(1+x)\vphantom{\frac{x^2}{2!}}}^{1\text{ E}}
\overbrace{\left(1{+}x{+}\frac{x^2}{2!}\right)}^{2\text{ D's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}\right)}^{3\text{ C's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3}{+}\frac{x^4}{4!}\right)}^{4\text{ B's}}
\overbrace{\!\left(1{+}x{+}\frac{x^2}{2!}{+}\frac{x^3}{3!}{+}\frac{x^4}{4!}{+}\frac{x^5}{5!}\right)}^{5\text{ A's}}\\[3pt]
&=\small1+5x+24\frac{x^2}{2!}+111\frac{x^3}{3!}+494\frac{x^4}{4!}+2111\frac{x^5}{5!}+8634\frac{x^6}{6!}+O\left(x^7\right)\tag{2}
\end{align}
$$
The coefficient of $\frac{x^5}{5!}$ in $(2)$ is $2111$.
|
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|
How to solve this pde equation: $(p^2 + q^2)y = qz$ My attempt at solution
$p^2 y = qz - q^2 = a... (I)$
This equation is of the form $f_1(x,p) = f_2(y,q)$.
Its solution is given by $dz=pdx + qdy$, upon integrating this we get value of $z$.
From (I)
$-yq_2 + zq - a = 0$, solving the quadratic equation for $q$, we get
$q=\frac{-z±\sqrt{z^2-4ay}}{-2y}$.
Taking the positive value only, $q=\frac{-z+\sqrt{z^2-4ay}}{-2y}$ .
Also, from (I), $p^2 y = a$, therefore $p=\frac{\sqrt{a}}{y}$.
Therfore $dz=\frac{\sqrt{a}{y}} dx + \frac{-z+\sqrt{z^2-4ay}}{-2y}dy$ .
I can't get any further. I know, how to solve by charpit's method, but my book mentions that I solve it without using charpit.
The answer is given: $z^2=(cx+a)^2 + c^2 y^2$ .
|
$let\ f(x,y,z,p,q)=(p^2+q^2)y-qz=0 \hspace{1cm} \to (1)$
Now,
$f_x=0$
$f_y=p^2+q^2$
$f_z=-q$
$f_p=2py$
$f_q=2qy-z$
The auxillary equations of the given DE are
$\dfrac{dp}{f_x+pf_z}=\dfrac{dq}{f_y+q.f_z}=\dfrac{dz}{-p.f_p-q.f_q}=\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}$
$\dfrac{dp}{-pq}=\dfrac{dq}{p^2+q^2-q^2}=\dfrac{dz}{-p(2py)-q(2qy-z)}=\dfrac{dx}{-2py}=\dfrac{dy}{z-2qy}\hspace{1cm} \to (2)$
From (2),
$\dfrac{dp}{-pq}=\dfrac{dq}{p^2}$
$-pdp=qdq $
On Integration,
$ - \dfrac{p^2}{2}=\dfrac{q^2}{2}+a^2$
$p^2+q^2=a^2\ (say)$
(1) becomes,
$a^2y=qz$
$q=\dfrac{a^2y}{z}$
$p=\dfrac{a}{2}\sqrt{z^2-a^2y^2}$
Since $dz=pdx+qdy$
$dz=\dfrac{a}{z}\sqrt{z^2-a^2y^2\ }dx+\dfrac{a^2y}{z}dy$
$zdz-a^2ydy=a\sqrt{z^2-a^2y^2}dx$
$\dfrac{zdz-a^2ydy}{\sqrt{z^2-a^2y^2}}=adx$
$\dfrac{d(z^2-a^2y^2)}{2\sqrt{z^2-a^2y^2}}=adx$
$\sqrt{z^2-a^2y^2}=ax+b$
$\implies z^2-a^2y^2=(ax+b)^2$
Answer is:
$\implies z^2-a^2y^2=(ax+b)^2$
|
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|
Simplifying $\tan100^{\circ}+4\sin100^{\circ}$ The answer is $-\sqrt3$.
I was wondering if this is just a coincidence?
Also, is there a relation between $$\tan(100^{\circ}+20^{\circ})=\frac{\tan100^{\circ}+\tan20^{\circ}}{1-\tan100^{\circ}.\tan20^{\circ}}=-\sqrt3$$ and the given expression? Or is there a more elegant method of solving the question?
|
Notice, $$\tan 100^\circ+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ}{\cos 100^\circ}+4\sin 100^\circ$$
$$=\frac{\sin 100^\circ+4\sin 100^\circ\cos 100^\circ}{\cos 100^\circ}$$
$$=\frac{\sin 100^\circ+2\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{(\sin 200^\circ+\sin100^\circ)+\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\sin \left(\frac{200^\circ+100^\circ}{2}\right)\cos \left(\frac{200^\circ-100^\circ}{2}\right)+\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\sin 150^\circ \cos 50^\circ +\sin 200^\circ}{\cos 100^\circ}$$
$$=\frac{2\frac{1}{2} \cos 50^\circ +\sin (270^\circ-70^\circ)}{\cos (90^\circ+10^\circ)}$$
$$=\frac{\cos 50^\circ -\cos 70^\circ}{-\sin 10^\circ}=\frac{\cos 70^\circ -\cos 50^\circ}{\sin 10^\circ}$$
$$=\frac{2\sin \left(\frac{70^\circ+50^\circ}{2}\right)\sin \left(\frac{50^\circ-70^\circ}{2}\right)}{\sin 10^\circ}$$
$$=\frac{2\sin 60^\circ(-\sin 10^\circ)}{\sin 10^\circ}=-2\sin 60^\circ=-2\times \frac{\sqrt 3}{2}=\color{red}{-\sqrt 3}$$
|
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|
Coefficient of the generating function $G(z)=\frac{1}{1-z-z^2-z^3-z^4}$ I am seeking the coefficient $a_n$ of the generating function
$$G(z)=\sum_{k\geq 0} a_k z^k = \frac{1}{1-z-z^2-z^3-z^4}$$
The combinatorial background of this question is to solve the recurrence
$$a_n=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4},\qquad (a_0,a_1,a_2,a_3)=(1,1,2,4).$$
My first idea was to use that $\frac{1}{1-x}=1+x+x^2+...$ which after some expansion leads to
$$a_n = \sum_{k_1+2k_2+3k_3+4k_4 = n}\binom{k_1+k_2+k_3+k_4}{k_1}\binom{k_2+k_3+k_4}{k_2}\binom{k_3+k_4}{k_3}$$
At this point I have no idea how to continue. Looking for a recurrence for $a_n$ would mean to run in circles. I feel that another combinatorial technique is needed, which I dont know. Any ideas?
|
Let $\left\{z_i\right\}_{i=1}^4$ be the roots of the polynomial in the denominator. Note that:
\begin{eqnarray}
\frac{1}{1-z-z^2-z^3-z^4} = \frac{-1}{\prod\limits_{i=1}^4 (z-z_i)} =
\sum\limits_{p=1}^4 \frac{-1}{z-z_p} \cdot \left(\prod\limits_{q\neq p} \frac{1}{z_p-z_q}\right)
\end{eqnarray}
Now clearly :
\begin{equation}
a_k = \frac{1}{k!} \frac{d^k}{d z^k}\left. \left( \frac{1}{1-\sum\limits_{j=1}^4 z^j} \right) \right|_{z=0} =
\sum\limits_{p=1}^4 \frac{1}{z_p^{k+1}} \cdot \left(\prod\limits_{q\neq p} \frac{1}{z_p-z_q}\right) = \left(1,1,2,4,8,15,29,56,108,208,401,\cdots\right)
\end{equation}
|
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|
How to solve this system of exponential equations? Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases}
3^{x+3y-2} + 6\cdot 3^{y^2+4x-2} &=5^{5y-3x} + 2\cdot 3^{y^2-2y+1}\\
1+2\sqrt{x+y-1} &=3\sqrt[3]{3y-2x}.
\end{cases}$$
I think about it but I still have no solution... :(
Since the second equation, I write $x+y \ge 1$ then $y \ge \dfrac{2}{3}x.$ So $y \ge \dfrac{2}{5}$.
I rewrite the 1st equation:
\begin{align*}
3^{x+3y-2} + 6\cdot 3^{y^2+4x-2} &=5^{5y-3x} + 2\cdot 3^{y^2-2y+1}\\
\iff 3^{x+3y-2} + 6\cdot 3^{y^2+4x-2}& \le 5^{5y-3x} -3^{5y-3x}+3^{5y-3x} + 2\cdot 3^{y^2+2y+1}\\
\iff (3^{x+3y-2}-3^{5y-3x})(1+2\cdot 3^{y^2+3x-3y+1})& \le 5^{5y-3x} -3^{5y-3x}\\
\iff (9^{2x-y-1}-1)\underset{>0}{\underbrace{(1+2\cdot 3^{y^2+3x-3y+1}})}&\le \left (\dfrac{5}{3} \right )^{5y-3x}-1.
\end{align*}
Now, I have trouble.... Can anyone post the roots of this system of exponential equations.
I really appreciate if some one can help me. Thanks!
|
The curves corresponding to the functions $y(x)$ , computed by numerical calculus, are drawn on the figure below.
Since they dosn't intersect, the system of equations has no real solution.
I guess that there is a mistake in the wording of the question or in copying the equations.
The more $x, y$ are large, the more the curves are distanced from one another :
|
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|
$a_1 =2$ and $a_{n+1}= \frac{2a_n +3}{a_n +2}$ the recursive sequence , converges?, and if yes, show it to which converges I do not know what to do, I try showing the first elements to see if they were behaving in some way, but no
|
Another less known way of solving this particular form of recurrence relations.
The key is transform the non-linear recurrence to a linear one.
Notice $$\begin{bmatrix}a_{n+1}\\1\end{bmatrix}
\propto \begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}
\begin{bmatrix}a_{n}\\1\end{bmatrix}, \forall n > 1
\quad\text{ and }\quad
\begin{bmatrix}a_{1}\\1\end{bmatrix} = \begin{bmatrix}2\\1\end{bmatrix} =
\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}
\begin{bmatrix}1\\0\end{bmatrix}$$
We have
$$\begin{bmatrix}a_{n}\\1\end{bmatrix}
\propto \begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}^n
\begin{bmatrix}1\\0\end{bmatrix}, \forall n > 1$$
The matrix $\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}$ has eigenvalues $\lambda_{\pm}$ and corresponding eigenvectors $u_{\pm}$ given by
$$\lambda_{\pm} = 2\pm \sqrt{3}\quad\leftrightarrow\quad u_{\pm} = \begin{bmatrix}\pm\sqrt{3}\\1 \end{bmatrix}$$
Since $\begin{bmatrix}1\\0\end{bmatrix} \propto u_{+} - u_{-}$, we have
$\begin{bmatrix}a_{n}\\1\end{bmatrix} \propto \lambda_{+}^n u_{+} - \lambda_{-}^n u_{+}$. Together with $\lambda_{+}\lambda_{-} = 1$ and $|\lambda_{-}| < 1$, we find
$$a_n = \sqrt{3}\frac{\lambda_{+}^n - \lambda_{-}^n}{\lambda_{+}^n - \lambda_{-}^n} =
\sqrt{3}\frac{1 + \lambda_{-}^{2n}}{1 - \lambda_{-}^{2n}}
= \sqrt{3}\frac{1 + (2-\sqrt{3})^{2n}}{1 - (2-\sqrt{3})^{2n}}
\to \sqrt{3}\quad\text{ as } n \to \infty$$
|
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|
Does the series $\sum\frac{(-1)^n cos(3n)}{n^2+n}$ converge absolutely? $\sum|\frac{(-1)^n \cos(3n)}{n^2+n}| \le \sum\frac{1}{n^2+n}$
since $-1 \le cos(3n) \le 1$
$\sum\frac{1}{n^2+n} = \sum\frac{1}{n(n+1)} = \sum(\frac{1}{n} - \frac{1}{n+1})$
$\int(\frac{1}{x} - \frac{1}{x+1}) dx = \log|x| + \log|x+1| + C$
which diverges.
Therefore by the integral test, the series diverges.
However, the answer is that the series converges absolutely. Where have I gone wrong?
|
In fact, it can be simple. If $n \geq 1$ then
$$
\bigg| \frac{(-1)^{n}\cos 3n}{n^{2}+n} \bigg| \leq \frac{1}{n^{2}+n} < \frac{1}{n^{2}}.
$$
But $\sum_{n} \frac{1}{n^{2}}$ converges.
|
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|
Problem with multivariable calculus: $\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$ Anyone can help me with this limit?
