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4

Consider the limit $$\lim_{n\to \infty} \frac{E_n}{n!}\left(\frac{\pi}{2}\right)^n$$ Using $E_{n} \sim \frac{(-1)^{(n-1)/2} 4^{n+1}}{n+1}B_{n+1}$ together with $B_{2n} \sim (-1)^{n+1}4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$ and Stirlings approximation we get \lim_{n\to \infty} \frac{E_{2n-1}}{(2n-1)!}\left(\frac{\pi}{2}\right)^{2n-1} = ... 3 By Abel's Theorem, if the power series converges, then it agrees with the limit of the function. This argues against -2/\pi. Instead, either the power series converges to 0 (by the limit Michael & Ewan give in the other answer) or it does not converge. Some messing around numerically: It looks like the terms of the sum (not the sum itself) ... 1 Leave out 1/x^2, to be reinserted later: \begin{align} \int xe^x\,dx &=\sum_{k=0}^{\infty}\frac{x^{k+2}}{(k+2) k!}\\ &=\sum_{k=0}^{\infty}\frac{(k+1)x^{k+2}}{(k+2)!}\\ &=\sum_{k=2}^{\infty}\frac{(k-1)x^k}{k!}\\ &=\sum_{k=2}^{\infty}\frac{kx^k}{k!}-\sum_{k=2}^{\infty}\frac{x^k}{k!}\\ ... 8 We have\int_0^t e^{\lambda x}dx = \dfrac{e^{\lambda t}-1}{\lambda}$$Differentiating with respect to \lambda, we obtain$$\int_0^t xe^{\lambda x}dx = \dfrac{\lambda t e^{\lambda t} - e^{\lambda t }+1}{\lambda^2}$$Set \lambda = 1, to obtain$$\int_0^t xe^{x}dx = t e^{t} - e^{t }+1$$EDIT To complete your approach, note that$$\sum_{k=0}^{\infty} ...

