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can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution using only high school methods? i guess i have to do an algebraic substitution to reduce $x^3+x$ to polynomial of degree 2?? i also know $$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$$ where $r_1,r_2$ are roots of polynomial of degree 2
|
we have $x^4-3x^3+2x^2-3x+1=0$ and we see that $x=0$ is not a solution, since we can write
$x^2+1/x^2-3(x+1/x)+2=0$
setting $t=x+1/x$ we get $x^2+1/x^2=t^2-2$ and we have a quadratic equation
$t^2-3t=0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$ Evaluate
$$\displaystyle \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx$$
$\bf{My\; Try::}$
Let $$\begin{align}I &= \int e^{x\sin x+\cos x}\left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx\\
&=\int e^{x\sin x+\cos x}\left(x^2\cos x+\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx\\
&= \int x\cdot e^{x\sin x+\cos x}\left(x\cos x\right)dx+\int e^{x\sin x+\cos x}\left(\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx\\
\end{align}$$
Now Let $x\sin x+\cos x = t\;,$ Then $x\cos x\,dx = dt$ and Integration by parts for $\bf{1^{st}}$ Integral
So $$\displaystyle I = x\cdot e^{x\sin x+\cos x}-\int e^{x\sin x+\cos x}dx+\int e^{x\sin x+\cos x}\left(\frac{\cos x-x\sin x}{x^2\cos^2 x}\right)dx$$
Now I do not understand how to solve after that.
|
Of course $\left( e^{x\sin x + \cos x}\right)' = x\cos x . e^{x\sin x + \cos x}$. But in the integrand we have terms such as $x^2\cos x.e^{...}$. So let's try
$$\left(x e^{x\sin x + \cos x}\right)' = (x^2 \cos x + 1)e^{...}$$
Hence the integral
$$I = x e^{x\sin x + \cos x} + \int e^{x\sin x + \cos x} \left( \frac{-\sin x}{x\cos^2 x} + \frac{1}{x^2\cos x} - 1 \right) \ dx$$
If we're lucky, the integrand of this new integral is exact in as much as it equals $$\left( f(x) e^{x\sin x + \cos x} \right)' = e^{x\sin x + \cos x} \left( f'(x) + x\cos x . f(x) \right) $$
Based on first two terms, something like $\displaystyle f(x) = -\frac{1}{x \cos x}$ looks like a good candidate. In fact that works. Hence
$$I = e^{x\sin x + \cos x} \left( x - \frac{1}{x\cos x} \right) + C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Optimization problem - Trapezoid under a parabola recently I've been working on a problem from a textbook about Optimization. The result that I get is $k = 8$, even thought the answer from the textbook is $k = \frac{32}{3}$
The problem follows:
--
The x axis interepts the parabola $12-3x^2$ at the points $A$ and $B$, and also the line $y = k$ (for $0 < k < 12$) at the points C and D. Determine $k$ in a way that the trapezoid $ABCD$ has a maximum area.
--
My solution was this
--
The trapezoid area is
$$A_{T} = \frac{(B+b) \cdot h}{2} = \frac{4+2 \cdot \sqrt{\frac{12-k}{3}} \cdot k}{2}$$
$$A_{T}' = 0 \therefore \frac{\sqrt{12-k}}{\sqrt{3}} - \frac{k}{\sqrt{3} \cdot 2 \cdot \sqrt{12-k}} = 0$$
$$2(12-k)-k=0 \therefore k = 8$$
--
Where did I go wrong on?
Thank you guys!
|
An alternative method:
Once you get to expressing $A$ in terms of $k$, substitute $k=12\sin^2\theta$ to help avoid messy surds.
$$\begin{align}\\
A&=k(2+\sqrt{4-\frac k3})\\
&=12\sin^2\theta(2+\sqrt{4-4\sin^2\theta})\\
&=24\sin^2\theta(1+\cos\theta) \\
\frac{dA}{d\theta}&=24[2\sin\theta\cos\theta(1+\cos\theta)+\sin^2\theta(\sin\theta)]\\
&=24\sin\theta (\cos\theta+1)(3\cos\theta-1)=0\;\text{when}\\
\cos\theta&=\frac 13\Rightarrow \cos^2\theta=\frac19\Rightarrow \sin^2\theta=\frac89\qquad \text{(NB: $\theta\neq0$ as $k\neq0$)}\\
\Rightarrow k&=\frac{32}3\qquad \blacksquare\end{align}$$
|
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|
Find the minimum value of the expression $P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$ Let $x,y,z$ be positive real number such that $xy+yz+zx=3$. Find the minimum value of the expression $$P=\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}$$
|
It is easy to guess from the symmetry that $x=y=z=1$ should give the minimum of $1$, and we will try proving this. By Cauchy-Schwarz Inequality:
$$\left(\sum_{cyc} \frac{x^2}{\sqrt{x^3+8}} \right)\left(\sum_{cyc} \sqrt{x^3+8} \right) \ge (x+y+z)^2 $$
We will use $\sum $ to denote cyclic sums. From the above, if we show that $(x+y+z)^2 \ge \sum \sqrt{x^3+8} $, then we have $P \ge 1$.
As you noted in comments, we can use $2\sqrt{x^3+8} \le (x+2)+(x^2-2x+4) = x^2-x+6$, so it is enough to show that:
$$2\sum x^2+4\sum xy \ge \sum x^2-\sum x+18 \iff \sum x^2+\sum x \ge 6$$
But $\sum x^2 \ge \sum xy = 3$ and $\left( \sum x\right)^2 \ge 3\sum xy = 9 \implies \sum x \ge 3$, so this is true.
|
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|
If the number $x$ is algebraic, then $x^2$ is also algebraic Prove that if the number $x$ is algebraic, then $x^2$ is also algebraic. I understand that an algebraic number can be written as a polynomial that is equal to $0$. However, I'm baffled when showing how $x^2$ is also algebraic.
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We have $P(x)=0$, where $P$ is some rational polynomial (that is, the coefficients are rational numbers). Break $P$ up into the terms with odd exponents and the terms with even exponents. For example, if $P(x)=x^4+x^3+5x^2+x+4$, then we would break it up as $(4+5x^2+x^4)+(x+x^3)$. The term with even exponents can be viewed as a polynomial in $x^2$: $4+5x^2+x^4=4+5x^2+(x^2)^2$ . The term with odd exponents can be viewed $x$ times a polynomial in $x^2$: $x+x^3=x(1+x^2)$. We now have an identity of the form:
$$P(x)=A(x^2)+xB(x^2)=0$$
Where $A$ and $B$ have rational coefficients. We almost have a rational polynomial which is $0$ at $x^2$, we just have to get rid of that $x$ infront of $B(x^2)$. We can do that like so:
$$(A(x^2)+xB(x^2))(A(x^2)-xB(x^2))=A(x^2)^2-x^2B(x^2)^2=0$$
And so the polynomial $A(X)^2 - XB(X)^2$ admits $x^2$ as a root, and of course has rational coefficients.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Approximate to whole number without calculator Approximate this number to the nearest whole number without a calculator:
$2^{\sqrt{5}}$
I don't know how to do this problem. Can you help me? My answer key says $5$.
Also, how do I approximate $2^\pi$ and $3^e$? The answers are $9$ and $20$, respectively.
|
i can get $4 < 2^{\sqrt 5} < 5.$ use the fact that $\sqrt x$ is concave down so the tangent approximation $\sqrt 5 = \sqrt {4 + 1)} < \sqrt 4 + {1 \over 2 \sqrt 4} = {5 \over 4}.$ so we have $2 < \sqrt 5 < {5 \over 4}$ and $2 = {512 \over 256} < {625 \over 256}$ gives us $2^{1/4} < {5 \over 4}.$
so far we have, $2 < \sqrt 5 < {5 \over 4}$ and $2 ^{1/4} < {5 \over 4}$ exponentiating shows $4 = 2^2 < 2^\sqrt 5 < 2^{5 \over 4} = 2^2 2^{1 \over 4} < 4 {5 \over 4} = 5.$
|
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|
Why is Implicit Differentiation needed for Derivative of $y = \arcsin (2x+1)$? my function is: $y = \arcsin (2x+1)$ and I want to find its derivative.
My approach was to apply the chain rule:
$$y' = \frac{dg}{du} \frac{du}{dx}$$
with $g = \arcsin(u)$ and $u = 2x+1$.
$$g' = \frac{1}{\sqrt{1-u^2}}.$$
${u}' = 2$.
My solution therefore was
$$\frac{1}{\sqrt{1-u^2}} \cdot 2 = \frac{2}{\sqrt{1-(2x+1)^2}}.$$
This seems to be wrong and the correct solution is given by:
$\frac{1}{\sqrt{-x^{2}-x}}$
I know that implicit differentiation should be used for this particular problem, but I do not really understand why. I appreciate any help!
|
Your solution is correct; they just simplified it further:
\begin{align*}
\frac{2}{\sqrt{1 - (2x + 1)^2}}
&= \frac{2}{\sqrt{1 - (4x^2 + 4x + 1)}} \\
&= \frac{2}{\sqrt{-4x^2 - 4x}} \\
&= \frac{2}{\sqrt{4(-x^2 - x)}} \\
&= \frac{2}{\sqrt{4}\sqrt{-x^2 - x}} \\
&= \frac{2}{2\sqrt{-x^2 - x}} \\
&= \frac{1}{\sqrt{-x^2 - x}} \\
\end{align*}
|
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|
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta
$$
After this, I thought of using the Taylor Expansion of
$\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.
Please Help!
|
By the way, there is a closed-form antiderivative (that could be proved by differentiation):
$$\int\arctan^2x\cdot
\ln\left(\frac{1+x^2}{2x^2}\right)\,dx=\\
\frac16\left[3 i \left\{\left(2 \operatorname{Li}_2(i
x)-2 \operatorname{Li}_2(-i x)+\operatorname{Li}_2\left(\frac{2
x}{x+i}\right)-\operatorname{Li}_2\left(\frac{2 x}{x-i}\right)\right)\cdot \ln \left(\frac{1+x^2}{x^2}\right)\\
+2 \left(2
\operatorname{Li}_3\left(\frac{x}{x-i}\right)-2
\operatorname{Li}_3\left(\frac{x}{x+i}\right)+\operatorname{Li}_3\left(\frac{2
x}{x+i}\right)-\operatorname{Li}_3\left(\frac{2
x}{x-i}\right)\right)\\
+\left(\operatorname{Li}_2\left(\frac{1}{2}-\frac{i
x}{2}\right)-\operatorname{Li}_2\left(\frac{i
x}{2}+\frac{1}{2}\right)\right)\cdot\ln2\right\}\\
+3 \left(2
\operatorname{Li}_2\left(-x^2\right)+\ln ^2\left(1+x^2\right)+\ln
\left(1+x^2\right)\cdot\ln2-2 \ln ^22\right)\cdot\arctan x\\
+6 x \ln \left(\frac{1+x^2}{2
x^2}\right)\cdot\arctan^2x+4 \arctan^3x\right]\color{gray}{+C}$$
This enables us to evaluate a definite integral over any region.
|
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|
Using mathematical induction to show that for any $n\ge 2$ then $\prod_{i=2}^n\bigl(1-\frac{1}{i^2}\bigr)=\frac{n+1}{2 n}$ I'm trying to work through some practice problems but I've been stuck on this for god knows how long now and I've no idea where to even start. Just wondering if it would be possible for someone to break this down and go over it with me step by step.
Using mathematical induction to show that for any $n \ge 2$ then $$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2 n}\tag{1}$$
|
Base case p(2):
LHS: $$\prod_{i=2}^2 \left(1-\frac{1}{i^2}\right) = 1-\frac{1}{2^2} = \frac{3}{4}$$
RHS: $$\frac{2+1}{2\cdot2} = \frac{3}{4}$$
Now assume p(k):
$$\prod_{i=2}^k\left(1-\frac{1}{i^2}\right)=\frac{k+1}{2k}$$
$$\Rightarrow \prod_{i=2}^{k+1}\left(1-\frac{1}{i^2}\right) = \frac{k+1}{2k} \left(1-\frac{1}{(k+1)^2}\right)$$
$$=\frac{k+1}{2k} - \frac{k+1}{2k(k+1)^2}$$
$$=\frac{k+1}{2k} - \frac{1}{2k(k+1)}$$
$$=\frac{2k(k+1)^2-2k}{4k^2(k+1)}$$
$$=\frac{2k(k^2+2k+1)-2k}{4k^2(k+1)}$$
$$=\frac{2k^3+4k^2+2k-2k}{4k^3+4k^2}$$
$$=\frac{2k^2(k+2)}{4k^2(k+1)}$$
$$=\frac{k+2}{2(k+1)}$$
$$\therefore p(k) \Rightarrow p(k+1) \mbox{, so} \prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n} \forall n\geq 2$$
|
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|
Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$ I have a function:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do it?
How would I go about solving it? This is what I did:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}} = y$$
$$y(1 + \sqrt{x}) = 1-\sqrt{x}$$
$$y + y\sqrt{x} = 1-\sqrt{x}$$
$$\text{Cancel out roots:}$$
$$y^2 + y^2x = 1+x$$
$$y^2 + y^2x - x = 1$$
$$y^2 + x(y^2 - 1) = 1$$
$$x(y^2 - 1) = 1 - y^2$$
$$x = \frac{1 - y^2}{y^2 - 1}$$
$$\text{Swap x and y and we get the inverse function:}$$
$$f^{-1}(x) = \frac{1 - x^2}{x^2 - 1}$$
$$\text{But the correct answer is supposed to be:}$$
$$f^{-1}(x) = \frac{(1 - x)^2}{(1 + x)^2}, -1< x \le 1$$
What am I doing wrong?
|
This is probably conceptually the simplest method.
For $\frac{a}{b}=\frac{c}{d}$, we always have $\frac{a+b}{a-b}=\frac{c+d}{c-d}$, provided that $a-b\not=0$ and $c-d\not=0$.
$$y = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
$$\frac{y}{1} = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
$$\frac{y+1}{y-1} = \frac{1-\sqrt{x}+(1+\sqrt{x})}{1-\sqrt{x}-(1+\sqrt{x})}=\frac{2}{-2\sqrt{x}}=\frac{1}{-\sqrt{x}}$$
$$\frac{\sqrt{x}}{1}=\frac{1-y}{1+y}$$
$$x=\frac{(1-y)^2}{(1+y)^2}$$
So
$$f^{-1}(x) = \frac{(1 - x)^2}{(1+x)^2}$$
|
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|
Estimating the modulus of the roots of $\sinθ_1z^3+\sinθ_2z^2+\sinθ_3z+\sinθ_4=3$ If $θ_1,θ_2,θ_3,θ_4$ are four real numbers, then any root of the equation
$$\sinθ_1z^3+\sinθ_2z^2+\sinθ_3z+\sinθ_4=3$$
lying inside the unit circle $\vert z\vert$=1, satisfies which inequality?
A)$\vert z\vert$ < $\frac{2}{3}$
B)$\vert z\vert$ > $\frac{2}{3}$
C)$\vert z\vert$ < $\frac{1}{2}$
D)None of these
We've recently passed equations of degree 3,4 and I know that it's maximum 4 roots that satisfy the equation. What does it mean for them to lie in the unit circle $\vert z\vert$=1? I guess it has to do something with complexical numbers?
|
I am assuming that by $\sin \theta_1 z^3$ our OP Jackie means $z^3 \sin \theta_1$ and so forth; with this understanding, we have the given equation
$z^3 \sin \theta_1 + z^2 \sin \theta_2 + z \sin \theta_3 + \sin \theta_4 = 3; \tag{1}$
taking absolute values and using the triangle inequality (several times) yields
$3 = \vert 3 \vert \le \vert z^3 \sin \theta_1 \vert + \vert z^2 \sin \theta_2 \vert + \vert z \sin \theta_3 \vert + \vert \sin \theta_4 \vert; \tag{2}$
now if we repeatedly apply the equality $\vert xy \vert = \vert x \vert \vert y \vert$, $x, y \in \Bbb C$ to (2) we obtain
$3 \le \vert z \vert^3 \vert \sin \theta_1 \vert + \vert z \vert^2 \vert \sin \theta_2 \vert + \vert z \vert \vert \sin \theta_3 \vert + \vert \sin \theta_4 \vert. \tag{3}$
We have $\vert \sin \theta_i \vert \le 1$, $ 1 \le i \le 4$; thus if $\vert z \vert \le 2/3$, (3) yields
$3 \le (\dfrac{2}{3})^3 + (\dfrac{2}{3})^2 + \dfrac{2}{3} + 1; \tag{4}$
however, the sum of the first three terms on the right of (4) is
$\dfrac{8}{27} + \dfrac{4}{9} + \dfrac{2}{3} =\dfrac{38}{27} < 2, \tag{5}$
so (4) becomes
$3 \le \dfrac{38}{27} + 1< 2 + 1 = 3. \tag{6}$
(6) is clearly not happening; this contradiction shows we must have
$\vert z \vert > \dfrac{2}{3}; \tag{7}$
thus (B) is the correct choice.
It is worth observing that with $\theta_i = \pi/2$, $1 \le i \le 4$, there is in fact a real solution $r$ to (1) satisfying $2/3 < r < 1$; this may be seen by applying the intermediate value theorem to the polynomial
$p(r) = \sum_0^3 r^i, \tag{8}$
which satisfies $p(2/3) = 65/27 < 3$, $p(1) = 4$. This shows that for at least some values of $\theta_i$, there is indeed a solution $z$ to (1) with $\vert z \vert < 1$; the problem is not vacuous.
As a closing comment, and to address Jackie's last two questions, we recall that for $z = a + bi$, $\vert z \vert = \sqrt{a^2 + b^2}$; saying $z$ lies within the unit circle thus means either $a^2 + b^2 < 1$ or $a^2 + b^2 \le 1$, depending on exaxtly how one parses the word "within"; this indeed has a lot to do with "complexical numbers".
Hope this helps. Cheers,
and as ever
Fiat Lux!!!
|
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|
Evaluation of $\lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\sqrt{x^2+x+1} -\lfloor \sqrt{x^2+x+1 }\rfloor\;,$ where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try}::$
$\bullet\; $If $x\in \mathbb{Z}\;,$ and $x\rightarrow \infty\;,$ Then $x^2<x^2+x+1<x^2+2x+1$
So $x<\sqrt{x^2+x+1}<(x+1)\;,$ So $\lfloor x^2+x+1 \rfloor = x\;,$ bcz it lies between two integers.
So $\displaystyle \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right) = \lim_{x\rightarrow \infty}\left(\sqrt{x^2+x+1}-x\right)\cdot \frac{\left(\sqrt{x^2+x+1}+x\right)}{\left(\sqrt{x^2+x+1}+x\right)}$
So $\displaystyle \lim_{x\rightarrow \infty} \frac{x+1}{\sqrt{x^2+x+1}+x} = \frac{1}{2}$
$\bullet\; $ If $x\notin \mathbb{Z}$ and $x\rightarrow \infty\;,$ Then How can i solve the above limit in that case,
Help me, Thanks
|
I think this problem is much more easily solved in slightly greater generality. If $f$ is a continuous function such that $\lim_{x \rightarrow \infty} f(x) = \infty$ then as $x \rightarrow \infty$, for arbitrarily large $x$, $f(x)$ is an integer and for any $\varepsilon < 1$, for arbitrarily large $x$, $f(x)$ is an integer minus $\varepsilon$. When $f(x)$ is an integer, $f(x) - \lfloor f(x) \rfloor = 0$ and when $f(x)$ is an integer minus $\varepsilon$, $f(x) - \lfloor f(x) \rfloor = \varepsilon$. So it is easy to see that $\lim_{x \rightarrow \infty} f(x) - \lfloor f(x) \rfloor$ doesn't exist.
Because $f(x) - \lfloor f(x) \rfloor$ is always non-negative and it is 0 for arbitrarily large $x$, its $\liminf$ is 0. Also, because $f(x) - \lfloor f(x) \rfloor$ is always less than 1 and is $\varepsilon$ for any $\varepsilon < 1$ for arbitrarily large $x$, its $\limsup$ is 1, as Yuval Filmus mentioned.
|
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|
Find the value of this infinite product $\prod_{n=1}^{\infty }\left [ 1+\frac{1}{a_{n}} \right ]$? Where $a_{1}=1$ and $a_{n+1}= \left ( n+1 \right )\left ( 1+a_{n} \right )$,
$\forall \left ( n\epsilon \mathbb{N} \right )$
I have found that for this infinite product to be convergent, the series $\sum \ln \left ( 1+\frac{1}{a_{n}} \right)$ must also be convergent, and that if it has a finite sum, let's name it s, the value of the infinite product is $ e^{s}$.