$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
I'm having trouble with proving that this limit really goes to $0$
thank you
|
$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$
Method 1: Using the two path test
Along the path of $y=0$, we have
$$\lim\limits_{(x,0)\to (0,0)} \frac{x^3 + 0^3}{x^2 + 0}=\lim\limits_{x\to 0} \frac{x^3}{x^2}=\lim\limits_{x\to 0} x=0$$
Along the path of $y=p(x)-x^2$, where $p(x)$ is a polynomial that passes through the origin, we have
$$\lim\limits_{(x,p(x)-x^2)\to (0,0)} \frac{x^3 + (p(x)-x^2)^3}{x^2 + p(x)-x^2}$$
$$=\lim\limits_{x\to 0} \frac{x^3 -x^6+ 3x^4p(x)-3x^2p(x)^2+p(x)^3}{p(x)}$$
$$=\lim\limits_{x\to 0} \left[\frac{x^3 -x^6}{p(x)}+ 3x^4-3x^2p(x)+p(x)^2\right]$$
Let $p(x)=x^3-x^6$, then
$$\lim\limits_{x\to 0} \left[1+ 3x^4-3x^2(x^3-x^6)+(x^3-x^6)^2\right]=1$$
Since we have different values along different paths, we can conclude that
$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$
Method 2: Using polar coordinates
$$\lim\limits_{r\to 0^+} \frac{r^3\cos^3\phi + r^3\sin^3\phi}{r^2\cos^2\phi + r\sin\phi}$$
$$=\lim\limits_{r\to 0^+} r^2\left(\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right)$$
Now lets attempt to find bounds that are independent of $\phi$. Since
$$\left|\cos^3\phi + \sin^3\phi\right|\leq 1$$
We have
$$r^2\left|\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right|\leq \frac{r^2}{\left|r\cos^2\phi + \sin\phi\right|}$$
Note that $\phi$ is a variable and we cannot treat it as a constant. Since the right hand side is dependent on $\phi$, then we can conclude that
$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$
|
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|
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
|
It should work if we start factoring consecutive integers out of the expression. You have it boiled down to this $$ 6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$
$$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$
This is a good start, but now we need an 8. we get an 8 by using our 4 and a factor of 2 from one of our 6s. This yields: $$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot(3\cdot6\cdot5)$$
We can get a 9 by using our 3 and a factor of 3 from our remaining 6. What remains:$$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot(2\cdot5)=10!$$
How about that?
|
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|
Breaking a determinant into eight pieces From Strang, Introduction to Linear Algebra, 4th ed. p. 250. This is an example whose answer is given in the text but I'm confused. I've paraphrased and corrected what I think is a typo in the first matrix (row 3 + row 3 changed to row 3 + row 2).
How is the determinant of M related to the determinant of A? Rows of M are combinations of the rows of A.
$$
M=
\begin{bmatrix} \text{row 1 + row 3} \\ \text{row 2 + row 1} \\ \text{row 3 + row 2} \\ \end{bmatrix}
$$
M can be split into eight matrices:
$$
\det M=
\begin{vmatrix} \text{row 1 + row 3} \\ \text{row 2 + row 1} \\ \text{row 3 + row 2} \\ \end{vmatrix}
= \begin{vmatrix} \text{row 1} \\ \text{row 2} \\ \text{row 3} \end{vmatrix}
+ \begin{vmatrix} \text{row 3} \\ \text{row 2} \\ \text{row 3} \end{vmatrix}
+ \begin{vmatrix} \text{row 1} \\ \text{row 1} \\ \text{row 3} \end{vmatrix}
+ \cdots
+ \begin{vmatrix} \text{row 3} \\ \text{row 1} \\ \text{row 2} \end{vmatrix}
$$
All but the first and last have repeated rows and zero determinant.
My question: By the rule that $$\begin{vmatrix} a + a^\prime & b + b^\prime \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a^\prime & b^\prime \\ c & d \end{vmatrix}$$ I'd have thought that the first three matrices (before the ellipsis) would suffice to produce the determinant of M. Why do we need the other bunch?
|
Let's apply the rule you quoted once:
$$
\det M=\left|
\begin{matrix}
\text{row 1}+\text{row 3}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|
=
\left|
\begin{matrix}
\text{row 1}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|+
\left|
\begin{matrix}
\text{row 3}\\
\text{row 2}+\text{row 1}\\
\text{row 3}+\text{row 2}\\
\end{matrix}
\right|
$$
Now you apply the rule to the second row, and it gives $2\cdot 2=4$ terms. Lastly apply it to the third row to get all $2\cdot 2\cdot 2=8$ terms.
Each time you apply the rule to a row, the number of terms doubles.
|
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|
Does the limit $\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}$ exist $$
\lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}
$$
Does this limit exist?
I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.
|
We have $$ \frac{|x|^3y^2}{ x^4 + y^6 } \leq c \sqrt{ |y|}$$
Indeed it suffices to show that
$$ |x|^3y^2\leq c \sqrt{ |y|} (x^4 + y^6) $$
Which we see it holds from AM - GM on $$ x^4/3 +x^4/3 +x^4/3+y^6 \geq C|x|^3\sqrt{|y|^3} $$
|
{
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"url": "https://math.stackexchange.com/questions/1442771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function. Solve the equation in interval $[0,\pi]:[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function.
How should i start this question,breaking it into intervals is difficult.Please guide me.
|
Clearly the equality holds true if one of $\sin x,\cos x=0$
This $\implies x=0,\dfrac\pi2,\pi$
$[\sin x]=\begin{cases} 1 &\mbox{if } x=\dfrac\pi2 \\
0 & \text{otherwise} \end{cases} $
$[\cos x]=\begin{cases} 1 &\mbox{if } x=0 \\
0 & \mbox{if } 0<x<\dfrac\pi2\\
-1 & \mbox{if } \dfrac\pi2<x\le\pi\\
\end{cases} $
For $0<x<\dfrac\pi2,[\cos x+\sin x]=0\implies0\le\cos x+\sin x<1$
But $\cos x+\sin x=\sqrt2\sin\left(\dfrac\pi4+x\right)$
$\implies0\le\sin\left(\dfrac\pi4+x\right)<\dfrac1{\sqrt2}$
In $0\le x\le\pi$ we need $\dfrac\pi2<x\le\dfrac{3\pi}4$
For $\dfrac\pi2<x<\pi,[\cos x+\sin x]=-1\implies-1\le\cos x+\sin x<0$
$\implies-\dfrac1{\sqrt2}\le\sin\left(\dfrac\pi4+x\right)<0$
In $0\le x\le\pi$ we need $\dfrac{3\pi}4<x\le\pi$
|
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"timestamp": "2023-03-29T00:00:00",
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|
figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.
|
What about this?
$$\frac{2^{n-1} (2 n-3)\text{!!}}{(n-1)!}$$
which gives
$$
\begin{array}{ccl}
f(1) & = & 1 \\
f(2) & = & 2 \\
f(3) & = & 6 \\
f(4) & = & 20 \\
f(5) & = & 70 \\
f(6) & = & 252 \\
f(7) & = & 924 \\
f(8) & = & 3432 \\
f(9) & = & 12870 \\
f(10) & = & 48620 \\
f(11) & = & 184756 \\
f(12) & = & 705432 \\
f(13) & = & 2704156 \\
f(14) & = & 10400600 \\
f(15) & = & 40116600 \\
f(16) & = & 155117520 \\
f(17) & = & 601080390 \\
f(18) & = & 2333606220 \\
f(19) & = & 9075135300 \\
f(20) & = & 35345263800 \\
\end{array}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can i factor out an $x^2$ while completing the square? Original equation $$\int\frac{x}{\sqrt{56+10x^2-x^4}}$$
This equation must be integrated by completing the square. My question is, when completing the square am i able to factor out $x^2$ like this $$-x^4+10x^2+56$$ $$-x^2(x^2-10)+56$$ would that be fine? and if is so how would I continue on about this with the $-x^2$ on the outside?
|
$$\int\frac{x}{\sqrt{56+10x^2-x^4}}dx$$
Using the substitution $x^2 = t$, $\,\,dt = 2xdx$
$$\frac{1}{2}\int\frac{dt}{\sqrt{-(t^2 -10t - 56)}}$$
$$\frac{1}{2}\int\frac{dt}{\sqrt{81-(t-5)^2}}$$
Substitute $s = t-5$, $\,\,ds = dt$
$$\frac{1}{2}\int\frac{ds}{\sqrt{81-s^2}}$$
$$\frac{1}{2}\int\frac{ds}{9\sqrt{1-\frac{s^2}{81}}}$$
Substitute $p = \frac{s}{9}$, $\,\,dp = \frac{ds}{9}$
$$=\frac{1}{2}\int\frac{dp}{\sqrt{1-p^2}}$$
$$=\frac{1}{2}\sin^{-1}(p)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{s}{9}\Big)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{t-5}{9}\Big)$$
$$=\frac{1}{2}\sin^{-1}\Big(\frac{1}{9}(x^2-5)\Big)$$
|
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|
Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital. I am trying to solve the limit
$$\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
Without using L'Hopital.
Evaluating yields $\frac{0}{0}$. When I am presented with roots, I usually do this:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \cdot \frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}$$
And end up with
$$\frac{9 - (5+x)}{(1-\sqrt{5-x}) \cdot (3+\sqrt{5+x})}$$
Then
$$\frac{9 - (5+x)}{3+\sqrt{5+x}-3\sqrt{5-x}-(\sqrt{5-x}\cdot\sqrt{5+x})}$$
And that's
$$\frac{0}{3+\sqrt{9}-3\sqrt{1}-(\sqrt{1}\cdot\sqrt{9})}= \frac{0}{3+3-3-3-3} = \frac{0}{-3} = -\frac{0}{3}$$
Edit: Ok, the arithmetic was wrong. That's $\frac{0}{0}$ again.
But the correct answer is
$$-\frac{1}{3}$$
But I don't see where does that $1$ come from.
|
I would rewrite the quotient as the quotient of two variation rates. Set $x=4+h\enspace(h\to 0)$:
\begin{align*}
\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}&=\frac{\sqrt{9}-\sqrt{9+h}}{\sqrt{1}-\sqrt{1-h}}\\
&=\frac{\sqrt{9+h}-\sqrt{9}}h\cdot\frac h{\sqrt{1-h}-\sqrt{1}}\to\bigl(\sqrt{x}\bigr)'_{x=9}\cdot-\frac1{\bigl(\sqrt{x}\bigr)'_{x=1}}\\
&=-\frac1{2\sqrt 9}\cdot2\sqrt{1}=-\frac13.
\end{align*}
|
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|
Find the determinant of $n\times n$ matrix Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc}
-x & a_2&a_3&a_4&\cdots &a_n\\
a_{1} & -x & a_3&a_4&\cdots &a_n\\
a_1&a_{2} & -x &a_4&\cdots &a_n\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
a_1&a_{2} & a_3&a_4&\cdots & -x\\
\end{array}\end{bmatrix}$, then how to find the $\det (M)$?
Proof: First I started by taking the $a_i$ from each $i$-th columns, then$ |M|=\prod_{i=1}^{n}{a_i} \begin{vmatrix}\begin{array}{ccccccc}
\frac{-x}{a_1} & 1&1&1&\cdots &1\\
{1} & \frac{-x}{a_2} & 1&1&\cdots &1\\
1&1 & \frac{-x}{a_3} &1&\cdots &1\\
\vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\
1&1 & 1&1&\cdots & \frac{-x}{a_n}\\
\end{array}\end{vmatrix}$. After this is there any easiest way to find the determinant.
|
Hint: $\det(M)$ is a polynomial in $x$ and $M$ is clearly singular if $x=-a_k$ for some $k$
|
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|
Is there anything wrong with this question? Find an equation of the plane that passes thru the points $p(1,0,1)$ and $q(2,1,0)$ which is parallel to the intersection of the two planes $x+y+z=5$ and $3x-y=4$?
I plotted the two points and the intersection of the two planes using a 3-d grapher and the points aren't parallel which means the equation of the plane requested wouldn't be parallel? Is that correct?
|
It's the plane that needs to be parallel, not the points.
The points just need to lie in the plane.
Let the equation of the plane be $ax+by+cz=d$. Without loss of generality we can let $d=1$.
Using the values of $(x,y,z)$ from the given points, we find that $a+c=d=1$ and that $2a+b=d=1$.
