All Questions
Tagged with taylor-series na.numerical-analysis
7 questions
4
votes
1
answer
225
views
Approximate the square root of (1-X) efficiently through (nested) products
Currently, I encountered a problem of approximating the following
series:
$$
(I-X)^{-\frac{1}{2}}=I+\frac{1}{2}X+\frac{1\cdot3}{2\cdot4}X^{2}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}X^{3}+\ldots
$$
where ...
- 133
2
votes
2
answers
6k
views
Numerical Computation of arcsin and arctan for real numbers [closed]
I'm coding some numerical methods and I do not know what the correct analysis would be for choosing the implementation for $arcsin$ and $arctan$ for real numbers. Here's what I know:
Both functions ...
- 205
2
votes
1
answer
1k
views
Can Convergence Radii of Padé Approximants Always Be Made Infinite?
I've found (as have others), that for some analytic functions, a Padé approximant of it has an infinite convergence radius, whereas its associated Taylor series has a finite convergence radius. $f(x)=...
- 165
2
votes
1
answer
78
views
An algebraic equation question [closed]
My question is this:
If $\frac{\sqrt[n]{\prod_{i=1}^n(p_i + 1)}}{\sqrt[n]{\prod_{i=1}^n(m_i + 1)}} = e ^\beta$
can I find an expression (either exact or approximate) for $\frac{\sqrt[n]{\prod_{i=1}^...
- 163
2
votes
1
answer
478
views
Approximation:- Algorithmic considerations
Hello
I want to approximate a function $f$ on $(a,b)$. The function is singular at the points $a$ and $b$, however I have asymptotic expansions at these points. I can also construct Taylor ...
- 579
0
votes
1
answer
58
views
Integration algorithm and analytic property
This question is the continuation of the previous one.
In the article about the integration of analytical polynomial - time computable function $f(x)$ with the Taylor series $$f(x) = \sum_{n=0}^{\...
- 81
0
votes
0
answers
136
views
Antiderivatives via Taylor series and the FT of Calculus
If $f$ is a real function on an interval $[a,b]$ such that
$f$ is computationally tractable on $[a,b]$: you can calculate $f(x)$ to $n$ bits of precision using an algorithm which is polynomial in $n$ ...
- 827