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0

As Julian Aguirre pointed, we are not allowed to use a comparison with an alternating series. However, Dirichlet's test is enough to ensure convergence, since the partial sums of $(-1)^n$ are bounded and $\frac{(n+a)}{(n+b)(n+c)}$ is eventually decreasing to zero as $n\to +\infty$. We may notice that: ...


6

For any continuous function $f$, if $\lim_{n\to\infty} a_n = L$ and $L$ is in the domain of $f$, then $$\lim_{n\to\infty} f(a_n) = f(L)$$


3

Your argument is incorrect. The comparison principle is for series of non-negative terms. The series $$ \sum_{n=2}^\infty\frac{(-1)^n}{\sqrt n+(-1)^n} $$ is alternating. Leibniz's test cannot be used because $1/(\sqrt n+(-1)^n)$ is not decreasing; in fact, the series diverges. But $$ \lim_{n\to\infty}\frac{\dfrac{1}{\sqrt n+(-1)^n}}{\dfrac{1}{\sqrt ...


3

Since if $b\neq c$ then there exists $A,B\in\mathbb{R}$ such that $$\frac{n+a}{(n+b) (n+c)}=\frac{A}{n+b} +\frac{B}{n+c} $$ the series is convergent by Leibniz test. If $b=c$ then $$\frac{n+a}{(n+b) (n+c)}=\frac{1}{n+b} +\frac{a-b}{(n+b)^2}$$ and the series is also convergent by Leibniz test.


1

Some new references [1] F. Ben Addaa J. Cresson, About Non-differentiable Functions, Journal of Mathematical Analysis and Applications, 263, (2), 2001, 721-737 [2] F. Ben Addaa J. Cresson, Fractional differential equations and the Schrödinger equation, Applied Mathematics and Computation, 161 (1), 2005, 323–345 [3] F. Ben Addaa J. Cresson, Quantum ...


2

No, the vanishing of the second derivative is a necessary condition for $c$ to be an inflection point (provided we are assuming $f''(c)$ is defined.) This condition is not sufficient, as can be seen with $c=0$ and $f(x) = x^4$.


0

I'm 14, and I started learning calculus when I was 12. From my experience, stay away from Khan Academy until later. I would recommend 2 books for you: Calculus Made Simple and after you have mastered that read Unified Calculus. How long it takes is up to you. Calculus is a broad subject. You can learn "calculus" in a day but to truly appreciate, understand, ...


1

What you linked does not really apply to taylor series. Taylor series have to converge in a disc on the complex plane. For example if you need $\log 3$ and you expand $\log x$ around $1$, it will only be valid in $[0,2]$ since log blows up at 0. Or consider $\sqrt x$ This is the $50^{th}$ order taylor expansion about $0.1$: Or consider $\frac{1}{1+x^2}$. ...


0

An equation like $$a^x-b^x=c$$ usually doesn't have closed-form solutions. Anyway, using the substitution $t^n=b^x$, it becomes $$t^{n\ln a/\ln b}-t^n=t^m-t^n=c.$$ For $n,m\in\{1,2,3,4\}$, you get a polynomial and analytical solutions are possible, though sometimes quite complicated. For instance, $$8^x-4^x=9$$yields by the Cardano formula ...


0

Question 1: you can not exchange limit and sum in this case because sum is bound by $n$ that is the index of limit. Question 2: in this case, sum is on index other than $n$. Or $\lim \limits_{n\to\infty}\sum \limits_{k=1}^{m}y_n^{(k)}=\sum \limits_{k=1}^{m}\lim \limits_{n\to\infty}y_n^{(k)}$ Sum must be finite in order to exchange order of $n$ and $k$.


1

The infinity symbol $\infty$ has at least a dozen different meanings according to context. In each context, one must guess which kind of infinity is intended by the author, if the author has not explicitly stated the meaning. In this case, I guess that what you are talking about is the elements $\infty$ and $-\infty$ of the extended real number system, which ...


2

Perhaps I don't get it right, but here is a try to simplify your question. $(x+y+3)^2>0$, hence $ \dfrac{-1}{(x+y+3)^2}$ is maximum when $(x+y+3)^2$ is too. But very clearly on $[-1,1]\times[-1,1]$, $|x+y|<3$. Thus $(x+y+3)^2$ is maximum on $[-1,1]\times[-1,1]$ when $(x+y+3)$ is maximum on $[0,1]\times[0,1]$ This is clearly for $x=y=1$ You have ...