The line of intersection has direction vector equal to the vector product of the normals to the two planes:
$\underline p=\left(\begin{array}{r}
1 \\
1 \\
1
\end{array}\right) \times
\left(\begin{array}{r}
3 \\
-1 \\
0
\end{array}\right)=\left(\begin{array}{r}
1 \\
3 \\
-4
\end{array}\right)
$
This direction vector is parallel to the required plane, so
$\left(\begin{array}{r}
1 \\
3 \\
-4
\end{array}\right) .
\left(\begin{array}{r}
a \\
b \\
c
\end{array}\right)=0
$
$a+3b-4c=0$
Now pull together these three equations:
$a+3b-4c=0$
$a+c=1$
$2a+b=1$
Let $c=1-a$, so that our equations are:
$a+3b-4+4a=0 \Rightarrow 5a+7b=4$
$2a+b=1 \Rightarrow 14a+7b=7$
These give $9a=3 \Rightarrow a=\frac 1 3$, $b= \frac 1 3$, $c=\frac 23$
Plane is $\frac13x+\frac13y+\frac23z=1$ or better $x+y+2z=3$.
|
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|
Problem with definite integral $\int _0^{\frac{\pi }{2}}\sin \left(\arctan \left(x\right)+x\right)dx$ Need to calculate this definite integral. It's seems very strange for me
$$\int _0^{\frac{\pi }{2}}\sin \left(\arctan\left(x\right)+x\right)dx$$
I dont see any reasonable way to calculate this integral. For instance, arctan of π/2 - it's incomprehensible value. I think there are some clever and special way.
|
$\int_0^\frac{\pi}{2}\sin(\tan^{-1}x+x)~dx$
$=\int_0^\frac{\pi}{2}\sin\tan^{-1}x\cos x~dx+\int_0^\frac{\pi}{2}\cos\tan^{-1}x\sin x~dx$
$=\int_0^\frac{\pi}{2}\dfrac{x\cos x}{\sqrt{x^2+1}}~dx+\int_0^\frac{\pi}{2}\dfrac{\sin x}{\sqrt{x^2+1}}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!\sqrt{x^2+1}}~dx$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!\sqrt{x^2+1}}~d(x^2+1)$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^n(-1)^{2n-k}(x^2+1)^k}{2(2n)!\sqrt{x^2+1}}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^n(-1)^{2n-k}(x^2+1)^k}{2(2n+1)!\sqrt{x^2+1}}~d(x^2+1)$
$=\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-\frac{1}{2}}}{2(2n)!k!(n-k)!}~d(x^2+1)+\int_0^\frac{\pi}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-\frac{1}{2}}}{2(2n+1)!k!(n-k)!}~d(x^2+1)$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k+\frac{1}{2}}}{2(2n)!k!(n-k)!\left(k+\dfrac{1}{2}\right)}\right]_0^\frac{\pi}{2}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k+\frac{1}{2}}}{2(2n+1)!k!(n-k)!\left(k+\dfrac{1}{2}\right)}\right]_0^\frac{\pi}{2}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(\pi^2+2)^{k+\frac{1}{2}}}{2^{k+\frac{1}{2}}(2n)!k!(n-k)!(2k+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(\pi^2+2)^{k+\frac{1}{2}}}{2^{k+\frac{1}{2}}(2n+1)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n+1)!k!(n-k)!(2k+1)}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $\frac{x^2+4x+7}{x^2+x+1}$ a one-to-one function or not? Is $\frac{x^2+4x+7}{x^2+x+1}$ a one-to-one function or not?
Leaving aside the graphical method is there any other way in which I can do this question?
|
An algebraic approach:
The function is one-one if the equation in $x$
$$
k=\dfrac{x^2+4x+7}{x^2+x+1}
$$
has only one solution for any $k$ that gives real solutions.
Since $x^2+x+1 \ne 0$ we have:
$$
x^2(k-1)+x(k-4)+k-7=0
$$
and we can easely see that the discriminant of this equation can be positive, so there is some $k$ (exactly: $4-2\sqrt{3}<k<4+2\sqrt{3} $) for which we have two values of $x$ and the function is not one to one.
|
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|
Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot x \cdot (1+\sqrt{\cos x})}$$
Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with
$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$
But hey, perhaps we can rationalize again?
$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$
Resulting in
$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$
Cancelling
$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$
Well that was hilarious. I ended up at the beginning! Dammit.
My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$
And then rationalize
$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$
$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:
$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Then
$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Well clearly that's not gonna work either. I will still get $0$ in the denominator.
The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.
I don't know how am I supposed to solve this without L'Hopital.
|
If L'Hopitals' Rule gives you $\frac{f'}{g'}\rightarrow\frac{0}{0}$, apply L'Hopital's Rule to it. :-)
$$\frac{f(x)}{g(x)}=\frac{1-\sqrt{cos(x)}}{x^{2}}
$$
$$f''(x)=-\frac{1}{2}\sqrt{cos(x)}-\frac{sin^{2}(x)}{4cos^{\frac{3}{2}}(x)}
$$
$$\frac{f''(x)}{g''(x)}\rightarrow-\frac{1}{4}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Thanks in advance
Update: I actually solved it, and this is the way that I wanted to:
$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$
Thanks for your two answers guys. I really appreciate it ! This community is awesome !
|
If ${\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0}$ and you are taking a limit of the form
$$\lim_{x \rightarrow \infty} \sqrt{x + f(x)} - \sqrt{x}$$
Then you can proceed by writing the limit as
$$\lim_{x \rightarrow \infty} \sqrt{x} \bigg(\sqrt{1 + {f(x) \over x}} - 1\bigg)$$
Since $\sqrt{1 + \epsilon} = 1 + {1 \over 2} \epsilon + O(\epsilon^2)$, the limit will be the same as
$$\lim_{x \rightarrow \infty} \frac{f(x)}{2\sqrt{x}}$$
In the case at hand, $f(x) = \sqrt{1 + \sqrt{x}}$, which is bounded by
for example $\sqrt{\sqrt{x} + \sqrt{x}} = \sqrt 2 x^{1 \over 4}$. Hence the limit will be zero.
|
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|
find all values of a for which the system has nontrivial solutions So i was given this question.
In each of the following, find all values of a for which the system has nontrivial solutions, and determine all solutions in each case.
a) x - 2y + z = 0
x + ay - 3z = 0
-x + 6y - 5z = 0
Here is my attempt:
$$\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 1 & a & -3 & 0 \\ -1 & 6 & -5 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ -1 & 6 & -5 & 0 \end{array}\right]\\ \sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 2 & -2 & 0 \end{array}\right]$$
I'm stuck on how to go about this, what really confuses me is the variable a, and how to subtract/add/multiply/divide when there is a variable between numbers.
|
You just need to go a little further with the row reduction.
$$\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 2 & -2 & 0 \end{array}\right]\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 1 & -1 & 0 \end{array}\right]\sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & a+6 & -8 & 0 \\ 0 & 0 & a-2 & 0 \end{array}\right].$$
So, in order for the system to have non-trivial solutions, we must have $a=2$. Then we have a row of zeros.
|
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|
compute the sum of a series $$
\sum_{n=0}^\infty \frac{x^n}{n(n+1)}
$$
what is the sum of this?
I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there
|
Your sum has an error in it because for n=0 you are dividing by 0. However, if you start instead with n=1, there is no error, so I will assume you meant that.
Notice that $$\int \int x^n dx dx = \frac{x^{n+2}}{(n+1)(n+2)} +Cx+D =x \cdot \frac{x^{n+1}}{(n+1)(n+2)} +Cx+D$$
Therefore $$\int \int \sum_{n=0}^{\infty} x^n dx dx = Cx+D+ \sum_{n=0}^{\infty} x \cdot \frac{x^{n+1}}{(n+1)(n+2)} =Cx+D+ x \cdot \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)(n+2)}$$
$$ =Cx+D+ x \cdot \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} $$.
Letting $C = D = 0$ and dividing by $x$ gives a closed form expression for the sum. We also have that first integral, $$\int \int \sum_{n=0}^{\infty} x^n dx dx = \int \int \frac{1}{1-x} dx dx =\int C + \log(x-1) dx = Cx + D + (x-1) \log(x-1) $$
Again letting $C = D = 0$ and dividing by $x$, we have $$ \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \frac{(x-1)\log(x-1)}{x} $$
|
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|
Find the limit of $x +\sqrt{x^2 + 8x}$ as $x\to-\infty$ $$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}$$
I multiplied it by the conjugate:
$\frac{-8x}{x - \sqrt{{x^2} + 8x}}$
I can simplify further and get:
$\frac{-8}{1-\sqrt{1+\frac{8}{x}}}$
I think there is an error with my math, because the denominator should probably be a 2.
I'm stuck on this one. I graphed it so I know the limit is -4, but I can't calculate it. Thanks a lot for the help!
|
We have
\begin{align}
\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}&=\lim_{x\to -\infty}\frac{(x +\sqrt{x^2 + 8x})(x -\sqrt{x^2 + 8x})}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{x^2 -(x^2 + 8x)}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{-8x}{x -\sqrt{x^2 + 8x}}\\
&=\lim_{x\to -\infty}\frac{\frac{-8x}{|x|}}{\frac{x}{|x|} -\sqrt{\frac{x^2}{|x|^2} + \frac{8x}{|x|^2}}}&& |\cdot |\text{ is needed since }x<0\\
&=\lim_{x\to -\infty}\frac{8}{-1 -\sqrt{1 - 8/x}}&&\text{since }|x|=-x\,\text{ for }x<0\\
&=\frac{8}{-1-\sqrt{1-0}}\\
&=\color{blue}{-4}
\end{align}
|
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|
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \cos^2x)(1) \longrightarrow\,
= \sin^2x - \cos^2x
\end{align}
Is that true?
|
Expand in terms of complex exponentials.
$$\sin^4 x + \cos^4 x = \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^4 + \left( \frac{e^{ix} + e^{-ix}}{2} \right)^4$$
Notice that $i^4 = +1$, so we get
$$\sin^4 x + \cos^4 x = \frac{1}{16} \left( 2e^{4ix} + 2 e^{-4ix} + 12 \right)$$
where we use the relation $(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 ab^3 + b^4$. The terms of the form $a^3 b$ and $ab^3$ all cancel by addition.
This leaves us with a final result:
$$\sin^4 x + \cos^4 x = \frac{4}{16} \left(\frac{e^{4ix} + e^{-4ix}}{2} \right) + \frac{12}{16} = \frac{3}{4} + \frac{1}{4} \cos 4x$$
|
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|
Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
|
Starting with
$$r^2=9\cos2\theta$$
Applying the double angle identity
$$r^2=9\cos^2\theta-9\sin^2\theta$$
Now multiply both sides by $r^2$, apply $x=r\cos\theta,y=r\sin\theta$ and $x^2+y^2=r^2$ to obtain
$$(x^2+y^2)^2=9x^2-9y^2$$
Which, when expanded gives
$${x}^{4}+2\,{x}^{2}{y}^{2}+{y}^{4}-9\,{x}^{2}+9\,{y}^{2}=0$$
|
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|
system of three equations with Cramer's rule First of - sorry for my English.
My math teacher gave me three tasks to complete, but I can't complete the second one.
Here is the system:
$$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$
We have to complete them using Cramer's rule. Determinant result is 43 for me, but today when I asked teacher she said there should be 1. I can't figure it out. How it can be 1? It would be great if someone could explain how to get to 1 step by step.
|
The coefficient matrix for the system $$\begin{cases} x+y-z=3 \\ x-3y+2z=1 \\ 7x-4y+z=7\end{cases}$$ is $$\pmatrix{1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1}$$
Let's find the determinant:
$$\begin{align}\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1\end{array}\right| &= \left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -4 & 3 \\ 0 & -11 & 8\end{array}\right| \\ &= \frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -12 & 9 \\ 0 & -11 & 8\end{array}\right| \\ &= \frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 1 \\ 0 & -11 & 8\end{array}\right| \\ &= -\frac 13\left|\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & -3\end{array}\right| \\ &= -\frac 13(1)(1)(-3) \\ &= 1\end{align}$$
EDIT: Here's the cofactor expansion method of finding the determinant. I'll expand across the top row:
$$\begin{align}\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -3 & 2 \\ 7 & -4 & 1\end{array}\right| &= (1)\left|\begin{array}{cc} -3 & 2 \\ -4 & 1\end{array}\right|-(1)\left|\begin{array}{cc} 1 & 2 \\ 7 & 1\end{array}\right|+(-1)\left|\begin{array}{cc} 1 & -3 \\ 7 & -4\end{array}\right| \\ &= (1)[-3\cdot 1 - 2\cdot(-4)] -(1)[1\cdot 1 - 2\cdot 7]+(-1)[1\cdot(-4)-(-3\cdot 7)] \\ &= 5+13-17 \\ &=1\end{align}$$
|
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|
Plotting $f(x) = \sin(x)+\cos(x)$ by converting it to another form Are there any trigonometric identity that can make $f(x) = \sin(x)+\cos(x)$ easier to plot? I have no idea how it becomes a sin graph shape in the end.
|
Special Case
It's not that hard. You should just use the summation formula for sines:
$$\sin (x + y) = \sin (x)\cos (y) + \cos (x)\sin (y)$$
This is how it works
$$\eqalign{
\sin (x) + \cos (x) &= \sqrt 2 \left( {{1 \over {\sqrt 2 }}\cos (x) + {1 \over {\sqrt 2 }}\sin (x)} \right) \cr
&= \sqrt 2 \left( {\sin ({\pi \over 4})\cos (x) + \cos ({\pi \over 4})\sin (x)} \right) \cr
&= \sqrt 2 \sin (x + {\pi \over 4}) \cr} $$
That's all you need to do for this case. If you are interested to tackle down the general case then read the sequel.
