# Tag Info

1

How about $a_n = \frac{1}{n}$ and $x_n = n$? Then $\sup a_nx_n = 1$ do we have $0 \times +\infty =1$, but then, what about $b_n = \frac{2}{n}$

0

For this $\lim\sup$ should not be $\infty$. I will do the lower bound and you should carry out the rest of the argument for the upper bound. By definition of limit and $\lim\sup$ given $\epsilon>0$ there exists $N$ such that for all $n>N$ $a_n>(1-\epsilon)a$ and for all $N$ there exists $m>N$ such that $x_m>(1-\epsilon)\lim \sup x_n$. ...

1

Other possibility: Use $x=1/y$, the integral becomes: $$I(a)=\int -\frac{1}{y^2}\frac{1}{\frac{1}{y^2} \sqrt{a^2-\frac{1}{y^2}}}dy=-\int \frac{y}{\sqrt{a^2y^2-1}}dy=-\frac{1}{a^2}\int \partial_y\left(\sqrt{a^2y^2-1}\right)dy=-\frac{\sqrt{a^2y^2-1}}{a^2}+C$$ Resubstitute: $$I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C$$

0

Covering is a standard technique to obtain bounds on the supremum of random processes. In Lemma 10 the author is essentially showing that the map $S$ is isometric restricted to the subspace. Isometry is ensured by the fact that $S$ is a JL transform. JL transform applies to a finite set of points however if you cover the subspace (rather the unit sphere over ...

1

I'd say you are on the right way, writing $$\frac{dx}{dy}=\frac{x}{y}+\frac{2y^2}{x}.$$ Let us view $x$ as a function of $y$, and introduce $z(y)=\frac{x(y)}{y}$. Then $x=yz$, so $\frac{dx}{dy}=z+y\frac{dz}{dy}$. Let us now replace all instances of $x$ and of $\frac{dx}{dy}$ in the differential equation, and simplify. Unless, I did something wrong, we end ...

0

Using your substition: $$\int\frac1{x^2\sqrt{a^2-x^2}}\;\rightarrow\;\int\frac{a\cos\theta\;d\theta}{a^2\sin^2\theta\cdot a\cos\theta}=-\frac1{a^2}\int\frac{d\theta}{-\sin^2\theta}=-\frac1{a^2}\cot\theta+C$$

3

We have, taking log and using L'Hopital, $$\lim_{n\rightarrow\infty}-t\sqrt{n}-n\log\left(1-\frac{t}{\sqrt{n}}\right)=\lim_{n\rightarrow\infty}\frac{\left(-\frac{t}{\sqrt{n}}-\log\left(1-\frac{t}{\sqrt{n}}\right)\right)}{1/n}=\frac{1}{2}t^{2}\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\sqrt{n}-t}$$ and so your limit.

0

Let $\sum_n a_n$ and $\sum_n b_n$ be series having positive terms and satisfying $$\frac{a_{n+1}}{a_{n}} \le \frac{b_{n+1}}{b_{n}}$$ for all sufficiently large $n$. If we rewrite the inequality as $$\frac{a_{n+1}}{b_{n+1}} \le \frac{a_{n}}{b_{n}},$$ we see that the sequence $(a_n/b_n)$ is monotone decreasing. Since $a_n, b_n \ge 0$, this sequence is ...

2

This is called the Ratio Comparison Test. You can find the proof on page 3 of these notes: http://www.westga.edu/~faucette/research/Raabe.pdf

1

Compose with the strictly increasing function $x=2\tan(z/2)$, $z\in(0,\pi)$. Such composition doesn't change the monotonicity. We get $$f(2\tan(z/2))=\frac{(1+\tan^2(z/2))z/2}{2\tan(z/2)}=\frac{z}{2\sin(z)}$$ And now $g(z)=\frac{\sin(z)}{z}$ is easy(er) to study in $(0,\pi)$. For this we can proceed similar to what you did, by looking at ...

1

Lets express $f(x)$ as $f_1(x)$ and $f_2(x)$, where $f_1(x)=ln(x$) and $f_2(x)=2x^2+1$, and so $f(x)=f_1(f_2(x))$ Ok. It is pretty obvious that $f(x)$ is defined over whole domain of real numbers. $f_1(x)$ is defined in $(0,\infty)$ domain, and it does happen so that $f_2(x)$ range is $[1,\infty)$ which fits into $(0,\infty)$ quite nicely. Only point of ...

