Can anyone explain to me the difference between the Taylor series of a function and Maclaurin Series of a function.

I thought it might have something to do with how the functions are centered.

Any explanation of these two concepts would be greatly appreciated.

a Maclaurin series is a taylor series centered at 0.

Aha, that is what I thought. Now, I have a question which I have been digging at for the last hour or so...

Find the Maclaurin Series for f(x) = sinh(x) and the radius of conversion. This is how I started...

f(x) = sinh(x) = (e^x - e^-x)/2

Taylor Series for e^x = 1 + x + (x^2)/2! + (x^3)/3! + ...
Taylor Series for e^-x = -1 + x - (x^2)/2! - (x^3)/3! + ...

So f(x) = [(1 + x + (x^2)/2! + (x^3)/3! + ...) - (-1 + x - (x^2)/2! - (x^3)/3! + ...)]/2

Where do I go from here (that is of course if so far this is correct)?? Or is there perhaps an easier way to do this?

Cheers. Ben.

i think you need to be more careful with your minus signs.

What you are doing is fine- now us a little "mathematical common sense" -x to an even power is the same as x to that power so that when you subtract e^(-x) from e^(x) the even powers will cancel. -x to an odd power is negative of x to that power so those terms from e^(-x) and e^(x) will add. Of course, the "2" in the denominator will cancel the "2" you get from summing the odd powers.

In other words: sinh(x)= (e^(x)- e^(-x))/2 is the "odd part" of e^x (and cosh(x) is the "even part").


The MacLaurin series for e^x is 1+ x+ ...+ (1/n!)x^n so the
MacLaurin series for sinh(x) is x+ (1/3!)x^3+...+ (1/(2n+1)!)x^(2n+1)- the odd power terms.

The MacLaurin series for cosh(x)= 1+ (1/2)x^2+ ...+(1(2n(!)x^(2n).- the even power terms.

Of course, both of these have radius of convergence (not radius of "conversion"!) the same as radius of convergence of the MacLaurin series for e^x.

Thankyou lethe and HallsofIvy. May I ask you both what you do for a living (or academically). You two always help me with my problems and you are very good at explaining things to me.

Originally posted by oxymoron
Thankyou lethe and HallsofIvy. May I ask you both what you do for a living (or academically). You two always help me with my problems and you are very good at explaining things to me.

i m a student.

If you wanted to find a Maclaurin Series for say sin(x). I would proceed in the following fashion...

f(x) = sin(x)......f(0) = 0
f'(x) = cos(x).....f'(0) = 1
f''(x) = -sin(x)....f''(0) = 0
f'''(x) = -cos(x)...f'''(0) = -1

This repeats 0, 1, 0, -1,...

Then using the Maclaurin Series I could define sin(x) as...

f(x) = sin(x) = f(0) + f'(0)(x-a)/1! + f''(0)(x-a)^2/2! + f'''(0)(x-a)^3/3! + ...
f(x) = sin(x) = x - x^3/3! + x^5/5! - ...

Now what if I changed a to say pi/4 or something. This would not be a Maclaurin Series because it is not centered at 0. However it would be centered at pi/4. The series turns out to be slightly different.

My question is what is the difference? Is it less accurate to determine the series at something other than 0 (possibly due to convergence/divergence of that function)?

Does it help doing the series at other centers so when you calculate a definite integral these functions are easier to handle?

I just don't understand why we do all the series for sin(x), cos(x), etc... centered at zero.

there is no particular reason why it has to be centered at zero. it just makes the formulas look a little nicer. for some taylor series, in fact you cannot expand around zero. for example log x.

if you want to use a taylor polynomial as a method of approximating a function, it will be more accurate with fewer terms if you choose to expand around a point close to where you will need to know approximate values.

for example, if you want to approximate the value of sin .5001, it makes sense to expand around .5, rather than 0.

O.K. Adding to this, how about a Laurent series for complex functions. You use these expansions to find residues, but fundamentally (I mean mathematically rigorous) is there any difference in the calculation of Laurent series or Taylor series. I mean you take derivatives which are evaluated at a particular point.

Originally posted by ryans
O.K. Adding to this, how about a Laurent series for complex functions. You use these expansions to find residues, but fundamentally (I mean mathematically rigorous) is there any difference in the calculation of Laurent series or Taylor series. I mean you take derivatives which are evaluated at a particular point.

well, taylor series only work for holomorphic functions, and laurents series are needed for more general functions.

Yes, but I can see no difference in the derivation of the 2, i.e. it is irrelevant if the variable is complex or not.