# Tag Info

25

Maybe I'm too late to be of much use to the original question-asker, but I was surprised to see that all of the previous answers seem to not quite address the real point in this question. While it is important to be aware of the dangers of rearranging conditionally convergent series, it not true that any rearrangement is invalid, in terms of changing the ...

15

Let's consider $$f(x) = \log (1+q^2+2q\cos x) = \log |1+qe^{ix}|^2 ,$$ which differs from your function only by the additive constant $2\log a_1$ if we take $q=a_0/a_1$. Since $|q|<1$, we can use the Taylor series of $\log(1+z)$ to write \begin{align} f(x) = 2\,\textrm{Re}\; \log (1+qe^{ix}) = 2\,\textrm{Re}\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} e^{... 14 Well, (a^2+x^2)^{-r}=(x+ai)^{-r}\cdot (x-ai)^{-r}. $$Differentiate this k times, we get$$ \left((a^2+x^2)^{-r}\right)^{(k)}=k!\sum_{s=0}^k {-r\choose s}{-r\choose k-s} (x+ai)^{-r-s}(x-ai)^{-r-(k-s)}. $$Estimate this by the triangle inequality, you get some constant times (x^2+a^2)^{-r-k/2}. But for a=0 we have equality (because the signs of all ... 11 In your question, the exponent of x_2 is a_2+k. I am assuming that is a typo, and you intended c_a(k) to be the coefficient of$$x_1^{a_1+k} x_2^{a_2} x_3^{a_3} \cdots x_{n-2}^{a_{n-2}} x_{n-1}^{a_{n-1}} x_n^{a_n-k}.$$If so, I can answer all of your questions. It will be convenient to switch the sign in your Vandermonde determinant, so I'll set \... 10 Robert Israel noted that p(n) is a e^a times a polynomial of degree n in a with a zero of order \lceil n/2 \rceil at a=0. We express this polynomial in terms of a Hermite polynomial H_n evaluated at an imaginary argument \alpha := \pm \sqrt{-a}. Start from the generating function$$ \exp(2xt-t^2) = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}, $$... 9 Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standard Lemma. If a bounded function f: [0,1]\to \mathbb{R} satisfies f(\frac{x+y}2)\leqslant \frac{f(x)+f(y)}2, then f is convex. Proof. At first, we prove that f is continuous on (0,1). If not, there exists a c\in (0,1/2)... 8 This is obvious, but the roots of the related polynomial (1+\frac{x}{n})^{n} -1 are of course much more manageable, all being of the form n(\omega -1) for \omega (not necessarily primitive) n-th root of unity. Clearly, 0 is the only root with zero real part, and when n is even, -2n is the only real root apart from 0, while 0 is the only ... 8 The coefficients of the series (6) can be found symbolically in \lambda and rational numbers. Consider some base b with 1 \le b \le e^{1/e} , the fixpoint t for f(x)=b^x  and f(t)=t. For bases b in this interval we have a real fixpoint t with 1 \le t \le e and \lambda = \log(t)  with 0 \le \lambda \le 1 with b^t = t (which is ... 8 The Bürmann-Lagrange theorem gives that$$\sum_{n\geq 0} {n\delta \choose n} t^n = \frac{1}{1-\delta t(1+z)^{\delta -1}}=\frac{1+z}{1+(1-\delta) z}$$where z=z(t)=\sum_{n\geq 1} \frac{1}{n}{n\delta \choose n-1}t^n is the solution of z=t(1+z)^\delta \big(i.e. z(t) is the local inverse at 0 of z \mapsto \frac{z}{(1+z)^\delta}\big). See e.g ... 7 This is follows directly from the Taylor development to order p-1 with integral remainder:$$ f(x) = 0+ \dots + 0+ x^p \int_0^1\frac{(1-t)^{p-1}}{(p-1)!}f^{(p)}(tx)dt. $$Your formulas are differentiated versions of this. By Whitney (Duke Math J. 10, 1943), if f is invariant under the \mathbb Z/(2)-action x\mapsto -x on \mathbb R, then f(x)=g(x^2)... 7 An operator performing the mapping is$$O= D/(e^D-1)=e^{B.(0)D},$$where D=d/dx  and (B.(0))^n=B_n(x)|_{x=0}, since the Bernoulli polynomials are an Appell sequence. Edit (6/20/2017): This operator is essentially the Todd operator. See the discussions on pg. 30 and Appendix B of "Permutohedra, associahedra, and beyond" by Postnikov of the Todd ... 6 Writing e^{-x} = t, your equation is t + t^b = 1. I'll assume 0 < b < 1 (for the case b > 1, write s = t^b and the equation becomes s + s^{1/b} = 1). Now the slightly more general equation t + \epsilon t^b = 1 has a nice series solution$$ t = 1 - \sum_{n=1}^\infty \left( \prod_{j=0}^{n-2} (j-nb)\right) \frac{\epsilon^n}{n!}$$... 6 Another way, somewhat related with the above answers, is the p-adic Volkenborn integral. You can find this, for example, in Schikhof's or in Alain Robert's books on p-adic calculus, or Henri Cohen vol. 2 of his books on number theory. This approach is useful because of the relation of Bernoulli numbers and L-functions: one can easily define good and ... 6 To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following: The formula implies a > 0 and |b| < a. To solve the problem, we reformulate the formula as:$$ \begin{align} r|1 + qe^{ix}|^2 &= r(1 + qe^{ix})(1 + qe^{-ix}) \\ &= r(1 + 2q\cos x + q^2) \\ & = a + b\cos x \\ \...

