- #1
jbowers9
- 89
- 1
I've been reading a paper at the following link:
www.cims.nyu.edu/~eve2/reg_pert.pdf
I have several questions:
In the first example they use the method to approximate the roots for
x^2 - 1 = "epsilon" x
I was under the impression - wrongly perhaps - that f(x) had to have continuous derivatives to use the Taylor series. When you go to f'''(a), the coefficient c3 vanishes. And, strictly speaking isn't this a Maclaurin series?
In the problem for x^2 -1 = "epsilon" e^x
The coefficient X2 for x = -1 is 3/8e^2.
I'm sure of this since I went through it several times.
I don't understand what the second figure - Figure 2 - is referring to at all (why two)??
And what "3" solutions? Doesn't a quadratic have 2 solutions? What is the solid line in the figure a plot of anyway?
I apologize that I can't post it verbatim but I'm unable to cut and paste from an Adobe file to this format and I'm really having problems using the LaTex editor. I'd appreciate any commentary. Thank you.
John
www.cims.nyu.edu/~eve2/reg_pert.pdf
I have several questions:
In the first example they use the method to approximate the roots for
x^2 - 1 = "epsilon" x
I was under the impression - wrongly perhaps - that f(x) had to have continuous derivatives to use the Taylor series. When you go to f'''(a), the coefficient c3 vanishes. And, strictly speaking isn't this a Maclaurin series?
In the problem for x^2 -1 = "epsilon" e^x
The coefficient X2 for x = -1 is 3/8e^2.
I'm sure of this since I went through it several times.
I don't understand what the second figure - Figure 2 - is referring to at all (why two)??
And what "3" solutions? Doesn't a quadratic have 2 solutions? What is the solid line in the figure a plot of anyway?
I apologize that I can't post it verbatim but I'm unable to cut and paste from an Adobe file to this format and I'm really having problems using the LaTex editor. I'd appreciate any commentary. Thank you.
John