Hot answers tagged

27

If $\mu=\lim_{n\to\infty} \frac{a_{n} }{b_{n} } $exists then $$ \mu=\lim_{n\to\infty} \frac{a_{n+1} }{b_{n+1} } = \lim_{n\to\infty} \frac{b_n+2a_n + 14}{9b_n+ 2a_n+70 }\\= \lim_{n\to\infty} \frac{\frac{b_{n} }{b_{n} }+2\frac{a_{n} }{b_{n} } + \frac{14 }{b_{n} }}{9\frac{b_{n} }{b_{n} }+ 2\frac{a_{n} }{b_{n} }+\frac{70 }{b_{n} } }\\ =\frac{1+2\lim_{n\to\infty}...


25

One idea to get the limit in closed form : Let $A=\begin{pmatrix} 2 & 1 \\ 2 & 9 \end{pmatrix}$, $X_n = \begin{pmatrix} a_n \\ b_n \end{pmatrix}$, and $b=\begin{pmatrix} 14 \\ 10 \end{pmatrix}$. You can write $X_{n+1} = AX_n + b$. The idea is to solve the equation $X=AX+b$ ($X$ being a two-dimensional vector) - this equation has a unique solution ...


12

It is sometimes better to take $x+a$ (for some constant $a$) as the antiderivative of $1$. In this case $$ \int \arctan\sqrt{x+2}\,dx=(x+a)\arctan\sqrt{x+2}-\frac{1}{2}\int(x+a)\frac{1}{1+x+2}\cdot\frac{1}{\sqrt{x+2}}\,dx, $$ so $a=3$ seems to be a good choice. I think you can continue from here.


11

Because $2\arctan\left(\frac{e^x-1}{e^x+1}\right)=2\left(\arctan(e^x)-\frac\pi4\right)=2\arctan(e^x)+C'$. The results differ by a constant.


10

Noting that $$ \phi=2\cos(\frac{\pi}{5})=2\sin(\frac{3\pi}{10}), \Phi=2\sin(\frac{\pi}{10})$$ from https://en.wikipedia.org/wiki/Golden_ratio, so one has \begin{eqnarray} {\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}&=&{2\sin(\frac{\pi}{10})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{10})\tan(\frac{3\pi}{20})\over \sin^2(\...


9

Do the substitution $x=2u$. $$\int \frac{dx}{1+\cos(x)} = 2\int \frac{du}{1+\cos(2u)} = \int \frac{du}{\cos^2(u)} = \tan(u)+C = \tan(x/2)+C.$$


8

The requested limit is: $$ \frac{4 \sqrt{57} - 20}{4 \sqrt{57} + 44} \approx 0.1374586 $$ This is $$ \frac{ \sqrt{57} - 5}{ \sqrt{57} + 11} $$ and rationalizing the denominator gives $$ \frac{ \sqrt{57} - 7}{ 4} $$ $$ a_{n+2} = 11 a_{n+1} - 16 a_n - 42 $$ $$ b_{n+2} = 11 b_{n+1} - 16 b_n - 42 $$ The separate linear recurrences are the result of the ...


6

Here is a rigorous and systematic way to analyze the existence and value of the limit. (miracle173's answer didn't prove existence but uses essentially the same method for finding the value if it exists.) Let $c_n = \frac{a_n}{b_n}$ (for each $n \in \mathbb{N}$). Then $b_{n+1} c_{n+1} = b_n + 2 b_n c_n + 14$. And $b_{n+1} = 9 b_n + 2 b_n c_n + 70$. Thus $...


6

Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get $$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} $$ ...


5

$$\int_{-1}^{1}\tan^{-1}(\sqrt{x+2})dx=\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx$$ if$f$ be differentiable, increasing and one-to-one on $[a,b]$ such that $f(a)=\alpha$ and $f(b)=\beta$ then $$\int_{a}^{b}f(x)dx+\int_{\alpha}^{\beta}f^{-1}(x)dx=b\beta-a\alpha$$ therfore $$\int_{1}^{3}\tan^{-1}\sqrt{x}\,dx+\int_{\large\frac{\pi}{4}}^{\large\frac{\pi}{3}}\tan^{2}x\,...


5

First you need $x \ne 0$ and $y \ne 0$ in order for your fraction $\dfrac{x^2+y^2}{2xy}$ to make sense. Then as $\sin \in [-1,1]$ you need $x^2+y^2 \leq 2xy$ which means: $x^2 -2xy+y^2 \leq 0$ ie $(x-y)^2 \leq 0$ Therefore you have solutions on $\theta$ only if $x = y \ne 0$, and those solutions are $\theta = \dfrac{\pi}{2} \pmod \pi$


4

The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, ...


4

Define $$ L = \lim_{n \to \infty} \frac{a_n}{b_n} $$ $$ r = \lim_{n \to \infty} \frac{a_{n+1}}{b_n} $$ $$ s = \lim_{n \to \infty} \frac{b_{n+1}}{b_n} $$ then it's not hard to see that $L = r/s$. Also, by substituting in the recursion, since we have $b_n \to \infty$ we can compute $$ r = \lim_{n \to \infty} \left(1 + 2 \frac{a_n}{b_n} + \frac{14}{b_n}\...


4

There is no mathematical reason for this. The reason for picking it this way (irritating people in other fields, as you can see from the comments!) is that rotating the [positive] $x$-axis onto the [positive] $y$-axis is about the simplest rotation you can think of, so we decide to call it "positive". And clearly rotating the positive $x$-axis onto the ...