I am not sure whether $a_{n}=n\cdot \left ( 1+a_{n-1} \right )$ also holds true.
The given series passes the n-th term test, ok but how do I apply another test to it?
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Note that
$$a_{n+1}=(n+1)(1+a_n)\Rightarrow \frac{a_{n+1}}{1+a_{n}}=n+1$$
Consider the finite product:
$$\mathcal{P}_n=\prod_{n=1}^{N}(1+\frac{1}{a_n})=\prod_{n=1}^{N}\frac{1+a_n}{a_n}=\prod_{n=1}^{N}\frac{a_{n+1}}{(n+1)a_n}=\prod_{n=1}^{N}\frac{1}{(n+1)}\cdot \prod_{n=1}^{N}\frac{a_{n+1}}{a_n}=\frac{1}{N!}\cdot\frac{a_N}{a_1}=\frac{a_N}{N!}$$
Now we need to find an expression for the general term $a_n$. Notice that
$$a_{n+1}=(n+1)(1+a_n)=(n+1)+(n+1)n(1+a_{n-1})=(n+1)+(n+1)n+(n+1)n(n-1)(1+a_{n-2})=...=(n+1)+(n+1)n+(n+1)n(n-1)+..+(n+1)n(n-1)(n-2)...2\cdot(1+1)$$
In other words
$$a_{n+1}=\frac{(n+1)!}{n!}+\frac{(n+1)!}{(n-1)!}+..+2\cdot\frac{(n+1)!}{1!}$$
Hence
$$a_N=\frac{N!}{(N-1)!}+\frac{N!}{(N-2)!}+..+2\cdot\frac{N!}{1!}$$
for all $N\geq1$ so
$$\mathcal{P}_N=\frac{a_N}{N!}=\frac{1}{(N-1)!}+\frac{1}{(N-2)!}+..+\frac{2}{1!}$$
One can see that
$$1\leq \mathcal{P}_N<\frac{1}{(N-1)(N-2)}+\frac{1}{(N-2)(N-3)}+..+2=3-\frac{1}{N-1}$$
then
$$1\leq\lim_{N\to\infty}\mathcal{P}_N\leq3$$
Indeed the limit is $e$. Check it with the Taylor expansion of $e^x$ at $x=1$.
|
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|
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
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What's wrong with just logging? Logarithm is a monotone increasing function, so the inequality sign stays the same.
First log, the multiply both sides by 2 and 3. LHS becomes $\log (\frac{1}{2})^2$, right $\log(\frac{1}{3})^3$. Now exponentiate. LHS is $\frac{1}{2} \cdot \frac{1}{2} \cdot 1$, RHS is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$. Each term on the LHSis greater than each term on the RHS, hence the inequality follows.
|
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|
$4$ dice sum probability What is the probability that, when $4$ dice are rolled, we can choose two of them such that the sum of the numbers on the upper faces is $7$?
I get $139 \over 216$ but I'm not sure I'm correct.
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It is convenient to assume that the dice were rolled one at a time, and that we recorded the results as they came in. So there are $6^4$ equally likely outcomes. We count the favourables.
First we count the number of strings that have $6$ and a $1$. It is easier to count the strings that don't. There are $5^4$ strings that don't have a $1$, and $5^4$ that don't have a $6$, and $4^4$ that are missing both. So the number of strings that have a $6$ and a $1$ is $6^4-2\cdot 5^4+4^4$. Call this number $a$.
There are also $a$ strings with a $2$ and a $5$, and $a$ with a $3$ and a $4$. So our first estimate for the number of favourables is $3a$.
However, we have double-counted the strings that have, for example, both $1$ and $6$ and $2$ and $5$ (or the other $2$ possibilities).
For both $1$ and $6$ and $2$ and $5$, we have $\binom{4}{2}\binom{2}{1}$. Multiply by $3$. We conclude that the number of favourables is $3a-3\binom{4}{2}\binom{2}{1}$.
We leave it to you to put the pieces together.
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|
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
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It is quite easy...
$\tan x + \cot x$
= $\tan x + \cot x - 2 \sqrt {\tan x \cot x} + 2 \sqrt{\tan x \cot x}$
=$(\sqrt {\tan x} -\sqrt {\cot x})^2+ 2$ [because $\tan x \cot x =1$]
Since square of any number is greater than or equal to zero, we can write,
$(\sqrt {\tan x} -\sqrt {\cot x})^2 \ge 0$
So, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 \ge 2$
But, $(\sqrt {\tan x} -\sqrt {\cot x})^2 + 2 = \tan x + \cot x$
So, $\tan x + \cot x \ge 2$
NOTE : We want any square number to make this problem easier, this is a standard technique for proving most of inequalities in mathematics, therefore we have added and subtract $2 \sqrt{\tan x \cot x}$ in the main expression to make the expression perfect square.
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|
Evaluating $\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$ $$\int\frac{4x^3-3x^2+6x-27}{x^4+9x^2}dx$$
this integral get very messy. Can I get a step by step breakdown of solving?
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No Step by step solution, but just a hint
By reducing to partial fractions we get
$$\frac{4x^3-3x^2+6x-27}{x^4+9x^2}=\frac{Ax+ B}{x^2+9}+\frac{C}{x^2}+\frac Dx$$
$$Ax^3+Bx^2+Cx^2+9C+Dx^3+9Dx=4x^3-3x^2+6x-27$$
$$9C=-27\implies C=-3$$
$$9D=6\implies D=\frac23$$
$$A+D=4\implies A=-\frac{10}{3}$$
$$B+C=-3\implies B=0$$
$$\frac{4x^3-3x^2+6x-27}{x^4+9x^2} =\frac{10x}{3(x^2+9)}-\frac{3}{x^2}+\frac{2}{3x}$$
|
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|
If $x$, $y$, $z$ are in arithmetic progression, show that $\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y. $ Show that if $x, y,$ and $z$ are consecutive terms of an arithmetic sequence, and $\tan y$ is defined, then
$$\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.
$$
I'm not sure what trig identities I would use and how to use them. Could I get some help? Thanks.
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Lets define the common ratio $r$
Using the 4 Trigonometric Conversion functions, we get:
\begin{eqnarray*} \sin(y+r) &=& \sin y \cos r + \cos y \sin r \\ \sin(y-r) &=& \sin y \cos r - \cos y \sin r \\ \cos(y+r) &=& \cos y \cos r - \sin y \sin r \\ \cos(y-r) &=& \cos y \cos r + \sin y \sin r
\end{eqnarray*}Since $x = y - r$, $y = y$, and $z = y+r$, we can rewrite the original equation prove:
\begin{align*}
\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.\ &= \frac{\sin (y-r) + \sin y + \sin (y+r)}{\cos (y-r) + \cos y + \cos (y+r)} \\
&= \frac{\sin y \cos r - \cos y \sin r + \sin y + \sin y \cos r + \cos y \sin r}{\cos y \cos r + \sin y \sin r + \cos y + \cos y \cos r - \sin y \sin r } \\
&= \frac{2 \sin y \cos r + \sin y}{2 \cos y \cos r + \cos y} \\
&= \frac{\sin y (2 \cos r + 1)}{\cos y (2 \cos r + 1)} \\
&= \frac{\sin y}{\cos y} \\
&= \boxed{\tan y}
\end{align*}
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|
How come $\frac{1}{\cos x} = 1 + \frac{x^{2}}{2} + o(x^{2})$ as $x \to 0$? Since
$$\cos x = 1 - \frac{x^{2}}{2} + o(x^{3})$$
as
$x \to 0,$
we have
$$\frac{1}{\cos x} = \frac{1}{1-\frac{x^{2}}{2} + o(x^{3})} = 1 + \frac{x^{2}}{2} + o(x^{3}) + o(\frac{-x^{2}}{2} + o(x^{3})).$$
But I do not see how to write the term on the extreme right as $o(x^{2})$.
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You may recall that
$$
\frac{1}{1-u}=1+u+O(u^2), \quad u \to 0.
$$
Thus
$$
\frac{1}{\cos x} = \frac{1}{1-\frac{x^{2}}{2} + o(x^{3})} = 1 + \frac{x^{2}}{2} + o(x^{3}) + O(\frac{-x^{2}}{2} + o(x^{3}))^2
$$ then use
$$
O(\frac{-x^{2}}{2} + o(x^{3}))^2 =o(x^{3}).
$$
|
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|
Can I get a little integral help? Find the integral $$\int_0^3 \sqrt {y+1}\, dy$$
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$$\int_0^{3} \sqrt{y+1} dy = \int_0^{3} (y+1)^{\frac{1}{2}}d (y + 1) = \frac{2}{3} \int_0^{3} \frac{3}{2}(y+1)^{\frac{1}{2}}d (y + 1) = \left. \frac{2}{3}(y+1)^{\frac{3}{2}} \right|_0^3 = \frac{2}{3} 4 ^{\frac{3}{2}} - \frac{2}{3}1^{\frac{3}{2}} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}$$
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|
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 3 } }{ 2 } }{ 2 } } =\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } $$
So, $\sin45+\sin15=\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 } } }{ 2 } $
2) Rewriting $\sin75$
$$\sin75=\sin(45+30)=\sin45\cos30+\sin30\cos45$$
$$=\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } $$
$$=\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } $$
So now I have these two expressions that are obviously equal (when I compute them) but how do I show them as being equal to each other. I assume the question is implying for the resulting expressions to look equal.
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We have:
$$\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 } } }{ 2 }$$
By squaring the numerator and denominator:
$$\frac {\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)^2}{4} = \frac{2+2-\sqrt{3}+ 2\sqrt{4-2\sqrt3}}{4} = \frac{4-\sqrt{3}+2\sqrt{4-2\sqrt3}}{4}$$
Next we must find:
$$\sqrt{4-2\sqrt3}$$
Let's call the answer $a-\sqrt{b}$
Now, we can say that:
$$\left(\sqrt{4-2\sqrt3}\right)^2 = 4 - 2\sqrt{3} = (a-\sqrt{b})^2 = a^2 - 2a\sqrt{b} + b$$
Therefore we can say:
$$a^2 + b = 4$$
and
$$2a\sqrt{b} = 2\sqrt{3}$$
We then get that: $a = 1$, $b = 3$
Therefore, $$\sqrt{4-2\sqrt3} = 1 - \sqrt{3}$$
Substituting:
$$\frac{4-\sqrt{3}+2\sqrt{4-2\sqrt3}}{4} = \frac{4-\sqrt{3}+2(1-\sqrt{3})}{4} = \frac{4-\sqrt{3}+2 - 2\sqrt{3}}{4} = \frac{6- 3\sqrt{3}}{4} = \frac{3(2- \sqrt{3})}{4}$$
Can you take the rest from here?
|
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|
Binomial Expansion with fractional or negative indices Question:
Expand the function $\frac{2}{(2x - 3)(2x+1)}$ in a series of powers of $x$ up to $x^2$. State the set of values of $x$ for which this expansion is valid.
I've come across this question and would like to ask how most of you would tackle it. I've never seen one of this form before as it's my first time tackling fractional or negative indices. Thus I kindly ask to correct me wherever I'm wrong in my attempt:
My Attempt:
I rewrote $\frac{2}{(2x - 3)(2x+1)}$ as $2(2x - 3)^{-1}(2x+1)^{-1}$ since it seemed to somehow make some more sense. I then followed the simple formula $$1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!}$$
And substituted it with one of my terms whilst leaving 2 outside: $$2[(2x-3)^{-1}] \equiv 2[1+(-1)(2x)+\frac{(-1)(-2)(2x)^2}{2(1)}]$$
As the question requests up to $x^2$. Once working all of that out I got:
$$2[1-2x-4x^2]$$
$\therefore$ The expansion is valid when $x$ is between $-\frac{1}{4}$ and $\frac{1}{4}$
And wrote it down as an inequality: $$-\frac{1}{4} < x < \frac{1}{4}$$
Second Attempt
I went ahead and worked it out using Partial Fractions. Seems to make much more sense.
$$\frac{2}{(2x - 3)(2x+1)} \equiv \frac{A}{(2x-3)}+\frac{B}{(2x+1)}$$
Rewrite as..
$$2 = A(2x+1)+B(2x-3)$$
$$2 = \frac{2}{(2x+1)}-\frac{2}{(2x-3)}$$
And once again rewrite as..
$$2(2x+1)^{-1} - 2(2x-3)^{-1}$$
$$2^{-1}(2x+1)^{-1}+6^{-1}(\frac{2}{3}x-1)^{-1}$$
From here I worked each expansion separately however got stuck at a certain point.
$$\frac{1}{2}[1+(-1)(2x)+\frac{(-1)(-2)(2x)^2}{2!} ] = \frac{1}{2}[1 - 2x - 4x^2 ]$$
And..
$$\frac{1}{6}[1 + (-1)(\frac{2}{3}x) + \frac{(-1)(-2)(\frac{2}{3}x)^2}{2!} ] = \frac{1}{2}[1 - \frac{2}{3}x - (\frac{2}{3}x)^2]$$
I'm not sure if I'm doing it right from here onwards
$$\frac{1}{2}[1-2x-4x^2] - \frac{1}{6}[1-\frac{2}{3}x-(\frac{2}{3}x^2)]$$
And my final answer..
$$\frac{1}{2} - x - 2x^2 - \frac{1}{6} - \frac{1}{9}x - \frac{2}{27}x^2 = \frac{2}{3} - \frac{8}{9}x - \frac{56x^2}{27}$$
I'm sorry if this is completely wrong, it's literally my first time trying it out however it makes sense to consult an experienced person before entering the topic thinking I understand it when I don't.
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The Binomial Theorem for negative powers says that for $|x| < 1$
$$(1+x)^{-1} = 1 - x + x^2 + \mathcal{o}(x^2)$$
Therefore we have:
$$\frac 2{(2x-3)(2x+1)} = \frac 1{2(2x-3)} - \frac 1{2(2x+1)} = -\frac 16\left(1-\frac 23x\right)^{-1} - \frac 12\left(1+2x \right )^{-1} = -\frac16\left(1 + \frac 23x + \frac 49x^2\right)-\frac 12\left(1 - 2x + 4x^2 \right ) = \boxed{-\frac 23 + \frac 89 x - \frac{56}{27}x^2}$$
This holds for $|x| < \frac12$.
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|
Solving $\sin\theta -1=\cos\theta $
Solve$$\sin\theta -1=\cos\theta $$
Steps I took to solve this:
$$\sin^{ 2 }\theta -2\sin\theta +1=1-\sin^2\theta $$
$$2\sin^{ 2 }\theta -2\sin\theta =0$$
$$(2\sin\theta )(\sin\theta -1)=0$$
$$2\sin\theta =0, \sin\theta -1=0$$
$$\quad \sin\theta =0, \sin\theta =1$$
$$\theta =0+\pi k, \theta =\frac { \pi }{ 2 } +2\pi k$$
Why is $\theta =0+\pi k$ wrong?
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When you square both sides of an equation, you may introduce extraneous solutions. Therefore, you must check that your solutions satisfy the original equations (a good idea in any case).
When you squared the equation $\sin\theta - 1 = \cos\theta$, you discovered that
the resulting equation was satisfied when $\sin\theta = 0$ or $\sin\theta = 1$.
In the interval $[0, 2\pi)$, the equation $\sin\theta = 0$ is satisfied when $\theta = 0$ or $\theta = \pi$. If $\theta = 0$, then
$$\sin(0) - 1 = 0 - 1 = -1 \neq 1 = \cos(0)$$
so $0$ is an extraneous solution. On the other hand, if $\theta = \pi$, then
$$\sin(\pi) - 1 = 0 - 1 = -1 = \cos(\pi)$$
so $\theta = \pi$ is a valid solution.
In the interval $[0, 2\pi)$, the equation $\sin\theta = 1$ is satisfied when $\theta = \dfrac{\pi}{2}$. When $\theta = \dfrac{\pi}{2}$,
$$\sin\left(\frac{\pi}{2}\right) - 1 = 1 - 1 = 0 = \cos\left(\frac{\pi}{2}\right)$$
Thus, $\theta = \dfrac{\pi}{2}$ is a valid solution.
Thus, the general solution of the equation $\sin\theta - 1 = \cos\theta$ is
$$
\theta =
\begin{cases}
\pi + 2n\pi, n \in \mathbb{Z}\\
\dfrac{\pi}{2} + 2n\pi, n \in \mathbb{Z}
\end{cases}
$$
|
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|
Show that this two variable limit don't exist Prove that the following limit doesn't exist: $$\lim_{(x,y)\to(1,0)}\frac{\sin(1-x^2+y^2)-y}{xy}$$
I guess I need to find different values for different directions.
|
First direction: $y=1-x,$ second direction: $y=-(1-x).$
If $y=1-x$ then
$$\begin{gather}\lim_{\substack{{(x,y)\to(1,0)}\\{y=1-x}}}\frac{\sin(1-x^2+y^2)-y}{xy}=
\lim_{x\to{1}}\frac{\sin((1-x)(1+x)+(1-x)^2)-(1-x)}{x(1-x)}=\\
=\lim_{x\to{1}}\frac{\sin{\left((1-x)(1+x+1-x)\right)-(1-x)}}{x(1-x)}=
\lim_{x\to{1}}\frac{\sin{\left(2(1-x)\right)-(1-x)}}{x(1-x)} =1\end{gather}$$
since $\lim\limits_{t\to{0}}\dfrac{\sin{t}}{t}=1.$
But for $y=-1+x$ we have
$$\begin{gather}\lim_{\substack{{(x,y)\to(1,0)}\\{y=-1+x}}}\frac{\sin(1-x^2+y^2)-y}{xy}=
\lim_{x\to{1}}\frac{\sin((1-x)(1+x)+(1-x)^2)+(1-x)}{x(x-1)}=\\
=\lim_{x\to{1}}\frac{\sin{\left((1-x)(1+x+1-x)\right)+(1-x)}}{x(x-1)}=\\
=\lim_{x\to{1}}\frac{\sin{\left(2(1-x)\right)+(1-x)}}{x(x-1)} =-3.\end{gather}$$
|
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|
Confusion about a Linear Transformation question. Let $\beta := [M_1, M_2, M_3, M_4]$ be the ordered basis of $R^{2×2}$ defined by:
$$ M_1 := \begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix},
M_2 := \begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix},
M_3 := \begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix},
M_4 := \begin{pmatrix}
0 & 0\\
0 & 1 \end{pmatrix}
$$
Define the linear transformation
$L := A → (A + A^T)$.
a) What is $L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$?
b) Find the matrix representation of $L$ with respect to the basis $\beta$.
c) Find all solutions to $L(A) = \begin{pmatrix}
4 & 0\\
0 & -6 \end{pmatrix}$.
Does the following look correct?
$L \left( \begin{pmatrix}
a & b\\
c & d \end{pmatrix} \right)$ = $ \begin{pmatrix}
2a & b + c\\
b+ c & 2d \end{pmatrix}$
$Ker(L) = span\{\begin{pmatrix}
0 & c\\
-c & 0 \end{pmatrix}\}$
$range(L) = span\{ \begin{pmatrix}
a & 0\\
0 & 0 \end{pmatrix}, \begin{pmatrix}
0 & s\\
s & 0 \end{pmatrix}, \begin{pmatrix}
0 & 0\\
0 & d \end{pmatrix}\}$ where s denotes the quantity $c + d$
Matrix Representation: $\begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 2\end{pmatrix} \begin{pmatrix}
a\\
b\\
c\\
d\end{pmatrix}=\begin{pmatrix}
2a\\
b + c\\
b + c\\
2d\end{pmatrix}$
Thanks for the help!
|
Your transformation is from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$; and from the given question, it looks like both the domain and the range spaces are represented using the same basis that is $\left\{b_1=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},b_2=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix},b_3=\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, b_4=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right\}$. Now, given a vector $x$ in the domain w.r.t. the basis of the domain, the actual vector in $\mathbb{R}^{2 \times 2}$ is $x_1 b_1 + x_2b_2+x_3b_3+x_4b_4$. The transformation $L:A\rightarrow A+A^T$ w.r.t. the given basis for the domain and the range spaces can be represented through a matrix, call it, $M_L$. $M_L$ consists of columns that are essentially $L(b_1), L(b_2), L(b_3),L(b_4)$ represented w.r.t. the basis (of the range space).
