taylor series question

[theoriginal169]

f(x + x[math]_{0}[/math])=[math]\sum[/math]1/n! x[math]_{0}[/math]^n(d/dx)^n f(x)

 

i saw this at my book but didnt understand .

[the tree]

What didn't you understand about it?

[ajb]

The Taylor series of an infinitely differentiable function [math]f(x)[/math] in the neighbourhood of a real number [math]a[/math] (complex also works) is

 

[math] f(a) + (x-a) f'(x)|_{x = a} + \frac{1}{2!}(x-a)^{2} f''(x)|_{x = a} + \frac{1}{3!}(x-a)^{3}f'''(x)|_{x = a} + \cdots .[/math]

 

Hope that is a little clearer.

[the tree]

Often, a is chosen to be 0, for simplicity's sake.

[ajb]

Often, a is chosen to be 0, for simplicity's sake.

 

Of course this depends on exactly what you are doing, in the case [math]a=0[/math] the series is also known as the Maclaurin series.

[theoriginal169]

The Taylor series of an infinitely differentiable function [math]f(x)[/math] in the neighbourhood of a real number [math]a[/math] (complex also works) is

 

[math] f(a) + (x-a) f'(x)|_{x = a} + \frac{1}{2!}(x-a)^{2} f''(x)|_{x = a} + \frac{1}{3!}(x-a)^{3}f'''(x)|_{x = a} + \cdots .[/math]

 

Hope that is a little clearer.

 

at my question a is x x-a is x0 why? what changes in f(x+x0)

[ajb]

Your original expression reads

 

[math]f(x + x_{0}) = f(x) + x_{0}f'(x) + \frac{1}{2}(x_{0})^{2} f''(x) + \cdots[/math].

 

This is just the Taylor expansion about "small" [math]x_{0}[/math].