0

$$\frac{\partial f}{\partial x}=\frac{2x(y-y^2)}{(1+x^2)^2}+\frac{8}{13}$$ $$\frac{\partial f}{\partial y}=\frac{2y-1}{1+x^2}+2y$$ Therefore $$f=\frac{y^2-y}{1+x^2}+\frac{8x}{13}+y^2$$ You can solve for the gradient function by integrating each equation with respect to its appropriate differential, then solve for the arguments which are exclusively of the ...

1

This is a heuristic argument commonly used to solve differential equations. Here is a more rigorous explanation: The differential equation you provided can be written as $$f'(x) = - \frac{1}{c}$$ where $A = e^{f(x)}$ and $B = e^x$. The general solution to this simple linear differential equation is $f(x) = -\frac{1}{c}x + K$, where $K$ is a constant of ...

1

We have $f(0)=f(1)=0$ and $f(x)$ is positive for $0\lt x\lt 1$. Thus there is a local (and global) maximum in the interval $(0,1)$. To show there cannot be two or more local maxima, note that (1) $f'(x)=0$ at local maxima, and (2) if we have a local maximum at $a$ and $b$, then $f'(x)=0$ somewhere between $a$ and $b$. But calculation shows that the first ...

1

The proposed method will not work, as $f''(x)$ is not always negative; for example try $p=3$. However, all is not lost. Taking just one derivative: $$f'(x)=\frac{x^{p-1}}{(2-x)^2}(px^2+(-3p-1)x+2p)$$ Note that $\frac{x^{p-1}}{(2-x)^2}>0$ for all $p>1$, $x\in(0,1)$. The second bit is an upward-facing parabola. It is positive for $x=0$ and negative ...

0

Thanks Bungo's comment, I've found the solution. As I've showed above I have $a_0=1$ and $a_n=0$ for $n \neq 2$. So I need to find $a_2$ to solve my problem: $$a_2 = \frac{2}{\pi}\int _0^\pi \cos^2 x \cos2x \,\text dx = \frac{1}{2}$$ Thus I've this Fourier series, that is equal to a famous trigonometrical formula: $$\cos^2x = \frac{1}{2} + \frac{1}{2} ... 1 For the Y Axis, we would want to use the shell method. The graph of the area you want to revolve can be seen here. The shell method states that the volume of revolution is equal to$$2\pi\int^4_2xf(x)$$We know f(x) = x, so we have$$2\pi\int^4_2x^2$$0 Let us denote the height of the ball at time t by x(t), the velocity by v(t) = x'(t) and acceleration by a(t) = v'(t) = x''(t) = -g (gravitational acceleration points downwards). Our initial conditions are then given by: x(0) = 8 \textrm{ft} and v(0) = 40 \textrm{ft}\ \textrm{s}^{-1}. We also have the following system of differential equations: ... 2$$\int(x^2+c)^{-3/2}dx=\int\frac{1}{\sqrt{(x^2+c)^3}}$$Let us use a new variable t such that \sqrt{x^2+c}=x+t. This is the first Euler substitution. From this formula we can solve for x to get x=\frac{c-t^2}{2t}, dx=\frac{-c-t^2}{2t^2}dt, \sqrt{x^2+c}=x+t=\frac{c+t^2}{2t}, and \sqrt{(x^2+c)^3}=\frac{(c+t^2)^3}{8t^3}. We put these into the ... 1 \textbf{HINT:} Try checking the secuence \{a_{2k}\}_{k\in\mathbb{N}} and \{a_{2k-1}\}_{k\in\mathbb{N}}. One of them is crecent and the other decrecent. But it converges iff \limsup=\liminf 1 Try trigonmetric substitution. Examples can be found here: http://en.wikipedia.org/wiki/Trigonometric_substitution In particular, try the substitution x=\sqrt{c}\tan\theta if c > 0. Try x=\sqrt{c}\sec\theta if c < 0 1 The result you are trying to prove is called Darboux's theorem. It says derivatives have the intermediate value property. http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29 The proof is on the page too. 1 Perhaps the sequence a_1,a_3,a_5,... is monotone, and also a_2,a_4,a_6,... 0 (1.) Its actually very easy. You can use algebra instead of integration. As the acceleration is constant, you can use one of the big five kinematics equations. You do not have the time, so it would be best to use: vf^2=2ax+vi^2 vf=velocity final vi=velocity initial a=acceleration x=distance traveled in air vf=0 vi=40 a=-32 ft/sec ... 1 First we can observe that n^{1/n}\to1. In fact n^{1/n}=e^{\frac{\ln(n)}{n}} and \frac{\ln(n)}{n}\to0. Therefore the denominator tends to 1 while the numerator tends to +\infty. 1 The derivative can be thought of as$$\frac{d^n}{dx^n}(f(x))=\underbrace{\frac{d}{dx}\big(\frac{d}{dx}\big(\cdots}_{n\text{ times}}f(x)\big)\cdots\big).$$Now, if you make the change of variables x\mapsto -x, you get the coefficient (-1)^n that you're looking for in your situation. 1 HINT Use induction. We have$$\dfrac{d^{m+1}}{d^{m+1}(-x)} = \dfrac{d}{d(-x)} \left(\dfrac{d^{m}}{d^{m}(-x)}\right) = -\dfrac{d}{dx} \left(\dfrac{d^{m}}{d^{m}(-x)}\right)$$6 No, the value \int_0^1 |f(x,x)|\,dx is not even well defined for f \in L^2([0,1]^2). Recall that the elements of L^2([0,1]^2) are strictly speaking not functions on [0,1]^2; they are equivalence classes of functions, mod equality almost everywhere. As Bungo's comment says, if you let D = \{(x,x) : x \in [0,1]\} be the diagonal of the square, then ... 1 Using Green's Theorem$$\int_C (Pdx+Qdy)=-\int_S \left(\frac{\partial P}{\partial y}-\frac {\partial Q}{\partial x} \right) $$Here, P=4y^3+\cos^2x and Q=-(4x^3+\sin^2y). Upon applying Green's Theorem, we find$$\int_C (Pdx+Qdy) =-\int_0^{2\pi}\int_0^2 12\rho^2\,\rho d\rho d\phi = -96\pi$$Now, we evaluate directly the line integral. To that ... 2 Note that the function inside the limit is not defined for x\le 0, so strictly the double-sided limit you wrote does not exist. However, you probably mean the right-hand limit, which is what I will work with here. (I see that @AndréNicolas beat me to this comment.) Split the limit into$$\lim_{x\to 0^+}\left(\frac{26x^3}{3}\ln x\right)-\lim_{x\to ...