For example, the first column of $M_L$ can be found out as the vector $M_{L_1}$ that is given by the components $[m_{11},m_{21},m_{31},m_{41}]^T$ that are found out by solving the following: $$L(b_1)=m_{11}b_1+m_{21}b_2+m_{31}b_3+m_{41}b_4.$$
Note that in the above I've used $b_1, b_2, b_3,b_4$ on the right hand side since the same basis is used to represent the range space as well. Were a different basis given for the range space then the elements of that basis should be plugged in instead. I hope you can take it from here and find the matrix $M_L$. The other two parts should be harmless; it's usually the matrix representation that's confusing at first looks.
|
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|
Determine when the system has a) no solution, b) 1 solution and c) infinitely many solutions This question is not for an assignment, it was on the midterm and I am interested in figuring out how to solve it before the final exam.
cheers,
Determine when the system has a) no solution, b) 1 solution and c) infinitely many solutions
Given the system
\begin{align*}
x + y + 7z & = -7\\
2x + 3y +17z & = -16\\
x +2y +(a^2+1)z & = 3a
\end{align*}
Attempted Solution
using matrix form $$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr1&2&a^2+1&3a\cr}$$ using gaussian elimination to isolate eqn 3 for variable z (sorry about the missing a^2+1 up there html doesn't like me)
line 1 - line 3
$$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr 0&1&7-(a^2+1)&-7-3a\cr}$$
2*line 1 - line 2
$$\pmatrix{1&1&7&-7\cr 0&-1&-3&2 \cr 0&-1&7-(a^2+1)&-7-3a\cr}$$
-1*line2
$$\pmatrix{1 & 1 &7 &-7\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 - (a^2 + 1) & -7 - 3a}$$
-1*line 2 + line 1
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 - (a^2 + 1) & -7 - 3a}$$
line 2 + line 3
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & 0 & 10 - (a^2 + 1) & -9 - 3a}$$
so we get equation 3
$10-(a^2+1)z=-9-3a$
rearranging,
$$(a^2+1)z=-19-3a \rightarrow$$
$$z=19/a^2+1 - 3a/a^2+1 \rightarrow $$
$$z=(19/a^2+1) - (-3/a+1)$$
$$z=19/a^2 + 1 - (-3/a) +1$$
$$z=19/a^2 + 2 + 3/a$$
multiply all terms by $a$
$$za=19/a + 2a + 3$$
$$za=(39/19)a + 3$$
$$za=3a+3$$
$$z=3+3/a$$
wut?
I am unsure how to properly answer the question, please help stack exchange!
My algebra skills are a little lacking, please point out helpfully where my thinking is wrong.
I can solve for z but what does that prove
|
You have arrived to this (which is followed by some not very reader friendly part, which I am ignoring EDIT: In the meantime, N. F. Taussig fixed it):
$$\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&-1&7-(a^2+1)&-7-3a
\end{array}\right)=
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&-1&6-a^2&-7-3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&9-a^2&-9-3a
\end{array}\right)=
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&(3-a)(3+a)&-3(3+a)
\end{array}\right)
$$
For $a=-3$ you have
$$\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&0&0
\end{array}\right)$$
This system has infinitely many solutions.
If $a\ne-3$, you can divide by $a+3$ to get
$$
\left(\begin{array}{ccc|c}
1&1&7&-7\\
0&-1&-3&2\\
0&0&3-a&-3
\end{array}\right)
$$
If $a=3$, then it has no solution. (The last equation is $0x+0y+0z=3$.)
For $a\ne3$ you can see that there is only one solution. (Maybe it would be better if I wrote for $a\ne\pm3$; just to remind that now we only consider the case when $a\ne-3$ and we got an additional condition $a\ne3$ to get unique solution.)
As a sanity check, you can plug $a=3$ in the original system. You can notice that if you subtract the first equation from the second, you get
$$x+2y+10z=-9.$$
For $a=3$ the third equation is
$$x+2y+10z=9.$$
These two equations are clearly incompatible.
Also for $a=-3$ you can see that the third equation becomes
$$x+2y+10z=-9.$$
So in this case the third equation is direct consequence of the first two equations.
I will also point out that if I was doing the Gaussian (or Gauss-Jordan) elimination, I would probably start by subtracting the first row from the second, since then I have two very similar rows, which makes things easier).
$$\left(\begin{array}{ccc|c}
1 & 1 & 7 &-7 \\
2 & 3 &17 &-16\\
1 & 2 &a^2+1& 3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 7 &-7 \\
1 & 2 &10 &-9\\
1 & 2 &a^2+1& 3a
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 7 &-7 \\
1 & 2 &10 &-9\\
0 & 0 &a^2-9& 3a+9
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 7 &-7 \\
1 & 2 &10 &-9\\
0 & 0 &(a-3)(a+3)& 3(a+3)
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 7 &-7 \\
0 & 1 & 3 &-2\\
0 & 0 &(a-3)(a+3)& 3(a+3)
\end{array}\right)$$
|
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|
Find the Viète formula I know that the Viète formula for the equation $ax^2+bx+c=0$ is:
$$x_1+x_2=-\frac{b}{a}$$
$$x_1x_2=\frac{c}{a}$$
But I didnt know which are the formula for the equation $ax^3+bx^2+cx+d=0$.
Please help me. Thanks.
|
Soppose that $x_1, x_2, x_3$ is zeros of equation $ax^3+bx^2+cx+d=0$. Now we have
$$ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3)$$
$$=a(x^2-xx_1-xx_2+x_1x_2)(x-x_3)$$
$$=a(x^3-x^2x_1-x^2x_2+xx_1x_2-x^2x_3+xx_1x_3+xx_2x_3-x_1x_2x_3)$$
$$=a(x^3-x^2(x_1+x_2+x_3)+x(x_1x_2+x_1x_3+x_2x_3)-x_1x_2x_3)$$
For the F. Viéte formula we have:
$$x_1+x_2+x_3=-\frac{b}{a}$$
$$x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}$$
$$x_1x_2x_3=-\frac{d}{a}$$
|
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|
deriving the sum of $x^n/(n+2)^2$ I am writing a research paper and I have stumbled upon an issue.
I have to evaluate
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2}$$
Here is what I did:
$$ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$
$$\sum_{n=1}^{\infty} x^{n+1} = \frac{x^2}{1-x}$$
Integrate once with respect to $x$.
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{n+2} = \frac{3}{2} - \frac{x^2}{2} - x - \log(1-x)$$
Divide by $x$
$$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+2} = \frac{3}{2x} - \frac{x}{2} - 1 - \frac{\log(1-x)}{x}$$
Integrate the expression again;
$$\sum_{n=1}^{\infty} \frac{x^{n+2}}{(n+2)^2} = \frac{3\log(x)}{2} - x - \frac{x^2}{4} + Li_2(x)$$
All we have to do is divide by $x^2$
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{3\log(x)}{2x^2} - \frac{1}{x} - \frac{1}{4} + \frac{Li_2(x)}{x^2} = \frac{4Li_2(x) - x^2 - 4x + 6\log(x)}{4x^2}$$
The issue is WolframAlpha returns it as:
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2} = \frac{4Li_2(x) - x^2 - 4x}{4x^2}$$
Why is the way I did, wrong? I have an extraneous, extra log term.?
Help is appreciated.
|
Somewhere in your question, you write "Integrate once with respect to $x.$". In the equation following that phrase, you have a $\frac{3}{2}$ on the r.h.s. That's superfluous. The derivatives of the l.h.s. and the r.h.s are equal, and setting $x=0,$ we find that the constant of integration is $0.$
|
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|
prove that $\sqrt{2} \sin10^\circ+ \sqrt{3} \cos35^\circ= \sin55^\circ+ 2\cos65^\circ$ Question:
Prove that: $\sqrt{2} \sin10^\circ + \sqrt{3} \cos35^\circ = \sin55^\circ + 2\cos65^\circ$
My Efforts:
$$2[\frac{1}{\sqrt{2}}\sin10] + 2[\frac{\sqrt{3}}{2}\cos35]$$
$$= 2[\cos45 \sin10] + 2[\sin60 \cos35]$$
|
Hints:
$$\begin{align}&\bullet\;\;\sin(x+y)=\sin x\cos y+\sin y\cos x\\{}\\&\bullet\;\;\sin 45^\circ=\frac1{\sqrt2}=\frac{\sqrt2}2=\cos45^\circ\\{}\\&\bullet\;\;\cos x=\cos(-x)\end{align}$$
|
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|
Proving that for all complex $z$, $\lim_{x\to0}\frac{1-\cos^{z}x}{x^2}=\frac{z}{2}.$ What do I need to study beforehand in order to prove it (not necessarily in only one way)? I found this sperimentally, at the moment we're beginning derivatives at school. By induction, I succeeded in proving $$\lim_{x\to0}\frac{1-\cos^{n}x}{x^2}=\frac{n}{2}.\tag{P(n)}$$
Basis: $P(0)$: $\displaystyle\lim_{x\to0}\frac{1-\cos^{0}x}{x^2}=0$ is true.
Inductive step: Assume $P(k)$ holds. This indeed implies $P(k+1)$ holds too: $$\lim_{x\to0}\frac{1-\cos^{k+1}x}{x^2}=\lim_{x\to0}\frac{\left(1-\cos^kx\right)\cos x-\cos x+1}{x^2}=\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lim_{x\to0}\frac{\left(1-\cos^kx\right)\cos x}{x^2}+\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{k}{2}+\frac{1}{2}=\frac{k+1}{2}.\ \ \ \ \ \ \ \ \ \ \ \ \ \blacksquare$$
I want to show it is true for all complex $z$ on my own.
|
$$\frac{1-\cos^z x}{x^2}=\frac{1-\exp(z\log \cos x)}{x^2}=\frac{1-\exp(z\log \cos x)}{\log\cos x}\cdot\frac{\log\cos x}{x^2}$$
But since $\log\cos x\to 0$ when $x\to 0$,
$$\lim_{x\to 0}\frac{1-\cos^z x}{x^2}=-z\cdot\lim_{x\to 0}\frac{\log\cos x}{x^2}\stackrel{H}{=}z\cdot\lim_{x\to 0}\frac{\tan x}{2x}=\frac{z}{2}.$$
|
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|
Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was able to show using the following method that $c_n$ is bounded:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n} < \overbrace{\frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{2}}}{n}}^{n-1 \text{ times}} = \frac{\large {n - 1}}{n\sqrt{2}} < \frac{1}{\sqrt{2}}$$
So now we know that $\large {0 < c_n < \frac{1}{\sqrt{2}}}$.
I know from testing for large values of $n$ that $c_n \to 0$.
What's left is actually finding a way to show this.
Any hints?
|
Hint : $c_n$ is Cesàro summation of sequence $\{\frac{1}{\sqrt{n}}\}_{n \in \Bbb N}$
Cesàro summation
|
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|
Forcing an absolute value of x after a square root operation Given the following two equations:
$$
f(x) = x \\
g^2(x) = 2x
$$
I need to find the $(x,y)$ coordinates for when they meet.
So after performing the square root operation, we have:
$$
f(x) = x \\
g(x) = \pm\sqrt{2\lvert x\rvert}
$$
So when trying to find the coordinates we get the following:
$$
f(x) = g(x) \\
x = \pm\sqrt{2\lvert x\rvert} \\
x \mp\sqrt{2\lvert x\rvert} = 0 \\
\sqrt{\lvert x\rvert} \cdot (\sqrt{\lvert x\rvert} \mp \sqrt{2} ) = 0 \\
\Rightarrow \sqrt{\lvert x\rvert} = \pm\sqrt{2} \quad\quad /(\quad)^2 \\
\Rightarrow \lvert x\rvert = 2 \\
\Rightarrow x = \pm2 \\
\Rightarrow f(\pm2) = \pm2
$$
Which doesn't make sense because that $g(x)$ isn't defined at $(-2,-2)$.
So what is wrong with the equations?
|
\begin{align}
f(x)&=g(x)\\
x&=\pm\sqrt{2x}\tag{1}\\
x^2&=2x\tag{2}\\
x^2-2x&=0\\
x(x-2)&=0\\
x&=0,\,2
\end{align}
Now check if these are indeed solutions to the original equation $(1)$. (They are.)
Back to your proposed solution set. Note that $x=-2$ is not a solution of $(1)$ since the left-hand side is $-2$, but the right-hand side is $\pm\sqrt{2(-2)}$ which not a real number.
More generally, for equations involving radicals, squaring both sides to obtain a different equation may introduce extraneous roots to the original equation. That is why you need to check to see if the roots of the new equation are indeed roots of the original equation.
For example, with $\sqrt{x-1}=x-7$, squaring both sides and rearranging leads to $x^2-15x+50=0$ which has solutions $x=10,5$. Now, $x=10$ is indeed a solution of the original equation, but $x=5$ is not.
|
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|
How to use complex analysis to find the integral $\int^\pi_{−\pi} \frac 1 {1+\sin^2(\theta)} d\theta$? How can I use complex analysis to solve the following:
$$\int^\pi_{−\pi} \frac 1 {1+\sin^2(\theta)} d\theta$$
|
Start with
$$
z=e^{i\theta} \\
\log z=i\theta \\
\theta=-i\log z \\
d\theta=-\frac izdz
$$
Now we substitute $z$ into our integral.
$$
\oint\frac{-i}{z(1+\sin^2(-i\log z))}dz \\
$$
Analyzing just the $\sin$ part for a moment:
$$
\begin{align}
\sin(-i\log z)&=-\frac i2\left(e^{i(-i\log z)}-e^{-i(-i\log z)}\right)\\
&=-\frac i2\left(e^{\log z}-e^{-\log z}\right) \\
&=-\frac i2\left(z-\frac 1z\right) \\
&=\frac{i(1-z^2)}{2z}
\end{align}
$$
Now we go back to the integral:
$$
\oint\frac{-i}{z\left(1+\left(\frac{i(1-z^2)}{2z}\right)^2\right)}dz \\
=\oint\frac{-i}{z\left(1-\frac{(1-z^2)^2}{4z^2}\right)}dz \\
=i\oint\frac{1}{\frac{1-6z^2+z^4}{4z}}dz \\
=i\oint\frac{4z}{1-6z^2+z^4}dz
$$
Now we can do partial fraction decomposition on the integrand to expose its poles:
$$
=\frac{i}{2\sqrt{2}}\oint\frac{1}{z-(-1-\sqrt{2})}-\frac{1}{z-(-1+\sqrt{2})}-\frac{1}{z-(1-\sqrt{2})}+\frac{1}{z-(1+\sqrt{2})}dz \\
$$
Two of these poles (the negative ones) are inside the unit circle, our integration contour. By Cauchy's integral formula, we get the value of the integral:
$$
2\pi i\left(\frac{i}{2\sqrt{2}}(-1-1)\right)=2\pi\frac{-2}{-2\sqrt{2}}=\sqrt{2}\pi
$$
|
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|
Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions Let
$$\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$$
How to show that it converges with no use of trigonometric functions?
(trivially, it is the anti-derivative of $\sin ^{-1}$ and therfore can be computed directly)
|
here is a way to show that $\int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}$ without the use of trigonometric functions. i will use the fact the area of unit circle is $\pi.$ that is $\int_0^1 \sqrt{1-x^2}dx = {\pi \over 4}$ twice and integration by parts.
\begin{eqnarray}
{\pi \over 4} & = & \int_0^1 \sqrt{1-x^2}\ dx \\
& = &
\int_0^1{1-x^2 \over \sqrt{1-x^2}} \ dx \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}}- \int_0^1{x^2 \over \sqrt{1-x^2}} \ dx \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} + \int_0^1 x\ d \sqrt{1-x^2} \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} +{x \sqrt{1-x^2}} |_0^1 -
\int_0^1\sqrt{1-x^2}\ dx \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} - {\pi \over 4}.
\end{eqnarray}
and that proves the claim $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}. $$
|
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|
How evaluate the following hard integrals? Prove:
$$\displaystyle\int_0^{\frac{\pi}{4}}{\,x}{\,\arctan\sqrt{\frac{\cos2x}{2\sin^2x}}}dx=\frac{\pi}{96}[{\pi^2}-6\ln^22]$$
And $$\displaystyle\int_0^{\frac{\pi}{4}}{\,x}{\,\arctan\sqrt{\frac{2\sin^2x}{\cos2x}}}dx=\frac{\pi}{192}[{\pi^2}+12\ln^22]$$
These integrals have been proposed by my friend,but I do not know how to proceed.
How do you evaluate these integral?
|
Denote the first integral by $I$ and the second by $J$. Then,
$$\begin{aligned}
J=&\int_0^{\pi/4} x\left(\frac{\pi}{2}-\arctan\sqrt{\frac{\cos 2x}{2\sin^2 x}}\right)\,dx \\
=&\frac{\pi^3}{64}-I \,\,\,\,\,\,\,(1)
\end{aligned}$$
$I$ can be simiplified to:
$$I=\int_0^{\pi/4} x\arccos(\sqrt{2}\sin x)\,dx=\left(\frac{x^2\arccos(\sqrt{2}\sin x)}{2}\right|_0^{\pi/4}+\frac{1}{2}\int_0^{\pi/4} \frac{x^2\left(\sqrt{2}\cos x\right)}{\sqrt{1-2\sin^2x}}\,dx$$
The first term is zero and with the substitution $2\sin^2x=\sin^2\theta$,
$$I=\frac{1}{2}\int_0^{\pi/2} \left(\arcsin\left(\frac{\sin \theta}{\sqrt{2}}\right)\right)^2\,d\theta$$
From here,
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!}x^{2n+1}$$
Integrate both sides within the limit $0$ to $\sin\theta/\sqrt{2}$, i.e:
$$\begin{aligned}
I &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}\frac{(2n)!!}{(2n+1)!!}\int_0^{\pi/2} \sin^{2n+2}\theta\,d\theta \\
&=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^{n+1}(n+1)}\frac{(2n)!!}{(2n+1)!!}\frac{(2n+1)!! \pi}{2^{n+2}(n+1)!}\\
&=\frac{\pi}{8}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}(n+1)^2} \,\,\,\,\,\,\,\,\,\left((2n)!!=2^nn!\right) \\
&=\frac{\pi}{8}\text{Li}_2\left(\frac{1}{2}\right) \\
\end{aligned}$$
Hence,
$$\boxed{I=\dfrac{\pi}{96}\left(\pi^2-6\ln^22\right)}$$
and from $(1)$,
$$\boxed{J=\dfrac{\pi}{192}\left(\pi^2+12\ln^2 2\right)}$$
|
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|
Discriminant of the polynomial $f(x)=4x^3-ax-b$
Definition. The discriminant of the polynomial $f(x)=4(x-x_1)(x-x_2)(x-x_3)$ is the product $16\{(x_2-x_1)(x_3-x_2)(x_3-x_1)\}^2$.
How to prove that the discriminant of $f(x)=4x^3-ax-b$ is $a^3-27b^2$.
Any help would be appreciated.
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You should read about the sylvester Sylvester matrix.
To be precise, $f(x) = 4x^3-ax−b$ gives $f'(x) = 12 x^2 - a$. So the discriminant is the determinant
$$ \Delta(a,b) = \left|\begin{array}{ccccc}
4 & 0 & -a & -b & 0 \\
0 & 4 & 0 & -a & -b\\
12 & 0 & -a & 0 & 0\\
0 & 12 & 0 & -a & 0\\
0 & 0 &12 & 0 & -a
\end{array}\right|
= -64(a^3 - 27b^2).$$
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|
In $\triangle ABC$ , find the value of $\cos A+\cos B$ The sides of $\triangle$ABC are in Arithmetic Progression (order being $a$, $b$, $c$) and satisfy
$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}=\dfrac{8^a}{(2b)!}$, Then prove that the value of $\cos A+\cos B$ is $\dfrac{12}{7}$.
It took me a long time to solve this question, if you have another way of solving, it is appreciated.
|
$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}$=$\dfrac{2}{10!}$($^{10}C_1$)+$\dfrac{2}{10!}$($^{10}C_3$)+$\dfrac{1}{10!}$($^{10}C_5$)
$\implies$$\dfrac{1}{10!}$(2$\cdot$$^{10}C_1$+2$\cdot$$^{10}C_3+^{10}C_5$)
$\implies$$\dfrac{1}{10!}$(2$\cdot$10+2$\cdot$120+252)=$\dfrac{512}{10!}$=$\dfrac{2^9}{10!}$
$\implies$$\dfrac{2^9}{10!}$$\implies$$\dfrac{8^3}{(2\cdot5)!}$=$\dfrac{8^a}{(2\cdot b)!}$(given) $\implies$$a=3, b=5$
If a,b,c are in AP $\implies$ $c=7$
$\cos A+\cos B$=$\dfrac{25+49-9}{90}$+$\dfrac{9+49-25}{42}$=$\dfrac{12}{7}$.
|
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|
Find all the prime numbers that satisfy the following conditions There was a brainteaser in the Science Magazine from University of Hong Kong which is as follow:
Find all the prime numbers $p$ such that $\sqrt{\frac{p+7}{9p-1}}$ is rational.
I tried a few numbers and it seems to suggest that $11$ is a suitable candidate.
Can I know the techniques to approach this question?