5

Let $$c_n(\lambda)=\frac{e^\lambda-{^n a}}{ \lambda^n},\quad d_n(\lambda)=e^{-\lambda}c_n(\lambda)$$ ($d_n$ seems to be simpler). In particular initial functions $c_0(\lambda)=e^\lambda-1$ and $d_0(\lambda)=1-e^{-\lambda}\sim\lambda$ (as $\lambda\to 0$) have good power series of the variable $\lambda$. We are intersted in the limits $$c_\lambda=\lim\limits_{... 5 The answer to the first question is yes, as explained by Liviu Nicolaescu. The answer to the second question is no. Consider a bounded analytic function in the strip \{z :|\Im z|<2/C\} for which the boundary of the strip is the natural boundary. Then the radius of convergence of the Taylor series at 0 is 2/C, and it cannot represent the function on ... 4$$\eqalign{G(s) &= \sum_{j=0}^\infty \dfrac{a^j (s - 1)^{2j}}{j!} = \sum_{j=0}^\infty \sum_{n=0}^{2j} {2 j \choose n} \dfrac{(-1)^{2j-n} a^j}{j!} s^n\cr & = \sum_{n=0}^\infty \sum_{j = \lceil n/2 \rceil}^\infty {2 j \choose n} \dfrac{(-1)^{n} a^j}{j!} s^n}$$so your coefficients are$$p(n) = (-1)^n \sum_{j=\lceil n/2 \rceil}^\infty {2j \choose n} \...

4

I am not sure how you are using Rouche's theorem, but if you make $f(z) = e^z,$ while $g(z)=e^z - P_k(z),$ then the minimum of $|e^z|$ on $|z|\leq R$ is $\exp(-R),$ so as long as this is larger than the remainder term, your polynomial has no zeros in the disk (by Rouche). Since the remainder term (the Lagrange formula is the easiest) decreases ...

4

The transfert operator describe an expression or formula of a transform that $\sum_{k=0}^\infty \frac{a_k x^k}{k!}$ into $\sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$ is the p-adic operator such that :The eigenvalues of the p-adic transfer operator are the Bernoulli polynomials,and are associated with the eigenvalues $p^{-n}$ , Try to check this paper by ...

4

You can use the Faa-di-Bruno formula, like it is done on page 7 of here or the proof of claim (b) on page 21 of here.

4

If the spectrum of $A$ is contained in a disk $\{z: |z - a| \le r\}$ where $|1-a| > r$, then the series $\sum_{n=0}^\infty (1-a)^{-1-n} (A - a I)^n$ converges to $(I-A)^{-1}$.

3

Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$ $$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left.\left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) + \left(\frac{\partial}{\partial s}\right)^j \log(\beta-e^s)\right|_{s=0}.$$ Now, $$\log(\alpha-e^s)... 3 \newcommand{\Catalan}{\operatorname{Catalan}} I made an attempt for \int_0^1 S^2\;dx/x, but with limited success. Let$$ q_1 := \frac{1}{16}\left( \ln \left( 1-i \right) {\pi }^{2}+16\,\zeta \left( 3 \right) -4\,i \ln \left( 1+i \right) \pi \,\ln \left( 2 \right) +i{\pi }^{3}+4\,i \ln \left( 1-i \right) \pi \,\ln \left( 2 \right) \\ -2\,\ln \...

3

What is your actual goal? To get a fixed accuracy? Arbitrary accuracy? An interesting method is CORDIC. It involves a limited amount of table lookup, addition and division by 2 (so bit shift if computations are in binary) but no multiplication. You seem willing to store $\pi$ so at least some lookup seems allowed. This is most directly used for $\tan$ or $\... 3 Let me sketch a solution as a three step process: For smooth function$f:M\to G_2$consider its left logarithmic differential$\delta^l f\in \Omega^1(M,\mathfrak g_2)$which satisfies the right Maurer-Cartan equation$d(\delta^l f) +\frac12 [\delta^l f,\delta^lf]=0$. It can be reconstructed on simply connected domains in$M$uniquely up to constant right ... 3 As mentioned already by Denis Serre, there is a rich literature investigating delay equations. If you make an experiment, and fix$\bar{x}=1$, then you see that you need as an initial value the complete past on$[-1,0]$. To play a bit, tak as an initial function the constant function$y(s)=1$for$s\in[-1,0]$. Then you can calculate the solution explicitly ... 3 This is not possible. Suppose e.g. that$e^\beta=2$. Then we have$p+1=2(m+1)$, i.e.$\frac{p}{m}=2+\frac{1}{m}$. Hence$\frac{p}{m}$can attain any real number$>2$, depending on what$m$is. In general the only thing one can say is$\frac{p}{m}>e^\beta$, if$\beta>0$, and$\frac{p}{m}<e^\beta$, if$\beta<0$. 3 It's been noted already in the comments that the problem is still too easy as stated, because one can easily find functions$f(z) = \sum_{n=0}^\infty c_n z^n$such as$f(z) = \sin(1 / (r-z))$, with infinitely many zeros inside the circle of convergence$|z|<r$, and as long as$r>1$the coefficients$c_n\$ decay exponentially, and thus satisfy ...

3

Taylor series are popular only because they are taught in calculus, but usually are quite bad for practical numerical approximation like you want here. Look for Padé approximants, Chebyshev series, or minimax approximations. Also, you should probably specify in which interval you want the approximation to be accurate.

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