28
votes
Why is $ \frac{\pi^2}{12}=\ln(2)$ not true?
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 ...
- 2,175
17
votes
Accepted
Fourier series of $\log(a +b\cos(x))$?
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 ...
- 24.2k
13
votes
Accepted
Bounding higher derivatives of $f(x) = 1/(1+x^2)^r$
Well,
$$
(a^2+x^2)^{-r}=(x+ai)^{-r}\cdot (x-ai)^{-r}.
$$
Differentiate this $k$ times, we get
$$
\left((a^2+x^2)^{-r}\right)^{(k)}=k!\sum_{s=0}^k {-r\choose s}{-r\choose k-s}
(x+ai)^{-r-s}(x-ai)^{-r-(...
- 109k
12
votes
Accepted
Estimating the growth of the Taylor coefficients given the growth of the function at the boundary
The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{...
- 24.2k
11
votes
What's the summation of formal series $\sum_{n\geq0}\binom{n\delta}{n}x^n$？
The Bürmann-Lagrange theorem gives that
$$\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}$$
where $z=z(t)=\sum_{n\geq 1} \frac{1}{n}{n\delta \...
- 3,255
9
votes
Accepted
Is there a bound for Lipschitz constant in terms of second differences?
Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standard
Lemma. If a bounded function $f$: $[0,1]\to \mathbb{R}$ satisfies $f(\...
- 109k
9
votes
A curious series related to the asymptotic behavior of the tetration
The coefficients of the series (6) can be found symbolically in $\lambda$ and rational numbers.
Consider some base $b$ with $1 \le b \le e^{1/e}$ , the fixpoint $t$ for $f(x)=b^x $ and $f(t)=t$...
- 5,342
9
votes
Accepted
Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable
Complex analysis can help. The rate of Taylor coefficients is determined by:
a) the radius of convergence, which is equal to the radius of the largest disk $|z|<r$
where your function is analytic. ...
- 91.8k
9
votes
Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable
Note that $g(1)=g'(1)=1$ and for real $x\in(-1,1)$
\begin{equation*}
g''(x)=\frac1{\pi\sqrt{1-x^2}}.
\end{equation*}
The map $z\mapsto1-z^2$ maps the set
\begin{equation*}
R:=\mathbb C\...
- 128k
9
votes
Accepted
Family of functions with prescribed derivatives
A counterexample:
$$f(z,t):=e^{tz}[1+(e^{|z|^2}-1)h(t)],$$
where $h(t):=e^{-1/|t|}$ for $t\ne0$, with $h(0):=0$.
Then all the assumptions on $f$ hold, but the conclusion
$$|f(z,t)|\le e^{c|z|}\ \;\...
- 128k
9
votes
Accepted
Taylor expansion of Stieltjes Transform
You certainly can't expand around $z=0$ if "$z$ has to be sufficiently large". Expand around $1/z =0$:
$$
\frac{1}{N} \sum_{n=1}^{N}\frac{1}{z-\lambda_{n} } =
\frac{1}{N} \frac{1}{z} \sum_{n=...
- 6,696
8
votes
Accepted
Transformation converting power series to Bernoulli polynomial series
An operator performing the mapping is
$$B(\partial_x) = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1},$$
with $\frac{\partial}{\partial x} = \partial_x$ and $(B.(0))^n=B_n(x)|_{x=0}= (b.)^n =...
- 10.5k
7
votes
Fourier series of $\log(a +b\cos(x))$?
To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following:
The formula implies $a > 0$ and $|b| < a$. To solve the problem, ...
6
votes
Transformation converting power series to Bernoulli polynomial series
Another way, somewhat related with the above answers, is the $p$-adic Volkenborn integral. You can find this, for example, in Schikhof's or in Alain Robert's books on $p$-adic calculus, or Henri Cohen ...
- 3,107
6
votes
Taylor series on a Riemannian manifold
If you fix $x\in M$, you can set up a Riemann normal coordinate chart $(z^i)$ centered at $x$. Then you can just do the usual Taylor series expansion in $z$. Semi-globally, in a normal convex ...