2

Just as in the expansion for $e^x$ you can expand $e^{f(x)}$ to $$e^{f(x)} \;\; =\;\; \sum_{n=0}^\infty \frac{(f(x))^n}{n!}.$$ For safety, I would just use functions $f(x)$ which are differentiable.

3

$$e^{x^2-1}=\frac1e\sum_{n=0}^\infty\frac{(x^2)^n}{n!}=\cdots$$

0

Not as pleasing as the AM-GM inequality, but since $f$ and its derivatives are all continuous except the discontinuity at $x=0$: you can differentiate $f$, then analytically show $f'$ is zero at x=1, as well as somewhere around 0.2 (here using the intermediate value theorem]. obtain $f''$ and show it is positive at $x\approx 0.2$ and negative at $x=1.$ ...

3

\begin{align} &\int_0^\pi\frac{x\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{1}\\ &=\int_0^\pi\frac{(\pi-x)\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{2}\\ &=\frac\pi2\int_0^\pi\frac{\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{3}\\ &=\frac\pi2\left[-\arctan(\cos(x))\vphantom{\int}\right]_0^\pi\tag{4}\\[3pt] ... 4 There is also the logarithmic differentiation approach: http://en.wikipedia.org/wiki/Logarithmic_differentiation Note that \ln((x+1)(x+2)^2(x+3)^3)=\ln(x+1)+2\ln(x+2)+3\ln(x+3). Then \frac{d}{dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)} using the chain rule. So take the derivative of the RHS (above) to get\frac{f^\prime(x)}{f(x)} = ...

6

I recommend the extended product rule which states that $$(fgh)'=(f')(gh)+(g')(fh)+(h')(fg)$$ For this particular problem, $f=(x+1), f'=1$, $g=(x+2)^2, g'=2(x+2)$, $h=(x+3)^3, h'=3(x+3)^2$. Assemble and win.

