Maclaurin series of the error function

[hobz]

Hi. I am trying to figure out what the Maclaurin series of the error function looks like.

 

I use the reference from Wikipedia to check my results, but I cannot seem to get it right.

 

with [math] f(x) = \int_0^x e^{-\frac1{2} t^2}~dt[/math]

 

I would expect [math]f(0) = 0[/math] and [math]f'(0)=-\frac1{2}[/math] and thus the first two terms of the Maclaurin series to be [math]\frac1{1!}x + \left(-\frac1{2}\right) \frac1{2!} x^2[/math]

but this does not seem to be correct.

 

What am I doing wrong?

 

Anyone?

[Bignose]

If you are just looking for the series, you can check out MathWorld's page: http://mathworld.wolfram.com/MaclaurinSeries.html

 

As to the specifics of how to calculate it, what steps did you do? Post some details and we may be able to diagnose where you made your mistakes.

[hobz]

With [math]f(x)=\int_0^x e^{-1/2 t^2} dt[/math] Taylor expanded around a=0 leading to the Maclaurin series.

 

[math]f(0)=\int_0^0 e^{-1/2 t^2} dt[/math] which must be equal to zero because of identical limits I presume.

 

[math]f'(0)= e^{-1/2\cdot 0^2} = 1[/math]

 

[math]f''(0)= -1/2 \cdot e^{-1/2\cdot 0^2} = -1/2[/math]

 

[math]f'''(0)= (-1/2)^2 \cdot e^{-1/2 \cdot 0^2} = (-1/2)^2[/math]

 

And thus I get the first three Maclaurin terms to be [math] f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x + \frac{f'''(0)}{3!}x = 0 + \frac{1}{1!}x + \frac{(-1/2)}{2!}x^2 + \frac{(-1/2)^2}{1!}x^3 [/math] which clearly differs from the series defined at MathWorld.

 

I realize that I exchange all [math]t[/math]'s with the value for [math]x=0[/math], but I have no idea what else to do.

[Bignose]

I think you need to be using the Leibniz rule for differentiating an integral:

 

http://mathworld.wolfram.com/LeibnizIntegralRule.html

 

I'll try to take a look at the problem a little later today...

 

Edited to add: Actually, a pretty neat trick would be to put in the series approximation of e^(-x^2) into the integral first, then do the integral, then do the differentiation.

[uncool]

The series for e^x is

1 + x + x^2/2 + ... + x^n/n! + ...

So the series for e^(-x^2/2) is

1 - x^2/2 + x^4/4 - x^6/(3!2^3) + ... + (-1)^n x^(2n)/(2^n n!) + ...

Therefore, the series for erf(x) is

C + x - x^3/6 + x^5/20 - x^7/(7*3!*2^3) + ... + (-1)^n x^(2n + 1)/(2^n n! 2n+1) + ...

C = 0, so the series is

erf(x) = x - x^3/6 + x^5/20 - x^7/(7*3!*2^3) + ... + (-1)^n x^(2n + 1)/(2^n n! 2n+1) + ...

=Uncool-

[hobz]

Thanks. Just what I needed.

Why is C = 0?

[D H]

with [math] f(x) = \int_0^x e^{-\frac1{2} t^2}~dt[/math]

 

What am I doing wrong?

 

That is not the error function. The standard definition is

 

[math]\mathrm{erf}(x) = \frac 2 {\surd \pi} \int_0^x e^{-t^2}dt[/math]

 

The Maclaurin series for erf(x) as reported at mathworld uses the defintion I reported, not the function in your original post.

 

BTW, wikipedia has the same definition for erf(x) as mathworld.

[hobz]

That is not the error function. The standard definition is

 

[math]\mathrm{erf}(x) = \frac 2 {\surd \pi} \int_0^x e^{-t^2}dt[/math]

 

The Maclaurin series for erf(x) as reported at mathworld uses the defintion I reported, not the function in your original post.

 

BTW, wikipedia has the same definition for erf(x) as mathworld.

 

Yes I am sorry. I did not mean erf(x).

[D H]

[math]f(0)=\int_0^0 e^{-1/2 t^2} dt[/math] which must be equal to zero because of identical limits I presume.

That's correct, but the I presume doesn't sound promising.

[math]f'(0)= e^{-1/2\cdot 0^2} = 1[/math]

That's correct too, but not quite right in formulation. This statement at the bottom of post #3 is at the bottom is what led you astray:

I realize that I exchange all [math]t[/math]'s with the value for [math]x=0[/math], but I have no idea what else to do.