- 21.5k
6
votes
Polynomial approximation for square root function with fast convergence and bounded coefficients
I have a strong impression that something like that has been asked before (perhaps, by somebody else) but it is easier to answer again than to find that old thread.
What you ask for is patently ...
- 61.9k
5
votes
A curious series related to the asymptotic behavior of the tetration
Let $$c_n(\lambda)=\frac{e^\lambda-{^n a}}{ \lambda^n},\quad d_n(\lambda)=e^{-\lambda}c_n(\lambda)$$ ($d_n$ seems to be simpler). In particular initial functions $c_0(\lambda)=e^\lambda-1$ and $d_0(\...
- 12.3k
5
votes
One-Sided Analyticity Condition Guarantees Analytic Function?
The answer to the first question is yes, as explained by Liviu Nicolaescu.
The answer to the second question is no. Consider a bounded analytic function in the strip
$\{z :|\Im z|<2/C\}$ for which ...
- 91.8k
5
votes
Accepted
Other expansion for positive Taylor expansion
I believe it is not possible. Here is an argument for this:
Disclaimer: All inequalities hereafter are meant elementwise
Let's consider a discrete version of this problem $x\in\{0,1,\dots,N-1\}$. Then ...
- 628
4
votes
Accepted
Asymptotic growth of the of Taylor coefficients of the inverse of a function
You can use the Faa-di-Bruno formula, like it is done on page 7 of here
or the proof of claim (b) on page 21 of here.
- 25.3k
4
votes
Accepted
Taylor expansion of determinant of Riemannian metric in normal coordinates up to higher order
After dropping the first order terms using the normal coordinate condition,
$$\partial^4_{ijkl} \det(g) = \partial^3_{ijk} (g^{-1} \partial_l g) = g^{-1} \partial^4_{ijkl} g + ( \partial^2_{ij} g^{-...
- 39k
4
votes
Are there any techniques that can be used in the case when a Neumann series doesn't converge?
Of course, one only has a chance if $1$ is not in the spectrum of $A$.
Robert Israel's answer gives a series that converges to the resolvent $(I-A)^{-1}$ if the spectrum of $A$ is, for instance, ...
- 12.5k
4
votes
Accepted
Are there any techniques that can be used in the case when a Neumann series doesn't converge?
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 $(I-A)^{-1}$.
- 54.2k
4
votes
Taylor expansion theorem for Gateaux differentiable functions
(There should be only one question in one post.)
Anyhow, concerning your first question: There is nothing here really to study, as the Gateaux derivative at a given point in a given direction is just ...
- 128k
3
votes
Accepted
Taylor-like expansion for a holomorphic function in non-simply-connected domain
OK, here goes the "annulus".
Let's say that a domain $\Omega$ has a good approximation property at a point $w\in\Omega$ if for every $\rho>1$ there is a compact set $K=K(\rho)\subset \Omega$ and a ...
- 61.9k
3
votes
Taylor $k$-differentiability of a real function at a point
It's called the “de la Vallée-Poussin derivative” or “Peano derivative” (both terms seem to have been used, sometimes with a difference between them; the latter is perhaps slightly more standard): see ...
- 32.5k
3
votes
Taylor $k$-differentiability of a real function at a point
I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this ...
- 60.5k
3
votes
One-Sided Analyticity Condition Guarantees Analytic Function?
The Taylor formula states that, under your assumptions, for any $x>0$ and any positive integer $n$ we have
$$
f(x)= f(0)+f'(0)x+\frac{1}{2!}x^2+\cdots +\frac{1}{n!} f^{(n)}(0)x^n+R_n(x),
$$
...
- 34.7k
3
votes
What's the summation of formal series $\sum_{n\geq0}\binom{n\delta}{n}x^n$？
Where there is abinomial coefficient in a summation, most of the time the result is a generalized hypergeometric function.
$$S_\delta=\sum_{n=0}^\infty\binom{n\delta}{n}x^n=\sum_{n=0}^\infty \frac{\...
- 1,440
3
votes
Accepted
Taylor expansion of cumulant generating function
Can something similar be said for the remainder of the cumulant
generating function $\log\mathbf E e^{itX}$ with an error bound in
terms of cumulants?
Yes and no.
Your question is whether the ...
- 8,071
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