1

This does not work for all $x,v$. The resulting matrix $I-xv^T$ needs to be invertible. If for instance $x=v= \pmatrix{1\\0\\ \vdots \\ 0 }$, then $I-xv^T$ is not invertible. On the other hand, if $I-xv^T$ is invertible, then the inverse is given by your formula $$(I-xv^T)^{-1} = I + \frac1{1 - x^Tv}x^Tv$$ since $$(I-xv^T)(I + \frac1{1 - x^Tv}xv^T)= ... 4 Using the hint of your prof we get$$\frac1{2^n}\tan\left(\frac x{2^n}\right)=\frac1{2^n}\cot\left(\frac x{2^n}\right)-\frac1{2^{n-1}}\cot\left(\frac x{2^{n-1}}\right)=u_n-u_{n-1}$$and then telescope. 2 To evaluate any infinite series, you need to first look at the partial sum. Use induction and the identity given by your Professor to prove that$$\sum_{n=1}^N \dfrac1{2^n} \tan\left(\dfrac{x}{2^n}\right) = \dfrac1{2^N} \cot\left(\dfrac{x}{2^N}\right)-\cot(x)$$1 When n=0, nc_nx^n is also zero. So the first sum just leaves that zero term out while the second one includes it. There is no change in the sum. 1 If \;z=-\frac i2\; then sequence's general term is$$\frac{(2i)^n}{n}\left(-\frac i2\right)^n=(-1)^n\frac{i^{2n}}n=(-1)^n\frac{(-1)^n}n=\frac1n$$and the series indeed diverges, so yes: it looks your work is fine. 1 For \;|z|>2\; :$$-\frac13\frac1{1+z}+\frac43\frac1{z-2}=-\frac1{3z}\frac1{1+\frac1z}+\frac4{3z}\frac1{1-\frac2z}==-\frac1{3z}\left[\left(1-\frac1z+\frac1{z^2}-\frac1{z^3}+\ldots+\right)-4\left(1+\frac2z+\frac{2^2}{z^2}+\ldots\right)\right]==-\frac1{3z}\left(-3-\frac9z-\frac{15}{z^2}-\ldots\right)=\frac1z+\frac3{z^2}+\frac5{z^3}+\ldots$$... 1 The result is true for a bounded sequence (y_k) i.e.$$\exists M>0,\quad \forall k\;\; |y_k|\le M$$and fails if (y_k) is just bounded below or bounded above. 1 Let x=y-\pi/2, then \cos x=\sin y and \sin x=-\cos y. So$$\sec x+\tan x=\frac{1-\cos y}{\sin y}and take the limit as y\to0 0 Nota Bene: Throughout this answer, I have deviated somewhat from the exact statement of the question as asked by allowing the initial condition f(x_0) = a for arbitrary x_0 \in \Bbb R; x_0 = 0 is of course then seen as a very special case. First of all, the relation 'twixt a and c as given in the text of the question has a small but significant ... 3 Denote \omega as the cube-root of unity. We then have 1+\omega+\omega^2 = 0. This gives us \begin{align} e^x & = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \cdots & \spadesuit\\ e^{\omega x} & = 1 + \dfrac{\omega x}{1!} + \dfrac{\omega^2 x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{\omega x^4}{4!} + ... 0 The number of ways you can color k identical objects with d colors is given by\binom{d+k-1}{d-1}$$To see this consider representing colorings as a string of the form oo|o|ooo|o|o|... where the "o" are the objects and the "|" are the separations between regions where the "o" get different colors. The number of different strings is then clearly given by ... 1 Use$$\begin{align} (1-t)\sum_{k=0}^\infty{d+k-1\choose d-1}t^k&=\sum_{k=0}^\infty\left({d+k-1\choose d-1}-{d+k-2\choose d-1}\right)t^k\\&=\sum_{k=0}^\infty{d+k-2\choose d-2}t^k\end{align}$$and$$ \sum_{k=0}^\infty{k\choose 0}t^k=\sum_{k=0}^\infty t^k=\frac1{1-t}$$0 An alternative approach using binomial coefficients:$$\begin{align} \sum_{i=1}^{N}(1+k)^i&=\sum_{i=1}^N\sum_{r=0}^i \binom ir k^r\\ &=k^0\sum_{i=1}^N\binom i0+\sum_{r=1}^Nk^r\sum_{i=r}^N\binom ir\\ &=N+\sum_{r=1}^Nk^r\binom{N+1}{r+1} \end{align}$$NB - the result is the same as$$\qquad N+\sum_{r=2}^{N+1}k^{r-1}\binom {N+1}r$$2 What am I doing wrong? Nothing, really. You're not using the fastest, least-work, solution, but that doesn't make yours wrong in any way. Well, provided your "et cetera" contains a proof, by induction or otherwise, of$$f^{(n)}(x) = 9e^x(n+x).$$If you merely spotted a pattern and concluded that it will continue without being able to say why it ... 2 The coefficient of x^n is \frac{i^n+(-i)^n}{2n!}. (We don't really need complex numbers here, but that's a convenient way of explcitly describing the coefficient) To ban the forbidden tricks, let us simply not use derivatives at all! We have$$ \sin 1 = 1-\frac1{3!}+\frac1{5!}\mp\ldots>1-\frac1{3!}=\frac56$$because the summands are decreasing in ... 1 Use the binomial theorem:$$(1+x)^n=\sum_{i=1}^n\binom nix^i\implies\;\text{the coefficient is}\;\;\binom nk:=\frac{n!}{k!(n-k)!}$$0 The power series y(x) = \sum_{n=0}^\infty c_n x^n is determined by c_0 and c_1. Plugging it into the equation and comparing the free terms on both sides yields$$ 2c_2=2 $$Hence c_2 = 1, as you got. You probably found it strange that this coefficient does not depend on c_0, c_1. But such is the nature of this nonhomogeneous equation: it forces ... 1 If F : \Omega\rightarrow E is a vector function on an open domain in \mathbb{C} into a Banach space E, then F is holomorphic iff it is weakly holomorphic. In that case (x^{\star}\circ F)'(\lambda)= x^{\star}\circ F'(\lambda) for all x^{\star}\in E^{\star}. Your power series is weakly holomorphic within the radius of convergence, with$$ ...

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If you know the result in the scalar case you can use Hahn-Banach to extend to the Banach case.

0

If you start with $$\log(1+y)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}y^n$$ and replace $y$ by $\frac{x^2}{4}$, you then have $$\log(1+\frac{x^2}{4})=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n~4^n}x^{2n}$$ So, integrating gives $$\int\log(1+\frac{x^2}{4})=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(2n+1)~n~4^n}x^{2n+1}$$ Using the given bounds ...

