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2

Consider the auxiliary function $$F(x):=\int_0^x|f'(t)|\>dt\qquad(x\geq0)\ .$$ By assumption, the $\lim_{x\to\infty} F(x)$ exists. By Cauchy's criterion it follows that for each $\epsilon>0$ there is an $M\geq0$ with $$|f(y)-f(x)|\leq\int_x^y|f'(t)|\>dt=F(y)-F(x)<\epsilon\qquad(M<x\leq y)\ .$$ The "essential part" of Cauchy's criterion then ...


3

$$I=\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$ Let $u=2012-x$ then $$I=\int_{1}^{2011} \frac{\sqrt{2012-u}}{\sqrt{2012 - u} + \sqrt{u}}du=\int_{1}^{2011} \frac{\sqrt{2012-x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$ Thus $$2I=\int_{1}^{2011} \frac{\sqrt{x}+\sqrt{2012 - x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$


4

HINT: As $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$ So, if $\int_a^bf(x)\ dx=I,$ $$2I=\int_a^b[f(x)+f(a+b-x)]\ dx$$


0

Actually there are solid arguments why $0/0$ can be defined as $0$. Here is a graphic of the function $f(x,y)=x/y$: The function is odd against both $x$ and $y$ variables. Along the axis $x=0$ it is constant zero. Along the axis $y=0$ it is unsigned infinity, but its Cauchy principal value is constant zero. Along the diagonal $y=x$ it is constant 1. ...


1

Hint: prove first that $f$ is bounded around $\infty$. Then assume that $x_n, y_n\to \infty$ with $x_n< y_n< x_{n+1}$ and that $f(x_n)\to X$, $f(y_n)\to Y$ and $X\neq Y$ and try to find a contradiction.


1

You are calculating the area in the following graph and your calculation is correct. But the question asks for the area "inside" both graphs which can be seen below: This area is then $$A=\int_0^{\pi/3}(1-\cos x)^2dx+\int_{\pi/3}^{\pi/2}\cos^2xdx$$ which is what the back of your book says.


1

There is a very general result which guarantees such substitutions. Let $\lim\limits_{x \to a}f(x) = L$ exist and let $\lim\limits_{t \to b}g(t) = a$ exist and also assume that $g(t) \neq a$ when $t$ is in a certain neighborhood of $b$ then $\lim_{t \to b}f(g(t)) = L$. Please understand that the theorem is valid only under the conditions given in the above ...


0

Hint:- $y=\left(\dfrac{a+x}{b+x}\right)^{a+b+2x} \implies \ln y=(a+b+2x)\ln (a+x)-(a+b+2x)\ln (b+x)$


0

Well, taking logarithm of both sides is not necessary. Just use a simple identity $$f=e^{\ln f}$$ For $f$ being a power: $f=B^E$ that makes $$f=e^{\ln(B^E)}=e^{E\ln B}$$ then by the chain rule $$f' = e^{E\ln B}\cdot (E\ln B)'=e^{\ln f}\cdot(E'\ln B + E(\ln B)')$$ $$=f\cdot(E'\ln B + \tfrac EB\,B')$$


1

Hint: Write the expression: $$f(x) = \left( \dfrac{a+x}{b+x}\right)^{a+b+2x}$$ as: $$\exp\left(\ln\left(\left(\dfrac{a+x}{b+x}\right)^{a+b+2x}\right)\right) = \exp\left((a+b+2x)\ln\left(\dfrac{a+x}{b+x}\right)\right)$$ We now have the form: $$\dfrac {d}{du} e^u = e^u \dfrac{du}{dx}$$ After you find the derivative, rewrite the expression without the ...


1

HINT: Using Trigonometric substitutions, set $2x=\sec\theta$ $$\implies4x^2-1=\tan^2\theta$$


5

To get a bound on $\dfrac{1}{x_{2015}}$, let $y_n = \dfrac{1}{x_n}$. Then, $\dfrac{1}{y_{n+1}} = \dfrac{1}{y_n}+\dfrac{1}{y_n^2} = \dfrac{y_n+1}{y_n^2}$. Hence, $y_{n+1} = \dfrac{y_n^2}{y_n+1} = y_n-\dfrac{y_n}{y_n+1}$. Rearrange to get $\left(1+\dfrac{1}{y_n}\right)(y_n-y_{n+1}) = 1$. Since $y_n$ is a decreasing sequence and $1+\dfrac{1}{y}$ is a ...


0

The result is correct. Don't forget $dx$ in the integrals.


