# Tag Info

0

I suspect that the problem is asking you to show that $2cos^2(\theta)- 1= cos(2\theta)$ is an identity- that is, that it is true for all $\theta$ That is simple if you know the "sum" identity $cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)$, taking $a= b= \theta$. If you do not know that identity it is somewhat harder and how you prove it depends upon exactly how ...

0

So we have that $f(0)=f(\pi/2)=0$, so by Rolle's theorem there is a (at least one) point $c$ in our interval $(0,\pi/2)$ such that $f'(c)=0$, and let's find it. Solve: $f'(x)=2e^{-2x}(\cos(2x)-\sin(2x))=0$. Now $\cos(y)=\sin(y)$ when $y=\pi/4 +2n\pi$, with $n\in\Bbb Z$. So letting $y=2x$, we have that $x=\pi/8$. We see that as Rolle's theorem claims, we ...

2

Perhaps this answers your question $$2\cos^2\theta−1$$ $$=2\cos^2\theta−\left(\sin^2\theta +\cos^2\theta\right)$$ $$=2\cos^2\theta−\sin^2\theta -\cos^2\theta$$ $$=\cos^2\theta−\sin^2\theta$$ $$=\cos\left(2\theta\right)$$

0

HINT $y=\sin(10 x^2), 0 \le x \le \sqrt{\frac{\pi}{10}}, y >= 0 \quad$ (see) $\displaystyle \Rightarrow dV = 2\pi x \cdot y\,dx \Rightarrow V = 2\pi \int_{0}^{\sqrt{\frac{\pi}{10}}}x\sin(10 x^2)\,dx$ I would say, the substitution $\quad 10x^2=t,\quad 20x\,dx = dt\quad\Rightarrow$ easy integration. I am sure that you can take from here.

1

There are multiple ways, but I'll give you one. From writing out the Taylor expansions of $f(x+\Delta x)$ and $f(x-\Delta x)$ you can show that we can approximate the second derivative in $\mathbb{R}^2$ by $$\frac{h(x+\Delta x,t)-2h(x,t)+h(x-\Delta x,t)}{(\Delta x)^2}=h_{xx}(x,t)+O(\Delta x^2)$$ So, substituting in we get $$\frac{h(x+\Delta ... 1 So you are trying to simplify this expression. 2\cos^2 \theta - 1 = \cos(2\theta) 0 Let f(z):D \to \mathbb{C} be holomorphic in a domain D of the complex plane. and suppose that f(z_0) \neq 0, then it is locally invertible. Because of the following, let f(z)=u(x,y)+iv(x,y), then we can think of this function as a function from f(x,y) : \mathbb{R}^2 \to \mathbb{R}^2 where f(x,y)=\left(u(x,y),v(x,y)\right). now using the inverse ... 7 My usual way is to use Cauchy's condensation test and recall that the harmonic series is divergent (by the same reason, if you like it). 0 It is, in general, very hard to prove there is NO closed form expression for the sum of a series when the terms of the series are given by a formula. It is much easier, for example, to prove that a closed form formula function has no closed form formula anti-derivative. On the other hand, there is no reason why we shouldn't be able to (eventually) come up ... 1 If f(x) is an analytic function in a neighbourhood of x=0, the radius of convergence of f(x) and f'(x) are the same. Now notice that, by differentiation under the integral sign:$$ f'(x) = \int_{0}^{1}\frac{dt}{x+\sqrt{1+t^2}}=\int_{0}^{1}\sum_{n\geq 0}\frac{(-1)^n x^n}{\sqrt{1+t^2}^{n+1}}\,dt $$hence:$$ \left|[x^n]\,f'(x)\right| = ...

0

HINT...the work done by a constant force $\underline{F}$ in moving a particle from $P$ to $Q$ is $$\underline{F}\cdot\overrightarrow{PQ}$$

0

The mere existence of the $n$'th derivative implies that all previous derivatives are differentiable, hence continuous.

0

It looks like you've made the following calculation xn-1 for n=1 x1-1 = x0 = 0 But x0 ≠ 0, x0 = 1 Now try the derivative.

1

You series is asymptotically equivalent to $$\sum_{n\geq 2}\frac{1}{n^{1+p/2}}$$ that is convergent for $\color{red}{p>0}$. That happens since $2\sqrt{n}<\sqrt{n}+\sqrt{n+1}<2\sqrt{n+1}$ and: $$\log\frac{n-1}{n+1} = \log\left(1-\frac{2}{n+1}\right) = -\frac{2}{n+1}+O\left(\frac{1}{(n+1)^2}\right).$$

2

Multiplying by $4R^2$ and exploiting the sine theorem we get: $$4R^2\sum_{cyc}\cos^2 A = 12R^2-(a^2+b^2+c^2)$$ hence the first inequality is equivalent to the trivial $OH^2\geq 0$, where $O$ is the circumcenter and $H$ is the orthocenter. On the other hand, if $ABC$ is an acute-angled triangle we have that $H$ lies inside $ABC$, hence $OH^2< R^2$ and ...