0

The first term of the expansion seems to be $$\dfrac{x^3}{3\cdot 4}.$$ But in your summation, you write $$\sum_{n=1}^\infty \frac{(-1)^{n+1}\cdot x^{2n+1}}{(2n+1)(n \cdot 2^n)},$$ which has as first term (when $n = 1$) $$\frac{x^3}{3\cdot1\cdot 2}.$$ So it seems like your error is in passing from the expansion to the $n$th term in the expansion.

1

Let $y(t)=\sum_{n=0}^\infty a_nt^n$, $f(t)=\sum_{n=0}^Ff_nt^n$ and $g(t)=\sum_{n=0}^Ga_nt^n$, where $F$ and $G$ are the degree of $f$ and $g$ respectively. Then \begin{align} y''(t)&=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}z^n\\ f(t)\,y'(t)&=\sum_{n=0}^\infty\Bigl(\sum_{i+j=n,\ j\le F}(i+1)a_{i+1}f_j\Bigr)z^n\\ ...

2

There is nowhere a converging series in sight. From the data given it seems that $N$ is large and $$p:=Nk$$ is very small. Therefore we may write $$S={1+k\over k}\bigl((1+k)^N-1\bigr)={1+k\over k}\left(\bigl(1+{p\over N}\bigr)^N-1\right)\doteq{1+k\over k}(e^p-1)\ .$$ If $N$ is not in the thousands use the first few terms of the binomial series: ...

2

Using $$\sum_1^N (1+k)^i = \frac{(k+1)\,((k+1)^N-1)}{k}$$ (which since this is a geometric series is not hard to prove) we get $$\sum_1^N (1+k)^i = \frac{\left(\sum_{j=0}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$ $$= \frac{1 + (N+1)k + \left(\sum_{j=2}^{N+1}\binom{N+1}{j}k^j\right) - (k+1)}{k}$$ $$= N + \sum_{j=2}^{N+1}\binom{N+1}{j}k^{j-1}$$

3

for $k<1$ we can use $$(1+k)^N \approx 1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3$$ thus \begin{align} \sum (1+k)^i &\approx&\frac{(k+1)(1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3-1)}{k}\\ &=&(k+1)\left(1+\frac{N-1}{2}k+\frac{(N-1)(N-2)}{3!}k^2\right) \end{align}

1

Note that $$\frac{1}{z}=\frac{1}{z-4+4}=\frac{1}{4}\frac{1}{\frac{z-4}4+1}$$ By hypothesis $|z-4|<4$, so indeed you want to expand this in positive powers of $z-4$, $$\frac 1 z=\frac 1 4\sum_{\nu \geqslant 0}(-1)^\nu (z-4)^{\nu}$$ This should be also evident from the fact that $1/z$ is holomorphic in $B_4(4)$ so represents the regular part of the ...

0

You need only find infinitely many points where it diverges. You may choose $S$ to be the set containing the infinitely many points that you find. By the wording, it is not necessary to find all points where it diverges. For example, depending on the functions the set could contain all numbers $1/n$ where $n$ is a positive integer.

1

I do not see what is the problem after Olivier Oloa's answer (use the generalized binomial theorem). Doing so, the series expansion you look for is, around $x=a$ $$\frac{1}{(1-x)^{3/2}}=\frac{1}{(1-a)^{3/2}}+\frac{3 (x-a)}{2 (1-a)^{5/2}}+\frac{15 (x-a)^2}{8 (1-a)^{7/2}}+\frac{35 (x-a)^3}{16 (1-a)^{9/2}}+O\left((x-a)^6\right)$$

7

Hint. You may use the generalized binomial theorem $$(1-x)^{\alpha}=\sum_{n=0}^{+\infty}(-1)^n{\alpha \choose n} x^n$$ for $x\in[0;1), \alpha \in\mathbb{R}$ and a change of variable from $x$ near $\pi$ to $x$ near $0$.

0

Why don't you just differentiate $f$ as given?