Thank you.
|
Let the rational number be $a/b$ in lowest terms. Rearrange the expression to $$p=\frac{a^2+7b^2}{9a^2-b^2}$$
Let $q$ be a prime factor of both numerator and denominator. Either $3a+b$ or $3a-b$ is a multiple of $q$, so $b=\pm3a+nq$ for some whole number $n$. Then $a^2+7b^2=a^2+7(9a^2+mq)$ for another whole number $m$, so $q$ is factor of $64a^2$.
If $q$ is a factor of $a$, and it is a factor of $3a+b$ or $3a-b$, then it is a factor of $b$ as well. That is a contradiction since $a/b$ is in lowest terms, so $q$ is a factor of 64.
$q$ was prime, so $q=2$, and the denominator is a power of 2.
$3a+b$ and $3a-b$ are both powers of $2$, so $a=(2.4^k+1)/3,b=2.4^k-1$. The denominator is $8.4^k$
Nine times the numerator is $$(2.4^k+1)^2+63(2.4^k-1)^2\\=256.4^{2k}-248.4^k+64=8.4^k(32.4^k-31)+64$$ This is a multiple of $8.4^k$, so $64$ is a multiple of $8.4^k$ and $k=0$ or $k=1$.
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|
Difference of the roots of quadratic formula I have a question to solve with roots quadratic formula that is ,
$$a^3 + b^3 = (a+b)(a^2-ab+b^2)$$
$$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$
but I didn't understand how the below formula is generated;
$$\alpha^3 - \beta^3 = (\alpha-\beta)^3+3\alpha\beta(\alpha-\beta)$$
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$(\alpha - \beta)^3 +3\alpha\beta(\alpha - \beta) = \alpha^3 -3\alpha^2\beta + 3\alpha\beta^2 -\beta^3+3\alpha^2\beta - 3\alpha\beta^2 =\alpha^3-\beta^3$
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Finding the residue of $\frac{e^{zt}}{z^2(e^{\pi z}-1)}$ using the laurent series $$f(z)=\frac{e^{zt}}{z^2(e^{\pi z}-1)}\tag{1}$$
I am doing it as follows, but my textbook is convinced that i am wrong:
$ \color{lime}{1.\space Residue\space at\space z\space =0}$
let $\pi z := q$
$e^{q}-1 = q+\frac{q^2}{2}+\frac{q^3}{6}+..=q(1+\frac{q}{2}+\frac{q^2}{6}+..)\tag{a}$
$e^{zt} =1+ zt +\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+.. \tag{b}$
using $(a)$ and $(b)$ we express $f(z)$ as follows:
$$f(z) = \frac{1}{qz^2}\{1+ z+\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+..\}\{1+\frac{q}{2}+\frac{q^2}{6}+..\}^{-1}$$
re substituting $q$ $f(z)$ becomes:
$$f(z) = \frac{1}{\pi z^3}\{1+ zt +\frac{(zt)^2}{2}+\frac{(zt)^3}{6}+..\}\{1-(\frac{\pi z}{2}+\frac{(\pi z)^2}{6}+..)+(\frac{\pi z}{2}+\frac{(\pi z)^2}{6}+..)^2+..\}$$
grouping coefficients of $z^2$ to find $c_1$:
$z^2\{\frac{\pi ^2}{12} -\frac{t\pi}{2}+\frac{t^2}{2}\}\frac{1}{z^3\pi}$
$\therefore$ $\operatorname{Res(f(z);0)}=\frac{\pi}{12}-\frac{t}{2}+\frac{t^2}{2\pi}$
i did however manage to find the $ \color{lime}{\space Residue\space at\space z\space = 2ki}$
->can someone please confirm that "hopefully" my book is wrong - Thanks.
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Since:
$$ e^{zt} = 1+zt+\frac{t^2}{2}z^2+O(z^3),$$
$$\frac{z}{e^{\pi z}-1}=\frac{1}{\pi}-\frac{1}{2}z+\frac{\pi}{12}z^2+O(z^3) $$
we have:
$$\begin{eqnarray*}\operatorname{Res}\left(\frac{e^{zt}}{z^2(e^{\pi z}-1)},z=0\right)&=&[z^2]\left(1+zt+\frac{t^2}{2}z^2\right)\cdot\left(\frac{1}{\pi}-\frac{1}{2}z+\frac{\pi}{12}z^2\right)\\&=&\color{red}{\frac{\pi}{12}-\frac{t}{2}+\frac{t^2}{2\pi}}\end{eqnarray*}$$
as you stated.
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Solve $\sin(x)=-\frac{1}{2}$ I have to solve the following equation for $x$
$$\cos(2x)+\sin(x)=0$$
After simplification i got
$\sin(x)=1$ or $\sin(x)=-\frac{1}{2}$
$\Rightarrow x=90^0$
But don't know how to solve for $x$ for $\sin(x)=-\frac{1}{2}$ ?
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You can observe that $-\sin x=\cos(\pi/2+x)$, so your equation becomes
$$
\cos2x=\cos\left(\frac{\pi}{2}+x\right)
$$
which is equivalent to
$$
2x=\frac{\pi}{2}+x+2k\pi
\qquad\text{or}\qquad
2x=-\frac{\pi}{2}-x+2k\pi
$$
and so, after simplifying,
$$
x=\frac{\pi}{2}+2k\pi
\qquad\text{or}\qquad
x=-\frac{\pi}{6}+2k\frac{\pi}{3}
$$
You can now obviate to the “strange” periodicity by dividing the second set of solutions by writing it as
$$
x=-\frac{\pi}{6}+2\cdot3k\frac{\pi}{3}
\quad\text{or}\quad
x=-\frac{\pi}{6}+2\cdot(3k+1)\frac{\pi}{3}
\quad\text{or}\quad
x=-\frac{\pi}{6}+2\cdot(3k+2)\frac{\pi}{3}
$$
(every integer is either a multiple of $3$ or $1$ more than a multiple of $3$ or $2$ more than a multiple of $3$) getting
$$
x=-\frac{\pi}{6}+2k\pi
\quad\text{or}\quad
x=\frac{\pi}{2}+2k\pi
\quad\text{or}\quad
x=\frac{7\pi}{6}+2k\pi
$$
Since these include the other set of solutions, the last sets are the whole sets of solutions.
When dealing with $\sin x=-\frac{1}{2}$, you can remember that $\sin(-x)=-\sin x$, so you can write the equation as
$$
\sin(-x)=\frac{1}{2}
$$
which gives
$$
-x=\frac{\pi}{6}+2k\pi
\qquad\text{or}\qquad
-x=\pi-\frac{\pi}{6}+2k\pi
$$
This becomes
$$
x=-\frac{\pi}{6}-2k\pi
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2k\pi
$$
Since $k$ is supposed to be an arbitrary integer, you can also write the last set of solutions as
$$
x=-\frac{\pi}{6}-2(-k-1)\pi
\qquad\text{or}\qquad
x=-\frac{5\pi}{6}-2(-k-1)\pi
$$
so that you get the “principal” solution in the interval $[0,2\pi)$:
$$
x=\frac{11\pi}{6}+2k\pi
\qquad\text{or}\qquad
x=\frac{7\pi}{6}+2k\pi
$$
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|
find x+y if $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $? Let $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $.
How find $x+y$? I tried a number of ways.
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There isn't just one possible value for $x + y$, which we can see as follows. Let $a = x + \sqrt{x^2 + 1}$ and $b = y + \sqrt{y^2 + 1}$. Then ${1 \over a} = -x + \sqrt{x^2 + 1}$ and ${1 \over b} = -y + \sqrt{y^2 + 1}$. As a result we have
$$x + y = {1 \over 2}\bigg(a - {1 \over a}\bigg) + {1 \over 2}\bigg(b - {1 \over b}\bigg)$$
$$= {1 \over 2}\bigg({a^2 - 1 \over a}\bigg) + {1 \over 2}\bigg({b^2 - 1 \over b}\bigg)$$
$$= {1 \over 2}{a^2b + ab^2 - a - b \over ab}$$
$$= {1 \over 2}{(ab - 1)(a + b) \over ab}$$
The condition given is that $ab = 42$ so this is the same as
$$ {41 \over 84} (a + b)$$
Hence if $x+ y$ is uniquely determined, $a + b$ must also be uniquely determined. But all we are given is that $ab = 42$, and there are many pairs $(a,b)$ which satisfy $ab = 42$, each of which will have a different value for $a + b$. Since $z \rightarrow z + \sqrt{z^2 + 1}$ is monotone with range $[1,\infty)$, each such $(a,b)$ with $a,b \geq 1$ will correspond to some value of $x$ and $y$.
Hence there is not just one answer to your question.
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|
Evaluating$ \int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx $using residues I need help to solve the next improper integral using complex analysis:
$$ \int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx $$
I have problems when I try to find residues for the function $ f = \displaystyle \frac{1}{(z^4+4)^2}$.
This is what I tried.
$$\displaystyle \text{res}(f,\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}) = \lim_{z\to \sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}} \left( \frac{\left(z-\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}\right)^2}{(z^4-4)^2}\right)'$$
with $k\in\{0,1,2,3\}$. What do you think about it?
I know there is a little more general problem involving this integral; for all $a>0$
$$ \int_{-\infty}^{\infty} \frac{x^6}{(a^4+x^4)^2} dx= \frac{3\pi\sqrt{2}}{8a} $$
Edit.
I've had an idea: from the integration by parts
$$\int u dv = uv - \int vdu$$
and if we let
$$ dv = \frac{4x^3 }{(4+x^4)^2}, \, u = \frac{x^3}{4}$$
with
$$dv = -\frac{d}{dx} \frac{1}{4+x^4} = \frac{4x^3 }{(4+x^4)^2}$$
we get finally
$$ \int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx = 0 + \frac{3}{4} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx $$
which I think is more easy to solve.
Anyway, if you know another idea or how to complete my first try will be welcome.
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If you are right, then $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$ can be easily done by a semi-circle contour, computing the residues at $(1+i)/\sqrt{2}$ and $(-1+i)/\sqrt{2}$. It is easy to check that the integral of $\frac{z^2}{1+z^4}$ on the arc of the semi-circle goes to 0.
$\frac{z^2}{1+z^4}=\frac{z^2}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})}$ shows that it is a simple pole at those residues.
$\text{ Res}(\frac{z^2}{1+z^4},e^{i\pi/4})=\lim_{z\to e^{i\pi/4}}\frac{z^2(z-e^{i\pi/4})}{1+z^4}=\lim_{z\to e^{i\pi/4}}\frac{3z^2-2ze^{i\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i\pi/4}$.
$\text{ Res}(\frac{z^2}{1+z^4},e^{i3\pi/4})=\lim_{z\to e^{i3\pi/4}}\frac{z^2(z-e^{i3\pi/4})}{1+z^4}=\lim_{z\to e^{i3\pi/4}}\frac{3z^2-2ze^{i3\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i3\pi/4}$.
I used L'Hospital's rule above because it seems simpler.
So the answer is $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx=2\pi i\frac{1}{4}\left(\frac{1-i}{\sqrt{2}}+\frac{-1-i}{\sqrt{2}}\right)=\frac{\sqrt{2}}2\pi$.
Edit: Something's wrong with the integration by parts; we do not seem to get the right answer. It is a 4 in the bottom and not 1. But we can do a further substitution to get it right.
$\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}}\frac{(\frac{x}{\sqrt{2}})^2}{1+(\frac{x}{\sqrt{2}})^4} dx=\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$.
So the final answer is $\int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx=\frac{3}{4}\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{3}{8}\pi$. You also have to argue that you are limiting $r\to\infty$ in $\left[\frac{x^3}{4(4+x^4)}\right]_{-r}^r$ which is why it is 0.
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|
Let $C$ be the curve of intersection of the plane $x+y-z=0$ with ellipsoid $\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$. Let $C$ be the curve of intersection of the plane $x+y-z=0$ and the ellipsoid $$\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$$ Find the points on $C$ which are farthest and nearest from the origin
When dealing with constraints I tried to consider the function $$F(x,y,z)=x^2+y^2+z^2-\lambda(x+y-z)-\mu\left(\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}-1\right)$$ However, I cannot solve this equation after differentiating respect to $x,y,z$ because it yields three equations with no common solution.
The system of equations are
$$2x=\lambda+\frac{\mu x}{2}$$
$$2y=\lambda+\frac{2\mu y}{5}$$
$$2z=-\lambda+\frac{2\mu z}{25}$$
How would I approach this problem, thanks.
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Since I am lazy to do it by hands, I give you the sage code.
$###lagrange method
sage:x, y, z, lam , mu= var('x, y, z, lam, mu')
sage:f=x^2+y^2+z^2
sage:g1 = x^2/4+y^2/5+z^2/25-1
sage:g2 = x +y -z
sage:h = f - g1 * lam -g2 * mu
sage:gradh = h.gradient([x, y, z, lam, mu])
sage:critical = solve([gradh[0] == 0, gradh[1] == 0, gradh[2] == 0, gradh[3] == 0, gradh[4] == 0],x, y, z, lam, mu, solution_dict=True)
sage:for i in range(0,len(critical)):
print show(critical[i]),float(f(critical[i][x], critical[i][y], critical[i][z]));
you shall get
$\left\{z : \frac{5}{646} \, \sqrt{2} \sqrt{17} \sqrt{19}, x :
\frac{20}{323} \, \sqrt{2} \sqrt{323}, y : -\frac{35}{646} \, \sqrt{2}
\sqrt{17} \sqrt{19}, \mbox{lam} : \frac{75}{17}, \mu : -\frac{70}{5491}
\, \sqrt{2} \sqrt{17} \sqrt{19}\right\} 4.41176470588$
$
\left\{z : -\frac{5}{646} \, \sqrt{2} \sqrt{17} \sqrt{19}, x :
-\frac{20}{323} \, \sqrt{2} \sqrt{323}, y : \frac{35}{646} \, \sqrt{2}
\sqrt{17} \sqrt{19}, \mbox{lam} : \frac{75}{17}, \mu : \frac{70}{5491}
\, \sqrt{2} \sqrt{17} \sqrt{19}\right\} 4.41176470588$
$\left\{z : \frac{5}{19} \, \sqrt{5} \sqrt{19}, x : \frac{2}{19} \,
\sqrt{5} \sqrt{19}, y : \frac{3}{19} \, \sqrt{5} \sqrt{19}, \mbox{lam} :
10, \mu : -\frac{6}{19} \, \sqrt{5} \sqrt{19}\right\} 10.0$
$\left\{z : -\frac{5}{19} \, \sqrt{5} \sqrt{19}, x : -\frac{2}{19} \,
\sqrt{5} \sqrt{19}, y : -\frac{3}{19} \, \sqrt{5} \sqrt{19}, \mbox{lam}
: 10, \mu : \frac{6}{19} \, \sqrt{5} \sqrt{19}\right\} 10.0
$
So you have two points where f has minimum 4.411 and two points where you have the maximum 10.
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|
Evaluating the sum : $\;\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$ How to evaluate this sum?
$$\frac{1}{3}+\frac{1}{4}.\frac{1}{2!}+\frac{1}{5}.\frac{1}{3!}+\ldots$$
Please give some technique. Binomial not working.
|
Some details regarding convergence aside, you have
$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$
so
$$x e^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$$
and
$$ \int_0^t x e^x \, dx = \sum_{k=0}^\infty \frac{1}{k+2} \frac{t^{k+2}}{k!}.$$
Thus
$$\sum_{k=1}^\infty \frac{1}{k+2} \frac{1}{k!} = \int_0^1 x e^x \, dx - 1.$$
The last integral is easy to evaluate.
|
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|
Prove that $5\mid x\,$ if $\,x, y \gt 1 $ satisfy $2x^2 - 1=y^{15}$ If $x\gt 1$ and $\,y\gt 1,$ with $ x, y \in \mathbb N$ so that $(x,y)$ satisfies the equation $$2x^2-1=y^{15},$$ then prove that $5\mid x$.
$\mod {10}$ gave me just what the last digit of y can be.
|
apply fermat, so $y^{15}\equiv y^3 \bmod 5$. So you have $2x^2\equiv y^3+1$. now cubes $\bmod 5$ are $1,3,2,4,0$. This tells us the statement is false, because when $y^3\equiv 1$ we get $2x^2\equiv 2$ and $x$ can be $1$ or $4 \bmod 5$
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|
Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?
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$$\int_{\mathbb{R}}\frac{dx}{x^4+x^2+1}=2\int_{0}^{+\infty}\frac{dx}{x^4+x^2+1} = 2\int_{0}^{1}\frac{dx}{x^4+x^2+1}+2\int_{0}^{1}\frac{x^2\,dx}{x^4+x^2+1}$$
so we just have to find:
$$ I=2\int_{0}^{1}\frac{1+x^2}{1+x^2+x^4}\,dx = 2\int_{0}^{1}\frac{1-x^4}{1-x^6}\,dx.$$
By expanding the integrand function as a geometric series we have:
$$ I = 2\sum_{n=0}^{+\infty}\left(\frac{1}{6n+1}-\frac{1}{6n+5}\right)=2\sum_{n=1}^{+\infty}\frac{\chi(n)}{n} $$
where $\chi(n)$ is the primitive non-principal Dirichlet character $\!\!\pmod{6}$. Since, by the residue theorem:
$$\frac{x^2+1}{x^4+x^2+1}=-\frac{i}{2\sqrt{3}}\left(\frac{1}{x-\omega}+\frac{1}{x-\omega^2}-\frac{1}{x-\omega^4}-\frac{1}{x-\omega^5}\right)$$
where $\omega=\exp\frac{2\pi i}{6}$, it follows that:
$$ I = \int_{0}^{1}\left(\frac{1}{1-x+x^2}+\frac{1}{1+x+x^2}\right)\,dx=\frac{2\pi}{3\sqrt{3}}+\frac{\pi}{3\sqrt{3}}=\color{red}{\frac{\pi}{\sqrt{3}}}. $$
As an alternative approach, we can just manipulate the series representation:
$$ I = 2\sum_{n\geq 0}\frac{4}{(6n+3)^2-4}=\frac{1}{9}\sum_{n\geq 0}\frac{8}{(2n+1)^2-\frac{4}{9}}\tag{1}$$
through the logarithmic derivative of the Weierstrass product for the cosine function:
$$ \cos z = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2 \pi^2}\right), $$
$$ \tan z = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 \pi^2 - 4z^2}$$
$$ \pi\tan(\pi z) = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 - 4z^2}\tag{2}$$
from which it follows that:
$$ I = \frac{\pi}{3}\tan\frac{\pi}{3} = \frac{\pi}{\sqrt{3}}.\tag{3} $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
What is the solution set of the given homogenous system? Write the solution set of the given homogenous system in parametric vector form:
\begin{align} 2x_{1}+2x_{2}+4x_{3} &= 0\\
-4x_{1}-4x_{2}-8x_{3} &= 0\\
0x_{1}-3x_{2}-3x_{3} &= 0\\
\end{align}
My attempt:
\begin{align} 2x_{1}+2x_{2}+4x_{3} &= 0\\
-4x_{1}-4x_{2}-8x_{3} &= 0\\
0x_{1}-3x_{2}-3x_{3} &= 0\\
\end{align}
Divided the second row by $-4$ and the third row by $-3$
\begin{align} x_{1}+x_{2}+2x_{3} &= 0\\
x_{1}+x_{2}+2x_{3} &= 0\\
0x_{1}+x_{2}+x_{3} &= 0\\
\end{align}
Row $2$ and $3$ plus the first row multiplied by $-1$
\begin{align} x_{1}+x_{2}+2x_{3} &= 0\\
0x_{1}+0x_{2}+0x_{3} &= 0\\
-x_{1}+0x_{2}-x_{3} &= 0\\
\end{align}
Thanks in advance.
|
That looks fine so far except that the bottom left $0$ in the very last equation should be a $-1$. The middle equation tells you nothing, so you can eliminate it. You'll end up with a whole line of solutions (all multiples of a single vector).
|
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|
How to proof that $\sum_{i=1}^{2^n} 1/i \ge 1+n/2$ I had troubles trying to prove that for every $n\ge1$
$$\sum_{i=1}^{2^n}\frac1i\ge 1+\frac n2$$
Can you give me a hint about the induction proof or show me in detail how can I prove it? I would appreaciate any help. Thanks!
|
$$A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2^n}=\\\frac{1}{1}\\+\frac{1}{2}\\+(\frac{1}{3}+\frac{1}{4})\\+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})\\+...\\+(\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n})$$now see
$$\frac{1}{1}\\+\frac{1}{2}\\+(\frac{1}{3}+\frac{1}{4})>(\frac{1}{4}+\frac{1}{4})\\+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})>(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})\\+...\\+(\frac{1}{2^n+1}+\frac{1}{2^n+2}+...+\frac{1}{2^n+2^n})>(\frac{1}{2^{n+1}}+\frac{1}{2^{n+1}}+....\frac{1}{2^{n+1}})$$so $$A>1+\frac{1}{2}+2\frac{1}{4}+4\frac{1}{8}+...\\A>1+(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}....)\\A>1+\frac{n}{2}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Folding a rectangular paper and finding the area of the triangle so formed. Given a rectangular sheet of paper ABCD such that the lengths of AB and AD are respectively 7 and 3 cms.Suppose B' and D' are two points on AB and AD respectively such that if the paper is folded along B'D' then A falls on A' on the side DC. Determine the maximum possible area of the triangle AB'D'.