2

Hint: Recall that $\csc^2 \theta = 1 + \cot^2 \theta$. Let us perform the substitution $\theta = 2x$. Then, $$\int \csc^6 2x ~dx = \frac{1}{2} \int \csc^6 \theta ~d\theta$$ $$ = \frac{1}{2} \int(1+\cot^2\theta)^2\csc^2\theta ~ d\theta$$ If $u = \cot \theta$, then $du = \cdots$?


1

Hint:- $$(n-1)I_n=-\dfrac{\cos2 x}{2\sin^{n-1} 2x}+(n-2)I_{n-2}$$ Where $I_n=\displaystyle\int\dfrac{dx}{\sin^n {2x}}$


2

If $f'(0) = 0$ there is nothing to prove, so suppose $f'(0)< 0$. If $x = \min\{1/2, -f'(0)\}$, then $0 < x < 1$ and $0 < x \le -f'(0)$. Thus $$1 - \frac{f'(0)^2}{2} = 1 + f'(0)\cdot [-f'(0)] + \frac{[-f'(0)]^2}{2} \ge 1 + f'(0)x + \frac{x^2}{2} \ge 0.$$ Consequently, $f'(0)^2 \le 2$, which implies $f'(0) = -|f'(0)| \ge -\sqrt{2}$.


6

By the use of the chain's rule you get $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ then $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\qquad(1)$$ Now ...


3

To help your intuitions: $f(x)=\sqrt{x}$ has no asymptote at $+\infty$, although $f'(x)$ tends to 0.


3

Yes, the gradient of $\ln x$ "at infinity" (more precisely, in the limit as $x$ becomes increasingly large) is zero. What this means is that the larger $x$ is, the less of an increase in $\ln x$ you get by a small change to $x$. It doesn't mean there has to be a horizontal asymptote ("adding up infinitely many infinitesimal things doesn't necessarily give ...


1

Use the transformation $g(\theta) = f(\cos\theta,\sin\theta)$ for $\theta \in [-\pi,\pi]$, as suggested in the question. Then, we need to show that $\displaystyle\lim_{t \to 1^-}\left[\dfrac{\sqrt{1-t^2}}{2\pi}\int_{-\pi}^{\pi}\dfrac{g(\theta)}{1-t\cos\theta}\,d\theta\right] = g(0)$. Since $f(x,y)$ is continuous on the unit circle, $g(\theta)$ is ...


5

Good question! :) The thing to remember is that differentiation is an operator. It isn't a scalar variable that can be tossed around. Here is an example: As you know, the definition of the first derivative would be: $$\frac{dy}{dx} \approx \frac{\Delta( y)}{\Delta x}$$ However, we don't know what the change in $y$ is for a given change in $x$ directly. This ...


0

$f(x) = x\ln x \to f'(x) = \ln x + 1 \to f''(x) = \dfrac{1}{x} > 0 \to f \text{ is convex} \to \displaystyle \sum_{i=1}^n x_i\ln x_i = \displaystyle \sum_{i=1}^n f(x_i) \geq nf\left(\dfrac{x_1+x_2+\cdots +x_n}{n}\right) = nf\left(\frac{1}{n}\right) = -\ln n \to \displaystyle \sum_{i=1}^n -x_i\ln x_i \leq \ln n$, and this maximum value is attained when ...


0

You can do this in many ways. The first is with Lagrange multipliers. The second is using the fact that your function is concave and then showing that nearby values to $x_i=1/n$ give smaller function values.


-1

"and is clear that for the same d∈(a,b) we chose f′(d)=0" is incorrect. It is not sufficient to deduce that $f^{\prime}(d)=0$ from $f(d)=0$. Suppose $f(x)=x$, then $f(0)=0$, but $f^{\prime}(0)=1$.


3

Consider $f(x) = x^2 + 1$. It is continuous and differentiable on $[-1, 1]$, and has no $0$ on this interval. Moreover, we have that $$(f(0) - f(-1))(f(1) - f(0)) = (1-2)(2-1) = -1 < 0$$ so this satisfies all the assumptions of the question with $a = -1, b=1$, and $c=0$. However, $f(x) \ne 0 $ for all $x\in [-1, 1]$, so that's an example of why your ...


2

It is clear that the radius of the shell is $r = y + 1$ (the height of the blue line minus the height of the red line). The height of the shell is given by $$ h = \sqrt y - (-\sqrt y) = 2 \sqrt y.$$ Using cylindrical shells, the desired quantity is $$ \int_0^4 2\pi r h \ dy = \int_0^42\pi(y+1)(2\sqrt y) \ dy = 4 \pi \int_0^4(y+1) \sqrt y \ dy \approx 228. ...