0

Hint: Consider for $n\ge1$, $$ a_n=2^{-n} $$ and $$ b_n=\left\{\begin{array}{} 3^{-(n+1)/2}&\text{if $n$ is odd}\\ 2\cdot5^{-n/2}&\text{if $n$ is even} \end{array}\right. $$ The discrete analog of L'Hôpital would be Stolz-Cesáro. See this question. What is suggested in the current question would be the analog of "if the limit of the ratio of two ...


1

How about calculating loan interests when taking out a loan at a bank? I would not consider that as something connected to science or engineering...


2

This symbol is used to indicate a line integral along a closed loop. if the loop is the boundary of a compact region $\Omega$ we use also the symbol $ \int_{\delta \Omega} $ we can generalize such notation to the boundary of a region in an n-dimensional space and, if $\Omega$ is an orientable manifold we have the generalized Stokes' theorem $$ \int_{\delta ...


2

Yes, so long as the infimum and supremum are understood to be taken in the extended reals.


0

The direction is irrelevant, except technically you will get a negative answer if you start at the outside and work inwards. The absolute value should be the same.


1

i think it is easier to do the shell method. break the region into thin strips of width $dx$ and length $(2-x^2) - x.$ if you rotate about the $y$-axis, then you cylindrical shell of radius $x.$ putting all these together, we have $$V = 2\pi\int_0^1 x\left(2-x^2 - x\right)\, dx $$ if you insist on the disk method, then break it into two solids: (a) ...


0

look at the first expression in you answer: you have $$\begin{align}\sin\left(2\sin^{-1}(2x)\right) &= \sin(2t)\\ & = 2\sin t \cos t \\ &=2 \times 2x\times \sqrt{1 -\sin^2 t}\\ &=4x\sqrt{1-4x^2}\end{align}$$ i have used $t = \sin^{-1}(2x), \sin t = 2x, \cos t = \sqrt{1-4x^2}.$


1

You need to appeal to an identity at the end, before using the triangle: $\int (1-4x^{2})^{1(/2)}dx=\int \cos ^{2}zdz=\int (\frac{1}{2}+\frac{1}{2}\cos 2z)dz=\frac{1}{2}z-\frac{1}{4}\sin 2z=\frac{1}{2}(z)-\frac{1}{4}(2\sin z\cos z)=\frac{1}{2}\arcsin x+\frac{1}{2}x(1-x^{2})^{1/2}+C$


0

hint: let $u = \sqrt{1-4x^2}, v = x$, can you proceed with integration by parts ?


0

You cannot find the primitive of solely $e^{x^3}$, simply because it doesn't have any. You can calculate it in terms of other integrals, but if you want to calculate $\int x^3e^{x^3} dx$, then I agree with kmitov.


2

Substitute $t=x^3$. Then $dt=3x^2dx$ and $x^2dx=\frac{dt}{3}$ You have $\int \frac{1}{3}\int e^t dt=\frac{1}{3}e^t+C=\frac{1}{3}e^{x^3}+C$


1

Yes, you have to find an antiderivative with respect to something. Since it is unspecified, the antiderivative of $\;\frac{\mathrm d y}{\mathrm d x}\;$ will implicitly be made with respect to the same variable of derivation. $x$. $$\int \dfrac{\mathrm d y}{\mathrm d x}\mathrm d x = y+c$$ What else can you do?


-1

$\frac{dy}{dx}$ is a function of $x$, so lets write it that way $\frac{dy}{dx}(x)$. Furthermore, lets change notation a little more. Lets write this as $y'(x)$. Then the antiderivative is: $$\int y'(x) dx$$ If we wanted to keep the original notation: $$\int \frac{dy}{dx}(x) dx$$ which should make sense, right?


0

Hint: For instance, try to visualize it here: https://graphsketch.com/ Add in the formula box one specific function, e.g. in "Enter graph equations", like $\frac{3x}{x-1}$, so you can see how it looks like, The horizontal asymptote is calculated by doing the limit of $f(x)$ when $x \to \infty$, so in your case is always f(x)=3 independently of $q$ (a ...