By the fundamental theorem of calculus,

[math]\frac d{dx} \int_a^x f(t)dt = f(x)[/math]

The Leibniz Integral Rule (see Bignose's post) generalizes the concept of differentiation under the integral sign. Applying this here,

[math]f'(x) = \frac d{dx}\int_0^x e^{-\,\frac {t^2} 2} dt = e^{-\,\frac {x^2} 2}[/math]

You don't simply substitute zeros for the ts. You substitute xs, and then set x=0 after performing all intermediate steps (i.e., calculating n derivatives).

A little more explicitly then,

[math]f'(0) = f'(x)|_{x=0} = e^{-\,\frac {x^2} 2}|_{x=0} = 1[/math]

You're shorthand started getting you in trouble here:

[math]f''(0)= -1/2 \cdot e^{-1/2\cdot 0^2} = -1/2[/math]

Everything that follows is off as well. Being explicit,

[math]f''(x) = \frac d{dx}f'(x) = \frac d{dx}e^{-\,\frac {x^2} 2} = e^{-\,\frac {x^2} 2}\frac d{dx}(-\,\frac {x^2} 2) = -xe^{-\,\frac {x^2} 2}[/math]

Thus [math]f''(0) = f''(x)|_{x=0} = 0[/math], not -1/2.

 

It looks like a pattern is developing here! I posit that the n^th derivative of f(x) is a product of a polynomial and f'(x):

[math]f^{(n)}(x) = p_n(x) e^{-\,\frac {x^2} 2}[/math]

This is obviously true for n=1, with p_1(x)=1. If this pattern is true, we should have

[math]f^{(n+1)}(x) = p_{n+1}(x) e^{-\,\frac {x^2} 2}\frac d{dx} f^{(n)}(x) = p_n^{\prime}(x) e^{-\,\frac {x^2} 2} - xp_n(x) e^{-\,\frac {x^2} 2} [/math]

or

[math]p_{n+1}(x) = p_n^{\prime}(x) - xp_n(x)[/math]

which gives a recursive relation for the n^th polynomial.

 

From above, we have [math]f''(x) = -xe^{-\,\frac {x^2} 2}[/math], from which [math]p_2(x) = -x[/math]. The recursive relation gives the same result. The relation holds for n=1 and n=2, and by induction, for all n>=1.

 

So what is [math]f^{(n)}(0)[/math]? Evaluating the general expression [math]f^{(n)}(x) = p_n(x) e^{-\,\frac {x^2} 2}[/math] at x=0 yields [math]f^{(n)}(0) = p_n(0)[/math]. We only need to evaluate the polynomials. The first few are

[math]p_1(x) = 1[/math]

[math]p_2(x) = -x[/math]

[math]p_3(x) = -1 -x(-x) = -1 + x^2[/math]

[math]p_4(x) = 2x -x(-1+x^2) = 3x - x^3[/math]

[math]p_5(x) = 3 - 3x^2 - x(3x-x^3) = 3 - 6x^2 + 3x^4[/math]

from which

[math]f'(0) = p_1(0) = 1[/math]

[math]f''(0) = p_2(0) = 0[/math]

[math]f'''(0) = p_3(0) = -1[/math]

[math]f''''(0) = p_4(0) = 0[/math]

[math]f'''''(0) = p_5(0) = 3[/math]

 

The Maclaurin series is

[math]f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}[/math]

which becomes

[math]f(x) = 0 + x - \frac 1 6 x^3 + \frac 1 {40} x^5 + \cdots[/math]

 

How does this compare to the series at mathworld? Using the standard definition of erf(x), you should be able to derive that

[math]\int_0^x e^{-\,\frac {t^2} 2} dt = \sqrt{\frac{\pi}2} \mathrm{erf}\left(\frac x {\surd 2}\right)[/math]

 

From mathworld,

[math]\mathrm{erf}(x) = \frac 2{\surd\pi} \left(x-\frac 1 3 x^3 + \frac 1 {10} x^5 + \cdots\right)[/math]

Evaluating [math]\mathrm{erf}(x/\surd 2)[/math],

[math]\mathrm{erf}\left(\frac x {\surd 2}\right) = \frac 2{\surd\pi} \left(\frac 1{\surd 2}x-\frac 1 3 \frac 1 {2\surd 2} x^3 + \frac 1 {10} \frac 1 {4\surd 2} x^5 + \cdots\right)[/math]

which simplifies to

[math]\mathrm{erf}\left(\frac x {\surd 2}\right) = \sqrt{\frac 2{\pi}}\left(x-\frac 1 6 x^3 + \frac 1 {40} x^5 + \cdots\right)[/math]

Multiplying by [math] \surd{\pi}/\surd 2[/math] recovers the series for f(x).

[hobz]

Thank you for your explanation! It is very insightful and helped me a lot.

Also thank you for the time you put into explaining this to me.