The Maclauren series of any given function allows us to get an increasingly accurate estimate of that function evaluated at any value within its' domain. What exactly is the purpose, or use, of a Taylor series centered at some nonzero value?

e.g. If I took the Taylor Series of e^x centered at x = 3, what use would that yield me? I only so far see Maclauren series as having any sensible conceptual application.

Question

The Maclauren series of any given function allows us to get an increasingly accurate estimate of that function evaluated at any value within its' domain. What exactly is the purpose, or use, of a Taylor series centered at some nonzero value? 

e.g. If I took the Taylor Series of e^x centered at x = 3, what use would that yield me? I only so far see Maclauren series as having any sensible conceptual application.

anonymous · Answer

centered at 3 simply means values near 3 are more accurately approximated than those that are far away from 3.

anonymous · Answer

if that makes sense at all lol

mendicant_bias · Answer

My phone is freaking out, one moment

mendicant_bias · Answer

For some reason I can't post replies in Firefox, I have to use Lynx, so now I'm incapable of seeing any LaTeX rendered images or anything. So, does that mean that a Maclauren Series will have its most accurate approximations of the function it's imitating near the origin or x = 0, assuming only two variables involved and considering y as a function of x?

anonymous · Answer

not exactly. Does not matter where you want the series to be centered at. What matters is the degree of error and how many terms are needed to approximate a given value with a given degree of error

anonymous · Answer

for example,

anonymous · Answer

say you want to approximate e^3.01

mendicant_bias · Answer

Wouldn't it be best to use a Taylor Series centered t and evaluated at 3.01, then?

anonymous · Answer

Depends on what you mean by best? However, i would say it's better to center at 3. Let me explain the example above

mendicant_bias · Answer

Clarification: If you took the infinite sum of the Taylor Series centered at and evalueated at 3.01, it would best approximate the value of the original function evaluated at 3.01 *as opposed to* centering a taylor series at 3.00 and evaluating it at 3.01 (Sorry, continue, but I really wanna check whether that line of thought is correct)

anonymous · Answer

again, say you want  approximate e^3.01.
if you want to use Maclauren series (which centered at 0), then you will need far more terms to approximate e^3.01 then you do if you center the series at 3 with a given degree of error

mendicant_bias · Answer

Oh, okay, so if we took the infinite sum of either series, we would still get the same answer, but one would more quickly approximate the correct value?

anonymous · Answer

yes, that's how taylor series is defined. Does not matter which one use use, when the number of terms become infinity, the end result is the same

mendicant_bias · Answer

Awesome! Also, I'm about to post another Taylor Series question (no numbers, just concept question) that you might be able to answer.

mendicant_bias · Answer

(Also, why are Maclauren series so special? I mean, *why* zero? It seems entirely arbitrary.)

anonymous · Answer

I don't think there's any thing special about being at centered at 0. As far as I can see, 0 is nice numbers to work with

mendicant_bias · Answer

Yeah. It's nice, I just don't...lol, yeah. Thanks, though. I'll post that next question in a sec.

anonymous · Answer

I would guess because being centered at 0, then it's easier to see the symmetry. If you now how one side behaves, you would now know how the other side behaves as well