0

Try the first one. Transform the triangle using linear transformations(thereby keeping angles invariant) to A(0,a),B(-b,0),C(c,0). What we are after is bb/(aa+bb) +cc/(aa+cc)+(aa/(sqrt(aa+cc)sqrt(aa+bb))-cb/(sqrt(aa+cc).sqrt(aa+bb)))^2>=3/4. Simplifying then we are after 4(3bbcc+aaaa+bbaa+aacc)>=3(aaaa+aabb+aacc+bbcc). This is clear.

1

You would get $\frac{d}{dx}x = 1\cdot x^0$, and $x^0=1$, so you get $\frac{d}{dx}x = 1\cdot 1 = 1$. Your calculation is allright to the point where you got $$n\cdot x^{n-1}$$ but then you seem to have made a mistake either in plugging in $n=1$ or in evaluating $x^0$ (which is equal to $1$, not $0$).

1

$$y=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$ Now $\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)=-\cos C\cos(A-B)$ $$\implies\cos^2C-\cos C\cos(A-B)+1-y=0$$ which is a Quadratic Equation in $\cos C$ $$\implies\cos^2(A-B)-4(1-y)\ge0\iff4y\ge4-\cos^2(A-B)=3+\sin^2(A-B)$$ $$\implies4y\ge3$$ The equality occurs if $\sin(A-B)=0\implies A=B\ \ \ \ (1)$ and ...

1

Eliminate $C$ from $A+B+C= \pi$ Next, recognize the bounds of $\cos A, \cos B.$

1

Given the notation, I would assume that $$r^{2} \cdot \vec{r}$$ means: $$(x^2 + y^2 + z^2) (x,y,z)$$ Yes, since $$\vec{dS} = \vec{\hat{n}}dS$$ where $dS$ is the surface element and $\vec{\hat{n}}$ is the unit normal of the surface, which assuming $R\equiv|\vec{r}|$ you have correct. The divergence theorem is in principle useful in problems like this, ...

2

The function is convex, so the graph is an upward pointing cup (for example, $f(x)=x^2$ is also convex and is an upward pointing cup. There is no ambiguity, $\sin^2\frac x2$ is convex. You may have doubts, but you can hardly argue with the proofs. Also, you can "see" here that the function is convex.