I discovered quite a few basic facts that everybody can, but cannot actually make any progress. Please help.
|
Rectangle $ABCD$ is drawn above with vertex $A$ at the origin. Let $DA' = x $ then slope of $AA' = \dfrac{3}{x} $ and therefore, slope of $B'D' = - \dfrac{x}{3} $ and we have that $B'D'$ passes through $( \dfrac{x}{2}, \dfrac{3}{2} )$, therefore, its equation is
$y' = \dfrac{3}{2} - \dfrac{x}{3} (x' - \dfrac{x}{2} ) $
$x'$ intercept is $\dfrac{x}{2} + \dfrac{9}{2x}$ and $y'$ intercept is $\dfrac{3}{2} + \dfrac{x^2}{6} $
We must have the $x'$ intercept in $[0, 7]$, i.e.
$ 0 \le x + \dfrac{9}{x} \le 14 $
So that
$ 0 \le x^2 + 9 \le 14 x $
And finaly $ x^2 - 14 x + 9 \le 0 $, which implies that $x \gt 0.67544 $
And we also have the condition of the $y'$ intercept, namely,
$ AD' = \dfrac{3}{2} + \dfrac{x^2}{6} \le 3 $
From which, we must have $ x \le 3 $
Area of $\triangle AB'D' = [AB'D'] = \dfrac{1}{2} \left(\dfrac{x}{2} + \dfrac{9}{2x} \right) \left(\dfrac{3}{2} + \dfrac{x^2}{6} \right) $
And this simplifies to
$ [AB'D'] = S(x) = \dfrac{1}{2} \left( \dfrac{3}{2} x + \dfrac{1}{12} x^3 + \dfrac{27}{4x} \right) = \dfrac{1}{8} \left( 6 x + \dfrac{x^3}{3} + \dfrac{27}{x} \right)$
Differentiating this expression with respect to $x$ and equating to zero, gives us,
$ 6 + x^2 - 27/x^2 = 0 $
Multiplying through by $x^2$,
$ x^4 + 6 x^2 - 27 = 0 $
$\Rightarrow (x^2 + 9)(x^2 - 3) = 0 $
And therefore, $x = \sqrt{3} $ gives the critical value for the area. To identify it as a local maximum or local minimum, we need to check the sign of second derivative.
$S''(x) = \dfrac{1}{8} ( 2 x + \dfrac{54}{x^3} ) \gt 0 $
So, at $ x = \sqrt{3} $ we have a local minimum.
In addition, we have to check the value of $S(x)$ at the end points $x = 0.67544 $ and $x = 3 $. We have
$S(0.67544) = 5.5161$
$S (\sqrt{3}) = 3.4641$
$S(3) = 4.5$
Therefore, $x = \sqrt{3} $ corresponds to the minimum area of $\triangle AB'D' $ where it is defined.
|
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|
Sequence pattern question I have the following question. Let $S_1$ be the sequence of positive integers $1,2,3,4,5 , \ldots$ and define sequence $S_{n+1}$ in terms of $S_n$ by adding $1$ to the integers of $S_n$ which are divisible by $n$. I need to find integers $n$ such that the first $n-1$ integers in $S_n$ are $n$.
So I wrote out some sequences
$S_2 : 2,3,4,5,6,7,8,9,10,11,12$
$S_3: 3,3,5,5,7,7,9,9,11,11,13,13$
$S_4: 4,4,5,5,7,7,10,10,11,11$
$S_5: 5,5,5,5,7,7,10,10,11,11$
$S_6: 6,6,6,6,7,7,11,11,11,11$
$S_7: 7,7,7,7,7,7,11,11,11,11$
and $S_2, S_3, S_5, S_7$ satisfies this property so I guess that the property is $n$ is prime but can anyone explain why?
|
Hint: $S_k$ is weakly increasing-it does not decrease because no number can pass another. A prime $p$ is not increased until $S_p \to S_{p+1}$
|
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|
Number of solutions to 3-variable sums with restrictions I came across the following problem
1) How many solutions does the equation $x_1+x_2+x_3=8$ have with integers $x_i\ge0$?
There are $9$ possible values for $x_1$. For each of those, there are $9-x_1$ possible values for $x_2$. And for each of those, there is only $1$ possible value for $x_3$. So To find the total number of solutions, we can sum the number of values for $x_2$ over the actual values of $x_1$.
$$
N=\sum_{x_1=0}^{8}(9-x_1)\cdot1=1+2+3+4+5+6+7+8+9=45
$$
The next two problems were identical, but with added restrictions.
2) How many solutions are there for problem (1) if $x_1\ge2$ and $x_3\ge3$?
3) How many solutions are there for problem (1) if $x_1\le2$ and $x_3\le3$?
Noticing that $N$ must not be too big, I found $N=10$ and $N=12$ by writing out tables of valid solutions. I am not a fan of brute force, however, and I am sure there is a better way of doing this, perhaps involving combinations. I did find this, which is a possible duplicate question, but I didn't quite follow the technique for counting solutions using generator functions. Got any hints to point me in the right direction?
|
Solving the equation $x_1 + x_2 + x_3 = 8$ involves determining how many ways two plus signs can be placed in a list of eight 1's. For instance, the list
$$1 1 1 + + 1 1 1 1 1$$
corresponds to the solution $x_1 = 3, x_2 = 0, x_3 = 5$, while
$$1 + 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 1, x_2 = 4, x_3 = 3$. The number of such lists that can be formed is
$$\binom{8 + 2}{2} = \binom{10}{2} = \frac{10!}{8!2!} = \frac{10 \cdot 9}{2} = 45$$
To solve the problem $x_1 + x_2 + x_3 = 8$ subject to the restrictions $x_1 \geq 2$ and $x_3 \geq 3$, let
\begin{align*}
y_1 & = x_1 - 2\\
y_2 & = x_2\\
y_3 & = x_3 - 3
\end{align*}
Then $x_1 = y_1 + 2$ and $x_3 = y_3 + 3$, so we obtain
\begin{align*}
y_1 + 2 + y_2 + y_3 + 3 & = 8\\
y_1 + y_2 + y_3 & = 3
\end{align*}
Applying the same method we used above to the equation $y_1 + y_2 + y_3 = 3$ gives
$$\binom{3 + 2}{2} = \binom{5}{2} = 10$$
solutions in the non-negative integers.
To solve the problem $x_1 + x_2 + x_3 = 8$ in the non-negative integers subject to the restrictions $x_1 \leq 2$ and $x_3 \leq 3$, we must subtract the number of solutions in which $x_1 \geq 3$ or $x_3 \geq 4$ from the $45$ solutions of the equation in the non-negative integers. To do so, we subtract the number of solutions in which $x_1 \geq 3$ and the number of solutions in which $x_3 \geq 4$, then add the number of solutions in which both $x_1 \geq 3$ and $x_3 \geq 4$ since those solutions would otherwise be subtracted from the total number of solutions twice.
The number of solutions of the equation $x_1 + x_2 + x_3 = 8$ in which $x_1 \geq 3$ is the number of solutions of the equation
\begin{align*}
y_1 + 3 + y_2 + y_3 & = 8\\
y_1 + y_2 + y_3 & = 5
\end{align*}
in the non-negative integers, which is
$$\binom{5 + 2}{2} = \binom{7}{2} = 21$$
The number of solutions of the equation $x_1 + x_2 + x_3 = 8$ in which $x_3 \geq 4$ is the number of solutions of the equation
\begin{align*}
y_1 + y_2 + y_3 + 4 & = 8\\
y_1 + y_2 + y_3 & = 4
\end{align*}
in the non-negative integers, which is
$$\binom{4 + 2}{2} = \binom{6}{2} = 15$$
The number of solutions of the equation $x_1 + x_2 + x_3 = 8$ in which $x_1 \geq 3$ and $x_3 \geq 4$ is the number of solutions of the equation
\begin{align*}
y_1 + 3 + y_2 + y_3 + 4 & = 8\\
y_1 + y_2 + y_3 & = 1
\end{align*}
in the non-negative integers, which is
$$\binom{1 + 2}{2} = \binom{3}{2} = 3$$
Hence, the number of solutions of the equation $x_1 + x_2 + x_3 = 8$ in the non-negative integers in which $x_1 \leq 2$ and $x_3 \leq 3$ is $45 - 21 - 15 + 3 = 12$.
|
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|
How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$
Any method for solving this problem? I've not achieved anything so far. Thanks in advance!
|
No formal answer, but a nice illustration:
The solution can exemplarically be shown by a matrix-multiplication-scheme. The following shows $S \cdot P = X$, where $X$ is simply a shifting of $P$ .
The left-bottom is matrix $S$ which performs the partial summations of the columns of matrix $P$ which is in the right-top-matrix, and the result $X$ is the right-bottom-matrix:
$$ \small \begin{matrix}
. & . & . & . & . & | & 1 & . & . & . & . & | \\
. & . & . & . & . & | & 1 & 1 & . & . & . & | \\
. & . & . & . & . & | & 1 & 2 & 1 & . & . & | \\
. & . & . & . & . & | & 1 & 3 & 3 & 1 & . & | \\
. & . & . & . & . & | & 1 & 4 & 6 & 4 & 1 & | \\
- & - & - & - & - & + & - & - & - & - & - & + \\
1 & . & . & . & . & | & 1 & . & . & . & . & | \\
1 & 1 & . & . & . & | & 2 & 1 & . & . & . & | \\
1 & 1 & 1 & . & . & | & 3 & 3 & 1 & . & . & | \\
1 & 1 & 1 & 1 & . & | & 4 & 6 & 4 & 1 & . & | \\
1 & 1 & 1 & 1 & 1 & | & 5 & 10 & 10 & 5 & 1 & | \\
- & - & - & - & - & + & - & - & - & - & - & +
\end{matrix} $$
From this a hypothetical result can easily be formulated and gives the direction of the attempt of a proof...
|
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|
Non-linear differential equation Let $ x \in \mathbb{R} $
Find all solutions of the following differential equation :
$$
y'=\tan(y+x)
$$
|
Let $u = y + x$. Then $u' = y' + 1$, so the ODE can be written $$u' = 1 + \tan(u),$$ which is a separable equation. We have $$\int \frac{du}{1 + \tan u} = \int dx,$$ which reduces to $$\frac{1}{2}(u + \log|\sin u + \cos u|) = x + C.$$ To see this, let $$I = \int \frac{du}{1 + \tan u}.$$ Then $$I = \int \left(1 - \frac{\tan u}{1 + \tan u}\right)\, du = u - \int \frac{\tan u}{1 + \tan u},$$ and
\begin{align}-\int \frac{\tan u}{1 + \tan u}\, du &= \int -\frac{\sin u}{\cos u + \sin u}\, du = \int \frac{-\sin u + \cos u}{\cos u + \sin u}\, du - \int \frac{\cos u}{\sin u + \cos u}\, du\\ &= \log|\cos u + \sin u| - \int \frac{1}{1 + \tan u}\, du\\
&= \log|\cos u + \sin u| - I.\end{align}
Therefore
$$I = \frac{1}{2}(u + \log|\sin u + \cos u|) + C.$$
Certainly $\int dx = x + C$, so we have $\frac{1}{2}(u + \log|\cos u + \sin u|) = x + C$. Now revert back to $y$.
|
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|
$ \lim_{n \to \infty} \frac {\sqrt{n+1}+\sqrt{n+2}+...+\sqrt{2n}}{n^{3/2}}$ Evaluate the limit
$$
\lim_{n \to \infty} \frac {\sqrt{n+1}+\sqrt{n+2}+...+\sqrt{2n}}{n^{3/2}}.
$$
Rearranging I can get
$$
\lim_{n \to \infty} \frac {\sqrt{\frac{n+1}{n}}+\sqrt{\frac{n+2}{n}}+...+\sqrt{\frac{2n}{n}}}{n}.
$$
but I do not see how that helps me. Perhaps I use L'Hopital's rule?
|
Using Stolz-Cesaro:
$$\eqalign{
\lim_{n\to\infty}\frac{\sqrt{n+1}+\cdots+\sqrt{2n}}{\sqrt{n^3}}
& =
\lim_{n\to\infty}\frac{(\sqrt{(n+1)+1}+\cdots+\sqrt{2(n+1))}-(\sqrt{n+1}+\cdots+\sqrt{2n})}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr
& =
\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr
&=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\cdots
}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
integration of $\large\int \frac{u^2}{(1-u^2)^2}$ $ du$ Is there a way to integrate $$\large\int \frac{u^2}{(1-u^2)^2} du$$ without using partial fraction decomposition?
|
Hint:
Let $f(u) = u$ and $g(u) = 1-u^2$. Hence,
$$ \left(\frac{f}{g}\right)'(u) = \frac{u^2+1}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} + \frac{1-u^2}{(1-u^2)^2} = \frac{2u^2}{(1-u^2)^2} + \frac{1}{2} \left(\frac{1}{1-u} + \frac{1}{1+u}\right).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a differential equation $F-y'F_{y'}=C$, with $F(y,y')= \frac{1+2y'^2}{3y^3\sqrt{1+y'^2}}$ If $$F= F(y,y')= \frac{1+2y'^2}{3y^3\sqrt{1+y'^2}},$$
where $y=y(x)$ and $y'= y'(x)=\frac{dy}{dx}$, then how to solve the differential equation:
$$F-y'F_{y'}=C, $$
that is:
$$F(y,y')-\frac{dy}{dx}\frac{\partial F}{\partial y'}=C,$$
where $C$ is some constant? What are the keys steps? I got a very hairy differential equation when I tried solving this directly by doing the calculations. Is there some substitution trick etc. that I should apply here?
|
We have that
$$\begin{align} \\
F(y,y') &= \frac{1 + 2y'^{2}}{3y^{3} \sqrt {1+y'^{2}}} \\
&= \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg]
\end{align} $$
As there is no explicit $x$ dependence, we can find a first integral of the form
$$\begin{align} \\
y'\frac{\partial F}{\partial y'} - F &= y'\cdot\frac{1}{3y^{3}}\bigg[4y'(1+y'^{2})^{\frac{-1}{2}} + (1 + 2y'^{2})\cdot\frac{-1}{2}\cdot2y'(1+y'^{2})^{\frac{-3}{2}}\bigg] - \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[(3y'^{2} + 2y'^{4})(1 + y'^{2})^{\frac{-3}{2}}\bigg] - \frac{1}{3y^{3}}\bigg[(1 + 2y'^{2})(1 + y'^{2})^{\frac{-1}{2}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[\frac{3y'^{2} + 2y'^{4} -(1 + 3y'^{2} + 2y'^{4})}{(1+y'^{2})^{\frac{3}{2}}}\bigg] \\
&= \frac{1}{3y^{3}}\bigg[\frac{-1}{(1+y'^{2})^{\frac{3}{2}}}\bigg] \\
&= C
\end{align} $$
(I omitted a few steps of algebra, if you would like me to put them in just write a comment below).
Hence we get
$$\begin{align} \\
(1+y'^{2})^{\frac{3}{2}} &= \frac{-1}{3y^{3}C} \\
\implies (1+y'^{2})^{3} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg) \\
\end{align} $$
Where in the last step we squared both sides.
Therefore,
$$\begin{align} \\
1 + y'^{2} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg)^{\frac{1}{3}} \\
\implies y'^{2} &= \bigg(\frac{1}{9y^{6}C^{2}}\bigg)^{\frac{1}{3}} - 1 \\
&= \alpha^{2}\cdot\frac{1}{y^{2}} - 1 \\
\implies y' &= \pm \bigg(\alpha^{2}\cdot\frac{1}{y^{2}} - 1\bigg)^{\frac{1}{2}} \\
&= \pm \bigg(\frac{\alpha^{2} - y^2}{y^{2}}\bigg)^{\frac{1}{2}} \\
&= \pm \frac{\sqrt{\alpha^{2} - y^2}}{y} \\
\end{align} $$
With
$$\alpha^{2} = \bigg(\frac{1}{9C^{2}}\bigg)^{\frac{1}{3}}$$
Separating and integrating we get
$$ \int \frac{y}{\sqrt{\alpha^{2} - y^2}} dy = \pm \int dx $$
where the LHS can be solved using a trig substitution.
For the future, a general method for solving maximising/minimising problems goes like this (this solution recipe is a grave oversimplification, but it may help you get used to these types of problems):
1) If there is no independent variable dependence ($x$ or $t$ or whatever other variable it might be), find a first integral using the formula
$$ y'\frac{\partial F}{\partial y'} - F = C $$
2) Substitute your functions $\frac{\partial F}{\partial y'}$ and $F$ into your equation.
3) Solve for $y'$
4) Separate and integrate
If you have any questions or you see a mistake in my algebra, just let me know in the comments.
|
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|
Evaluating$\int\frac{1}{(x^2-1)^2}$ This is the integral:
$\int\frac{1}{(x^2-1)^2}$
I have tried several ways to solve this but I always end up that last parameter equals 1 and all others equals 0 so I end up where I started. Examples over the internet with similar fraction have more than $1$ in the numerator which makes the example simplier.
Here's what I tried
$\frac{A}{(x-1)^2}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^2(x-1)^2}$ as ${(x^2-1)^2}=(x+1)(x-1)(x+1)(x-1)$
from there I got to point where
$A+B+C=1\\4B+C=1\\4A+C=1$
but this leads to $A=0,\ B=0$ and $C=1$ so to nowhere
I also tried this way:
$\frac{Ax+B}{(x^2-1)}+\frac{Cx+D}{(x^2-1)^2}$
but this resulted into
$A = 0\\B=0\\-A+C=0\\-B+D=1$
so again I got $A=0,\ B=0,\ C=0$ and $D=1$ which led to nowhere.
How to tackle this problem?
|
Since:
$$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right),$$
then:
$$\frac{1}{(x^2-1)^2}=\frac{1}{4}\left(\frac{1}{(x-1)^2}-\frac{2}{(x-1)(x+1)}+\frac{1}{(x+1)^2}\right)$$
or:
$$\frac{1}{(x^2-1)^2}=\frac{1}{4}\left(\frac{1}{(x-1)^2}-\frac{1}{x-1}+\frac{1}{x+1}+\frac{1}{(x+1)^2}\right).$$
Better now?
|
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|
Evaluating $\int\frac{\sqrt{x^2-1}}x\mathrm dx$ How can one evaluate the integral
$$\int\frac{\sqrt{x^2-1}}x\mathrm dx$$?
I tried substituting $x = \cosh t$ but got stuck at
$$\int\frac{\sinh^2t}{\cosh t}\mathrm dt$$
Any hints?
|
$$
\begin{aligned}\int \frac{\sqrt{x^{2}-1}}{x} d x =& \int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
=& \int \frac{x^{2}-1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) \\
=& \int\left(1-\frac{1}{x^{2}}\right) d\left(\sqrt{x^{2}-1}\right) \\
=& \sqrt{x^{2}-1}-\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} \\
=& \sqrt{x^{2}-1}-\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
|
{
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|
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.
$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!
|
As I said we need to solve a system of equations.
The system of equations:
$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$
the solutions have the form: $$x=4p^4-s^4$$ $$y=4p^2s^2$$ $$q=4ps(2p^2-s^2)$$ $$t=4p^4-8p^2s^2+s^4$$ $$z=4p^4+s^4$$
$p,s,k$ - integers.
Formulas you can write a lot, but will be limited to this. Will make a replacement.
$$a=p^2+s^2-k^2$$
$$b=p^2+s^2+k^2-2pk-2ks$$
$$c=p^2+k^2-s^2+2ps-2kp$$
$$r=s^2+k^2-p^2+2ps-2ks$$
The solution then is.
$$x=2ab$$
$$y=a^2-b^2$$
$$q=2cr$$
$$t=c^2-r^2$$
$$z=a^2+b^2$$
|
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|
Combinatorial Proof For Counting All Binary Strings The Question
Provide a combinatorial proof for the following:
For $n \ge 1$,
$$2^n = \binom{n+1}1+\binom{n+1}3+\ldots+\begin{cases}
\binom{n+1}n,&\text{if }n\text{ is odd}\\\\
\binom{n+1}{n+1},&\text{if }n\text{ is even}
\end{cases}$$
My Work
Parts a,b of the question involved counting strings. This lead me to notice that $2^n$ is actually all $n$-bit binary strings. Which means the series on the right is somehow counting all total binary strings of length $n$. I'm kind of stumped as to how the series works though. I tried writing out some examples of it.
