# Tag Info

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


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), thenz''(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] ... 81 - \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.

Top 50 recent answers are included