0

You can substitute $x \rightarrow x^2$ (or rather: use $x^2$ as an argument in the sine function) in Taylor series since it's correct for every real number. However, the formula changes so you can't say that $\sin x$ is identical with $\sin x^2$. $$\sin x = x - \frac{x^3}{3!} + \frac {x^5}{5!} + \ldots,$$ $$\sin x^2 = x^2 - \frac{(x^2)^3}{3!} + \frac ... 1 Let's say we calculate the gradients at point (x_0,y_0,z_0). We are interested in the rate of increase of f in the direction of some vector (x_1, y_1, z_1), and let's say that the vector has norm 1. Let's have a single variable function$$g(t)=f(x_0 + tx_1, y_0 + ty_1, z_0 + tz_1)$$and let's find g'(t). Using the chain rule, you can verify ... 0 Using the addition theorems you get$$\cos(x+y)+\cos(x-y)=2\cos(x)\cos(y).$$So in our case we are looking for a solution of$$\cos\left(\frac{2\pi j t}{m}\right) \cos\left(\frac{2\pi j}{3m}\right)=\frac{1}{4}.1 You can try checking out Problems in Real Analysis: Advanced Calculus on the Real Axis. 1 In the classical setting a field is a vector valued function f:D\to R^n defined on some subset D\subset R^m. So the answer to both your questions is "yes". 1 Its much better to change \sec into \cos and then we get \begin{align} L &= \lim_{x \to \pi/2}\frac{\sec x}{\sec^{2}3x}\notag\\ &= \lim_{x \to \pi/2}\frac{\cos^{2}3x}{\cos x}\notag\\ &= \lim_{h \to 0}\dfrac{\cos^{2}3\left(\dfrac{\pi}{2} + h\right)}{\cos \left(\dfrac{\pi}{2} + h\right)}\text{ (by putting }x = (\pi/2) + h)\notag\\ &= ... 0 Do it by induction. Base case: N=1.\sum_{i=1}^N i^3 = 1^3 = 1^2 = \left(\sum_{i=1}^N i\right)^2$$Now let M = N+1. Then you have$$\sum_{i=1}^M i^3 = \sum_{i=1}^N i^3 + M^3$$and$$\left(\sum_{i=1}^M i\right)^2 = \left(\sum_{i=1}^N i + M\right)^2 = \left(\sum_{i=1}^N i\right)^2 + 2M\sum_{i=1}^N i + M^2$$Then recall the summation formula for an ... 0 Hint: you could try induction on N. It might help if you knew the formula for \sum_{i=1}^N i. 1 Another approach is to write$$\begin{align} \sum_{i=0}^{\infty}\frac{i}{4^i}&=\sum_{i=1}^{\infty}\frac{1}{4^i}\left(\sum_{j=1}^{i}1\right)\\\\ &=\sum_{j=1}^{\infty}\sum_{i=j}^{\infty}\frac{1}{4^i}\\\\ &=\sum_{j=1}^{\infty}\frac{1}{4^j}\frac{1}{1-\frac14}\\\\ &=\frac{1/4}{(1-\frac14)^2}\\\\ &=\frac49 \end{align}$$0 Let me try. Set$$S = \sum_{i\geq 0}\frac{i}{4^i}.$$Then we have$$4S = \sum_{i \geq 0} \frac{i+1}{4^i} = \sum_{i\geq 0}\frac{i}{4^i} + \sum_{i\geq 0}\frac{1}{4^i} = S + \frac{1}{1-\frac{1}{4}}$$So, 3S = \frac{4}{3}. It implies that$$S = \frac{4}{9}.$$1 For -1<x<1, the series \sum_{i=0}^{\infty}x^i converges absolutely to \frac{1}{1-x}$$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}Then \begin{align*} \sum_{i=0}^{\infty}ix^{i-1} &= \frac{d}{dx}\left(\frac{1}{1-x}\right)\\ &=\frac{1}{(1-x)^2}\\ \sum_{i=0}^{\infty}ix^i&=\frac{x}{(1-x)^2} \end{align*} Now, by plugging x=1/4 into the last ... 0 This is a straight-forward integral. \theta is fixed for the projectile. If \theta=0 then the problem does not make much sense. If \theta=\pi/2 then the area is undefined. So it makes sense to assume \theta\in(0,\pi/2)\cup(\pi/2,\pi)f(x)=0 \iff x=0 \ \ \vee \ \ x=\dfrac{2v^2\cos^2\theta\tan\theta}{g}$$And so,$$A(\theta) = ...

3

If this is a quartic then you're given the three roots of its derivative. You can integrate this cubic to recover the quartic and use the known point to find the leading coefficient and the constant of integration. Note that $x=0$ is a zero of both the quartic and the cubic. Solution:

2

Solution With out Using $\bf{D-Lhopital \; Rule}$ Let $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{\sec x}{(\sec 3x)^2} = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(\cos 3x)^2}{\cos x}.$$ Now Using $$\bullet\; \cos 3x = 4\cos^3 x-3\cos x$$ We get $$\displaystyle y = \lim_{x\rightarrow \frac{\pi}{2}}\frac{(4\cos^3 x-3\cos x)^2}{\cos x} = ... 1 Let$$\displaystyle y=\lim_{x\rightarrow 0^{+}}\left(\frac{1}{x}\right)^{\sin x}\;,$$Now Let x=0+h\;, Then \displaystyle y=\lim_{h\rightarrow 0}\left(\frac{1}{h}\right)^{\sin h} So$$\displaystyle \ln(y) = \lim_{h\rightarrow 0}\sin (h)\cdot \ln\left(\frac{1}{h}\right) = -\lim_{h\rightarrow 0}\sin h\cdot \ln(h) = -\lim_{h\rightarrow ...

0

Usually we tell people not to just copy-paste HW problems onto here, but also show their efforts. Now since this is your first time, here we go: We are looking at a triangle with $H=6$ and $R=4$. Let's partition the height, and, by similar triangles, we get that the "radius" for a height $h$ is $\frac{2h}{3}$. Now we know that the approximate volume of hot ...