$2^1 = \binom{1+1}{1} = \binom{2}{1} = 2$
$2^2 = \binom{3}{1}+\binom{3}{3} = 4$
$2^3 = \binom{4}{1}+\binom{4}{3} = 8$
$2^4 = \binom{6}{1}+\binom{6}{3}$ This is where I realized I must be doing it wrong because $2^4 = 16$ and $\binom{6}{3} = 20$
My Question
I'm trying to understand the problem. How does the series work? How would I write out $2^5$ as the $RHS$ series for example? And then how does it count the total amount of $n$ length binary strings.
|
Your $n=4$ line should read
$$2^4=\binom51+\binom53+\binom55=5+10+1=16\;.$$
The upper number in each of the binomial coefficients is $n+1$, which is $5$ in this case, and the lower number runs through all of the odd numbers less than or equal to $n+1$.
HINT:
$$\binom{n+1}1+\binom{n+1}3+\ldots+\begin{cases}
\binom{n+1}n,&\text{if }n\text{ is odd}\\\\
\binom{n+1}{n+1},&\text{if }n\text{ is even}
\end{cases}\tag{1}$$
is the number of subsets of $\{1,2,\ldots,n+1\}$ that have an odd number of elements. Show that exactly half of the subsets of $\{1,2,\ldots,n+1\}$ have an odd number of elements. Then use your observation that a set of $m$ elements has $2^m$ subsets altogether to complete the proof that $(1)$ is equal to $2^n$.
|
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|
Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a(a+1)(a-1) = 3k$
since $a(a+1)(a-1)$ is the product of 3 consecutive integers, the expression is divisible by 3.
Is this ok?, just a check. I'm not really all that good at number theory.
|
$a^3-a=(a-1)a(a+1)$ is always true and therefore is always the product of three consecutive integers.
Another way is to see that $a$ can be congruent to $0$, $1$ or $2$ $mod 3$. In each case we get that $a^3-a$ is congruent to $0$.
|
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|
Laurent series expansion, can one simplify this? I have to expand $f(z)=\frac{z-1}{(z^2+1)z}$ in an annulus $R(i,1,2)$.
$$f(z)=\frac{1}{z-i}\frac{1}{z+i}-\frac{1}{z-i}\Big(\frac{i}{z+i}-\frac{i}{z}\Big)$$
$$\frac{1}{z-i}\frac{1}{z+i}=\frac{1}{z-i}\cdot\frac{1}{2i}\cdot\frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n$$
$$\frac{i}{z+i}=\frac{1}{2}\cdot \frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n}\frac{1}{2}(z-i)^n$$
$$\frac{i}{z}=\frac{i}{z-i}\cdot\frac{1}{1-(-i/(z-i))}=\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+1}$$
Finally we got:
$$f(z)=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n-\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}=\frac{1}{2i}\cdot\frac{1}{z-i}+\sum_{n=1}^{\infty}\Big( (-1)^{n-1}(\frac{1}{2i})^{n}-(-1)^n (\frac{1}{2i})^{n+1}\Big)(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}$$
My questions are: is this expansion correct? And can one simplify it?
|
Hi this series as generalitation of Laurent Series in three poles
$$\sum _{k=-1}^n \left(\frac{1}{2}+\frac{i}{2}\right) \left((-i)^k-i i^k\right) x^k+\frac{(-1)^{n-1} \left(\left(\frac{1}{2}-\frac{i}{2}\right) x^{n+1}\right)}{x-i}+\frac{(-1)^{n-1} \left(\left(\frac{1}{2}+\frac{i}{2}\right) x^{n+1}\right)}{x+i}$$
|
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|
$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$ $$I=\int \frac{\cos^3(x)}{\sqrt{\sin^7(x)}}\,dx$$
I tried to write it as
$$I=\int \sqrt{\frac{\cos^6(x)}{\sin^7(x)}}\,dx$$
And $$I=\int \sqrt{\frac{1}{\tan^6(x)\sin(x)}}\,dx$$ but it seems to go nowhere , how can I manipulate it so that its becomes solvable?
|
Here are the steps
$$\int \frac{\cos^3 x}{\sqrt{\sin^7 x}}dx= \int \frac{(1-\sin^2 x)(\cos x)}{\sqrt{\sin^7 x}}dx $$
Let $u=\sin x$, then $du=\cos x\ dx$. So now
$$ \int \frac{1-u^2}{\sqrt{u^7}}du = \int \frac{1}{\sqrt{u^7}}-\frac{u^2}{\sqrt{u^7}}du $$
$$= \int u^{-\frac72}du-\int u^{-\frac32}du $$
$$= \frac{u^{-\frac72+1}}{-\frac72+1}- \frac{u^{-\frac32+1}}{-\frac32+1}+C $$
$$= \frac{u^{-\frac52}}{-\frac52}- \frac{u^{-\frac12}}{-\frac12} +C $$
$$= 2u^{-\frac12} -\frac{2}{5}u^{-\frac52}+C$$
$$= \frac{2}{\sqrt{u}}-\frac{2}{5\sqrt{u^{5}}}+C$$
$$= \frac25\left(\frac{5\sqrt{u^5}-\sqrt{u}}{\sqrt{u}\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5u^2-1}{\sqrt{u^{5}}}\right) +C$$
$$= \frac25\left(\frac{5\sin^2(x)-1}{\sqrt{\sin^{5}(x)}}\right) +C$$
|
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|
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?
Thanks.
|
Recall that
$$\text{arsinh } x := \ln\left(x + \sqrt{1 + x^2}\right),$$
and that it has a nice Taylor series expansion:
$$\text{arsinh } x \sim x - \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^5}{5} - \cdots$$
(it's not too hard to write the coefficients in a closed form, but we only need the first terms here).
Accounting for the subtracted $x$ in the numerator of the original ratio, the Taylor series of the numerator is
$$- \frac{1}{6} x^3 + O(x^5).$$
Now, the Taylor series of $\tan x$ is$$\tan x \sim x + \frac{1}{3} x^3 + \frac{2}{15} x^5 + \cdots.$$ So, multiplying gives that the Taylor series of the denominator $\tan^3 x$ is $$x^3 + O(x^5).$$
The leading terms are both third-order, so the limit is the ratio of the coefficients of those terms, that is $$\lim_{x \to 0} \frac{\ln(x + \sqrt{1 + x^2}) - x}{\tan^3 x} = \frac{\left(-\frac{1}{6}\right)}{(1)} = -\frac{1}{6}.$$
|
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|
Commutative matrices Knowing that $AB=BA$, find the matrices that commute with the matrix
\begin{pmatrix}
1 & 2 \\
3 & 4 \\
\end{pmatrix}
I have assumed that multiplying matrix
$\begin{pmatrix}
a & b \\
c & d \end{pmatrix}
$ by the first one should be equal to multiplying the first one by $\begin{pmatrix}
a & b \\
c & d \end{pmatrix}
$
Alright:
$\begin{pmatrix}
a+2c & b+2d\\
3a+4c & 3b+4d\end{pmatrix}
$
equals
$\begin{pmatrix}
a+3b & 2a+4b\\
c+3d & 2c+4d\end{pmatrix}
$
so:
$\begin{pmatrix}
a+2c = a+3b => 2c = 3b = 0\\
3a+4c = c +3d => a+c - d=0\\
b+2d = 2a+ 4b => -2a - 3b + 2d = 0\\
3b + 4d = 2c + 4d => 3b - 2c =0\end{pmatrix}
$
I was left with equations from which I have formed another matrix and used Gauss algorithm to evaluate it:
$\begin{pmatrix}
1 & 0 & 1 & -1 & 0 \\
-2& -3 &0 &2& 0\\
0 &3 &-2& 0 &0\\
0 &-3& 2& 0 &0\end{pmatrix}
$
I made some changes (r3:=r3+r4; r4:=r4+r3; r2:=r2+2r1) and the upper diagonal matrix is:
$\begin{pmatrix}
1 & 0 & 1 & -1 & 0 \\
0& -3 &2 &0& 0\\
0 &0 &0& 0 &0\\
0 &0& 0& 0 &0\end{pmatrix}
$
I was left with a set of equations
\begin{eqnarray}
a+ c + d &=& 0 \\
-3b + 2c &=& 0
\end{eqnarray}
and now I'm stuck. Any help would be highly appreciated!
EDIT: A is the given matrix.
|
You should be left with a little more than two equations. If $$A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$$
and $$B=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$
then $$AB=\begin{bmatrix}a+2c & b+2d\\ 3a+4c & 3b+4d\end{bmatrix}$$
and $$BA=\begin{bmatrix}a+3b & 2a+4b\\ c+3d & 2c+4d\end{bmatrix}$$
which gives you $4$ equations.
|
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|
Find all positive integers $n$ for some given condition. Find all positive integers $n>1$ such that $n^2$ divides $2^n+1$
I found that $n$ is of the form $6k+3$.
|
The only answer is $n=3$
(1) Since $n$ divides $2^n + 1$, $n$ is odd. Let $p$ be the smallest prime divisor of $n$
(2) Let $a$ be the smallest positive integer such that $2^a \equiv -1 \pmod p$. $a$ must exist since $2^n \equiv -1 \pmod p$
(3) Let $b$ be the smallest positive integer such that $2^b \equiv 1 \pmod p$. $b$ must exist and $b < p$ since $2^{p-1} \equiv 1 \pmod p$
(4) There exists $q,r$ such that $n = qb+r$ and $0 \le r < b$ so that $-1 \equiv 2^n \equiv (2^b)^q2^r \equiv 2^r \pmod p$ and $a \le r < b$ and $r > 0$
(5) There exists $h,k$ such that $n = ha+k$ and $0 \le k < a$ so that $-1 \equiv 2^n \equiv (2^a)^h2^k \equiv (-1)^h2^k \pmod p$
(6) $k=0$ since if $k > 0$ then $h$ cannot be even since then $2^k \equiv -1 \pmod p$ and $k < a$ and $h$ cannot be odd since $2^k \equiv 1 \pmod p$ and $k < a \le r < b$
(7) Since $k=0$, $a | n$. Since $a < p$, $a=1$ but if $2^1 \equiv -1 \pmod p$, $p=3$
(8) From the article cited by Shivang, we have:
Let $v_p(x)$ be the greatest power in which prime $p$ divides $x$.
Theorem 2: (Second form of LTE) Let $x,y$ be two integers, $n$ be an odd positive
integer and $p$ an odd prime such that $p|(x+y)$ and none of $x$ and $y$ is divisible by
$p$, we have:
$$v_p(x^n + y^n) = v_p(x+y) + v_p(n)$$
(9) Let $x=2$, $y=1$
(10) By Theorem 2: $v_3(2^n + 1) = v_3(2^n + 1^n) = v_3(2+1) + v_3(n) = 1 + v_3(n)$
(11) So, there exists $w$ with odd $r,s$ such that $3^wr = n$ and $3 \nmid r$ and $3^{w+1}s = 2^n + 1$ and $3 \nmid s$. Since $n^2 | 2^n+1$, it follows that $3^{2w}r^2 |\, (3^{w+1}s)$ but this is only possible if $2w = w+1$ so that $w=1$.
(12) So, we have $n=3r$. Asume $r > 1$ Let $p$ be the least prime that divides $r$. Since $r | (2^n+1)(2^n - 1) = 2^{2n} - 1$, so $2^{2n} \equiv 1 \pmod p$ and using Fermat's Little Theorem gives us $2^{p-1} \equiv 1 \pmod p$. Therefore $2^{\gcd(2n,p-1)}\equiv 1 \pmod p$. Since $p$ is the least prime, $(p-1) \nmid n$, $\gcd(2n,p-1)=2$
(13) But if $2^2\equiv 1\pmod p$, then $p=3$ which is impossible since $3 \nmid r$ so $r=1$.
(14) $3^2s = 2^3 + 1$ is only true if $s=1$. So, the only solution is $n=3$.
|
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|
Calculate this infinite sum $$s= \sum_{n=1}^\infty \frac{n+3}{(2^n)(n+1)(n+2)}$$
Any method to calculate this type of infinite sums?
|
Using Partial Fraction Decomposition,
$$\text{let }\frac{n+3}{(n+1)(n+2)}=\frac A{n+1}+\frac B{n+2}$$
$$n+3=n(A+B)+2A+B\implies A+B=1,2A+B=3\implies A=2, B=-1$$
$$\implies\frac{n+3}{(n+1)(n+2)}=\frac 2{n+1}-\frac1{n+2}$$
$$\implies\frac{n+3}{2^n(n+1)(n+2)}=\frac{(1/2)^{n-1}}{n+1}-\frac{(1/2)^n}{n+2}$$
If $T_m=\dfrac{(1/2)^{m-1}}{m+1},$
$$\frac{n+3}{2^n(n+1)(n+2)}=T(n)-T(n+1)$$ which is clearly Telescoping
|
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|
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
|
Note
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{1}$$
$$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{2}$$
$(1)+(2)$
$$\Longrightarrow x+y=-(x+y)$$
$$\Longrightarrow x+y=0$$
|
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|
How to integrate $1/\sqrt{(1+x^2)^3}$? Normally I use WolframAlpha pro to help me with problems I don't know however wolfram wont/cant show me the steps only the final solution to this integration problem.
Is anyone able to assist me with a walk through of atleast the start if not all of the steps to solving this equation?
$$\int\frac{dx}{\sqrt{(1+x^2)^3}}$$
|
$$\int\frac{dx}{\sqrt{(1+x^2)^3}}$$
make a trigonometric substituition
$$
x=\tan\theta\\
dx=\sec^2\theta d\theta\\
1+x^2=\sec^2\theta\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta$$
if $\theta\in(0|\frac{\pi}{2})\cup(\frac{3\pi}{2}|2\pi)$ then $\cos\theta>0$ and then
$$\begin{align}
\tan\theta&=x\\
\sec\theta&=\sqrt{1+x^2}\\
\cos\theta&=\frac{1}{\sqrt{1+x^2}}\\
\sin\theta&=\frac{x}{\sqrt{1+x^2}}\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta&=\int\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\int\frac{1}{\sec\theta}d\theta\\
&=\int\cos\theta d\theta\\
&=\sin\theta+C\\
&=\frac{x}{\sqrt{1+x^2}}+C
\end{align}$$
if $\theta\in(\frac{\pi}{2}|\frac{3\pi}{2})$ then $\cos\theta<0$ and then
$$\begin{align}
\tan\theta&=x\\
\sec\theta&=-\sqrt{1+x^2}\\
\cos\theta&=-\frac{1}{\sqrt{1+x^2}}\\
\sin\theta&=-\frac{x}{\sqrt{1+x^2}}\\
\int\frac{\sec^2\theta}{\sqrt{(\sec^2\theta)^3}}d\theta&=\int-\frac{\sec^2\theta}{\sec^3\theta}d\theta\\
&=\int-\frac{1}{\sec\theta}d\theta\\
&=\int-\cos\theta d\theta\\
&=-\sin\theta+C\\
&=\frac{x}{\sqrt{1+x^2}}+C
\end{align}$$
|
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|
How to integrate $(x^2 - y^2) / (x^2 + y^2)^2$ How do I integrate
$$\int \int \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dx dy?$$
The WolframAlpha page gives
$$
c_1 + c_2 + \tan^{-1}(x/y).
$$
And I kind of specifically need
$$
\int_{0}^{x} \frac{(x^2 - y^2)}{(x^2 + y^2)^2} dy.
$$
Note
*
*I want to know integration technique to solve this without using $F' = f$.
*For the double integral above, what I'm interested is Lebesgue integral, but I guess what Wolfram gave is in the Riemann sense.
|
By the quotient rule $$\frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right) = \frac{x^2 - y^2}{(x^2 + y^2)^2}.$$
Therefore
$$\int_0^x \frac{x^2 - y^2}{(x^2 + y^2)^2}\, dy = \int_0^x \frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right)\, dy = \frac{y}{x^2 + y^2}\bigg|_{y = 0}^{y = x} = \frac{1}{2x}.$$
|
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|
Study the convergence of $\sum_{n=1}^\infty \frac{(-1)^n \cos^2(n)}{n}$
Study the convergence of $\sum_{n=1}^\infty \frac{(-1)^n \cos^2(n)}{n}$
Abel's/Dirichlet's tests cannot be applied here.
I guess it's something more tricky involving integration maybe (?)
|
First write
$$\sum_{n = 1}^\infty \frac{(-1)^n\cos^2 n}{n} = \sum_{n = 1}^\infty \left(\frac{(-1)^n}{2n} + \frac{(-e^{2i})^n}{4n} + \frac{(-e^{-2i})^n}{4n}\right)$$
using the identities $$\cos^2 x = \frac{(e^{ix} + e^{-ix})^2}{4} = \frac{1}{2} + \frac{e^{2ix}}{4} + \frac{e^{-2ix}}{4}.$$ The series $\sum_{n = 1}^\infty (-1)^n/(2n)$ converges by the alternating series test, and series $\sum_{n = 1}^\infty (-e^{2i})^n/(4n)$ and $\sum_{n = 1}^\infty (-e^{-2i})^n/(4n)$ converge by Dirichlet's test. This shows that $\sum_{n = 1}^\infty (-1)^n\cos^2(n)/n$ converges. It's value can be computed:
\begin{align}&\sum_{n = 1}^\infty \frac{(-1)^n}{2n} + \sum_{n = 1}^\infty \frac{(-e^{2i})^n}{4n} + \sum_{n = 1}^\infty \frac{(-e^{-2i})^n}{4n}\\
& = -\frac{\log 4 + \log(1 + e^{2i}) + \log(1 + e^{-2i})}{4}\\
&= -\frac{\log 4 + \log[(1 + \cos 2)^2 + \sin^2 2]}{4}\\
&= -\frac{\log 4 + \log|2 + 2\cos 2|}{4}\\
&= -\frac{\log 4 + \log 2 + \log(1 + \cos 2)}{4}\\
&= -\frac{\log[8(1 + \cos 2)]}{4}.
\end{align}
Note. In general, series of the form $$\sum_{n = 1}^\infty \frac{e^{in\theta}}{n},\quad \theta\in \Bbb R\setminus 2\pi \Bbb Z$$ are convergent, which I'll show by Dirichlet's test. Let $\theta \in \Bbb R\setminus 2\pi \Bbb Z$. The sequence $\left\langle \frac{1}{n}\right\rangle_{n = 1}^\infty$ decreases to $0$, and for every positive integer $N$,
$$\left\lvert \sum_{n = 1}^N e^{in\theta}\right\rvert = \left\lvert e^{i\theta}\frac{1 - e^{iN\theta}}{1 - e^{i\theta}}\right\rvert = \dfrac{\left\lvert\sin \frac{N\theta}{2}\right\rvert}{\left\lvert\sin \frac{\theta}{2}\right\rvert} \le \frac{1}{\left\lvert \sin \frac{\theta}{2}\right\rvert}.$$ Hence, by Dirichlet's test, the series $\sum_{n = 1}^\infty e^{in\theta}/n$ converges.
|
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|
Integral $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx$, where $0I am trying to evaluate the following integral:
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(1+c\sin x) dx,$$
where $0<c<1$.
I can't really think of a way to find it so please give me a hint.
|
Express the integrand as a series:
$$\begin{align}I(c) &= \int_{-\pi/2}^{\pi/2} dx \, \log{(1+c \sin{x})} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} c^k \int_{-\pi/2}^{\pi/2} dx \, \sin^k{x} \\ &= - \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{c^{2 k}}{k} \frac1{2^{2 k}} \binom{2 k}{k} \end{align} $$
$$I'(c) = -\pi \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k-1} = -\frac{\pi}{c} \sum_{k=1}^{\infty} \frac1{2^{2 k}} \binom{2 k}{k} c^{2 k}= -\frac{\pi}{c} \left (\frac1{\sqrt{1-c^2}}-1\right )$$
$$\implies I(c) = -\pi \int dc \frac{1}{c} \left (\frac1{\sqrt{1-c^2}}-1\right ) = \pi \int dt \, (\sec{t}-\tan{t}) = \pi \log{(1+\sin{t})} + K$$
or
$$I(c) = K +\pi \log{\left ( 1+\sqrt{1-c^2} \right )} $$
$$I(0) = 0 \implies K=-\pi \log{2}$$
Therefore
$$I(c) = \pi \log{\left ( \frac{1+\sqrt{1-c^2}}{2}\right )} $$
|
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|
To show two matrices are conjugate to each other Given two matrices A and B
$$
A =
\begin{pmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ 1 & 2 & 1 \end{pmatrix}, \quad
B =
\begin{pmatrix} 3 & 0 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
$$
I have checked basic properties such as trace, charatersitic polynomial and are same .