0

The Weierstrass-Caratheodory formulation asserts that $f\colon E\rightarrow\mathbb{R}$ is differentiable at a point $x_0\in E$ if and only if there exists a function $\phi$ continuous at $x_0$ such that $$f(x)=f(x_0)+\phi(x)(x-x_0)$$ and the derivative is given by $f^{\prime}(x_0)=\phi(x_0)$. Take $f(x)=\sin(x)$, then suppose we have the following ...


1

Since you want a physical interpretation of the differential elements, I'll assume you know what mechanical work is. The differential elements are not really infinitesimals. That may have been how Newton and Leibniz thought about them, but in standard analysis today they are just linear functions. Think about it this way. Say you have a constant force ...


1

You want: $$\frac{A(x+3)+B}{(x+3)^2}=\frac{3-2x}{(x+3)^2}$$ You lost the denominator on the right hand side. Much, then, to solve: $$A(x+3)+B = 3-2x$$


-1

$m=f'(2) = \dfrac{g'(2)h(2) - g(2)h'(2)}{(h'(2))^2}= \dfrac{5\cdot 6-18\cdot 2}{2^2}= -\dfrac{3}{2}.$, and also $f(2) = \dfrac{g(2)}{h(2)} = \dfrac{18}{6} = 3$. The slope of the normal line is $m' = -\dfrac{1}{m} = \dfrac{2}{3}$, thus the normal equation is: $y - 3 = \dfrac{2}{3}\left(x-2\right)$


1

First you get the derivative of $f(x) = e^{2x}$ which is $f^\prime(x) = 2e^{2x}$ by the chain rule. Now the line is tangent to $f(x)$ at $(1,e^2)$ then they have the same slope. Now the slope of $f(x)$ at $(1,e^2)$ is $f^\prime(1) = 2e^{2}$. and now you should use the formula $$m = \frac{y_2 - y_1}{x_2 - x_1}$$, Your $m$ is the slope that you found which is ...


0

Since $$\lim_{n \to \infty} \frac{(-3)^{n-1}}{n^5} = \infty \neq 0,$$ the series diverges by the $n$th term test.


1

The general formula for a function $f$ at $(x_0,y_0)$ is: $$y=f(x_0)+f'(x_0)(x-x_0).$$


0

If you wanted to check the answer you get from the Lagrange multipliers, you could look for "extrema" along the boundary of the region by looking at the map: $$F: \mathbb R \rightarrow \mathbb R^{2} \rightarrow \mathbb R$$ $$F(t) = f(\gamma(t)) = f(\sqrt{2} \cdot cos(t), \sqrt{2} \cdot sin(t)), t \in [0, 2\pi]$$ Which is a function on one variable relating ...


-1

Looks good, I have just a few comments: "So the extras are the points..." at this point we don't know which of these is an extremum. These are possible extreme points. It is only after we plug in all the values that we know which are extrema. "$f$ has a maximum and minimum" I would phrase as "$f$ achieves its maximum and minimum"


0

$lim_{n \to \infty}$ $C/n^2=0$ <=> $C*lim_{n\to\infty}1/n^2=0$ <=> $lim_{n\to\infty} 1/n^2=0$ Given any $\epsilon \gt 0$, there is some N such that for any $n \gt N$,∣∣$1/n^2−0$∣∣$\lt\epsilon$. $∣∣1/n^2−0∣∣\lt\epsilon <=> ∣∣1/n^2∣∣\lt\epsilon <=> 1/n^2\lt\epsilon <=> n^2 > 1/\epsilon <=> n \gt \sqrt{1/\epsilon} $ Let ...


1

Look at the definition they have given. You want $\ell=0$ in this case. Given any $\epsilon>0$, you have to find some $N$ such that for any $n>N,\; \left|\frac C{n^2} - \ell\right|<\epsilon$. See if you can work from there.


1

You want to prove that $(C/n^2)\to 0$ as $n\to\infty$. You are required to use the definition that: $$ \lim_{n\to\infty}a_n = l \quad\iff\quad \forall \epsilon>0:\exists N:\forall n>N: \big(\lvert a_n-l\lvert < \epsilon \big) $$ Since $a_n=C/n^2$ and $l=0$ you must show that $$ \lim_{n\to\infty}\frac C{n^2} = 0 \quad\iff\quad \forall ...


0

Danielle, First, the presence of the constant $C$ is immaterial to the limit. You can factor is out and reduce the question to $C \lim_{n \to \infty} \dfrac{1}{n^2} = 0 \implies \lim_{n \to \infty} \dfrac{1}{n^2} = 0$.Then your $|a_n - \ell|$ is fairly straightforward, as $\ell = 0$ means that $|a_n - 0| = a_n$. Choose such an $\epsilon = \dfrac{1}{N} > ...