0

$x^{'} (t) = 3(x (t)-2) ; \quad x(0)=-1$ The differential equation can be written equiavalently as $$x^{'} (t) -3x(t)=-6$$ Now multiplying both sides of the above with the "integrating factor" $e^{-3t}$ we obtain: $$x^{'} (t) e^{-3t} -3e^{-3t}x(t)= -6e^{-3t}$$ $$(x(t) e^{-3t})^{'} = (2 e^{-3t})^{'}$$ Hence, $x(t)e^{-3t} = 2e^{-3t} +c$, where $c ... 2 Hint. What you have done is not correct. Observe that $$\int 3 \:dt= 3t+C, \quad C \, \text{is a constant}.$$ Here you have $$\int\frac{dx}{x-2}=\int 3 \:dt \tag1$$ or $$\ln|x-2|=3t+C$$$$x(t)-2=Ke^{3t}. \tag2$$ You obtain$K$by putting$t=0$in$(2)$, $$x(0)-2=K$$$$-1-2=K.$$ Then $$x(t)=2-3e^{3t}.$$ 3 First integral: Let us multiply the numerator and denominator of the integrand by$\sqrt{1+x^n}$, so that the initial integral becomes $$\mathcal{I}=\underbrace{\int_0^1\frac{dx}{\sqrt{1-x^{2n}}}}_{\mathcal{I}_1}+ \underbrace{\int_0^1\frac{x^ndx}{\sqrt{1-x^{2n}}}}_{\mathcal{I}_2}.$$ After the substitution$x=\sin^{\frac1n}\theta$the integrals ... 3 You've lost an$r$:$3x^2+3y^2 = 3r^2$, rather than$3r$. The$r$just before the differentials should be distributed to both term in the integrand as well. Response to edit: the$z$is fine now, but that$r$at the end isn't making it into the first term, I think. $$\int_{0}^1 \int_{0}^{2\pi}\int_0^2 (3r^2+3z^2)r\,dz\,d \varphi \,dr =\int_{0}^{1} ... 0 Based on the answer of Jack D'Aurizio: The parameterization u(\theta) = \frac{v(\theta)}{||v(\theta)||_{\infty}} will work since: ||u(\theta)||_{\infty} = ||\frac{v(\theta)}{||v(\theta)||_{\infty}}||_{\infty} = \frac{1}{||v(\theta)||_{\infty}} ||v(\theta)||_{\infty} = 1. (I.e we normalize v(\theta) with respect to the || \bullet ||_{\infty} norm) 0 like the Weierstrass function but monotonic The Weierstrass function is quite unlike any monotone functions. It is nowhere differentiable, while every monotone function is differentiable at almost every point. Specifically, your function f has f'(x)=0 for almost every x. Indeed, by definition f is the cumulative distribution function for a ... 1$$u(\theta)=\frac{v(\theta)}{\max\left(\left|\sin\theta\right|,\left|\cos\theta\right|\right)}=\frac{1}{\max\left(\left|\sin\theta\right|,\left|\cos\theta\right|\right)}\left(\cos\theta,\sin\theta\right).$$3 Ok maybe not the most elegant solution but nevertheless: I concentrate on the integral first integral and call it I_1. Using a substitution x^n=\cos^2(y),\,dx=-\frac{2}{n}\sin(x)\cos^{\frac{2}{n}-1}(x) we obtain$$ I_1=\frac{2}{n}\int_{0}^{\pi/2}\cos^{\frac{2}{n}-1}(y)\sqrt{1+\cos^2(y)} $$Expanding the square root as a Taylor series allows us to ... 5 As long as there is some neighborhood U of 0 so that f(x) \ne 0 for all x \in U\setminus\{0\}, this is correct - and this assumption just needs to be made so the limit is well-defined. You can see this directly from the definition of the limit. It is pretty easy if you work with sequences. Let x_n be an arbitrary sequence so that x_n \to 0. Then ... 2 For any x,$$\sin x = x + \frac{\cos(\zeta_x)\cdot x^3}{3!},$$for some \zeta_x between 0 and x. Then$$\frac{\sin f(x)}{f(x)} = \frac{f(x) + \frac{\cos(\eta_x)\cdot f(x)^3}{3!}}{f(x)} = 1 + \frac{\cos(\eta_x)\cdot f(x)^2}{3!}$$for some \eta_x between 0 and f(x). As x \to 0, f(x) \to 0 and \eta_x \to 0, so the limit is 1. This does not ... 5 Hint:$$\frac{\Bigl(2+\dfrac{1}{k}\Bigr)^k}{2^k}=\Bigl(1+\dfrac{1}{2k}\Bigr)^k=\sqrt{\Bigl(1+\dfrac{1}{2k}\Bigr)^{2k}}.$$14 Hint:$$\frac{a^k}{b^k}=\left(\frac ab\right)^k,$$and in general,$$\lim_{k\to\infty}\left(1+\frac xk\right)^k=e^x.$$1 By point-slope form of the secant line, we have$$y-y_0=m(x-x_0)$$for any given point (x_0,y_0) on the line and any (x,y) on the line distinct from the given point. Well, one point on the secant line is (a,f(a)), so$$y-f(a)=m(x-a)$\$ is an equation for the line. To calculate the slope, we need two points (which, fortunately, we have), giving us ...

Top 50 recent answers are included