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$$y(t) = \sum_{n=0}^{\infty}a_n t^n$$ thus \begin{align} y' &=&\sum_{n=1}^{\infty}a_n n t^{n-1}\\ y'' &=&\sum_{n=2}^{\infty}a_n n(n-1) t^{n-2} = \sum_{n=0}^{\infty}a_{n+2} (n+2)(n+1) t^{n} \end{align} thus your equation is $$\sum_{n=0}^{\infty}a_{n+2} (n+2)(n+1)t^n + \sum_{n=1}^{\infty}a_n n ... 0 The power series must be convergent when x satisfies that -R <x+2< R, and be divergent when x <-R and x>R and be unkown when x+2=-R or x+2=R, where 2\le R \le 3. So (a) is convergent; b is divergent; c and d are unknown. 2 I think that one want that you use power series to find f. a) if \displaystyle f(x)=\sum_{n\geq 0} c_n x^n, then you have c_0=f(0)=a. b) You have \displaystyle f^{\prime}(x)=\sum_{n\geq 1} nc_n x^{n-1}. Now \displaystyle f(x)^2=\sum_{n\geq 0} (\sum_{k=0}^n c_kc_{n-k})x^n. So you get that for n\geq 0$$(n+1)c_{n+1}=-\sum_{k=0}^n c_kc_{n-k}$$... 1 I think f(x)=1/(x+1/a)=a/(1+ax) Recall the formula for the geometric series$$\sum_{k=0}^{\infty}y^k=\frac1{1-y}$$1 Hint: In the interesting case where the coefficients are not all 0, we have a geometric series. Added: Fix x, with |x|\lt 1. Then our series is a+ar+ar^2+ar^3+\cdots, where a=c_0+c_1x+c_2x^2+c_3x^3 and r=x^4. 1 What we are looking for is the Taylor series at z=\pi/2 for g(z). The Taylor series is defined as$$ f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n Therefore, for g(z), we have \begin{align} g(z) &= g(\pi/2) + g'(\pi/2)(z - \pi/2) + \frac{g''(\pi/2)}{2!}(z - \pi/2)^2 + \cdots\\ &= 1 + (z - \pi/2) + \frac{1}{2!}(z - \pi/2)^2 + ... 2 Hint: Consider the case a_n=2^n 2 Hint: Apply the root test to the series. If we let a_n be the sequence starting x,\frac{1}2x^2,3x^3,\ldots, then we wish to calculate \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}, knowing that if this is less than 1, the series converges, and if it is greater than one, it diverges. Notice that the coefficients of odd powers are always the greatest so ... 2 There's a formula (Cauchy) for computing the radius of convergence of arbitrary series:\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{k\to\infty}\sqrt[2k]{2k}=\lim_{n\to\infty}\sqrt[n]n=1$$It's rather clear we are only interested in odd terms because of \limsup and \sqrt[n]n\to1 is well known, therefore the radius is 1/1=1. 0 I would recommend splitting the x^k*k and the \frac{x^k}{k} terms and seeing where the two intervals intersect. (∑(2k+1)x^{2k+1}) + (∑\frac{x^{2k}}{2k}) 2 Let u = z - 1. Then$$ g(z) = \frac{\ln(u + 1)}{u^3} $$The series expansion of$$ \ln(u + 1) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}u^n $$Now divide the expanded terms by u^3 and make the substitution u = z - 1. 2 You need the Laurent series for \ln z = (z-1) - \frac{1}{2} (z-1)^2 + \frac{1}{3} (z-1)^3 -\cdots. The rest should be easy. 1 We have for all 0\le a<1$$\frac {a^n}{\ln^2n}=_\infty o\left(\frac1{n^2}\right)$$and the Riemann series \sum\frac1{n^2} is convergent so R\ge1. Moreover for a=1 we have$$\frac1{\ln^2n}\ge\frac1n,\quad \text{for $n$ large enough}$$and the harmonic series \sum \frac1n is divergent so R\le1 hence R=1. 2 The easiest way to look at this problem is to convert the numerator and denominator from the presented sums to ratios using:$$\Sigma_{k=0}^n x^k = {{1 - x^{n+1}} \over {1 - x}}$$Using some standard manipulations, use can then replace your sequence with:$$a_n(x) = {{1-x^{[t_n]+1}} \over {x^{n+1}(1-x^{[t_n]-n})}} = {{x^{[t_n]+1}-1} \over ...

6

This is the binomial identity $\sum_{m=0}^n\binom{m}{j}\binom{n-m}{k-j} = \binom{n+1}{k+1}$ with $j = 1$, $n = a+ b$ and $k = b+1$.

1

Some experimentation gives $\dbinom{a+b+1}{b+2}$. This is the correct answer for $1\le a, b\le 5$.

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