I'm trying to figure out how to calculate the power series for exp(exp(x)) using exp(x) and then to write down the first few terms. I have the answer for the terms but I don't know how they arrived at it   Thanks in advance  For the first few terms we have: exp(exp(x)) = e(1 + x + x^2 + . . .)
69806
I need to find terms up to degree $5$ of $e^{e^{z}}$ at $z=0$.    I tried letting $\omega = e^{z} \approx 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots$, and then substituting these first few terms into the Taylor series expansion for $e^{z}$ as follows:     $e^{\omega} = \sum_{n=0}^{\infty}\frac{\omega^{n}}{n!} = 1 + \omega + \frac{\omega^{2}}{2!}+\frac{\omega^{3}}{3!}+\cdots = 1 + (1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots) + \frac{1}{2}\left(1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots \right)^{2}+ \frac{1}{3!}\left( 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots\right)^{3}+\cdots$   But, then when I try to multiply this all out (using Maple), I find that going even further out than I have here, the terms of degree $\leq 5$ do not terminate. So, then I tried putting the command into Maple, and it came out to be something long and horrible with $e$ coefficients for each term.  I must be missing something here.  Could somebody please tell me how to finish this problem? I would ask for a complete solution, not hints or leading questions. I've been working on this for hours and am beyond frustrated. The most edifying thing for me at this point would be to see the correct, full way it's supposed to be done, work backwards, take it apart, and then be able to apply it to other situations.  Thank you.
47213
$\lim n^n/(e^nn!)=0$ using Stirling approximation it is obvious. But can we do it without using Stirling approximation.  Now series with terms $x^n n^n/n!$ has ROC $1/e$. What we can say about behaviour of power series at endpoints?
120834
A few days ago, I was wondering about a series expansion for $$f(x) = \  x^x$$ so I tried taking a couple derivatives. From these I was able to guess at some formulas for the general coefficients of the first few terms of the nth derivative as well as the last term (where I put the terms of the derivative in the order  $$f^{(n)}(x) = c_1x^x + c_2x^{x-1} + c_3x^{x-2}\ + \ ... \ +\ c_nx^{x-(n-1)}$$  Please let me know if there is a general formula for the coefficients as a function of n. In my own fooling around, I found that $$\ln(x)+1$$ appears quite a bit, so I called this $a$ in my work. Since I would like to center my expansion at $1$, and this expression evaluated to $1$, at $x=0$, I was interested in writing my formulas as functions of $n$ and $a$. I found that you get something of the form (this could very well be completely incorrect):  $$f^{(n)}(x) = ax^x + a^{n-2}{\binom n2}x^{x-1} + a^{n-4}(3{\binom n4}-a{\binom n3})x^{x-2}\\ + a^{n-6}(15{\binom n6}-10{\binom n5}a+2{\binom n4}a^2)x^{x-3}\ + \ ... \ +\ (-1)^n(n-2)! \ x^{x-(n-1)}$$  An interesting property that seemed to recur for the various coefficients was  that once I found a recursion formula for the coefficient in question and plugged in my initial value, it always seemed to be the same as taking the integral of the previous coefficient with respect to $a$. That is to say, if I plugged in the nth coefficient (a polynomial of $a$) into my recursion formula for the $(n+1)$th coefficient, this gave me the same result as integrating. In general, as reflected by the few coefficient formulas above,  the formula for the coefficient of $$x^{x-c}$$ where c is an integer such that $$0\le c&lt;n$$ "Stabilized" to a polynomial of a with c terms after the first few applications of the recursion formula. This stabilization occurred when $$n=2c$$ Also, as a direct result of the tentative and incomplete formula I have written above, the first nonzero-value for the coefficient of $$x^{x-c}$$ is $$(-1)^{c+1}(c-1)!$$ and appears in $$f^{(c+1)}(x)$$ If these various properties are indeed true, can anyone explain why this would be the case? I am aware that this function can be expanded using the expansion for $$e^x$$ with $$x\ln(x)$$ in the exponent. This does not strike me as particularly enlightening (let me know if I'm missing something). I am most interested in the coefficients on the binomial coefficients, since these are the only things that stand between me and a general formula for the coefficients (since the form is quite standard). These are: $$1$$ then $$3,-1$$ then $$15, -10,2$$The sum of these coefficients should also give the nth derivative evaluated at zero since $$f(1) = 1$$ and $$a(1) = 1$$  Anyway, please let me know what you think! I am currently stuck due to an inability to brute force my way any further and a complete lack of cleverness. My knowledge of mathematics is quite elementary, so please explain any higher level concepts if they are necessary to explain this problem.
10033
I'd like to determine the function corresponding to the following power series: $$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1}, $$ where $|x|&lt;1$.