4

We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of $$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$ $$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$ $$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$ simplify giving the coefficients of $x^...


4

$\sin x=0$ is rejected because when $\sin x=0$, the value $\csc x$ does not exist, so $(\sin^2 x)(1+\csc x)$ does not exist, so it can't be equal to $0$ or to anything else either.


4

Differentiate it: $$\left(2\arctan x+\arcsin\frac{2x}{1+x^2}\right)'=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac1{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}=$$ $$=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac{1+x^2}{\sqrt{(1-x^2)^2}}=\frac2{1+x^2}+\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$$ We can thus see that if $\;|1-x^2|=-(1-x^2)\iff 1-x^2<0\iff |x|>1\;$ , then the ...


4

The author of the question may have been thinking purely geometrically, picturing the situation below, with $x$ the (non-negative) length of a segment: For completeness, here's the argument one can make ... Writing $\alpha := \angle BAC = \angle BAD$ and $\beta := \angle CBD$, we have $$\tan \alpha = x \qquad\text{and}\qquad \sin\beta = \frac{2x}{1+x^2} =...


4

Hint. One may write $$ \cos \phi =\cos (\phi/2+\phi/2)=\cos^2( \phi/2)-\sin^2 (\phi/2)=\cos^2( \phi/2)(1-\tan^2(\phi/2))=\frac{1-\tan^2(\phi/2)}{1+\tan^2(\phi/2)} $$ where we have used the standard identity $$ \cos (a+b)=\cos a \cos b-\sin a \sin b. $$


3

Suppose that we define the function $\def\RR{\mathbb R}f:\mathbb R\to\mathbb R$ putting $$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$ for each $x\in\mathbb R$, after checking that the power series converges for all real $x$. It is easy to compute derivatives of functions defined by power series, and therefore it is trivial to show that $$f''(x)=...


3

$$(a-c)(a+c)^2+bc(a+c)=ab^2$$ $$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$ $$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$ $$\implies (a-b+c)(a(a+b)-c^2)=0$$ Dividing the both sides by $a-b+c\gt 0$ gives $$c^2=a^2+ab$$ from which we have $c\gt a$. Case 1 : $...


3

Hint: Simply write $$ \sin x+\sin 5x=\sin(3x-2x) + \sin(3x+2x)=2\sin 3x\cos 2x$$ and similarly fot $\cos x+\cos 5x$.


3

If you want an expression that holds for all $n \geq 2$ then it might be a tad ugly in Cartesian, because: $$\cos(n\theta) = \sum_{\text{even }k} (-1)^{k/2}{n \choose k}\cos^{n-k} \theta \sin^k \theta \\ = (x^2+y^2)^{-n/2}\sum_{\text{even }k} (-1)^{k/2}{n \choose k} x^{n-k}y^k.$$ But this will get you there.


3

In A Note on Trigonometric Algebraic Numbers by D. H. Lehmer (and also here with a different notation) we find that $$ z^{-d}\Phi_n(z) = \psi_n(z+z^{-1}) $$ where $\Phi_n$ is the $n$-th cyclotomic polynomial and $d=\frac{\phi(n)}{2}$ is the degree of $\psi_n$, which is half the degree of $\Phi_n$. Lehmer proves that $\psi_n$ is irreducible. The roots of $\...


3

HINT: Find $$\cos(45^\circ-17^\circ)$$ and $$\sin(45^\circ-17^\circ)$$ and use $\cos45^\circ=\sin45^\circ=\text{?}$


3

$$x=e^{\pm i\alpha},y=?$$ If we take the same sign, $$xy=e^{\pm i(\alpha+\beta)},\dfrac1{xy}=e^{\mp i(\alpha+\beta)}$$ Now $e^{iA}+e^{-iA}=2\cos A$ But if we take the opposite sign, $$xy=e^{\pm i(\alpha-\beta)},\dfrac1{xy}=e^{\mp i(\alpha-\beta)}$$


2

$$\frac { 2\sin { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) -\sin { \left( 3x \right) } } } }{ 2\cos { \left( \frac { x+5x }{ 2 } \right) \cos { \left( \frac { x-5x }{ 2 } \right) } -\cos { \left( 3x \right) } } } =\frac { \sin { \left( 3x \right) \left( 2\cos { 2x-1 } \right) } }{ \cos { \left( 3x \right) \...


2

As mentioned, the limiting case is when the small circle touches the larger circle internally. Then, we have the figure below:- Applying Pythagoras theorem to the red triangle, we have $OM^2 = 1 – (o.5L)^2$. Applying Pythagoras theorem to the green triangle, we have $(R + d)^2 = (1 – R)^2 – OM^2$ Eliminating $OM^2$ from the two equations, we get $d$ in ...


2

Hint: $$\cot{\omega}=\cot{A}+\cot{B}+\cot{C} \Longleftrightarrow \sin{(A-\omega)}\sin{(B-\omega)}\sin{(C-\omega)}=\sin^3{\omega}\tag{1}$$ Now using sine rule, we get $$\dfrac{\sin{(A-\omega)}}{\sin{\omega}}=\frac{CP}{AP}\tag{2}$$ $$\dfrac{\sin{(B-\omega)}}{\sin{\omega}}=\frac{AP}{BP}\tag{3}$$ $$\dfrac{\sin{(C-\omega)}}{\sin{\omega}}=\frac{BP}{CP}\tag{4}$...



Only top voted, non community-wiki answers of a minimum length are eligible