I also realise that i need to find the P
Such that $A=P^{-1}BP $
But how do i find P ? Thanks
|
Compute the eigenvalues of $A$ and $B$. These are of course $1, 1, 3$.
Compute eigenvectors of $A$. These are
$$
\begin{bmatrix}-2\\1\\0\end{bmatrix},
\begin{bmatrix}0\\0\\1\end{bmatrix},
\begin{bmatrix}0\\1\\1\end{bmatrix},
$$
where the first two are relative to $1$, and the third to $3$.
Simlarly for $B$ one has
$$
\begin{bmatrix}0\\1\\0\end{bmatrix},
\begin{bmatrix}-2\\0\\1\end{bmatrix},
\begin{bmatrix}1\\0\\0\end{bmatrix}.
$$
Then you build matrices with the eigenvectors as columns
$$
\alpha
=
\begin{bmatrix}-2&0&0\\1&0&1\\0&1&1\end{bmatrix},
$$
and
$$
\beta
=
\begin{bmatrix}0&-2&1\\1&0&1\\0&1&0\end{bmatrix}.
$$
You will have
$$
\alpha^{-1} A \alpha
=
\begin{bmatrix}1\\&1\\&&3
\end{bmatrix}
=
\beta^{-1} B \beta,
$$
where I have omitted zeroes.
Therefore
$$
B = (\alpha \beta^{-1})^{-1} A \alpha \beta^{-1}.
$$
|
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|
Algebra QF $f(x)=x^2-18x-4$ I have the problem to find the zero of the function:
$$f(x)=x^2-18x-4$$
I have it mostly worked out as $a=1, b= -18, c= -4$
worked out I have:
$$\frac{18\pm \sqrt{340}}{2}$$
I know the answers are
$$9+ \sqrt{85}\\
9- \sqrt{85}$$
The question is when working the problem out I end up with:
$$\frac{18\pm \sqrt{340}}{2}$$
I am not sure how that step (with √ 340) becomes $9\pm\sqrt{85}$
Any help would be awesome!
|
Note that:
$$\frac{18\pm\sqrt{340}}{2}=\frac{18\pm\sqrt{4\cdot85}}{2}=\frac{18\pm\sqrt{4}\cdot\sqrt{85}}{2}=\frac{18\pm2\sqrt{85}}{2}=\frac{2\cdot(9\pm\sqrt{85})}{2}=9\pm\sqrt{85}$$
|
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|
Recurrence Relations Closed Form So, the question is to derive the closed form solution to the recurrence relation $$T(n) = 3T(n-1) + 5,\hspace{5mm} T(0) = 0.$$
$\begin{align}T(n) &= 3T(n-1)+5
\\&= 3(3T(n-2)+5)+5
\\&= 3(3(3T(n-3)+5)+5)+5\end{align}$
I'm just struggling to go from here. I understand that it's $3$ to the power of something but I'm lost on what I have to do after this.
|
Calculating the first few values: $0,5,20,65,200$ , dividing by $5$ this is $0,1,4,13,40$ , multiplying by $2$ we get $0,2,8,26,80$ adding $1$ we get $1,3,9,27,81$. Otherwise you can use generating functions:
Let $A=\sum\limits_{n=0}^\infty t(n)x^n$.
Then $A=3xA+5\sum\limits_{n=1}^\infty x^n$ (since $f(0)=0$).
Then $(1-3x)A=\frac{5}{1-x}-5$ so $A=\frac{5}{(1-3x)(1-x)}-\frac{5}{1-3x}=\frac{5}{2(x-1)}-\frac{15}{2(3x-1)}+\frac{5}{3x-1}=\frac{5}{2(x-1)}-\frac{5}{2(3x-1)}=\frac{1}{2}(\sum\limits_{n=0}^\infty 5\cdot 3^{n} -5)$
therefore $f(n)=\frac{5(3^{n}-1)}{2}$
|
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|
What general function can I use to represent the next sequence: 2,2,2,2,2,3,3,2,2,2,2,2,3,3,2,2,2,2,2,3,3…? Look at this sequence:
2,2,2,2,2,3,3,2,2,2,2,2,3,3,2,2,2,2,2,3,3...
It is defined as follows:
$$f(n)=\begin{cases}
3 &\text{if $n \bmod 7=6,0$}\\
2 &\text{otherwise}\\
\end{cases}$$
David found a good representation for $f(n)$:
$$g(n)=\frac{\sin\dfrac{2\pi(n+1)}{7}}{\sin\dfrac{2\pi}{7}}
\frac{\sin\dfrac{2\pi(n+2)}{7}}{\sin\dfrac{4\pi}{7}}\cdots
\frac{\sin\dfrac{2\pi(n+6)}{7}}{\sin\dfrac{12\pi}{7}}\ ;$$
$$f(n)=2+g(n)+g(n+1)\ ;$$
What is a general representation of $f(n)$ that defined as fallow:
$$f'(n)=\begin{cases}
a &\text{if $n \bmod c=0$ or in the range of $\{c-d,c-1\}$}\\
b &\text{otherwise}
\end{cases}$$
*
*$a,b,c,d$ are known positive integeres
To be more specific I have
$$h(n)=y - \sum_{0}^{n}{f'(n)}$$
*
*$y$ is a know positive integer
And $h(n)$ is defined as fallow:
$$h(n) = \frac{e}{n}$$
*
*$e$ is a known positive integer
How do I find $n$?
|
If I understand what you are asking, we can define
$$
\newcommand{\flfrac}[2]{\left\lfloor\frac{#1}{#2}\right\rfloor}
f(n)=b+(a-b)\left(\flfrac{n+d}c-\flfrac{n-1}c\right)\tag{1}
$$
then
$$
f(n)=\left\{\begin{array}{}
a&\text{if }n\equiv c-d\dots c\pmod{c}\\
b&\text{if }n\equiv1\dots c-d-1\pmod{c}
\end{array}\right.\tag{2}
$$
Next we compute
$$
\begin{align}
\sum_{k=0}^n\flfrac km
&=\sum_{k=0}^{qm-1+r+1}\flfrac km\\
&=\overbrace{m\frac{(\color{#0000FF}{q}-1)\color{#0000FF}{q}}2}^{\text{$m$ copies of $0\dots q-1$}}+\overbrace{(\color{#C00000}{r}+1)\color{#0000FF}{q}\vphantom{\frac{()}2}}^{\text{$r+1$ copies of $q$}}\\
&=\left(\frac{m\left(\color{#0000FF}{\flfrac nm}-1\right)}2+\color{#C00000}{n-m\flfrac nm}+1\right)\color{#0000FF}{\flfrac nm}\\
&=\left(n+1-\frac{m\left(\flfrac nm+1\right)}2\right)\flfrac nm\tag{3}
\end{align}
$$
Now, apply $(3)$ to
$$
\begin{align}
\sum_{k=0}^n\left(\flfrac{k+d}c-\flfrac{k-1}c\right)
&=\sum_{k=0}^{n+d}\flfrac kc-\sum_{k=0}^{d-1}\flfrac kc-\sum_{k=0}^{n-1}\flfrac kc+\sum_{k=0}^{-2}\flfrac kc\\
&=\sum_{k=0}^{n+d}\flfrac kc-\sum_{k=0}^{d-1}\flfrac kc-\sum_{k=0}^{n-1}\flfrac kc+1\tag{4}
\end{align}
$$
|
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|
finding $\int\frac{1}{(t^2+25)^2} dt$ without trig substitution Our calculus book covers partial fractions but not trig substitution, so I would like to find out the most elementary way to evaluate
$$\displaystyle\int\frac{1}{(t^2+25)^2}\;dt$$
without using trig substitution (or partial fractions over the complex numbers).
|
$$
\begin{aligned}
\int \frac{1}{\left(t^{2}+25\right)^{2}} d t &=-\frac{1}{2} \int \frac{1}{t} d\left(\frac{1}{t^{2}+25}\right) \\
&=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{2}\left(\frac{1}{t^{2}\left(t^{2}+25\right)} d t\right.\\
&=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50} \int\left(\frac{1}{t^{2}}-\frac{1}{t^{2}+25}\right) d t \\
&=-\frac{1}{2 t\left(t^{2}+25\right)}-\frac{1}{50}\left[-\frac{1}{t}-\frac{1}{5} \tan ^{-1}\left(\frac{t}{5}\right)\right]+C \\
&=\frac{1}{250}\left[\frac{5}{t^{2}+25}+\tan ^{-1}\left(\frac{t}{5}\right)\right]+C
\end{aligned}
$$
|
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|
Given some of the roots of the function $f(x) = x^3+bx^2+cx+d$, how do I find the coefficients of that function? Two of the roots of $f(x) = x^3+bx^2+cx+d$ are $3$ and $2+i$. How do I find b+c+d? The answer choices are -7, -5, 6, 9, and 25.
|
x^3+bx^2+cx+d = 0 has 3 solutions 3 and 2+i are given. third solution z is not.
this polynomial can decomposed as (x-3)(x-2-i)(x-z)=0 from here follows
1. 3*(2+i)*z = -d
2. 3z + 3(2+i) + z(2+i) = c
3. 3 + 2+i + z = -b
these admits a solution:
b = -5 - i - z;
c = 6 + 3 i + 5 z + i z,
d = -3 z(2 + i)
in order b to be real z = y-i
in order d to be real y = 2;
so b = -7;
c = 17;
d = -15;
|
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|
Prove that a recurrence relation (containing two recurrences) equals a given closed-form formula. Prove that $a_n = 3a_{n-1} - 2a_{n-2} = 2^n + 1$ , for all $n \in \mathbb{N}$ , and $a_1 = 3$ , $a_2 = 5$ , and $n \geq 3$
Basis:
$a_1 = 2^1 + 1 = 2 + 1 = 3$ $\checkmark$
$a_2 = 2^2 + 1 = 4 + 1 = 5$ $\checkmark$
Inductive Hypothesis:
$a_k = 2^k + 1$ , where $n = k$
$a_k = 3a_{k-1} - 2a_{k-2}$
Inductive Step:
$$\begin{align}
\ a_{k+1} & = 3a_k - 2a_{k-1} \\
& = 3(2^k + 1) - 2(2^{k-1} + 1) \\
& = \color{red}{6^k} + 3 - \color{red}{4^{k-1}} - 2 \\
& = \color{red}{6^k} - \color{red}{4^{k-1}} + 1
\end{align}$$
Now, the $ + 1$ looks very promising, but $6^k - 4^{k-1}$ makes me sick. Anyone have some good hints? That second recurrence ( $2a_{n-2}$ ) seems to be stumping me.
|
$$3\cdot 2^k - 2\cdot 2^{k-1}= 3\cdot 2^k -2^k=2^k(3-1)=2^k\cdot2=2^{k+1}$$
Notice that $$ 3\cdot 2^k = 2^k + 2^k + 2^k$$
While
$$ 6^k=(3 \cdot 2)^k=3^k\cdot 2^k=2^k + 2^k ....... + 2^k$$
The last line shows $2^k$ being added $3^k$. There is a way to show this better in latex (a bracket under the ellipses) but I don't know how.
|
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|
Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.
|
Put $a=6k^2-2$, $b=6k$ and $c=6k^2+1$. Then
$$\left(6k^2-2\right)^2+(6k)^2=\left(6k^2+1\right)^2+3.$$
Behind these solutions is the observation that $$\left(\frac{x+y}{2}\right)^2-\left(\frac{x-y}{2}\right)^2=xy$$ We can re-write the given equation as $c^2-a^2=b^2-3$ and choose $b$ such that $b^2-3$ is the product of two odd numbers $xy$. This allows us to choose $a=\frac{x-y}{2}$ and $c=\frac{x+y}{2}$.
|
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|
$f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ For all $x,y\in\mathbb{R}$. also $f : \mathbb{R} → \mathbb{R}$ and $x+y\not=0$.
My attempt:
I restated it as
$a[x^2 y^2 (\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2})] +
b[xy(x+y-\frac{1}{y}-\frac{1}{x})] + c [x+y-2]=0$
because of $f(xy) - \frac{f(x)+f(y)}{x+y}=0$
(we know that $f$ is identically equal to $0$)
and later tried to prove that
$\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2}$
$x+y-\frac{1}{y}-\frac{1}{x}$
$x+y-2$
are all equal to 1. I eventually got to that
$x^2 -2x -1,5 =0$ And i don't like this
Later I tried ;
$\frac{a(x^2+y^2)+b(x+y)+2c}{x+y} = \frac{f(x+y)}{x+y}$
because if
$(x+y)^2=x^2+y^2$
$xy=0$ but that implies that $c$ is equal to $0$ too. Both of my trials are propably wrong.
|
Let $y=1$
$f(x)(x+1)=f(x)+f(1)$ Hence $$f(x)=\frac{f(1)}{x}$$
You can verify that : $$f(xy)=\frac{f(1)}{xy}=\frac{\frac{f(1)}{x}+\frac{f(1)}{y}}{x+y}$$
As Peter Taylor explained in the comment :
$$f(0)=f(1)+f(0)$$
So if you want $f$ to be defined on $\mathbb R$, you need to have $f(1)=0$. Hence $$\forall x\neq 0\quad f(x)=\frac{f(1)}{x}=0$$
Then $\forall x\neq 0\quad x.f(0)=f(0)$ implies $f(0)=0$
|
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|
Surface Area by Integration $$2\pi\int_{3}^6\left(\frac{1}{3}x^\frac{3}{2}-x^\frac{1}{2}\right)\left(1+\left(\frac{1}{2}x^\frac{1}{2}-\frac{1}{2}x^\frac{-1}{2}\right)^2\right)^\frac{1}{2}dx$$
I've managed to simplify this down to the equation below (not sure if it'll help), but I still can't integrate it.
Please help.
$$\frac{\pi}{3}\int_{3}^6(x^4-4x^3-2x^2+12x+9)^\frac{1}{2}dx$$
PS. The answer you should get is 9$\pi$.
|
HINT
I would say
$x^4-4x^3-2x^2+12x+9=(x^2-2x-3)^2$
|
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|
Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then the trinomial is factorized
$$(x-3)(x-1) \le 0 $$
Therefore
$$1\lt x \lt 3 $$
Can someone tell me if I make a mistake, or if the process is good enough. Will you have done something different?
|
Never multiply an inequality by a variable since when multiplying by a negative number you have to change the sign.This is how this kind of problems should be solved
$$\frac{x^2+3}{x}\leq 4\\\frac{x^2+3}{x}-4\leq 0\\\frac{x^2-4x+3}{x}\leq 0\\\frac{(x-3)(x-1)}{x}\leq 0$$
Now you should spit it into cases
*
*$x<0$ Then you have that $x,(x-1),(x-3)$ are all negative so everything is negative
*$0<x<1$ Then you have that $x$ is positive and $(x-1),(x-3)$ negative hence the expression is positive since 2 negatives give positive
*$1<x<3$ Then you have that $x,(x-1)$ are positive and $x-3$ is negative hence the expression is negative
*$x>3$ everything is positive hence expression is positive
Now you see that $x<0$ and $1<x<3$ are solutions,now to check the boundaries,$x=0,1,3$ you can see that for $x=0$ it's undefined and that $x=1,3$ fit so the solution is $x\in(-\infty,0)\cup [1,3]$
|
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|
Hockey pucks and parameters There is one hockey puck with a diameter of $3$ inches. The puck is spinning around its center at a speed of $3$ counterclockwise rotations per second. At the center, the puck is traveling at a speed of $24$ inches per second at an angle of $45^\circ$ to the positive $x$-axis.
(a) At time $t=0$, the center of the puck is at the origin. Find the location of the center of the puck at time $t$? (Note that time $t$ is measured in seconds.)
(b) A point on the outer edge of the puck begins at the point $\left(0,\frac32\right)$. Find its location at time $t$?
I am not sure how to find part (a), but this is what I have for part b).
The puck has diameter of $3$ inches, a radius of $\frac{3}{2}$.
I used the formula where that in order to find a point on the circle's circumference, given a circle with center $(a,b)$ and radius $r$,
$x(t) = r\cos (t) + a$
$y(t) = r \sin (t) + b$
the circle's center is at $(0,0)$ because of part a) statement.
Therefore $x(t) = r \cos (t)$ and $y(t) = r \sin (t)$. I plugged in $x(t) = 0$ and $y(t) = \frac{3}{2}$ because of what was given in part b).
From there I found that $t = \frac{\pi}{2}$ and that $r = \frac{3}{2}$,
so now the formula is $x(t) = \frac{3}{2} \cos (t)$ and $y(t) = \frac{3}{2} \sin (t)$. But now I am stuck. What do I do next for part b), and how do I solve part a)?
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(a) The hockey puck is traveling along the line $y=x$, traveling 24 inches every second. Since the puck moves 24 inches when $t=1$ and the puck travels at a constant pace, we can graph:
When $t=1$, $x = y = 12\sqrt{2}$. Therefore, the center of the puck at time $t$ is at:
\begin{align*}
x &= 12\sqrt{2}t \\
y &= 12\sqrt{2}t
\end{align*}
(b) Let's make a parametric equation for the puck using variables:
\begin{align*}
x &= a + r \cos \theta \\
y &= b + r \sin \theta
\end{align*}
We know that the center of the puck at time $t$ is at $12 \sqrt{2} t$. We also know that the radius is $\frac{3}{2}$. Filling in the variables, we get:
\begin{align*}
x &= 12 \sqrt{2}t + \frac{3}{2} \cos \theta \\
y &= 12 \sqrt{2}t + \frac{3}{2} \sin \theta
\end{align*}
But what is $\theta$? Since the puck spins $6\pi$ radians every second and starts at $\frac{\pi}{2}$, $\theta = 6\pi t + \frac{\pi}{2}$
\begin{align*}
x &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t + \frac{\pi}{2}) \\
y &= 12 \sqrt{2}t + \frac{3}{2} \sin (6\pi t + \frac{\pi}{2})
\end{align*}
Since $\sin(\theta+90) = \cos(\theta)$ and $\cos(\theta + 90) = -\sin \theta$
\begin{align*}
x &= 12 \sqrt{2}t - \frac{3}{2} \sin (6\pi t) \\
y &= 12 \sqrt{2}t + \frac{3}{2} \cos (6\pi t)
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
exercise: ordinary differential equations I am struggling with an exercise. Can you please give me a hint?
Exercise:
Show that the solution curves of the differential equation:
$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.
I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.
Hence I get:
$u+x\frac{du}{dx}=-\frac{u(2-u^3)}{2u^3-1}$
This gives:
$2u^4-u+(2u^3-1)x\frac{du}{dx}=-2u+u^4$
$(2u^3-1)x\frac{du}{dx}=-(u^4+u)$
$\frac{(2u^3-1)}{(u^4+u)}\frac{du}{dx}=-\frac{1}{x}$
So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?
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$\bf hint:$ so you need to integrate $\int \frac{2u^3 - 1}{u^4 + u}$ we can do this by using partial fraction. here is how it goes
$$ \frac{2u^3 - 1}{u^4 + u} =
\frac{1}{2}\left(\frac{4u^3 +1}{u^4 + u} - \frac{3}{u(u+1)(u^2-u+1)}\right)$$
now you need to find the constants $A, B, C$ and $D$ so that $$ \frac{3}{u(u+1)(u^2-u+1)} = \frac{A}{u} + \frac{B}{u+1} + \frac{C(2u-1)}{u^2-u+1} + \frac{D}{u^2 - u + 1} $$ is an identity.
finding $A, B$ are easier. just by looking at the behaviour near $u = 0, u = -1.$ they turn out to be $A = 3, B = -1.$
|
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|
Partial Fraction Decomposition Clarification I'm just looking for some overall clarification for the following cases. Now, to the extent of my knowledge, the following examples of partial fractions would be split up in the following way:
\begin{align}
\frac{1}{x^2+3x-4}&=\frac{1}{\left(x+4\right)\left(x-1\right)}=\frac{A}{x+4}+\frac{B}{x-1},\tag{1}\\
\frac{1}{x^3+x^2}&=\frac{1}{x^2\left(x+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1},\tag{2}\\
\frac{1}{x^2\left(x+1\right)^3}&=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{\left(x+1\right)^2}+\frac{E}{\left(x+1\right)^3},\tag{3}\\
\frac{1}{x^2\left(x^2+x+1\right)}&=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+x+1},\tag{4}\\
\frac{1}{x^2\left(x^3+x^2+x+1\right)^2}&=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx^2+Dx+E}{x^3+x^2+x+1}+\frac{Fx^2+Gx+H}{\left(x^3+x^2+x+1\right)^2},\tag{5}
\end{align}
have I made a mistake anywhere?
