# Tag Info

1

This means that the radius of convergence of the given power series is infinite, we have $$\left|\sum_{n=0}^{\infty}\frac{n^{1000}}{\sqrt{n!}}x^n\right|<\infty$$ for any complex value of $x$.

4

Your mistake $$f'(0)=2 \quad \text{and} \quad f''(0)=4$$ Another way to proceed: if you know the series expansion of both $e^x$ and $\sin(2x)$: $$e^x=1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots$$ $$\sin(2x)=2x-\frac {4x^3}3+\cdots$$ Then $$e^x\sin(2x)=\left(1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\right)\left(2x-\frac {4x^3}3+\cdots\right)$$ $$=2x-\frac {4x^3}3+... 1 Based on your work above f''(0) = 4 check this. Where did the x^1 term go? f(x) = \frac {0}{0!} + \frac {2}{1!}x + \frac {4}{2!}x^2 + \frac {-2}{3!} x^3 \cdots I would say 3^{rd} degree polynomial, and not "grade." But that would be correct, the highest exponent of x = 3. And, the highest derivative you would need is 3. What about the ... 1 You miscalculated the first and second order terms, remember that sin(0) = 0 and  cos(0) = 1. 1 Yes, it's false as others have noted. I think that there's a more closely related claim, namely that if$$ Df(a) \ne 0 $$and f is C^1, then$$ Df(x) \ne 0 $$for all x in some neighborhood of a, i.e., I've switched inequality for equality. This fact is useful in the proofs of the inverse and implicit function theorems. 1 Yes, the claim is false, and your example disproves the claim. Another, somewhat simpler, counterexample is$$ f(x) = {1 \over 2} x^2. $$The closest-sounding claim to yours that is true (and that I know of) is the statement that if an analytic function of a complex variable is zero in a neighborhood of a point in its connected domain, then the function is ... 3 Note that X has the form X(x,y) = \varphi(x,y) Y(x,y) with \varphi(x,y) = \frac{1}{x^2+y^2} and Y(x,y) = (x,y) and the solutions to the system \dot{\alpha}(t) = Y(\alpha(t)) are radial line segments. Hence, the solutions of \dot{\gamma}(t) = X(\gamma(t)) will also be radial line segments with a different parametrization. Fix p = (x_0,y_0) \neq (... 1 We may compute the whole Taylor series from the Weierstrass product:$$ \cos(x)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \tag{1}$$and the Taylor series of \log(1-x), leading to:$$ \log\cos(x) = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\,\zeta(2m)}{m\,\pi^{2m}}\,x^{2m}.\tag{2}$$Since ... 1 For the derivatives one get:$$f'(x_0)=-\tan(x)f''(x_0)=-\frac{1}{\cos^2(x)}$$(Use the chainrule with u=\log(x), v=\cos(x)) Then you get:$$T_{f,0,2}(x)=\sum\limits_{j=1}^2 \frac{f^{(j)}(0)}{n!}\cdot x^nT_{f,0,2}(x)=0-\tan(0)(x-0)-\frac{1}{2}x^2T_{f,0,2}(x)=-\frac{1}{2}x^2$$0 Hint for the case k=1: If c \in \mathbb R and \mu is a finite real Borel measure on [0,1], then$$\varphi(f) = cf(0) + \int_0^1 f'\, d\mu$$is a bounded linear functional on C^1([0,1]). 0 Here is a graph: (Large version) 1 The ratio test will prove that the series is absolutely convergent if the absolute value of the result is less than 1. This is true because the series will eventually grow smaller than an arbitrary convergent geometric series. Therefore, since we know that series a and b converges with |x|<r_a,r_b respectively, we also know that as lim_{k>\infty},... 1 In general$$x\in f^{-1}(X)\iff f(x)\in X\tag1$$Observe that x\in X\implies f(x)\in f(X)\implies x\in f^{-1}(f(X)) so it is always true that:$$X\subseteq f^{-1}(f(X))\tag2$$Essential is the information f^{-1}(f(X))\subseteq X for every X\subseteq A. Applying this on X=\{a\} where a\in A we get f^{-1}\left(f\left(\left\{ a\right\} \right)\... 