General Case
Consider a linear combination of $\sin \alpha x$ and $\cos \alpha x$ as follows
$$y = A \cos \alpha x + B \sin \alpha x$$
where $A$ and $B$ are some real constants. Then we rewrite $y$ in this way
$$y = \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}} \cos \alpha x + \frac{B}{\sqrt{A^2+B^2}} \sin \alpha x \right)$$
Now, the magic comes in! We can find a unique angle $\phi$ in the interval $[0,2\pi)$ such that
$$\begin{array}{}
\sin \phi = \dfrac{A}{\sqrt{A^2+B^2}} \\
\cos \phi = \dfrac{B}{\sqrt{A^2+B^2}}
\end{array}$$
and hence
$$y = \sqrt{A^2+B^2} \left( \sin \phi \cos \alpha x + \cos \phi \sin \alpha x \right) \\$$
Finally, using the summation formula for sines we get
$$y = \sqrt{A^2+B^2} \sin(\alpha x+\phi)$$
|
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|
What is the probability that the number $3^a+7^b$ has a digit equal to $8$ at the units place? The number $a$ is randomly selected from the set $\left\{0,1,2,3,....,98,99\right\}$.The number $b$ is selected from the same set.What is the probability that the number $3^a+7^b$ has a digit equal to $8$ at the units place?
I only know this much that $3^a+7^b(mod 10)=8$.When $3^a+7^b$ is divided by $10$,its remainder is $8$ because its unit place is $1.$Total number of cases$=100\times 100$,but i cannot count favorable number of cases.Hence could not find the probability.Please help me.Thanks.
|
Use the Chinese theorem. $\equiv 8\pmod{10}$ means even and $\equiv 3\pmod{5}$. Now both $3$ and $7\equiv 2$ are generators for $\mathbb{F}_5^*$, hence the working cases are $a\equiv 2\pmod{4}$ and $b\equiv 2\pmod{4}$, $a\equiv 3\pmod{4}$ and $b\equiv 0\pmod{4}$, $a\equiv 0\pmod{4}$ and $b\equiv 1\pmod{4}$. $3^a$ and $7^b$ are always odd, hence $3^a+7^b$ is always $\equiv 0\pmod{2}$. Since the residue classes $\pmod{4}$ are equidistributed in $\{0,1,2,\ldots,98,99\}$, the wanted probability is:
$$ \frac{3\cdot 25\cdot 25}{100\cdot 100}=\color{red}{\frac{3}{16}}.$$
|
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|
Implicit equation of Semicircle and ellipse May I know what is the Implicit equation to define the upper bound of a circle of radius 1?
Is it $ y^2 - \sqrt{1-x^2} = 0$?
and for the lower bound, $ y^2 + \sqrt{1-x^2} = 0$?
What is the implicit equation of the upper bound of an ellipse of radius 0.3 then?
|
The implicit equation of a circle of radius $1$ centered at the origin is $x^2+y^2=1$, so you can find: $y^2=1-x^2$ and the upper esmicircle ($y\ge 0$) is given by: $y=\sqrt{1-x^2}$ ( that you can write as $ y-\sqrt{1-x^2}=0$ if you want). The other semicircle ($y\le 0$) is $y=-\sqrt{1-x^2}$.
For an ellipse with center in the origin and symmetry axes parallel to the coordinate axis you can start from the equation
$$
\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1
$$
where $a$ and $b$ are the semiaxis, and, in the same way, you find
$$
y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}=\pm \dfrac{b}{a}\sqrt{a^2-x^2}
$$
that with the $+$ sign represent the upper ($y\ge 0$) semiellipse.
Finally, note has no sense to talk of an ellipse of radius $0.3$
|
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|
Taylor series of arctan(x) (Spivak) At p. 388 of Calculus, Spivak gives a formula:
$$\frac{1}{1+t^2} = 1 - t^2 + t^4 - ... + (-1)^nt^{2n} + \frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$$
Which can be integrated to find $\arctan(x)$.
I don't understand where this formula comes from, but I found it up to $(-1)^nt^{2n}$ by considering the geometric series for $\frac{1}{1-x}$ and replacing $x$ by $-x^2$ to get the series for $\frac{1}{1+x^2}$. I don't see the term $\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}$ though, because the series I got this way is $\frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^nx^{2n}$.
|
Consider
$$
1+x+x^2+\dots+x^n=\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}-\frac{x^{n+1}}{1-x}
$$
that can also be written
$$
\frac{1}{1-x}=1+x+x^2+\dots+x^n+\frac{x^{n+1}}{1-x}
$$
Now substitute $x=-t^2$, that gives
$$
\frac{1}{1+t^2}=1+(-t^2)+(-t^2)^2+\dots+(-t^2)^n+\frac{(-t^2)^{n+1}}{1+t^2}
$$
and not it's just a matter of observing that $(-t^2)^k=(-1)^kt^{2k}$.
|
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|
Help in proof involving matrices
Let $$ P=\begin{pmatrix}11&-6\\18&-10\end{pmatrix}. $$ Show that
there is no $2\times2$ matrix $Q$ such that $Q^2=P$.
I don't know how to approach this question. Is brute force the only way to go? That is, take a general matrix
$$
Q=\begin{pmatrix}a&b\\c&d\end{pmatrix}
$$
and get
$$
Q^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+c)&bc+d^2\end{pmatrix}.
$$
Then equate element by element and try to find a contradiction? This seems tedious and not very instructive.
In a previous part of the question, we established that $P^2+P=2I$, $|P|=-2$, and $$P^{-1}=\frac12\begin{pmatrix}10&-6\\18&-11\end{pmatrix}.$$
But I don't see how these would be useful...
Any hint would be appreciated.
|
The Eigen values of $P$ are $2$ and $-1$. Now there exists a Matrix $Q$ such that $Q^2=P$. So $Det(Q^2)=Det(P)$ $\implies$ $Det(Q)=i\sqrt{2}$ or $-i\sqrt{2}$.
Now by Cayley hamilton theorem the characteristic equation of $Q$ is of the form
$$Q^2-Tr(Q)Q+Det(Q)I=0$$ $\implies$ Since $Q^2=P$
$$Tr(Q)Q= P+Det(Q)I=\begin{pmatrix}
11+i\sqrt{2} & -6\\
18 & -10+i\sqrt{2}
\end{pmatrix}$$
Taking Determinant on both sides we get
$$Det(Tr(Q)Q)=-4+i\sqrt{2}$$ Using $Det(kA)=k^nDet(A)$ we get
$$Tr(Q)^2Det(Q)=-4+i\sqrt{2}$$ So
$$Tr(Q)^2=\frac{-4+i\sqrt{2}}{i\sqrt{2}}=1+2i\sqrt{2}$$ So
$$Tr(Q)=\sqrt{1+2i\sqrt{2}}=\sqrt{2}+i$$ Finally
$$Q=\frac{1}{\sqrt{2}+i}\begin{pmatrix}
11+i\sqrt{2} & -6\\
18 & -10+i\sqrt{2}
\end{pmatrix}$$
$$Q=\begin{pmatrix}
4\sqrt{2}-3i& -2\sqrt{2}+2i\\
6\sqrt{2}-6i & -3\sqrt{2}+4i
\end{pmatrix}
$$
|
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|
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$,
Since 5 is a prime number, we get that $5\mid (a+\sqrt2b)$or $5\mid (a-\sqrt2b)$,
If $5\mid (a+\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
If $5\mid (a-\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
Thus $5\nmid a^2-2b^2$.
|
You just have to make a table for the function $a^2-5b^2$ in $\mathbf Z/5\mathbf Z$ above the horizontal line, the possible values for $a^2$; to the left of the vertical line, those for $b^2$):
$$\begin{array}{r|rrr}
&0&1&-1\\
\hline
0&0&1&-1\\1&-2&-1&2\\-1&2&-2&1
\end{array}$$
which shows the only case with $a^2-5b^2\equiv 0\mod5$ is when $a\equiv b\equiv 0\mod5$.
|
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|
In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be equal to
$(A)15^\circ\hspace{1cm}(B)75^\circ\hspace{1cm}(C)115^\circ\hspace{1cm}(D)135^\circ$
Let AD be the median.Angle $ADC=45^\circ,ACD=30^\circ,CAD=105^\circ$.and $BD=CD$Then i am stuck,how to find angle $B?$Please help me.Thanks.
|
here is way to compute the angle $\angle BAD$ using the rule of sines where $D$ is mid point of $BC.$
let me use $t = \angle BAD.$
suing the rule of sin on $\Delta ABD, \Delta ADC$ we have $$\frac{BD}{\sin t} = \frac{AD}{\sin(t + 45^\circ)}, \frac{CD}{\sin(15^\circ)} = \frac{AD}{\sin 30^\circ} $$
from the two equations, we get $$\frac{AD}{BD} = \frac{sin(t+45^\circ)}{\sin t}=\frac{\sin 30^\circ}{\sin 15^\circ}=2\cos 15^\circ \to \tan t =\frac 1{2\sqrt 2 \cos 15^\circ - 1} = \frac 1 {\sqrt 3}. $$
therefore $t = 30^\circ, \angle B = 105^\circ$
|
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|
How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$? How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?
I found the derivative to be $f'(x)= (2\sin 4\theta)(1+\cos 2\theta) + (\sin 2\theta)^2 (-2\sin 2\theta)$.
Now I know that I have to set this equal to zero and solve for $\theta$ but that is where I get stuck, can I solve for $\theta$ or is there a different way that is easier?
|
I’d start by rewriting $f(x)$, using the identities $\sin2\theta=2\sin\theta\cos\theta$ and $\cos^2\theta=\frac12(1+\cos2\theta)$:
$$f(x)=\sin^22\theta(1+\cos2\theta)=8\sin^2\theta\cos^4\theta\;.$$
In fact, let’s go a step further and get rid of the sine:
$$f(x)=8\sin^2\theta\cos^4\theta=8(1-\cos^2\theta)\cos^4\theta=8\cos^4\theta-8\cos^6\theta\;.$$
Differentiating and factoring yields
$$f\,'(x)=16\sin\theta\cos^3\theta(3\cos^2\theta-2)\;.$$
Now set this to $0$ and solve. If you want, you can further simplify a little by noting that
$$3\cos^2\theta-2=3(1-\sin^2\theta)-2=1-3\sin^2\theta\;.$$
|
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|
Solution of quadratic diophantine equations Is there any algorithm so that solution to the following equation can be found?
$(x+a)^2-y^2=c$ where $c$ and $a$ is a constant.
It is similar to Pells eqution with a variation where $D=1$.
I am new to this number theory problem. So can anyone please help me?
|
I.e. you want to solve in integers, where $c$ is given (we can let $x$ be $x+a$):
$$x^2-y^2=c$$
The typical solution is factoring $(x-y)(x+y)=c$, noticing the factors of $c$ and checking cases, when $x-y=d$ and $x+y=\frac{c}{d}$. Note $\begin{cases}x-y=d\\x+y=\frac{c}{d}\end{cases}$ is solvable in integers if and only if $d$ and $\frac{c}{d}$ are of the same parity.
