# Conceptual: Are all MacLaurin Series = to their Power Series?

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1. Apr 15, 2015

1. The problem statement, all variables and given/known data
To rephrase the question, given a power series representation for a function, like ex , and its MacLaurin Series, when I expand the two there's no difference between the two, but my question is: Is this true for all functions? Or does the Radius of Convergence have to do with anything?

On many questions in my Calculus course I am asked to find MacLaurin Polynomial for certain functions, so I don't understand why I should go to all the trouble and find their derivatives and use the definition of a Taylor/MacLaurin Series to find the MacLaurin/Taylor Polynomial, when I can just manipulate the power series and smack down the Nth amount of terms from the power series representation.

2. Apr 15, 2015

### Staff: Mentor

The MacLaurin series is, by definition, the power series expanded at x=0.
If you know this power series already you are done, but using the final answer as solution step is probably not the intention of those homework problems.

The series can be different from the expansion around other points. ln(x+2) is an example where the terms depend on the point of the expansion.

3. Apr 15, 2015

Hi, Thanks for the response this clears thing up a bit. When you said:
I'm a little confused, so it would be correct for me to say that this means for ln(x+2) depending on where I choose to centre the Series, say I want a Taylor Series for ln(x+2) centred at x=5, then the terms of the Taylor Series would be different than if I wanted ln(x+2) the Taylor series, centred at x=0 (i.e. MacLaurin). So I have two questions:

Does this mean that if I centre a Taylor Polynomial somewhere other than x=0, the Taylor Polynomial no longer equals the power series?

And, what role exactly does the Radius of convergence play, if I had a finite radius of convergence say for 1/(1-x) it's power series has a radius of convergence of 1, what role exactly, if any, does that play on the MacLaurin Series?

4. Apr 15, 2015

### LCKurtz

To answer your first question, the Taylor series of a function may converge for all $x$ and not equal the function except at the expansion point. The function given by $f(x) = e^{-\frac 1 {x^2}}$ for $x\ne 0$ and defined as $f(0)=0$ has derivatives of all orders $f^{(n)}(0)=0$, so its Taylor expansion is identically $0$, only equaling the function at $x=0$.

Even in cases where the Taylor series equals the function, that can only happen where the series converges. So if the radius of convergence is finite, the series diverges outside that range and can't represent the function or anything else.

5. Apr 15, 2015

### Tarpie

Hey I don't mean to hijack but let me just add to this inquiry.

If a function is equal to its mclaurin series and the mclaurin series has a radius of convergence of infinity, then the function is equal to to that series at all X right?

ex equal to it's mclaurin series ∑Xn/n! for every X.

But this can also be written centred at another value, say, 2, as ∑e2/n! ⋅ (x-2)^n

So here we have two different power series representations for ex, one centred at 0 and the other centred at 2.

Now understandably a function will be more accurately estimated using a taylor series about the point you're interested in, but if the mclaurin series has a radius of infinity then
will expanding that function about a different point other than 0 actually provide a better estimate?

Or is this expansion about different points only required with taylor(mclaurin) series with a finite interval of convergence.

I don't know how clear this question is but to summarize: if a fxn is equal to it's mclaurin series for all x∈ℝ (b/c the remainder approached 0 and the interval of convergence of the series itself is (-inf,inf) ), then does taylor expansion about a point other than 0 give a closer estimate about certain point than the mclaurin at 0?

6. Apr 15, 2015

### Ray Vickson

Expansion about a different point can lead to a significantly faster rate of convergence, hence be better for numerical evaluation. For example, if you want the values of $e^x$ for $x$ near 10, you would get faster convergence (needing fewer terms for a given level of accuracy) if you expanded about 10 instead of about 0. (However, in this case the coefficients would involve $e^{10}$, which you would somehow need to know. One possibility would be to take $e^{10} = 1/e^{-10}$ and to spend a lot of time and effort getting an accurate estimate of $e^{-10}$, using the expansion about $x = 0$.)

7. Apr 16, 2015

### Staff: Mentor

You can calculate those expansions and see how they look like. They will all converge down to x=-2 if I remember correctly, but the upper limit of convergence changes so clearly you get different things.

@Ray Vickson: If you expand around x=0, the convergence at x=-10 is horrible - the absolute deviation is similar to the one at x=10, but you take the inverse value of something small and you need ~35 terms just to get a value that stops changing its sign - more if you want a result with a meaningful precision. Better calculate e as precisely as possible and multiply it with itself until you reach e10.