105347
Let $$ S = \sum_{n=0}^\infty \frac{1}{n!(n+2)} $$  Integrate the Taylor Series of $xe^x$ to show that S = 1  Also, Differentiate the Taylor series of $\frac{e^x - 1}{x}$ to show that S = 1  For the integration one, I got $\int xe^x dx$ = $x^2\sum_{n=0}^\infty \frac{1}{n!(n+2)} x^n$ and pretty much stuck. I can see the S term but i have no idea how to move on from there
38814
How can one solve the following differential equation   $$f'' + f =0$$   with the usage of power series?  Writing:   $$f(x) = \displaystyle\sum_{k=0}^{\infty} a_k x^k$$  $$f'(x) = \displaystyle\sum_{k=1}^{\infty} a_k k x^{k-1}$$  $$f'' (x) = \displaystyle\sum_{k=2}^{\infty} k (k-1) a_k x^{k-2}$$  How to continue?
29072
Bernoulli stated sum of series of powers as:      I had a doubt in the given formula in the picture! What if $n &lt; p$  i.e. $1^4 + 2^4 + 3^4$  here $n = 3$ and $p = 4$ so outer summation runs ($i = 1$ to $p$) from $1$ to $4$. And we have to find $(n-i)!$ [factorial]. Won't it lead to factorial of negative number when $i &gt; n$ ?
49994
I am trying to see where this relationship comes from: $\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x$ Does anyone have any special knowledge that me and my summer math teacher doesn't know about? He went over generating functions a few days ago and I did not know exactly what he was talking about. In the middle of his lecture he wrote this generating function. He explained that in this case $a_{n} = \frac{1}{n!}$ for $\displaystyle \sum_{n=0}^{\infty} a_{n} x^n$.   Is the relationship a definition or can it be derived using $\displaystyle \sum_{n=0}^{\infty} x^n $?
90142
I would like to evaluate in a closed form the integral  $$\int_{\mathbb{R}}\frac{\exp(-x^2)}{1+x^2}\,\mathrm{d}x$$  I tried various methods :    integration by parts some changes of variables ($y=x^2$, $x=\tan$) residue calculus (but the factor $\exp(-x^2)$ forbid to send the contour to infinity) developping $\exp(-x^2)$ or $\frac{1}{1+x^2} $ in power series (but in both cases one cannot exchange sum and integral)   Does anyone knows if this integral is known, or how to evaluate it ? At least I would like to find an exact expression.  The reason for me to believe that a closed form exists is that this integral arose in a problem of probability where I expect - if I haven't made any mistake previously - a very simple expression.
16750
I need help with how to compute the radius of convergence of the power series $\sum_{n=0}^{\infty} n!x^n$?  I was thinking of using the ratio test but am unsure of how to go about it. Any help would be greatly appreciated, thanks
106805
So I just started learning series, and I am having trouble understanding them. I have a series below:  $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)2^n}$$  Can someone explain to me how I should go about this, from start to finish? I couldn't find any online websites that explained how to find sums, so if you have that, it would be greatly appreciated!
43493
The Taylor expansion of $\exp(-x)$ is $$e^{−𝑥} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!}  ....$$ What I want to prove is that $$ \frac{x^2}{2!} - \frac{x^3}{3!}  .... $$  is greater than or equal to $0$.  I would really appreciate some help.
49871
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.
111831
What i've tried to do is define a formal power series such that $$A(x)=\sum_{n=0}^{\infty}a_n \dfrac{x^n}{n!}$$ Using my recurrence relation I find that $A(x)-1-2x = x(A(x)-1)+ \displaystyle\sum_{n=2}^{\infty}na_{n-2} \dfrac{x^n}{n!}$  I do not know how to proceed after this.
115525
when I try to show that the derivative of exp function is  $$\frac{\partial e^t}{\partial t}=e^t$$ I need to prove firstly that this limit is equal to 1.  $$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$  If you try to use L'Hospital's rule then you need firstly to know what is the derivative of exp function that we want to prove it.  If you want to use approximation of $e^\epsilon \simeq1+\epsilon$ you well found the answer '$1$' but I don't want to use this approximation because it based on the Taylor series that required also to know the derivative of exp function that we don't have it yet   The only way that I can prove it, is with the definition of $e$ number  $$ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e $$ so I ask if there are another way to show this limit.
67178
I'm wondering about how to show compute this series: $$\sum_{n=0}^{\infty}(-1)^{n-1}\binom{1/2}{n}$$ My approach was to use the general formula of the binomial series, which is: $$(1+z)^r=\sum_{k=0}^{+\infty}z^{k}\binom{r}{k}$$ Yet this can't be used because in this case, we have $|z|=1$. Thus, is there any method that I can use for this?
113204
How could the series below be computed ? $$\sum_{k=0}^{\infty}\frac{1}{2^{k!}}$$ It's not a series from a book, but a series I thought of many times, and I didn't manage to figure out what I should do here. I'm just curious to know if there are some known ways for approaching such a series. Thanks!