E.g., let's solve $x^2-y^2=36$. $(x-y)(x+y)=2^2\cdot 3^2$. Both $x-y, x+y$ must be even (since they're of the same parity, like I said). Cases:
$\begin{cases}x-y=2\\ x+y=2\cdot 3^2\end{cases}$, $\begin{cases}x-y=2\cdot 3^2\\x+y=2\end{cases}$,
$\begin{cases}x-y=-2\\ x+y=-2\cdot 3^2\end{cases}$, $\begin{cases}x-y=-2\cdot 3^2\\x+y=-2\end{cases}$
|
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|
If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ My question:
If $0<|z|<1$, show that $\frac{1}{4}|z|<|1-e^z|<\frac{7}{4}|z|$ ($z$ is complex)
what I have tried:
I tried to expand the middle term in its Taylor series but I can't get the appropriate bound.
|
The Maclaurin expansion of $e^z - 1$ is
$$z + \frac{z^2}{2!} + \cdots + \frac{z^N}{N!} +\cdots \tag{*}$$
Since $0 < |z| < 1$ and $k! \ge 2\cdot 3^{k-2}$ for all $k \ge 2$, then by the triangle inequality, (*) is strictly bounded above in modulus by
$$\left[1 + \frac{1}{2} + \frac{1}{2}\left(\frac{1}{3}\right) + \frac{1}{2}\left(\frac{1}{3}\right)^2+\cdots\right]|z| = \left(1 + \frac{1/2}{1-1/3}\right)|z| = \frac{7}{4}|z|.$$
By a similar argument, (*) is strictly bounded below in modulus by
$$\left[1 - \left(\frac{1}{2} + \frac{1}{2}\left(\frac{1}{3}\right) + \cdots\right)\right]|z| = \left(1 - \frac{1/2}{1 - 1/3}\right)|z| = \frac{1}{4}|z|.$$
Thus,
$$\frac{1}{4}|z| < |e^z - 1| < \frac{7}{4}|z|\qquad (0 < |z| < 1).$$
Since $|e^z - 1| = |1 - e^z|$, the above inequality is the same as your inequality.
|
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|
Use the PMI to prove the following for all natural numbers $3^n≥1+2^n$ Use the PMI to prove the following for all natural numbers
$3^n≥1+2^n$
Base Case: $n=1$
$3^1≥1+2^1$
$3 ≥ 3$, which is true
Inductive Case:
Assume $3^k ≥ 1+2^k$
[Need to Show for k+1]
$3^{(k+1)} \ge 1+2^{(k+1)}$
Now from here i always get stuck trying to show this next step. I already saw that this same question is asked on this website but i still don't understand why the steps are correct.
for example,
$3^{(k+1)}=3^k⋅3 \ge (1+2^k)3$
how does the RHS turn into that? is there an algebra step that i just dont remember learning?
|
Assume $$3^k\ge 1+2^k$$
Add $2^k$ to both sides:
$$3^k+2^k\ge 1+2^k+2^k=1+2^{k+1}$$
Now notice that $3^{k+1}=3\cdot 3^k=3^k+2\cdot 3^k\ge 3^k+2^k$
So, we combine $3^k+2^k\ge 1+2^{k+1}$ with $3^{k+1}\ge 3^k+2^k$ and get
$$3^{k+1}\ge 1+2^{k+1}$$
|
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|
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series?
I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:
\begin{equation}
y(x) = \tan x
\end{equation}
\begin{equation}
y^{\prime} = \frac{1}{\cos^{2} x}
\end{equation}
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x}
\end{equation}
This can be rewritten as:
\begin{equation}
y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime}
\end{equation}
Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$:
\begin{equation}
y^{\prime} = y^{2}
\end{equation}
And therefore
\begin{equation}
\frac{1}{\cos^{2} x} = \tan^{2} x
\end{equation}
Now, evalueting this equation at $x = \pi / 4$
\begin{equation}
\frac{1}{(\sqrt{2}/2)^{2}} = 1^{2}
\end{equation}
\begin{equation}
2 = 1
\end{equation}
|
$x = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\underbrace{\left(x + x + x + \ldots + x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x$
|
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|
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
|
$3\sin^2{\theta}-2\cos^2{\theta}+5\sin{\theta}\cos{\theta}=0$
Transfer the $\sin{\theta}\cos{\theta}$ term to the RHS.
$3\sin^2{\theta}-2\cos^2{\theta}=-5\sin{\theta}\cos{\theta} $
Then square both sides.
$(3\sin^2{\theta}-2\cos^2{\theta})^2=(-5\sin{\theta}\cos{\theta})^2$
$9\sin^4{\theta}-12\sin^2{\theta}\cos^2{\theta}+4\cos^4{\theta}=25\sin^2{\theta}\cos^2{\theta}$
Then convert all occurrences of $\sin^2{\theta}$ to $1-\cos^2{\theta}$.
$9(1-\cos^2{\theta})^2-12(1-\cos^2{\theta})\cos^2{\theta}+4\cos^4{\theta}=(1-\cos^2{\theta})\cos^2{\theta}$
This will give an equation which can be solved for $\cos^2{\theta}$,
$50\cos^4{\theta}-55\cos^2{\theta}+9=0$
then for $\cos{\theta}$,
then for $\theta$.
|
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|
How to prove this inequality about $xyz=1$ Let $x,y,z>0$,and such $xyz=1$, show that
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3$$
I tried use $AM-GM$ inequality
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3\sqrt{\dfrac{(x+y)(y+z)(z+x)}{(x^3+x)(y^3+y)(z^3+z)}}$$
which shows that $$(x+y)(y+z)(z+x)\ge (x^3+x)(y^3+y)(z^3+z)$$
Am I on the right lines?but I don't have any idea how to start proving it
|
firstly we can rewrite the terms like : $$\frac{x+y}{x^3+x}=\frac{1}{1+x^2}+\frac{y}{x+x^3}=1-\frac{x^2}{1+x^2}+\frac{y}{x}-\frac{yx^2}{x+x^3}$$ then we can apply $AM\geq GM$ for the denominators and we get $$\frac{x+y}{x^3+x}\geq 1+\frac{y}{x}-\frac{x^2}{2x}-\frac{yx^2}{2x^2}=1+\frac{y}{x}-\frac{x}{2}-\frac{y}{2}$$ similarly we get the other terms and after adding these 3 inequalities we have $$\frac{x+y}{x^3+x}+\frac{y+z}{y^3+y}+\frac{z+x}{z^3+z}\geq 3 +\frac{y}{x}+\frac{z}{y}+\frac{x}{z}-x-y-z $$ now it is suffice to show that $\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\geq x+y+z$ and this is well known ineqality when $xyz\le 1$ we are done
|
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|
When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
I do not have a specific idea how to solve this one
|
If $n$ is odd then $2^n+3^n+6^n\equiv -1\pmod{3}$, so $2^n+3^n+6^n$ cannot be a square.
For similar reasons $\!\!\pmod{5}$, $n$ has to be a number of the form $4k+2$.
For similar reasons $\!\!\pmod{7}$, $n$ has to be a number of the form $12k\pm 2$, so $2^n+3^n+6^n$ is a multiple of $7$. $n=2$ gives the solution:
$$ 2^2+3^2+6^2 = 7^2. $$
Let we set $a_n=2^n+3^n+6^n$ and $\nu_7(m)=\max\{n\in\mathbb{N}:7^n\mid m\}$.
Since $\frac{1}{4}+\frac{1}{9}+\frac{1}{36}=\frac{7}{18}$, if $n=12k-2$ then $\nu_7(a_n)\equiv 1\pmod{2}$, so $a_n$ cannot be a square and it is enough to study the case $n=12k+2$ and the sequence given by $b_n=a_{12n+2}$.
My bet now is on some variation of Zsigmondy's theorem. Numerical evidence gives that every $b_n$, for $n>0$, has a big prime divisor that appears with multiplicity one. If we manage to prove the last claim, we have that the previous solution ($n=2$) is the only one.
|
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|
How do I express 108 as the product of powers of its prime factors? How do I express $108$ as the product of powers of its prime factors?
I've got $2^2 \times 3^3$ is this correct?
|
First notice that $108$ is even so $2$ is a factor:
$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &54\end{matrix}$$
$54$ is also even:
$$\begin{matrix} & & 54 \\ & \large\swarrow & & \large\searrow \\ 2 & & & &27\end{matrix}$$
$27$ is not even so let's check the next prime: $3$. We can see that $3$ divides $27$ because $2+7=9$ is a multiple of $3$. So:
$$\begin{matrix} & & 27 \\ & \large\swarrow & & \large\searrow \\ 3 & & & &9\end{matrix}$$
And then hopefully you can see that $9$'s prime factorization is just $3\times 3$. So putting all of these steps together we see that
$$\begin{matrix} & & 108 \\ & \large\swarrow & & \large\searrow \\ \color{red}2 & & & & 54 \\ & & & \large\swarrow & & \large\searrow \\ & & \color{red}2 & & & & 27 \\ & & & & & \large\swarrow & & \large\searrow \\ & & & & \color{red}3 & & & & 9 \\ & & & & & & & \large\swarrow & & \large\searrow \\ & & & & & & \color{red}3 & & & & \color{red}3\end{matrix}$$
Thus
$$108 = 2\times 2 \times 3 \times 3\times 3 = 2^2\times 3^3$$
|
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|
Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$
|
$$\begin{align}
z^2+\sqrt{z^4+2z^3}-2\sqrt{z^4+z^3}
&=(\sqrt{z^4+2z^3}-\sqrt{z^4+z^3})+(z^2-\sqrt{z^4+z^3})\\
&={z^3\over\sqrt{z^4+2z^3}+\sqrt{z^4+z^3}}-{z^3\over z^2+\sqrt{z^4+z^3}}\\
&={z^3(z^2-\sqrt{z^4+2z^3})\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})}\\
&={z^3(-2z^3)\over(\sqrt{z^4+2z^3}+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+z^3})(z^2+\sqrt{z^4+2z^3})}\\
\end{align}$$
Can you take it from here?
|
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|
Prove , there exists $\theta , \phi \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$ and $f'(\phi)\neq 0$ Let the function
$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) & \tan(\frac{\pi}{6}) & \\ \sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) & \tan(\frac{\pi}{3}) \end{vmatrix} $$
where
$\theta \in \left[ \frac{\pi}{6},\frac{\pi}{3} \right]$ and $f'(\theta)$ denote the derivative of $f$ with respect to $\theta$. Which of the following statements is/are TRUE?
(I) There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$
(II) There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta)\neq 0$
*
*I only
*II only
*Both I and II
*Neither I Nor II
I try to explain $:$
$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{3}} & \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & \sqrt{3} \end{vmatrix} $$
$f(\theta) =$$\left(\frac{3}{2}-\frac{1}{2\sqrt{3}}\right)sin\left(\theta\right)-\left(\frac{1}{2\sqrt{3}}-\frac{1}{2}\right)cos\left(\theta\right)+\left(\frac{1}{4}-\frac{3}{4}\right)tan\left(\theta\right)$
Since ,
$f'(\theta) =$$\frac{1}{6}\left(-3\sec ^2\left(\theta\right)-\left(\sqrt{3}-9\right)\cos \left(\theta\right)+\left(\sqrt{3}-3\right)\sin \left(\theta\right)\right)$
I know option (3) is true , since it is possible for both statements .
Can you prove in formal way ?
|
Without explicitly expanding the determinant we see that
$$f(\pi/6) = f(\pi/3) = 0$$
since two of the row are equal in these cases giving us a zero determinant. The mean value theorem then tells us that there exists a $\theta\in(\pi/6,\pi/3)$ where $f'(\theta) = 0$.
If II is false then $f’(\theta) \equiv 0 \implies f(\theta) \equiv 0$ on $(\pi/6,\pi/3)$. Thus a single non-zero value of $f$ on the interval is enough to show that II is true. Take for example $\theta = \pi/4$.
|
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|
Factorial modular arithemetic
Find $50!/(10^{12}) \pmod{10}$.
Let $N$ represent the number then,
$N \equiv 0 \pmod{2}$
For mod 5 it is harder.
$N^{4} \equiv 1 \pmod{5}$.
How to solve this?
|
\begin{align}
\lfloor 50/2 \rfloor &= 25\\
\lfloor 25/2 \rfloor &= 12\\
\lfloor 12/2 \rfloor &= 6\\
\lfloor 6/2 \rfloor &= 3\\
\lfloor 3/2 \rfloor &= 1\\
\lfloor 1/2 \rfloor &= 0\\
25 + 12 + 6 + 3 + 1 &= 47
\end{align}
So $2^{47} \mid 50!$. It follows that $\dfrac{50!}{2^{12}} \equiv 0 \pmod 2$
\begin{align}
\lfloor 50/5 \rfloor &= 10\\
\lfloor 10/5 \rfloor &= 2\\
\lfloor 2/5 \rfloor &= 0\\
10+ 2 &= 12
\end{align}
So $5^{12} \mid 50!$.