0

I know you already have an answer but you could divide top and bottom by $\cos^2(x)$ So you have $\int \frac{\sec^2(x)}{2+\tan^2(x)} dx$ Let $u=\tan(x)$ But this looks similar to @math 's idea. Oh by the way I used $\sec^2(x)=\tan^2(x)+1$

5

Without reference to a "graph," note that if we have $f(a)=c$ and $f(b)=d$, then \begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b f(x)+x f'(x) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ ... 1 The optimization problem is\max_{x,y} 1-rx^2-y^2$$subject to$$ x+y=m$$. So the decision varibles are \mathbf{x}=(x,y) and the parameters are \mathbf{r}=(r,m). That's how we connect the symbols in the definition to the symbols in the problem. It will be useful to define the objective f(\mathbf{x},\mathbf{r})=1-rx^2-y^2 and constraint function ... 4 Not true! Take f(x)=x - it is a bijection. Then f^{-1}(x)=x, and$$ \int_c^d f^{-1}(y)\,dy+\int_a^bf(x)\,dx=\frac{1}{2}(d^2-c^2)+\frac{1}{2}(b^2-a^2) \ne bd-ac. $$However, if c=f(a), and d=f(b), then it holds! Simply draw the figure of the graph of f. Then f^{-1} is also visible in the same graph (reflection along y=x). Then bd-ac is the ... 1 Setting$$ t=\tan(x) $$we get$$ \frac{1}{1+\cos^2x}=\frac{\cos^2x+\sin^2x}{2\cos^2x+\sin^2x}=\frac{1+t^2}{2+t^2}. $$Since$$ t=\tan x\iff x=\tan^{-1}t, $$and$$ dx=\frac{dt}{1+t^2}, $$it follows that \begin{eqnarray} ... 2 Let u=\tan x then du=(1+u^2)dx so$$\int \frac{dx}{1+\cos^2x}=\int\frac{du}{(1+u^2)(1+(1+u^2)^{-1})}=\int\frac{du}{2+u^2}$$Can you take it from here? 0 If you read it in the other direction, then for a function g, the anti-derivative is$$G(x) = \int_a^x g(t) dt + c.0 The decay formula is A(t) = A_0 2^{-\lambda t} with \lambda = \dfrac{1}{h} the decay rate. Half-time t_{1/2} being the time such that A(t_{1/2}) = \dfrac{A_0}{2} one has : \begin{align} A(t_{1/2}) = \dfrac{A_0}{2} &\Leftrightarrow A_0 2^{-\lambda t_{1/2}} = \dfrac{A_0}{2}\\ &\Leftrightarrow \lambda t_{1/2} = 1\\ &\Leftrightarrow ... 3 If f is an arbitrary continuous funciton, then you can define a function g via the integral and by the fundamental theorem, the derivative of g is f. That is: g is an antiderivative of f. That's all. 2\int \frac{1}{1+\cos x+\sin x}dx=\int \frac{1}{\frac{2}{2}(1+cos x)+2\cos(x/2)\sin(x/2)}dx=\int \frac{1}{2\cos^2(x/2)+2\cos(x/2)\sin(x/2)}dx=\int \frac{1}{2\cos^2(x/2)(1+\tan(x/2))}dx=\int \frac{.5\sec^2(x/2)}{1+\tan(x/2)}dx=\log(1+\tan(x/2))+C

0

This might help. I will put some more steps in a bit. $\frac{1}{1+(\sin(x)+\cos(x))} \cdot \frac{1-(\sin(x)+\cos(x))}{1-(\sin(x)+\cos(x))}=\frac{1-(\sin(x)+\cos(x)}{1-(\sin(x)+\cos(x))^2}=\frac{1-\sin(x)-\cos(x)}{1-(\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x)}=\frac{1-\sin(x)-\cos(x)}{1-(1+2 \sin(x) \cos(x))}=\frac{1-\sin(x)-\cos(x)}{-2 \sin(x) \cos(x)} \\ \int ... 3 We take$u=\tan\left(\frac{x}{2}\right) $,$du=\frac{1}{2}\sec^{2}\left(\frac{x}{2}\right)dx $, then we use the substitutions$\sin\left(x\right)=\frac{2u}{u^{2}+1} $,$\cos\left(x\right)=\frac{1-u^{2}}{u^{2}+1} $and$dx=\frac{2du}{u^{2}+1} $. So we have ... 0 Quite a number of sequences$(a_n)$will be analogous to the one in question 1243909, and any way to compute the limit will be similar. If you can show that your sequence$a_n$is bounded below by$x^{f(n)}$and above by$k x^{f(n)}$, then the limit of$a_n ^ {(1/n)}$is the limit of$x^{f(n)/n}\$ if it exists.

1

Taylor series don't always converge. The Taylor series method can be very good if the interval of integration is inside the radius of convergence of the series, otherwise it may not converge at all. But as Yves Daoust mentioned, you should be comparing this to more sophisticated numerical methods, not to Riemann sums. In practice, Riemann sums are almost ...

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