0

You do not need to solve the integral because you can use the Second Fundamental Theorem of Calculus. With that the integral is cancelled by the derivative but you need to use the composition of functions. If we call the function that is inside the integral f(t) and the upper limit v(x) and the lower limit u(x). The solution is ((derivative v(x)) f(v(x)) ...


1

Hint: $$ F(x)=\int_{x^2}^{4x^2} \sin \sqrt t\;\;dt = \int_{0}^{4x^2} \sin \sqrt t\;\;dt - \int_{0}^{x^2} \sin \sqrt t\;\;dt $$ We could define $$G(x) = \int_0^ x \sin \sqrt t\;\;dt$$ and rewrite as $$ F(x) = G(4x^2) - G(x^2). $$ By Chain Rule, $$ F^\prime(x) = G^\prime(4x^2)(8x) - G^\prime(x^2)(2x). $$ By FTC, $$G^\prime(x) = \sin \sqrt x.$$ Is it clear ...


0

You'll need to be able to sketch or analyze functions without Google. :) You have hints. The curve $y=\sqrt{2x}$ suggests that it isn't straight. At this point, you'd either picture what a square root plot looks like, or graph out a few points on paper to see it. Where it crosses the $x$ axis is important because that's also one of your bounds. This ...


2

Looks like you need to show that $$\prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right) = \frac{\pi}{2}.$$ This Wikipedia article uses Euler's infinite product for the sine function: $$\frac{\sin x}{x} = \prod_{k=1}^{\infty}\left(1 - \frac{x^2}{k^2 \pi^2}\right).$$ The result follows quickly by applying $x = \pi/2.$


1

It is rather strange, because Wolfram Alpha is perfectly happy to return a Laurent series for e.g. series of 1/(x+x^2) at x = 0 Somehow, $1/x$ is treated differently.


0

Let $x=a\cos t, y=b\sin t$ to get $S=\displaystyle2\int_0^{\frac{\pi}{2}}2\pi(b\sin t)\sqrt{(-a\sin t)^2+(b\cos t)^2} dt=4\pi b\int_0^{\frac{\pi}{2}}\sin t\sqrt{a^2-(a^2-b^2)\cos^{2}t}dt$. Now let $u=\cos t, du=-\sin t dt$ to get $\;\;\;\displaystyle S=4\pi b\int_0^1\sqrt{a^2-(a^2-b^2)u^2} du$. Then let $u=\frac{a}{\sqrt{a^2-b^2}}\sin\theta,\;\; ...


1

If you calculate the curvature tensor of a cylinder you come up with $0$. That means that the $2D$ cylinder living in $3D$ space does not really have any curvature. You can form a cylinder by wrapping around a flat sheet of paper without stretching it. Therefore the geodesics on the cylinder are just geodesics on flat $2D$ space wrapped around. And we all ...


2

First, a quick but very important warning: In basic calculus, the symbol combination $$ydy - x dx = 0$$ does not really make sense. It is a dirty trick because it is the quickest way of transforming $$y\frac{dy}{dx} - x = 0,$$ which is a differential equation, into $$\int ydy - \int xdx = 0,$$ which is the next step in solving the differential ...


3

The locus of points that see an ellipse under a right angle is a circle having radius $\sqrt{a^2+b^2}$: Since the tangent to the ellipse in the point $P=(a\cos\theta,b\sin\theta)$ is given by: $$ P+\lambda(-a\sin\theta,b\cos\theta),$$ by solving: $$ a^2(\cos\theta - \lambda\sin\theta)^2 + b^2(\sin\theta+\lambda\cos\theta)^2=a^2+b^2$$ with respect to ...


2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


3

From the Euler's identity: $$\begin{eqnarray*}z!=\Gamma(z+1) &=& \lim_{n\to +\infty}\frac{n! n^z}{(z+1)\cdot\ldots\cdot(z+n)}=\lim_{n\to +\infty}n^{z}\prod_{k=1}^{n}\left(1+\frac{z}{k}\right)^{-1}\\&=&\prod_{k=1}^{+\infty}\left(1+\frac{z}{k}\right)\left(1+\frac{1}{k}\right)^z\end{eqnarray*}$$ it follows that: $$\frac{1}{2}! = ...


1

You may recall that $$ \sum_{k=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|<1. \tag1 $$ Just substitute $r \rightarrow 1-x $ in $(1)$, you get, for the right hand side $$ \frac{1}{1-r}=\frac{1}{1-(1-x)}=\frac 1 x $$ as long as $|1-x|<1$ ($0$ being excluded).