Note
\begin{align}
50!
&= \prod_{n=0}^{9} (5n+1)\times(5n+2)\times(5n+3)\times(5n+4)\times(5n+5)\\
&= \left( \prod_{n=0}^{9} (5n+1)\times(5n+2)\times(5n+3)\times(5n+4)\right)
\left( \prod_{n=0}^{9} (5n+5)\right)\\
\end{align}
So
\begin{align}
\dfrac{50!}{5^{12}}
&= \left( \prod_{n=0}^{9} (5n+1)\times(5n+2)\times(5n+3)\times(5n+4)\right)
\dfrac{\prod_{n=0}^{9} (5n+5)}{5^{12}}\\
&\equiv \left( \prod_{n=0}^{9} (1\times 2\times 3\times 4)\right)
(1\times 2\times 3\times 4) \times 1 \times
(1\times 2\times 3\times 4) \times 2
\pmod 5\\
&\equiv (-1)^{12} \times 1 \times 2 \pmod 5\\
&\equiv 2 \pmod 5\\
\end{align}
That is, $\dfrac{50!}{5^{12}} \equiv 2 \pmod 5$
If $x \equiv a \pmod 2$ and $x \equiv b \pmod 5$, then $x \equiv 5a + 6b \pmod {10}$. So $\dfrac{50!}{5^{12}} \equiv 2 \pmod{10}$
|
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|
A sum including binomial coefficients I would like to prove the following equality:
$$\sum_k (-1)^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{k}=\sum_{k=0}^{n-2}\binom{2n-k-2}{n-1}\binom{n-2}{k}$$
but the power over two and the switch on the number of sums bothers me.
Any help would be welcome.
(the equality it part of Note 1.41 in the book "Analytic Combinatorics")
|
The following answer is purely algebraic. We transform both sides of OPs expression to finally obtain the same representation. We also use the coefficient of operator $[z^n]$ to denote the coefficient $a_n$ of $z^n$ of a series $\sum_{k=0}^{\infty}a_kz^k$.
At first we transform the right-hand side. It's the easier one.
\begin{align*}
\sum_{k=0}^{n-2}&\binom{2n-k-2}{n-1}\binom{n-2}{k}\\
&=\sum_{k=0}^{n-2}\binom{n+k}{n-1}\binom{n-2}{k}\tag{1}\\
&=\sum_{k=0}^{n-2}[z^{n-1}](1+z)^{n+k}\binom{n-2}{k}\tag{2}\\
&=[z^{n-1}](1+z)^n\sum_{k=0}^{n-2}\binom{n-2}{k}(1+z)^{k}\\
&=[z^{n-1}](1+z)^n(2+z)^{n-2}\tag{3}\\
&=\sum_{k=0}^{n-1}\left([z^k](1+z)^n\right)\left([z^{n-1-k}](2+z)^{n-2}\right)\\
&=\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-2}{n-1-k}2^{(n-2)-(n-1-k)}\\
&=\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-2}{k-1}2^{k-1}\tag{4}\\
\end{align*}
Comment:
*
*In (1) we change the order of summation by transforming the index $k$ with $n-2-k$
*In (2) we represent the binomial coefficient $\binom{n+k}{n-1}$ as the coefficient of $z^{n-1}$ of $(1+z)^{n+k}$
*In (3) we write the sum as polynomial $(2+z)^{n-2}$.
And here's the left-hand side.
\begin{align*}
\sum_{k=0}^{n-1}&(-1)^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{k}\\
&=(-1)^{n-1} \sum_{k=0}^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{n-1}\\
&=(-1)^{n-1} \sum_{k=1}^{n}(-2)^{k-1}\binom{n}{k}\binom{n+k-2}{n-1}\\
&=\frac{1}{2}(-1)^{n-1} \sum_{k=1}^{n}(-2)^{k}\binom{n}{k}[z^{n-1}](1+z)^{n+k-2}\\
&=\frac{1}{2}(-1)^{n-1} [z^{n-1}](1+z)^{n-2}\sum_{k=1}^{n}\binom{n}{k}(-2)^k(1+z)^{k}\\
&=\frac{1}{2}(-1)^{n-1} [z^{n-1}](1+z)^{n-2}\left\{(-1-2z)^n-1\right\}\\
&=\frac{1}{2} [z^{n-1}](1+z)^{n-2}\left\{(1+2z)^n+1\right\}\tag{5}\\
&=\frac{1}{2} [z^{n-1}](1+z)^{n-2}(1+2z)^n\\
&=\frac{1}{2}\sum_{k=0}^{n-1}\left([z^k](1+2z)^n\right)\left([z^{n-1-k}](1+z)^{n-2}\right)\\
&=\frac{1}{2}\sum_{k=0}^{n-1}\binom{n}{k}2^k\binom{n-2}{k-1}\tag{6}\\
\end{align*}
Since the expressions (4) and (6) are equal, the claim follows.
The summand with $k=0$ does not contribute anything. So we can start with $k=1$ and after an index transformation we obtain the somewhat more convenient representation
\begin{align*}
\sum_{k=0}^{n-2}\binom{n}{k+1}\binom{n-2}{k}2^k
\end{align*}
Comment:
*
*In (5) we need not to consider the summand $+1$ in the rightmost expression. It does not contribute anything to the coefficient $[z^{n-1}]$.
|
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|
If $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $ then find $a+b$ $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $
if I use this $$\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right| $$
I find $a=1,b=4$ but if I try to multiple by its conjugate
$$\lim _{x\rightarrow -\infty }\dfrac {x^{2}+6x+3-\left( ax+b\right) ^{2}} {\left| x\right| \sqrt {1+\dfrac {6} {x}+\dfrac {3} {x^{2}}}-ax-b}=\lim _{x\rightarrow -\infty } \dfrac {x\left( x+6+\dfrac {3} {x}-a^{2}x-2ab-\dfrac {b^{2}} {x}\right) } {x\left( -\sqrt {1+\dfrac {6} {x}+\dfrac { {3}} {x^{2}}}-a-\dfrac {b} {x}\right) }$$
in order to lim of this equal to $1$ $$x\left(\underbrace { 1-a^{2}}_0\right) +6-2ab=-1-a$$
if $a=1$ then $ b=4$
if $a=-1$ so $b=-3$ but wolfram says $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}-x-3=\infty$
can u tell me where is my mistake?
|
Here is a slightly different approach.
First note that
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b}{x^1}&=\lim_{x\to \infty}\sqrt {1-\frac6x+\frac3{x^2}}-a+\frac bx\\
&=1-a
\end{align}
and
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b-(1-a)x}{x^0}&=\lim_{x\to \infty}\sqrt {x^{2}-6x+3}+b-x\\
&=\lim_{x\to \infty}\frac{2(b-3)x+3-b^2}{\sqrt {x^{2}-6x+3}-b+x}\\
&=b-3
\end{align}
and finally
\begin{align}
\lim_{x\to \infty}\frac{\sqrt {x^{2}-6x+3}-ax+b-(1-a)x-(b-3)}{x^{-1}}&=...\\
&=-3
\end{align}
Putting all this together we see that $$\sqrt {x^{2}-6x+3}-ax+b=(1-a)x+(b-3)-\frac3x+O(\frac1{x^2}).$$
The conclusion is yours!
|
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|
Verifying partial order relation I have the following question where i have to verify if the relation is partial order:
$A=\{1,2,3,\ldots,100\}$, relation $x\mathrel{R}y \leftrightarrow \frac{y}x=2^k$, where $k\ge 0$ is an integer.
This how i am thinking :
Reflexive: $x\mathrel{R}x\implies\frac{x}x=1$, therefore the relation is reflexive !
Symmetric: $x\mathrel{R}y\implies\frac{y}x=2^k \Leftrightarrow\frac{x}y=2^k$? $\frac42=2$ while $\frac24= 0.5$, thefore not symmetric !
AntiSymmetric: $\implies\frac{y}x=2^k$, $\frac{x}y=2^k\implies \frac{y}x=\frac{x}y\implies x=y$. eg $\frac21=\frac12$? clearly not, therefore not AntiSymmetric !
Transitive: $\implies\frac{y}x=2^k,\frac{z}y=2^k\implies \frac{z}x=2^k$? not really sure how to check the transitivity here
|
Your argument for reflexivity is correct: if $x\in A$, then $\frac{x}x=1=2^0$, so $x\mathrel{R}x$. Your example showing that $R$ is not symmetric is correct, though you’ve not explained it adequately: $\frac42=2=2^1$, so $2\mathrel{R}4$, but $\frac24=\frac12$ is not of the form $2^k$ for any integer $k\ge 0$, so $4\not\mathrel{R}2$, and therefore $R$ is not symmetric.
However, $R$ is antisymmetric. Suppose that $x,y\in A$, $x\mathrel{R}y$, and $y\mathrel{R}x$. Then by definition $\frac{y}x=2^k$ for some integer $k\ge 0$, and $\frac{x}y=2^\ell$ for some integer $\ell\ge 0$. But then
$$2^\ell=\frac{x}y=\left(\frac{y}x\right)^{-1}=(2^k)^{-1}=2^{-k}\;,$$
so $\ell=-k\le 0$. Thus, $0\le\ell\le 0$, so $\ell=0$, and $x=y$. This shows that if $x\mathrel{R}y$ and $y\mathrel{R}x$, then $x=y$, which is exactly what it means for $R$ to be antisymmetric.
To show that $R$ is transitive, start by assuming that $x,y,z\in A$, $x\mathrel{R}y$, and $y\mathrel{R}z$; you need to use this information to prove that $x\mathrel{R}z$. Use the definition of $R$ to translate these into more basic statements about $x,y$, and $z$. For instance, since $x\mathrel{R}y$, there is an integer $k\ge 0$ such that $\frac{y}x=2^k$. Similarly, since $y\mathrel{R}z$, there is an integer $\ell\ge 0$ such that $\frac{z}y=2^\ell$. Now combine these equations to find one involving $\frac{z}x$.
|
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|
A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$ Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.
In my view, maybe we can use
$${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$
And then
$${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$
By Stolz formula, we have
\begin{align*}
&\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\
= &\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\
= &\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right].
\end{align*}
And how can we continue?
|
We have:
$$ a_n(a_{n+1}-a_n) = n $$
so:
$$ a_{n+1}^2-a_{n}^2 = a_{n+1}(a_{n+1}-a_n) + n = n\left(1+\frac{a_{n+1}}{a_n}\right)=2n+\frac{n^2}{a_n^2}$$
and:
$$ a_{N+1}^2-a_1^2 = N(N+1)+\sum_{n=1}^{N}\frac{n^2}{a_n^2} $$
from which $a_{N+1}\geq \sqrt{N(N+1)}$ and $a_n\geq \sqrt{(n-1)n}$.
If we plug this inequality back into the previous line, we get:
$$\begin{eqnarray*} a_{N+1}^2 &\leq& N(N+1)+a_1^2+\frac{1}{a_1^2}+\sum_{n=2}^{N}\frac{n}{n-1}\\&=& (N+1)^2+\left(a_1-\frac{1}{a_1}\right)^2+H_{N-1}.\end{eqnarray*} $$
The process continues by keep turning lower/upper bounds into tighter upper/lower bounds.
Can you check it proves your statement?
|
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|
Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ I've seen this identity on examsolutions, but I'm unsure on how to prove it.
$$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
|
Notice,
$$RHS=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$
$$=2\cos \left(\frac{A}{2}+\frac{B}{2}\right)\cos \left(\frac{A}{2}-\frac{B}{2}\right)$$
$$=2 \left(\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\right)\left(\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\sin^2\frac{A}{2}\sin^2\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-\left(1-\cos^2\frac{A}{2}\right)\left(1-\cos^2\frac{A}{2}\right)\right)$$
$$=2 \left(\cos^2\frac{A}{2}\cos^2\frac{B}{2}-1+\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-\cos^2\frac{A}{2}\cos^2\frac{B}{2}\right)$$
$$=2 \left(\cos^2\frac{A}{2}+\cos^2\frac{B}{2}-1\right)$$
$$=2\cos^2\frac{A}{2}+2\cos^2\frac{B}{2}-2$$
$$=\left(2\cos^2\frac{A}{2}-1\right)+\left(2\cos^2\frac{B}{2}-1\right)$$
$$=\cos A+\cos B=LHS$$
|
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|
multiplication over gf(16) Can some one show me how to do multiplication over gf(16) step by step
I found this example online, http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/12-13_Finite_Fields.html#An_Example.