0 The general idea is great, except I'd be careful about the last \iff, since the converse isn't necessarily true. It's safer to use a \Longrightarrow, since that's all we really need. Indeed, if c = 7 and \square = 1, then:$$ 13 < x + 1 < 15 \implies -15 < x + 1 < 15 \iff |x + 1| < 15 $$But the converse is not true, since if x = 4, ... 0 You have picked \delta>0 such that \delta=-c+\sqrt{c^2+\varepsilon} This is a problem. Let c=-5. so \delta=5+\sqrt{25+\varepsilon} substitute \varepsilon=0.0001. So \delta is approximately equal to 10. This is where the problem lies. So x can range from -15 to +5. Let x=0. So x^2 =0. But 0 does not into (25-\varepsilon,25+\varepsilon) or (... 3 The brakes reduce it's velocity by 10 m/s in every second. So the acceleration is -10 \, \text{m/s}^2. So$$y(t)=v-10\,\text{ms}^{-2}tx(t)=100\,\text{ms}^{-1}t+\frac{1}{2}(-10\,\text{ms}^{-2})t^2$$0 Following notation and terminology as in Terence Tao's "Analysis I", as it seems this question is taken from Exercise 6.1.8 (p. 132) where the reader is asked to prove Theorem 6.1.19 (p. 130). We need to show that for any \epsilon > 0, the sequence \left\{\max(a_n, b_n)\right\}_{n=m}^\infty is eventually \epsilon-close to \max(x, y). Let \... 1 The summand is equivalent to: \frac{1}{m} \cdot [\frac{1}{n}-\frac{1}{m+n}] Assuming finite integer m, we can begin telescoping when we are at the m+1th term. Hence this is just an expression for the mth harmonic number. (If you are unfamiliar with telescoping series: https://en.m.wikipedia.org/wiki/Telescoping_series) 0 I think there is a simpler way to do this by introducing some more machinery. Let f_n=\sum_{m=1}^\infty\frac{m}{2^n}\chi_{E_{n,m}}. Show that f_n\rightarrow f pointwise and then that f_n is monotonically increasing. Then apply the Lebesgue monotone convergence theorem. 1 We want there to be three appropriate solutions to$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 2y = 0$$Consequently, we will have y = 0 and so$$\frac{\partial f}{\partial x} = \frac{dp}{dx} = 0$$at each critical point. Since there are three critical points of p, it will have a degree of at least 4. Lastly, we need to ensure that ... 1 Since f_y=2y and f_x=p^{\prime}(x), you need to find p(x) so that p^{\prime}(x)=0 for 3 values of x, and since D=f_{xx}f_{yy}-(f_{xy})^2=2p^{\prime\prime}(x), you also need p^{\prime\prime}(x)<0 at two of the critical points and p^{\prime\prime}(x)>0 at the other critical point. 2 The main term goes to zero so fast that any crude comparison gives absolute convergence. Moreover, the value of the series is simple to compute:$$ \sum_{n\geq 0}\frac{n^2+1}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 0}\frac{n(n-1)+n}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 1}\frac{1}{(n-1)!}+\sum_{n\geq 2}\frac{1}{(n-2)!} = \color{red}{3e}.$$1 You know that you can use \frac 0 0 and \frac \infty \infty in L'Hôpital's rule. Thus, if you have 0\cdot\infty where f(k)\rightarrow0 and g(k)\rightarrow\infty, you can rewrite \lim_{k\rightarrow c}{f(k)\cdot g(k)} as \lim_{k\rightarrow c}{\frac {f(k)}{\frac 1 {g(k)}}}. Now the problem is no longer in the form 0\cdot \infty but \frac 0 0, ... 10 Just to provide a concrete underpinning to this general reasoning:$$\sum_{n=0}^{\infty}\frac{1}{n!}x^n=e^x$$Differentiate and then multiply by x:$$\sum_{n=0}^{\infty}\frac{n}{n!}x^n=xe^x$$Differentiate and then multiply by x:$$\sum_{n=0}^{\infty}\frac{n^2}{n!}x^n=x(x+1)e^x$$Adding the first and last equations:$$\sum_{n=0}^{\infty}\frac{n^2+1}{n!}x^...