0

You can use the chain rule on $y(x)$ to get $$y'(x) = e^{z(x)}\cdot z'(x)$$ and then use the chain and product rule on $y'(x)$ to get $$y''(x) = e^{z(x)}\cdot (z'(x))^2+z''(x)e^{z(x)}$$ Now you can solve for $z''$ and should get $$z''(x) = \frac{y''(x)- e^{z(x)}\cdot (z'(x))^2}{e^{z(x)}}$$ Next you can substitute $y(x) = e^{z(x)}$ and $z'(x) = ...


2

You are starting with $$ e^{-x} \sum_{k = 1}^\infty \frac{kx^k}{(k+1)!} = e^{-x}F(x),$$ which is almost very easy. Let's ignore the $e^{-x}$ piece, because it's not interesting. So we just consider $F(x)$. Notice that $\dfrac{F(x)}{x} = \displaystyle\sum_{k = 1}^\infty \frac{kx^{k-1}}{(k+1)!}.$ If we integrate this, we see that $$\begin{align} \int_0^x ...


1

Once you have $z'(x) = y'(x)/y(x)$, then $$z''(x) = \frac{d}{dx}\left[ \frac{y'(x)}{y(x)} \right],$$ and you evaluate this via the quotient rule: $$z''(x) = \frac{(y'(x))' y(x) - y'(x)y'(x)}{y(x)^2} = \frac{y'' y}{y^2} - \frac{(y')^2}{y^2}.$$ I don't know how you went from the fourth expression to the fifth; i.e., the fourth equality seems dubious to me.


3

Here is one line proof $$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$ Q.E.D.


3

For any $\epsilon > 0$, choose $\delta = \text{min}\left(1,\frac{7\epsilon}{2}\right)$, then: if $0 < |x| < \delta$ then $\left|\dfrac{x^2-8}{x-8} - 1\right| = \left|\dfrac{x^2-x}{x-8}\right| \leq \dfrac{|x^2-x|}{8-|x|} < \dfrac{|x^2-x|}{7} < \dfrac{|x^2| + |x|}{7} < \dfrac{|x|+|x|}{7} = \dfrac{2|x|}{7} < \dfrac{2}{7}\cdot ...


2

Let $$ F(n)=\prod_{k=1}^\infty\left(\frac{k+1}k\right)^n\frac{k}{k+n} $$ Then, using Stirling's Formula, $$ \begin{align} F\left(\frac12\right)^2 &=\prod_{k=1}^\infty\frac{k+1}k\frac{k^2}{\left(k+\frac12\right)^2}\\ &=\lim_{n\to\infty}\prod_{k=1}^n\frac{k+1}k\frac{4k^2}{(2k+1)^2}\\ ...


4

\begin{align} \int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\mathrm dx&=\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\left(1-\sin^2x\right)^2}\mathrm dx\\[7pt] &=\int_0^{\pi/2}\frac{\sin x\cos x}{2\sin^4x-2\sin^2x+1}\mathrm dx\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{t^2-t+\frac12}\qquad\color{blue}{\implies}\qquad t=\sin^2x\\[7pt] ...


8

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$


1

First note $$dH(p,q)(x,y) = q\cdot y + dV(p)(x)$$ for all $(p,q), (x,y) \in \Bbb R^n \times \Bbb R^n$. Suppose $c$ is a regular value of $H$. Let $p \in \Bbb R^n$ such that $V(p) = c$. Then $H(p,0) = c$. Since $dH(p,0)$ is surjective, given $r \in \Bbb R$, there exists $(x,y)\in \Bbb R^n \times \Bbb R^n$ such that $dH(p,0)(x,y) = r$, i.e., $dV(p)(x) = r$. ...


1

with the hint in the coment i find that the equality holds for $$|z+iw|\rightarrow z=iw\\ |z-i\overline{w}|\rightarrow z=-i\overline{w}$$ then $$iw=-i\overline{w}\\ w^2=-|w|^2=-1\\ w=\pm i\\ z=iw=\pm i^2=\mp 1$$


1

For simplicity consider the case n=1. $H(x,y)=y^2/2+V(x)$, $dH=ydy+V_xdx$ is a linear map in the tangent space whose matrix representation is the gradient: $$\operatorname{grad}(H): (u,v)\mapsto V_xu+yv=\begin{bmatrix}V_x & y\end{bmatrix} \begin{bmatrix}u \\ v\end{bmatrix}$$ For general n $$dH=\sum_{i=1}^ny^idy^i+\frac{\partial{V}}{\partial x^i}dx^i$$ ...


2

If $z=a+bi$, then $e^z=e^a(cos(b)+isin(b))$ Since the cos- and sin- function have periodicity $2\pi$, you get what you want.



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