0

Consider how you would write a general equation of a disk in $\mathbb C$. Suppose it has centre $a\in \mathbb C$ and radius $r\in \mathbb R$. The one thing we know about points in this disk is that the distance between those points and the centre is less than to $r$. Let $z\in\mathbb C$ be an arbitrary point in the complex plane. Think of the vector from $a$ ...


0

$D$ is derivative wrt $t$ $$\left( \begin{array}{cc} D & 0 \\ 0 & D \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ \end{array} \right)-\left( \begin{array}{cc} -11 & 15 \\ -30 & 31 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ \end{array} \right)=\left( \begin{array}{c} 18 e^t \\ 30 e^t \\ \end{array} \right)$$ ...


0

Let $$ f_a(x)=\frac{1}{\sqrt{|x|}} \chi_{[-a,a]} $$ Observe that $f_a \in L^1$ but not in $L^2$. Now just compute the fourier transform and play with $a$. \begin{align} \hat{f_a}(y) &= \int_{-a}^a \frac{1}{\sqrt{|x|}}e^{-iyx} \, dx \\ &=\int_0^a\frac{1}{\sqrt{x}}e^{-iyx}+\frac{1}{\sqrt{x}}e^{iyx} \,dx \\ & =2 \int_0^{\sqrt{a}}e^{-iys^2} + ...


0

Consider all $z\in \mathbb{C}$ with $|z|<5$. Thats the set of all points with distance less than five from the origin, so it's the open disc around zero of radius 5. By now going to $z+1$ you shift the entire space so that -1 gets mapped to zero while preserving relative distances. So now the disk is centred around -1.


1

There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps. Notice $$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right] = \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right] $$ and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have $$\mathcal{I} ...


2

Observe that $$\int_0^x\sin\bigg(\frac{\pi}{t}\bigg)dt = \int_0^x {t^2 \over \pi}\bigg({\pi \over t^2}\sin\bigg(\frac{\pi}{t}\bigg)\bigg)dt$$ Integrating the right-hand side by parts gives $$-{x^2 \over \pi}\cos\bigg({\pi \over x}\bigg) + \int_0^x {2t \over \pi} \cos\bigg({\pi \over t}\bigg)\,dt$$ So $${\int_0^x\sin(\frac{\pi}{t})dt \over x} = -{x \over ...


0

The change of variables $t= 1/s$ gives $$(1)\,\,\,\,\frac{1}{x}\int_{1/x}^\infty \frac{\sin \pi s}{s^2}\, ds.$$ Integrate by parts to see (1) equals $$ \frac{1}{x}\left [ x^2 \cos (\pi/x)/\pi - 2\int_{1/x}^\infty \frac{\cos \pi s}{\pi s^3}\,ds\right ]$$ In absolute value, the last integrand is no larger than $1/(\pi s^3),$ which is simple to integrate. It ...


0

Let's start from hint $(2)$, and take the natural log of both sides:- $$\sum_{r=1}^n\log\left(\cos\left(\frac{\theta}{2^r}\right)\right)=\log(\sin \theta)-\log\left(\sin\left(\frac{\theta}{2^n}\right)\right)-\log(2^n)$$ Differentiating both sides, we use the identity that $\frac{d}{d\theta}\log(f(\theta))=\frac{f'(\theta)}{f(\theta)}$. For the left hand ...


2

We may just exploit: $$\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sin\left(\frac{\pi}{t}\right)\,dt=\int_{n}^{n+1}\frac{\sin(\pi u)}{u^2}\,du=\frac{2(-1)^n}{\pi n^2}+O\left(\frac{1}{n^3}\right) $$ that follows from integration by parts.


3

Conveniently, your integration limits are $(0,4)$ so the inside absolute value can be ignored. You are left with the following: $$\int_0^4 |x^2 - 2x| \ dx = \int_0^4x\cdot|x-2|\ dx$$ $$= \int_0^2 x\cdot(2-x)\ dx + \int_2^4x\cdot(x-2) \ dx$$ I leave the rest to you.


1

Another bijection from $(0,1)$ to $\mathbb{R}$ to is given by $\displaystyle f(x)=\frac{2x-1}{x^2-x}$.



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