1

The limit of a rational function (quotient of two polynomials) at $\infty$ is the limit of the ratio of the highest degree terms, namely $$\lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{n^{2}}{n^{3}}=\lim_{n\to\infty}\frac{1}{n}=0.$$

1

HINT Observe that $$\lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{\frac{1}{n}+\frac{2}{n^2}+\frac{2}{n^3} }{ 1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}}=\frac{0}{1}=0$$

1

You made a mistake at the final step to make a conclusion. :) Since $\lim_{n\rightarrow \infty}\left |\frac{a_{n+1}}{a_{n}} \right |=0$ then the series converges absolutely. Also, we don't know the value of the sum so it is still wrong to say it converges absolutely to $0$! The ratio test just tells us that the limit $$\lim_{m \to \infty} \sum_{n=0}^{m}... 2 Because the question is not in indeterminate form ie it's not 0/0 7 l'Hopital's rule only works for the limits "0/0" and "\infty/\infty". To compute this limit try this:$$\frac{\sin x+\cos x}x=\frac{\sin x}x+\frac{\cos x}x2 You can rewrite the product as a quotient, but you have to do this before you do the derivatives. So if you've got a limit \lim_{x\to x_0}(f(x)g(x)) and f(x)\to 0 and g(x)\to\infty, then you can for example decide to move g to the denominator giving f(x)/(g(x))^{-1}. And now both numerator and denominator go to zero, thus if also the other ... 3 Hint: The binomial formula with the Cauchy product \begin{align*} (x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k} \end{align*} does not use falling factorials x^{\underline{k}} resp. y^{\underline{n-k}}. Here is a step by step answer similar to that by @MarkoRiedel. It's convenient to use the coefficient of operator [z^k] to denote the coefficient ... 1 I am not sure about the geometric interpretation, but this might be a way to find your D^2. Similarly to your intuition, think of D^2f(x)(v,w) as the derivative ofx\mapsto Df(x)v=:G_v(x)$$in direction w. Then you can easily check by hand that the two definitions agree. Of course this easily generalizes to maps f:\mathbb{R}^n\rightarrow \mathbb{R}... 3 If g(x) is a differentiable function, the derivative of g(x)e^{-x} is given by \left(g'(x)-g(x)\right)e^{-x}. If g(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!} we have that g'(x)-g(x) is a monomial and an even function, hence f(x)=g(x)e^{-x} has a unique stationary point at the origin and is a decreasing function. Since \lim_{x\to +\infty}g(x)e^{-x}=0 ... 2 Let f(x) = e^{-x} (1 + x + x^2/2) then$$f'(x) = -\frac{1}{2}x^2e^{-x}$$via the product rule, this shows that (since x^2 and e^{-x} are non-negative for all real x) that f'(x) \leq 0  and so f is a decreasing function. It is obviously continuous and tends to (positive) infinity as x \to -\infty and tends to 0 as x \to \infty. 0 Here is some information pertaining to P(j,s): I get the impression that if this function is used at all for anything the values that matter s\in(0,1). Now for the actual functions: P(1,s)=-1 P(2,s)=-s^6+s^5+2 s^4-2 s^2-2 s+1 P(3,s)=-s^{30}+s^{29}+2 s^{28}-3 s^{25}-3 s^{24}-4 s^{23}+s^{22}+6 s^{21}+7 s^{20}+7 s^{19}+2 s^{18}-4 s^{17}-13 s^{16}-7 ... 8 Expand the cube and divide top and bottom by n to get$$\begin{align*}\lim_{n \to \infty} \frac{n\sqrt{n} +3n + 5\sqrt{n} + 2}{n + \sin n} &= \lim_{n \to \infty} \frac{\sqrt{n} + 3 + \frac{5}{\sqrt{n}} + \frac{2}{n}}{1 + \frac{\sin n}{n}} \end{align*}$$from which you can see the numerator diverges off to \infty whilst the denominator tends to 1. ... 1 To answer your question you should consider what L. Hospitals rule says. I will break up the theorem to two parts: condition and conclusion. I will highlight conditions only. Conditions: 1.  g(x) and f(x) should be continuously differentiable in the open neighborhood of the real number a. 2. g'(x) should not be zero for all values of x in the ... 3 Consider the integral$$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$Since$$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$and$$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$we have$$\...