An Example:
$x^4 = x+1$
$x^5 = x^2+x$
$(x^2+x+1) (x^3+x^2+1)$
$= x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1$
$= x^2 + 1$
I know $x^4 = x+1$, because $x^4 mod (x^4 + x + 1)$, but how does $x^5 = x^2+x$?
|
A better way to do the multiplication (not necessarily more efficient, but a better way to think about it) is by using remainders of polynomial division. So for example, $x^{5} = x(x^{4}+x+1) + x + 1$, and since $x^{4}+x+1 = 0$, $x^{5} = x+1$. You really are looking at polynomials $\bmod{(x^{4}+x+1)}$.
|
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|
Probability to identify faulty machines.
There are four machines and it is known that exactly
two of them are faulty. They are tested, one by one,
in a random order till both the faulty machines are
identified. Find the probability that exactly $3$ tests will
be required to identify the $2$ faulty machines.
$a.)\ \dfrac{1}{2} \\
b.)\ 1 \\
\color{green}{c.)\ \dfrac{1}{3}} \\
d.)\ \dfrac{2}{3} $
I did $\dbinom{3}{2} \times \dfrac{2}{4} \times \dfrac{1}{3}= \dfrac{1}{2} $
But the answer given is option $c.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.
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The probability for faulty-not faulty-faulty is
$\frac{1}{2}\times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The probability for not faulty-faulty-faulty is
$\frac{1}{2} \times \frac{2}{3} \times \frac{1}{2}=\frac{1}{6}$
The sum is $\frac{1}{3}$.
|
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|
Integration with substitution of trig: $\int\frac1{(x^2-4)^{3/2}}dx$ $$\int\frac1{(x^2-4)^{3/2}}dx $$
I'm unsure how to go about this integral after subbing $x=2\cosh(\theta)$; to the power of $3/2$ is confusing me when trying to simplify.
Thanks.
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Let $u^2x^2=x^2-4$. Then $x=\pm\sqrt{\frac{-4}{u^2-1}}$, $dx=\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du$.
$$\int\frac{1}{\left(x^2-4\right)^{3/2}}\, dx=\int\frac{\pm(-4)\sqrt{\frac{u^2-1}{-16}}\left(\frac{-2u}{\left(u^2-1\right)^2}\right)\,du}{u^3\cdot \pm\left(\frac{-4}{u^2-1}\right)\sqrt{\frac{-4}{u^2-1}}}$$
$$=\frac{1}{4}\int\frac{1}{u^2}\, du=\frac{x}{-4\sqrt{x^2-4}}+C$$
|
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|
$-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$. Show that $-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$.
Here its plotting.
By first derivative it involves natural log and expression become more difficult.
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Note that $a^x$ is a decreasing convex function of $a>0$ for $x<0$. Hence
$$-3\cdot7^x+2\cdot 6^x+5^x>0$$
$$5^x-2\cdot 4^x+3^x>0$$
Add these up to get the needed inequality.
|
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Find the value of $(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots$. Show that: $$(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots = \dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}.$$
By the duplication theorem and $\Gamma(z+1)=z\,\Gamma(z)$, we have: $$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=\dfrac{\Gamma\left(-\dfrac{z}{2}\right)}{z\,\Gamma\left(-z\right)\Gamma\left(\dfrac{z}{2}\right)} \cdot 2^{-z}.$$
Using the definition $\Gamma\left(z\right) = \dfrac{1}{z}\prod_{k=1}^\infty \left(1+\dfrac{1}{k}\right)^z \left(1+\dfrac{z}{k}\right)^{-1}$, and simplifying terms, we obtain:$$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=2^{-z}(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots .$$
How can I get rid of the $2^{-z}$, and am I going about this the right way?
Cheers!
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I would do the following: consider a finite product
$$\mathcal{P}_N:=(1-z)\left(1+\frac z{2}\right)\ldots \left(1-\frac{z}{2N-1}\right)\left(1+\frac{z}{2N}\right)$$
and rewrite it as
\begin{align*}\mathcal{P}_N&=\frac{\prod_{k=0}^{N-1}\left(2k+1-z\right)\prod_{k=0}^{N-1}\left(2k+2+z\right)}{(2N)!}=\\
&=\frac{2^{2N}}{(2N)!}{\color{red}{\prod_{k=0}^{N-1}\left(k+\frac{1-z}{2}\right)}}{\color{blue}{
\prod_{k=0}^{N-1}\left(k+1+\frac{z}{2}\right)}}=\\&=
\frac{2^{2N}}{(2N)!}{\color{red}{\frac{\Gamma\left(N+\frac{1-z}{2}\right)}{
\Gamma\left(\frac{1-z}{2}\right)}}}{\color{blue}{\frac{\Gamma\left(N+1+\frac{z}{2}\right)}{
\Gamma\left(1+\frac{z}{2}\right)}}}=\\
&=\frac1{\Gamma\left(\frac{1-z}{2}\right)\Gamma\left(1+\frac{z}{2}\right)}
\times\underbrace{
\frac{2^{2N}\Gamma\left(N+\frac{1-z}{2}\right)\Gamma\left(N+1+\frac{z}{2}\right)}{\Gamma(1+2N)}}_{\mathcal{Q}_N}.
\end{align*}
Now it suffices to use Stirling formula to show that $\mathcal{Q}_{\infty}=\sqrt\pi$. Indeed, as $N\to\infty$, one has
\begin{align*}
\mathcal Q_N&= 2^{2N}\frac{\sqrt{\frac{2\pi}{N+\frac{1-z}{2}}}\left(\frac{N+\frac{1-z}{2}}{e}\right)^{N+\frac{1-z}{2}}\cdot \sqrt{\frac{2\pi}{N+1+\frac{z}{2}}}\left(\frac{N+1+\frac{z}{2}}{e}\right)^{N+1+\frac{z}{2}}}{\sqrt{\frac{2\pi}{2N+1}}\left(\frac{2N+1}{e}\right)^{2N+1}}\left[1+O\left(N^{-1}\right)\right]=\\
&= \sqrt{\pi}\cdot e^{-\frac12}
\frac{\left(1+\frac{1-z}{2N}\right)^N\left(1+\frac{1+\frac z2}{N}\right)^N}{\left(1+\frac1{2N}\right)^{2N}}\left[1+O\left(N^{-1}\right)\right]=
\sqrt{\pi}\left[1+O\left(N^{-1}\right)\right].
\end{align*}
|
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|
Show that $\lim \limits_{x \to \infty} (x + x^2 \log \frac{x}{x+1}) = \frac{1}{2}$ I am self studying real analysis and I found this problem in a book.
We have to show that if $x > 1$ then
$$\lim_{x \to \infty} \left[ x + x^2 \log \frac{x}{x+1} \right] = \frac{1}{2}$$.
I can prove that the sequence is bounded.
It is bounded below by 0.
Suppose not, then there exists a $x > 1$ such that
$$x + x^2 \log \frac{x}{x + 1} < 0$$.
Which in turn implies
$$1 + \frac{1}{x} > e^{\frac{1}{x}}$$,
a contradiction.
It is bounded above by 1.
Suppose not, then there exists a $x > 1$ such that
$$x + x^2 \log \frac{x}{x + 1} > 1$$.
Which means $1 + \frac{1}{x} < e^{\frac{1}{x} - \frac{1}{x^2}}$.
Which in turn implies
$$\frac{1}{x^2} < \sum_{n = 2}^{\infty} \frac{\left( \frac{1}{x} - \frac{1}{x^2} \right)^n}{n!}$$.
Now,
$$\sum_{n = 2}^{\infty} \frac{\left( \frac{1}{x} - \frac{1}{x^2} \right)^n}{n!} < \sum_{n = 2}^{\infty} \left( \frac{1}{x} - \frac{1}{x^2} \right)^n = \left( \frac{1}{x} - \frac{1}{x^2} \right)^2 \sum_{n = 0}^{\infty} \left( \frac{1}{x} - \frac{1}{x^2} \right)^n = \frac{1}{x^2} \frac{x^2 - 2x + 1}{x^2 - x + 1} < \frac{1}{x^2}$$a contradiction.
However the sequence is not monotonic.
Let $u(x) = x + x^2 \log \frac{x}{x+1}$.
Then $u(183073) \approx 0.5000000077125151$, $u(183074) \approx 0.4999997131235432$ and $u(183089) \approx 0.5000000357104$
Any help on this?
Thanks.
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Set $1/x=h$
to get $$\lim_{h\to0}\dfrac{h-\ln(1+h)}{h^2}$$
Now use $\ln(1+h)=h-\dfrac{h^2}2+O(h^3)$
|
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|
Finding first three terms of a GP whose sum and sum of squares is given? We have to find three numbers in G.P. such that their sum is $\frac{13}{3}$ and the sum of their squares is $\frac{91}{9}$.
Here's what I have tried to do:-
$$a+ar+ar^2=\frac{13}{3}$$
$$a^2+a^2r^2+a^2r^4=\frac{91}{9}$$
$$\frac{a^2+a^2r^2+a^2r^4}{a+ar+ar^2}=\frac{91}{9}*\frac{3}{13}$$
$$\frac{a+ar^2+ar^4}{1+r+r^2}=\frac{7}{3}$$
We have not been taught how to solve a biquadratic equation in school, so how should I proceed from here on?
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$a+ar+ar^2=\frac{13}{3}$ [equation 1]
$a^2+a^2r^2+a^2r^4=\frac{91}{9}$ [equation 2]
Square the first equation:
$a^2(1+r+r^2)^2 = \frac{169}{9}$
$a^2(1 + r^2 + r^4 + 2r + 2r^2 + 2r^3) = \frac{169}{9}$ [equation 3]
Now take equation 3 minus equation 2:
$2a^2r(1 + r + r^2) = \frac{26}{3}$
Using equation 1 again,
$2ar \frac{13}{3} = \frac{26}{3}$
So $ar = 1 \implies r = \frac{1}{a}$
Now put that back into equation 1 to get:
$a + 1 + \frac 1a = \frac{13}{3}$
Multiply throughout by $a$, rearrange:
$a^2 - \frac{10}{3}a + 1 = 0$
Solving that will give you the nice solutions $a = 3$ or $a = \frac 13$, which are both admissible, and the respective values of $r$ are the reciprocals of $a$, $\frac 13$ and $3$.
So the possible sequences are $3,1,\frac 13$ and $\frac 13, 1, 3$. Both possibilities are admissible because they haven't told you if the GP is increasing or decreasing.
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|
Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$ Evaluate the Integral: $\int^{0.6}_0\frac{x^2}{\sqrt{9-25x^2}}dx$
I believe my work is correct until I try to change the bounds according to theta, $\theta$.
How do I do that? Also, please comment freely on my work, that is, is my method correct and cogent.
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Notice, you let $$x=\frac{3}{5}\sin \theta\implies \theta=\sin^{-1}\left(\frac{5x}{3}\right)$$ then
upper limit at $x=0.6$ : $\theta=\sin^{-1}\left(\frac{5\times 0.6}{3}\right)=\sin^{-1}\left(1\right)=\frac{\pi}{2}$
lower limit at $x=0$ : $\theta=\sin^{-1}\left(\frac{5\times 0}{3}\right)=\sin^{-1}\left(0\right)=0$
after substitution, we get
$$\int_{0}^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\ dx=\int_{0}^{\pi/2}\frac{\left(\frac{3}{5}\sin\theta\right)^2}{\sqrt{9-9\sin^2\theta}}\frac{3}{5}\cos\theta\ d\theta$$
$$=\frac{9}{25}\cdot \frac{3}{5}\cdot \frac{1}{3}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{\sqrt{\cos^2\theta}}$$
$$=\frac{9}{125}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{|\cos\theta|}$$
we know $|\cos\theta|=\cos \theta\ \ \forall \ \ 0\le \theta \le\pi/2$
$$=\frac{9}{125}\int_{0}^{\pi/2}\frac{\sin^2\theta\cos \theta d\theta}{\cos\theta}$$
$$=\frac{9}{125}\int_{0}^{\pi/2}\sin^2\theta\ d\theta$$
|
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|
Solve for $x$: $x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$ Solve for $x$: $$x\lfloor x + 2 \rfloor +\lfloor 2x - 2 \rfloor +3x =12$$
My attempt: I have changed this equation into the fractional part function. so that we know $0 \leq \{x\}<1$. I have final equation $x^2 +7x-x\{x+2\}-\{2x-2\}=14$. How to proceed now?
|
Since
$$\lfloor x+2\rfloor=\lfloor x\rfloor+2,\quad \lfloor 2x-2\rfloor=\lfloor 2x\rfloor-2$$
the equation can be written as
$$x(\lfloor x\rfloor+2)+\lfloor 2x\rfloor-2+3x=12,$$
i.e.
$$x\lfloor x\rfloor+5x+\lfloor 2x\rfloor=14\tag 1$$
Now let us separate it into cases :
Case 1 : $x=m+\alpha$ where $m\in\mathbb Z$ and $0\le\alpha\lt \frac 12$.
$$\begin{align}(1)&\Rightarrow (m+\alpha)m+5m+5\alpha+2m=14\\&\Rightarrow (m+5)\alpha=-m^2-7m+14\\&\Rightarrow \alpha=\frac{-m^2-7m+14}{m+5}\qquad (\text{since $m+5\not=0$})\\&\Rightarrow 0\le \frac{-m^2-7m+14}{m+5}\lt \frac 12\end{align}$$
There is no such $m$.