EDIT: I should have factored $\left(5\right)$ further to get
\begin{align}
\frac{1}{x^2\left(\left(x+1\right)\left(x^2+1\right)\right)^2}.
\end{align}
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$$
x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1)
$$
So
$$
\frac{\cdots\cdots\cdots}{x^3+x^2+x+1} = \frac{Cx+D}{x^2+1} + \frac{E}{x+1}.
$$
|
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|
Why is there different results when calculating a double sum? Consider:
$$
\begin{matrix}
-1 & 0 & 0 & 0& 0& \ldots\\
1/2 & -1 & 0 & 0& 0& \ldots\\
1/4 & 1/2 & -1 & 0& 0& \ldots\\
1/8 & 1/4 & 1/2 & -1& 0& \ldots\\
1/16&1/8 & 1/4 & 1/2 & -1& \ldots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{matrix}
$$
When I calculate the sum of each column first, I got $0,0,0\ldots$ then sum them up the answer should be $0$. But when I calculate the sum of each row, I got $-1, -1/2, -1/4, -1/8,\ldots$, then the sum will be $-2$. Why is there different results when I calculate the sum?
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The sum is not absolutely convergent (http://en.wikipedia.org/wiki/Convergent_series) so we can make the sum equal whatever we want.
With divergent series the way that we order the series may change the sum. As @DuAravis said the series can be written as
\begin{align*}
S_n &=(-1) +(\frac{1}{2} -1) +(\frac{1}{4}+\frac{1}{2} -1)+(\frac{1}{8}+\frac{1}{4}+\frac{1}{2} -1) +\ldots=-2 \quad (1)
\end{align*}
or
\begin{align*}
S_n &=(-1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\ldots) +\ldots (0-1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\ldots) +\ldots=0 \quad (2)
\end{align*}
but we can also rearrange the matrix as follows,
$$
\begin{matrix}
-1 & 0 & 0 & 0& 0& \ldots\\
1/2 & -1 & 0 & 0& 0& \ldots\\
1/2 & 1/4 & -1 & 0& 0& \ldots\\
1/2 & 1/4 & 1/8 & -1& 0& \ldots\\
1/2 &1/4 & 1/8 & 1/16 & -1& \ldots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{matrix}
$$
now if we sum the columns we get $\infty$.
Also, we need to be careful about multiplying. For example, if multiply each term by $x$. Then by (2) this shouldn't change the total sum,
\begin{align*}
S_n &=x \times (-1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\ldots) +\ldots x \times (0-1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} +\ldots) +\ldots=x \times 0=0
\end{align*}
but now if we use (1),
\begin{align*}
S_n &=x \times (-1) +x \times (\frac{1}{2} -1) +x \times (\frac{1}{4}+\frac{1}{2} -1)+(\frac{1}{8}+\frac{1}{4}+\frac{1}{2} -1) +\ldots=-2x \quad (1)
\end{align*}
The problem I think is that you're working with $\infty$'s so when you manipulate the sum it's like adding, subtracting, or multiplying by $\infty$. Here are two famous examples http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF and http://en.wikipedia.org/wiki/Grandi%27s_series.
Here are some links on mathexchange Sum of divergent series and Sum of infinite divergent series
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|
Proving $1+\frac{4}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{4}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{4}{10^2}+\frac{1}{11^2}+\cdots=\frac{\pi ^2}{4}$ Proving $$1+\frac{4}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{4}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{4}{10^2}+\frac{1}{11^2}+\cdots=\frac{\pi ^2}{4}$$
Firstly, I thought to prove it by comparison the terms with the terms of $1/n^2$ , but the problem with the missing terms, so I couldn't reach to the proving . Can anybody help? Best regards.
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This sum is explicitly:
$$\sum_{n=1}^\infty\frac{1}{n^2}+3\frac{1}{(2n)^2}-4\frac{1}{(4n)^2}$$
You can check for yourself that this eliminates $n$'s which are a multiple of four, while multiplying by four terms with even $n$ but which are not a multiple of four. In other terms:
$$\sum_{n=1}^\infty\frac{1}{n^2}\left(1+\frac{3}{4} - \frac{1}{4}\right)=\frac{\pi^2}{6}\cdot\frac{3}{2}=\frac{\pi^2}{4}$$
|
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|
How to solve $y′′′+y'=2-\sin(x)$ I have tried to solve this but with no luck.
So far I just get,
$$y_p(x) = A \sin x + B \cos x \\
y'_p(x) = A \cos x - B \sin x \\
y''_p(x) =-A \sin x - B \cos x \\
y'''_p(x) =-A \cos x + B \sin x $$
$$y'''+ y' = -A \cos x + B \sin x + A \cos x - B \sin x \\ \\
= \cos x (-A+A) + \sin x (B-B) = 2 - \sin x $$
I would really appreciate help in solving this
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here is a way to why $x$'s are added. we will find a particular solution to $$y'' + y = \sin x \tag 1$$ the equation is forced by a solution of the homogeneous problem(called the resonant case). first you look at a more general problem of the solution of
$$y'' + y = \sin kx$$ a particular solution is $$y_p = \frac{\sin kx}{1-k^2}$$ so is $$y_p(x) = \frac{\sin kx - \sin x}{1-k^2}= -\frac{1}{1+k}\frac{\sin kx - \sin x}{k - 1} $$ in the limit as $k \to 1, $ we have
$$y_p = -\frac{1}{2}\left( \frac{d}{dk}\sin kx \right)|_{k = 1}
=-\frac{1}{2}x\cos x $$
you can see that a particular solution of $(1)$ is $$y_p = -\frac{1}{2}x\cos x. $$
|
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|
How to compute $\int_0^{\infty} \ln (1 + e^{-x})\, dx$ and $\int_0^{\infty} \ln (1 - e^{-x})\, dx$? Bierenes de Haan's book (page 377) shows that $\int_0^{\infty} \ln (1 + e^{-x})\, dx = \frac{\pi^2}{12}$, and $\int_0^{\infty} \ln (1 - e^{-x})\, dx = -\frac{\pi^2}{6}$. Anybody know how to compute them? Thanks.
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Using the Maclaurin series for $\ln(1 + x)$, we have
\begin{align}\int_0^\infty\ln(1 + e^{-x})\, dx &= \int_0^\infty \sum_{n = 1}^\infty (-1)^{n-1}\frac{e^{-nx}}{n}\, dx \\
&= \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}\int_0^\infty e^{-nx}\, dx\\
& = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}\cdot\frac{1}{n}\\
& = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n^2}\\
& = \sum_{n = 1}^\infty \frac{1}{(2n-1)^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2}\\
&= \sum_{n = 1}^\infty \frac{1}{n^2} - 2\sum_{n = 1}^\infty \frac{1}{(2n)^2}\\
&= \frac{1}{2}\sum_{n = 1}^\infty \frac{1}{n^2}\\
&= \frac{1}{2}\cdot\frac{\pi^2}{6}\\
&= \frac{\pi^2}{12}.
\end{align}
A similar method works for the second integral.
\begin{align}
\int_0^\infty \ln(1 - e^{-x})\, dx &= \int_0^\infty -\sum_{n = 1}^\infty \frac{e^{-nx}}{n}\, dx\\
&= - \sum_{n = 1}^\infty \frac{1}{n}\int_0^\infty e^{-nx}\, dx\\
&= -\sum_{n = 1}^\infty \frac{1}{n}\cdot\frac{1}{n}\\
&= -\sum_{n = 1}^\infty \frac{1}{n^2}\\
&= -\frac{\pi^2}{6}.
\end{align}
The interchange of series and integral are justified for the first integral since $$\sum_{n = 1}^\infty \int_0^\infty \left|\frac{(-1)^{n-1}e^{-nx}}{n}\right|\, dx = \sum_{n = 1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\ dx = \sum_{n = 1}^\infty \frac{1}{n^2} < \infty,$$ and for the second integral since similarly $$\sum_{n = 1}^\infty \int_0^\infty \left|\frac{e^{-nx}}{n}\right|\, dx = \sum_{n = 1}^\infty \frac{1}{n^2} < \infty.$$
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Derive a Recurrence Could really use some help with this.
For an integer $m \geq 1$ and $n \geq 1$, consider $m$ horizontal lines and $n$ non-horizontal lines, such that no two of the non-horizontal lines are parallel and no three of the $m+n$ lines intersect in one single point. These lines divide the plane into regions (some of which are bounded and some of which are unbounded). Denote the number of these regions by $R_{m,n}$. For example, $R_{4,3} = 23$.
Derive a recurrence for the numbers $R_{m,n}$ and use it to prove that
$$R_{m,n} = 1 + m(n+1) + \binom{n+1}{2}$$
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Think about what happens each time you add a line of each type. If you add a horizontal line, it will intersect each of the $n$ non-horizontal lines, but none of the $m$ horizontal lines. On the other hand, if you add a non-horizontal line, it will intersect all of the existing $m + n$ lines. After drawing a couple pictures of small cases, you will see that $l + 1$ regions are added when a new line intersects $l$ lines at $l$ distinct points. Since no three lines meet at the same point by assumption in the problem statement, we know each new intersection intersects at a new, distinct point.
Thus we have our recurrences:
$\begin{align}
R_{m,n} &= R_{m-1,n} + (n + 1)\\
&= R_{m,n-1} + (m + (n-1)) + 1
\end{align}
$
We will now prove that $$R_{m,n} = 1 + m(n+1) + \binom{n+1}{2}$$ via double induction.
Let $\mathbf{P}(m,n)$ be the statement that the assertion holds for $m, n$
Base Case $\mathbf{P}(1,1):$
It is not hard to see that we have $R_{1,1} = 1 + 1(1 + 1) + {1 + 1 \choose 2} = 1 + 2 +1=4$ regions.
$\mathbf{P}(m-1,1) \Rightarrow \mathbf{P}(m,1):$
$$\begin{align}
R_{m,n} &= R_{m-1,n} + (n + 1)\hspace{3mm}\text{ gives us,}\\
R_{m,n} &= 1 + (m-1)(n+1) + \binom{n+1}{2} + (n+1)\hspace{3mm}\text{so we show,}
\end{align}$$
$$\begin{align}
R_{m-1,1} + (1+1) &= 1 + (m-1)2 + \binom{2}{2} + 2=1 + (m-1 + 1)2 + \binom{2}{2}\\&=1 + m(1+1) + \binom{1+1}{2} \hspace{3mm}\text{as desired.}
\end{align}
$$
$\mathbf{P}(m,n-1) \Rightarrow \mathbf{P}(m,n):$
$$\begin{align}
R_{m,n} &= R_{m,n-1} + (m + (n-1)) + 1 \hspace{3mm}\text{gives us,}\\
&= 1 + m(n) + \binom{n}{2} + (m + (n-1) + 1)\\
&= 1 + m(n+1) + {n(n-1) \over 2} + n\\
&= 1 + m(n+1) + {n(n-1+2) \over 2}\\
&= 1 + m(n+1) + \binom{n+1}{2}
\end{align}$$
Thus by induction the formula holds for all $m,n\in\mathbb{N}$
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|
What's wrong with how I calculated the inverse of this matrix? $\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
2 & 0 & 0 & | & 0 & 1 & 0\\
2 & 2 & 0 & | & 0 & 0 & 1\end{array} \right) $ is the original matrix. Now, add 2*R1 to R2 and R3.
$\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
0 & -2 & 4 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -R3 to R2.
$\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
0 & -2 & 0 & | & 0 & 1 & -1\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -.5*R3 to R1.
$\left( \begin{array}{ccc}
-1 & -1 & 0 & | & 0 & 0 & -.5\\
0 & -2 & 0 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -.5*R2 to R1.
$\left( \begin{array}{ccc}
-1 & 0 & 0 & | & -1 & -.5 & -.5\\
0 & -2 & 0 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, simple division yields
$\left( \begin{array}{ccc}
1 & 0 & 0 & | & 1 & .5 & .5\\
0 & 1 & 0 & | & -1 & -.5 & 0\\
0 & 0 & 1 & | & .5 & 0 & .25\end{array} \right) $
|
It seems you made a small computational error in rewriting the right hand inverse matrix after step 3. You gotta be really careful about that,it's like a complicated integration-it can really shaft you on an exam or important homework problem. If a row reduction to yield an inverse matrix looks wrong to you but you're not sure if you made a mistake, you can always multiply the matrix by the obtained inverse. If you made a mistake, you won't get the identity matrix and row where the mistake is is where you won't have a 1 or a 0 usually. Even better,if you've got access to a computer algebra system like Mathematica, that'll make checking really quick and simple.
|
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|
finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min?
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$. (long division)
$x_{min}=\frac{-b}{2a}=\frac{1}{2}$. (when $ax^2+bx+c=0$)
So $f(0.5)=0.5^2-0.5+1=\frac{3}{4}$
Thanks.
|
$$x^2-x+1=\frac{4x^2-4x+4}4=\frac{(2x-1)^2+3}4\ge\frac34$$
The equality occurs if $2x-1=0\iff x=\dfrac12$
|
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|
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
|
As noted, induction is a more difficult way to prove this. Here it is.
Claim:
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}
$$
for $n=2,3,4,\cdots$. First, when $n=2$ we have
$$
\frac{1}{1}+\frac{1}{4} = \frac{5}{4} < \frac{3}{2} = 2-\frac{1}{2}
$$
which is correct.
Now, suppose $n \ge 2$ and
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}
$$
Then
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}+\frac{1}{(n+1)^2} < 2-\frac{1}{n}+\frac{1}{(n+1)^2}
$$
so it suffices to prove
$$
2-\frac{1}{n} + \frac{1}{(n+1)^2} < 2-\frac{1}{n+1}
$$
This is equivalent to
$$
\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}
$$
which holds iff
$$
n+n(n+1) < (n+1)^2
$$
or
$$
2n+n^2 < 1+2n+n^2
$$
which is true. This completes the induction.
|
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Integrate $ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$ Evaluate the integral:
$$\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$$
I have no clue on how to attack it.
The only thing I noticed is that there exists a symmetry around $\pi/8$, meaning that from $\pi/8$ to $\pi/4$ is the negative of zero to $\pi/4$. But, there exists a root of the integrand at $\pi/6$ and the limit of the integrand at $\pi/4$ is $-\infty$.
Conjecture: The integral is $0$ for the reason of symmetry I mentioned above.
However I cannot prove that. I would appreciate your help.
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By replacing $\phi$ with $\arctan(t)$, then using integration by parts, we have:
$$ I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$
Now comes the magic. Since:
$$\int \frac{3\sqrt{2}\,t}{(3-t^2)\sqrt{1-t^2}}\,dt = -3\arctan\sqrt{\frac{1-t^2}{2}}\tag{1}$$
integrating by parts once again we get:
$$ I = \frac{\pi^2}{8}-3\int_{0}^{1}\frac{1}{1+t^2}\arctan\sqrt{\frac{1-t^2}{2}}\,dt \tag{2}$$
hence we just need to prove that:
$$ \int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan\sqrt{1-2t^2}}{1+t^2}\,dt=\color{red}{\frac{\pi^2}{24}}\tag{3}$$
and this is not difficult since both
$$\int_{0}^{1}\frac{dt}{1+t^2}(1-t^2)^{\frac{2m+1}{2}},\qquad \int_{0}^{\frac{1}{\sqrt{2}}}\frac{(1-2t^2)^{\frac{2m+1}{2}}}{1+t^2}\,dt $$
can be computed through the residue theorem or other techniques. For instance:
$$\int_{0}^{1}\frac{(1-t)^{\frac{2m+1}{2}}}{t^{\frac{1}{2}}(1+t)}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{1}(1-t)^{\frac{2m+1}{2}} t^{n-\frac{1}{2}}\,dt=\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(m+\frac{3}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(m+n+2)}$$
or just:
$$\int_{0}^{1}\frac{\sqrt{\frac{1-t^2}{2}}}{(1+t^2)\left(1+\frac{1-t^2}{2}u^2\right)}\,dt = \frac{\pi}{2(1+u^2)}\left(1-\frac{1}{\sqrt{2+u^2}}\right)\tag{4}$$
from which:
$$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\frac{\pi}{2}\int_{0}^{1}\frac{du}{1+u^2}\left(1-\frac{1}{\sqrt{2+u^2}}\right) =\color{red}{\frac{\pi^2}{24}} $$
as wanted, since:
$$ \int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/1151817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
how to evaulate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ How do I evaluate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ ?
according to Taylor's series, I did like this:
$$\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)}=\lim \limits_{x \to 0} \frac{x^5 - \frac{x^{10}}{2}+ O_3(x)}{(x^3+ \frac{x^6}{2!} +O_4(x))(x^2- \frac{x^6}{3!} +\frac{x^{10}}{5!} +O_4(x))} $$
but how do I continue from here?
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We have
\begin{align}
&\lim_{x\to 0} \frac{x^5 - \frac{x^{10}}{2} + O(x^{15})}{(x^3 + \frac{x^6}{2!} + O(x^9))(x^2 - \frac{x^6}{3!} + O(x^{10}))}\\
&= \lim_{x\to 0}\frac{1 + O(x^5)}{(1 + O(x^3))(1 + O(x))}\\\\
&= \frac{1}{(1)(1)}\\
&= 1.
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1153010",
"timestamp": "2023-03-29T00:00:00",
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|
Implicit derivative switching $y$ and $x$ Regard $y$ as the independent variable and $x$ as the dependent variable and use implicit differentiation to find $dx/dy$ if
$y \sec x = 4x \tan y$
I got $(\sec x-4x\sec^2 y)/4\tan y-y\sec \tan x$ but it was wrong.
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You almost got it.
$\begin{align}
y\sec x & = 4x \tan y
\\[1ex]
y' \sec x + y \tan x \sec x & = 4 \tan y + 4 x y' \sec ^2 y & \text{implicit derivation using the chain rule}
\\[1ex]
y' (\sec x - 4 x \sec ^2 y) & = 4 \tan y - y \tan x \sec x & \text{associate like elements}
\\[1ex]
\frac{\mathrm d y}{\mathrm d x} & = \frac{4 \tan y - y \tan x \sec x}{\sec x - 4 x \sec ^2 y} & \text{cross division}
\\[1ex]
\therefore \frac{\mathrm d x}{\mathrm d y} & = \frac{\sec x - 4 x \sec ^2 y}{4 \tan y - y \tan x \sec x} & \text{inverting}
\end{align}$
Note though: You wrote: $(\sec x−4 x \sec^2 y)/4\tan y−y\sec\tan x$
Not only is that $\dfrac{\sec x - 4 x \sec ^2 y}{4 \tan y} - y \sec \tan x$
But also You have a typo: $\;\sec \tan x \;$ is not the same as $\;\sec x \tan x\;$
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{
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"url": "https://math.stackexchange.com/questions/1153611",
"timestamp": "2023-03-29T00:00:00",
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|
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
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$a, b, c$ are sides of a triangle iff there exists positive reals $x, y, z$ s.t. $a=x+y, b=y+z, c = z+x$. In terms of these variables, the inequality is
$$\sum_{cyc} \frac{a}{b+c-a} = \sum_{cyc} \frac{x+y}{2z} \ge 3$$
Now the last is easy to show with AM-GM of all $6$ terms.
$$\sum_{cyc} \frac{x+y}{2z} = \frac12\left(\frac{x}z+\frac{y}z+\frac{y}x+\frac{z}x+\frac{z}y+\frac{x}y \right) \ge \frac12\left(6\sqrt[6]{\frac{x}z\cdot\frac{y}z\cdot\frac{y}x\cdot\frac{z}x\cdot\frac{z}y\cdot\frac{x}y} \right) = 3$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
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|
Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid?
Proof
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
We have $n \ge 1 \implies 6n -1 > n + 3 \implies \frac{n+3}{n(n+3)} < \frac{6n-1}{n^2+3n}.$
Let $\epsilon > 0$ be given.
Note that $ \frac{6n-1}{n^2+3n}< \epsilon \iff \frac{n+3}{n(n+3)} < \epsilon \iff n > \frac{1}{\epsilon}.$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (**)$
By the Archimedean Property of $\mathbb R$, $\exists N \in \mathbb N^+$ such that $N > \frac{1}{\epsilon}.$
If $n \ge N$, then $n > \frac{1}{\epsilon}$, and from $(*)$ and $(**)$ it follows that $\left|\frac{2n^2+1}{n^2+3n} - 2\right| < \epsilon$.
Therefore $\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$
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From $\lim_{n\to\infty} \dfrac{2n^2+1}{n^2+3n}$, divide top and bottom by $n^2$ to get:
$\lim_{n\to\infty} \dfrac{2+1/n^2}{1+3/n}$, which, as $n\to\infty$ becomes
$\dfrac{2+0}{1+0}=\dfrac{2}{1}=2$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1158558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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.