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That's a fairly standard rewrite. If you need to find something like $$\lim_{x\to a} f(x)g(x)$$ where $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=\infty$, you can define $h(x)=\frac{1}{g(x)}$ and consider $\lim_{x\to a} \frac{f(x)}{h(x)}$. You should be able to use the form of l'Hôpital you know on that. Of course that requires $g$ to be well-behaved ...

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\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{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 3 Using Negative Binomial Coefficients and Vandermonde's Identity, we get \begin{align} \sum_{k=0}^n\binom{x+n-k-1}{n-k}\binom{y+k-1}{k} &=\sum_{k=0}^n(-1)^{n-k}\binom{-x}{n-k}(-1)^k\binom{-y}{k}\tag{1}\\ &=(-1)^n\sum_{k=0}^n\binom{-x}{n-k}\binom{-y}{k}\tag{2}\\ &=(-1)^n\binom{-x-y}{n}\tag{3}\\ &=\binom{n+x+y-1}{n}\tag{4} \end{align} ... 2 We have $$\sum_{k=0}^n {y-1+k\choose k} {x-1+n-k\choose n-k} = \sum_{k=0}^n [z^{k}] \frac{1}{(1-z)^y} [z^{n-k}] \frac{1}{(1-z)^x} \\ = [z^n] \frac{1}{(1-z)^y} \frac{1}{(1-z)^x} = [z^n] \frac{1}{(1-z)^{x+y}} ={x+y-1+n\choose n}.$$ -1 Geometric Solution: Assume that the based field is\mathbb{K}$, which is of characteristic$0$. Prove that the equality holds when$x$and$y$are integers (for example, via combinatorial arguments). Now,$\mathbb{Z}^2$(as well as$\mathbb{N}^2$) is a Zariski-dense subset of$\mathbb{K}^2$. Therefore, the equality holds for all$(x,y)\in\mathbb{K}^2$, ... -1 The multiset numbers$(\!\tbinom{m}{r}\!)=\binom{m+r-1}{r}$count the multisets of cardinality$r$with elements drawn from a set of size$m$. Using the multiset numbers, the identity becomes: $$\left(\!\tbinom{x+y}{n}\!\right)=\sum_{k=0}^n \left(\!\tbinom{x}{n-k}\!\right) \left(\!\tbinom{y}{k}\!\right)$$ This has more or less the same combinatorial ... 1 Hint: consider$g:R^2\times R\rightarrow R^3$defined by$g(x,y,z)=(f(x,y),z)$shows that the rank of the differential is 3 and deduce that it is a local diffeomorphism by using the local inversion theorem, then consider the composition of$g$with the projection (which is an open map) on$R^2$which is$f$. 3 If$A$is bounded above there is a real number$M$so that if$x>M$then$x \in A^{c}$. Therefore no$N$exists so that if$x>N$then$ x\not\in N$( since$\max(N+1,M+1)\in A^c$). So$A^c$is not bounded above. 5 No, you can't do it like that: when you have a product where one factor has limit$\infty$and the other has limit$0$, you cannot apply a theorem on products of limits. The theorem helps when both limits are finite and you just multiply them; it also works when one limit is$\infty$(or$-\infty$) and the other one is either infinite or *finite and not$0$”... 0 Definition of upper bound k, states that$k \ge s$for all s in S. A supremum is an upper bound. So$u $is upper bound of S. If$n $is a natural number then$n >0$. Therefore$1/n >0$and therefore$u + 1/n > u \ge s $for all s in S. So$u+ 1/n $is an upper bound. A supremum is the least upper bound and by definition if$k < u $then$k $... 8 $$n-n\sqrt{1-\frac{5}{n}}=\frac{(n-n\sqrt{1-\frac{5}{n}})(n+n\sqrt{1-\frac{5}{n}})}{n+n\sqrt{1-\frac{5}{n}}}=\frac{5}{1+\sqrt{1-\frac{5}{n}}}\rightarrow \frac{5}{2}, \space \text{as} \space \space n \rightarrow \infty.$$ 16 In short: no, you did not do it correctly. The reason is that$\infty\cdot 0$is one of the indeterminate forms, you cannot conclude it's equal to zero. The rationale being: it can be anything. For instance,$\$\begin{align} \underbrace{n}_{\to \infty}\cdot \underbrace{\frac{1}{n}}_{\to 0} &\xrightarrow[n\to\infty]{} 1\\ \underbrace{n^2}_{\to \infty}\...

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