Case 2 : $x=m+\alpha$ where $m\in\mathbb Z$ and $\frac 12\le\alpha \lt 1$.
$$\begin{align}(1)&\Rightarrow (m+\alpha)m+5m+5\alpha+2m+1=14\\&\Rightarrow (m+5)\alpha=-m^2-7m+13\\&\Rightarrow \alpha=\frac{-m^2-7m+13}{m+5}\qquad (\text{since $m+5\not=0$})\\&\Rightarrow \frac 12\le \frac{-m^2-7m+13}{m+5}\lt 1\\&\Rightarrow m=1,\alpha=\frac 56\end{align}$$
Hence, $\color{red}{x=\frac{11}{6}}$ is the only solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1510756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Solving the equation $z^7=-1$
Solve the equation $z^7=-1$
My attempt:
$$z=x+yi$$
$$(x+yi)^7+1=0$$
$$(x^2+2yi-y^2)^3(x+yi)+1=0$$
but now it's start to look ugly.
I'm sure that there is a simple way
|
$$z^7=-1\Longleftrightarrow$$
$$z^7=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$z^7=1\cdot e^{\pi i}\Longleftrightarrow$$
$$z^7=e^{\pi i}\Longleftrightarrow$$
$$z=\left(e^{\left(\pi+2\pi k\right)i}\right)^{\frac{1}{7}}\Longleftrightarrow$$
$$z=e^{\frac{1}{7}\left(\pi+2\pi k\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-6$
So the solutions are:
$$z_0=e^{\frac{1}{7}\left(\pi+2\pi\cdot 0\right)i}=e^{\frac{\pi}{7}i}$$
$$z_1=e^{\frac{1}{7}\left(\pi+2\pi\cdot 1\right)i}=e^{\frac{3\pi}{7}i}$$
$$z_2=e^{\frac{1}{7}\left(\pi+2\pi\cdot 2\right)i}=e^{\frac{5\pi}{7}i}$$
$$z_3=e^{\frac{1}{7}\left(\pi+2\pi\cdot 3\right)i}=e^{\pi i}=-1$$
$$z_4=e^{\frac{1}{7}\left(\pi+2\pi\cdot 4\right)i}=e^{-\frac{5\pi}{7}i}$$
$$z_5=e^{\frac{1}{7}\left(\pi+2\pi\cdot 5\right)i}=e^{-\frac{3\pi}{7}i}$$
$$z_6=e^{\frac{1}{7}\left(\pi+2\pi\cdot 6\right)i}=e^{-\frac{\pi}{7}i}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1511178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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|
Linear Algebra Matrix and Inverse I have 4x4 matrices:
$$A\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix} =
\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}
$$
and
$$\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix}
B = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$$
and I need to find $A^{-1}$ and $B^{-1}$.
Now, there's the long way of doing it (augmented matrix with identity on the right side and solving for A, then finding A inverse, B, then B inverse), but I thought this was too much work and there must be an efficient way of doing it.
I've noticed that the right side of the equation for both A[] and []B look like they are just another arrangement of identity matrix, which must provide some useful clue to this problem.
I have a gut feeling that we must do something with elementary row matrices but don't have a concrete idea to get me started..
Can anyone give me some insight or introduce efficient way of solving this question?
(Without determinant or long mechanical computation)
Thank you.
|
The right hand matrix is known as permutation matrix and its inverse is its transpose, i.e.
$$
P^{-1}=\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}^T=\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}=[e_2\: e_4\: e_3 \:e_1]=\begin{bmatrix}e_4'\\e_1'\\e_3'\\e_2'\end{bmatrix}
$$
where $e_i$ is elementary column vector and $e_i'$ is elementary row vector. Also $Ae_i$ is the $i$th column of $A$ and $e_i'A$ is the $i$th row of $A$.
So
$$
A^{-1}=\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix}[e_2\: e_4\: e_3 \:e_1]=\begin{bmatrix}
1&-4&3&3\\
4&10&8&6\\
2&-1&5&3\\
5&-4&-2&-9
\end{bmatrix}
$$
And
$$
B^{-1}=\begin{bmatrix}e_4'\\e_1'\\e_3'\\e_2'\end{bmatrix}\begin{bmatrix}
3&1&3&-4\\
6&4&8&10\\
3&2&5&-1\\
-9&5&-2&-4
\end{bmatrix}=\begin{bmatrix}
-9&5&-2&-4\\
3&1&3&-4\\
3&2&5&-1\\
6&4&8&10
\end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1512344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Pythagorean Triple Inequality When finding the Pythagorean triple where $a+b+c=1000$,
Wolfram alpha shows me that $a< -500\left(\sqrt{2} - 2\right)$
When I input $a^2+b^2=c^2, a<b<c$ and $a+b+c=1000$
How does wolfram arrive at that inequality:
$a< -500\left(\sqrt{2} - 2\right)$
Here is the link: Wolfram
|
For the inequality, we observe that since $b > a$, we must have
$$
c^2 = a^2+b^2 > 2a^2
$$
or
$$
c > a\sqrt{2}
$$
Thus
$$
1000 = a+b+c > a+a+a\sqrt{2} = a(2+\sqrt{2})
$$
or
$$
a < \frac{1000}{2+\sqrt{2}} = \frac{1000(2-\sqrt{2})}{2^2-(\sqrt{2})^2}
= 500(2-\sqrt{2})
$$
As regards the original problem: Primitive Pythagorean triples can be obtained using the well-known schema
$$
a, b = u^2-v^2, 2uv
$$
in some order, with $u > v$ both integers, and
$$
c = u^2+v^2
$$
Note that in this case,
$$
a+b+c = 2u^2+2uv = 2u(u+v)
$$
In order for $ka+kb+kc = 1000$ for some integer $k$, we need $u(u+v) \mid 500$. For example, with $u = 20, v = 5$, we obtain $a = 2uv = 200, b = u^2-v^2 = 375, c = u^2+v^2 = 425$. Note that $a+b+c = 200+375+425 = 1000$, and $a^2+b^2 = 40000+140625 = 180625 = c^2$.
That this is the only solution can be demonstrated by noting first that $500 = 2^2 \times 5^3$, and observing that for $u$ and $u+v$, we need two disjoint subsets of factors whose separate products differ by less than a factor of $2$.* This only happens for the above case with $20$ and $25$ (yielding $u = 20, v = 5$), and also with $4$ and $5$ (yielding $u = 4, v = 1$, and requiring us to scale the resulting triple by a factor of $25$). Since these two pairs are in direct proportion to each other, they produce the same unique solution.
*ETA: Only if we require $a > 0$. If $a$ can be less than $0$, then we only need disjoint factor subsets $u$ and $u+v$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1516450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:
Prove that $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $
where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma function.
I've tried reflection formula and some Feynmann method on the integrals in the wikipedia but nothing works.
(I've copied this from AoPS because there's no response there)
|
A little late, but here is yet another solution not shown previously.
I will first show that
$$\int_0^1 x \ln\left(\Gamma(x)\right)\,dx=\frac{1}{4} \ln \frac{2 \pi}{A^4} \tag{1}$$
And then from $(1)$ derive the desired result.
Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{2}$$
Then, if we plug $(2)$ in $(1)$
$$
\begin{aligned}
\int_0^1 x \ln\left(\Gamma(x)\right)\,dx&=\frac{\ln 2 \pi}{2}\int_0^1 x\,dx+ \sum_{k=1}^{\infty}\frac{1}{2k}\int_0^1 x \cos\left(2 \pi k x\right)\,dx + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\int_0^1 x \sin\left(2 \pi k x \right)\,dx\\
&=\frac{\ln 2 \pi}{4}+ \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\left(-\frac{1}{2 \pi k} \right)\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln\left(2 \pi k\right)}{k^2}\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{2\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln k}{k^2}\\
&=\frac{\ln 2 \pi}{4}-\frac{\gamma}{12}-\frac{\ln 2 \pi}{12}-\frac{1}{2\pi^2}\sum_{k=1}^{\infty}\frac{\ln k}{k^2}\\
&=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{1}{2\pi^2}\zeta^\prime(2)\\
&=\frac{\ln 2 \pi}{6}-\frac{\gamma}{12}+\frac{\zeta(2)}{2\pi^2} \left(-12 \ln A + \gamma + \ln2 \pi \right)\\
&=\frac{\ln 2 \pi}{6}- \ln A +\frac{\ln2 \pi}{12} \\
&=\frac{\ln 2 \pi}{4}- \ln A \\
&=\frac{1}{4} \ln \frac{2 \pi}{A^4} \qquad \blacksquare\\
\end{aligned}
$$
We used that $\zeta^\prime(2) =\zeta(2)\left(-12 \ln A + \gamma + \ln2 \pi \right)$
On the other hand, if we integrate by parts the L.H.S. of $(1)$ we have that ($du=\psi(x)$ and $v=\frac{x^2}{2}$):
$$\int_0^1 x^2 \psi(x)\,dx =-2\int_0^1 x \ln\left(\Gamma(x)\right)\,dx\tag{3}$$
Which imiediatly gives the desired result
$$\boxed{\int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right)}$$
Appendix
$$
\begin{aligned}
\int_0^1 x \cos(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{2 \pi k} x \cos(x)\,dx \qquad (2 \pi kx \to x)\\
&=\frac{1}{(2 \pi k)^2}\left(x\sin(x)\Big|_0^{2 \pi k} -\int_0^{2 \pi k}\sin(x)\,dx\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-\cos(x)\Big|_0^{2 \pi k}\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-1+1\right)\\
&=0 \qquad \blacksquare
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 x \sin(2 \pi kx)\,dx&=\frac{1}{(2 \pi k)^2}\int_0^{2 \pi k} x \sin(x)\,dx \qquad (2 \pi kx \to x)\\
&=\frac{1}{(2 \pi k)^2}\left(-x\cos(x)\Big|_0^{2 \pi k} +\int_0^{2 \pi k}\cos(x)\,dx\right)\\
&=\frac{1}{(2 \pi k)^2}\left(-2 \pi k\right)\\
&=-\frac{1}{2 \pi k}\qquad \blacksquare
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1517085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Suppose ${3^x} + {4^x} = {5^x}$. What is zero of this equation? Suppose ${3^x} + {4^x} = {5^x}$.
What is zero of this equation?
|
$x=2$
Since
${\left( {\frac{3}{5}} \right)^x} + {\left( {\frac{4}{5}} \right)^x} = 1 = {(\sin x)^2} + {(\cos x)^2} \Rightarrow x = 2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1518591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to integrate a 4th power of sine and cosine? I'm having some trouble figuring out the right substitutions to make to integrate
$$\int \sin^4(\theta)d\theta$$
and
$$\int \cos^4(\theta)d\theta$$
Any hints or suggestions are welcome.
Thanks,
|
HINT (using partial integration):
$$\int\sin^4(x)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\sin^2(x)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)+\frac{3}{4}\int\left(\frac{1}{2}-\frac{1}{2}\cos(2x)\right)\space\space\text{d}x=$$
$$-\frac{1}{4}\sin^3(x)\cos(x)-\frac{3}{8}\int\cos(2x)\space\space\text{d}x+\frac{3}{8}\int 1\space\space\text{d}x$$
$$\int\cos^4(x)\space\space\text{d}x=$$
$$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{4}\int\cos^2(x)\space\space\text{d}x=$$
$$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{4}\int\left(\frac{1}{2}+\frac{1}{2}\cos(2x)\right)\space\space\text{d}x=$$
$$\frac{1}{4}\sin(x)\cos^3(x)+\frac{3}{8}\int\cos(2x)\space\space\text{d}x+\frac{3}{8}\int 1\space\space\text{d}x$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1520224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.