Question

I don't understand why, when evaluating a Maclauren Series, the term to the right of the coefficient doesn't disappear.

I am thinking this is a sort of algebraic workaround that just demonstrates how to get the value of the coefficient in the Taylor Series terms in the first place, but I don't understand why if we earlier set x equal to a in order to get the coefficients for each term why each generalized term doesn't disappear (better explanation in a sec.)

Answer 1

you mean the (x-a)^n term?

Answer 2

Yeah, I closed the question because on my tablet I'm not able to use the LaTeX equation editor, but I'll go for it anyways. Why, if x is set equal to a, does that term not always cancel out, no matter what kind of series it is? I'm guessing it's because the "set x = a" is sort of a "suppose we do this..." approach to finding the coefficient while not actually applying it outside of finding that coeffcient.

I am having a very, very, very hard time ifferetiating between what x and a are in effect becase I constantly see x and a being used almost interchangeably; I'm quite aware a is a constant, I'm quite aware x is a variable, but I don't understand why we selectively evaluate the x found in the derivative part of the coefficient while we don't plug in a number for the x in the (x-a) part.

Answer 3

i.e. in the maclauren series we evaluate the derivative at x = 0, we set a= 0, and then we just.leave the other x alone and don't evaluate it?

Answer 4

so you mean, of all values x can be, why is it that x has to be the value at which the series is centered at? (i.e, if centered at 2, then we have to find f'(2)?)

Answer 5

f''(2), f'''(2)... and so on

Answer 6

No, no, I get why that's the case from what we discussed a minute ago; the value of x that we choose to evaluate with is equal to what it's centered around because that value of x will most quickly (with accuracy) approximate the original function evaluated at the same x value.

What I mean is I literally just don't understand why we choose NOT to evaluate (x-a) by plugging in both x and a; the result would be zero because the two equal each other IF we're seeking the approx. value of the function at the exact same point as a.

Answer 7

e.g. I have a taylor series centered at 3 (so a = 3) and I choose to evaluate it at x = 3. Up until we choose to evaluate it, it totally makes sense to me why we wouldn't plug the value of x into the (x-a) part because up until that point the series is generalized so we could approximate any x-value with enough terms, but hen we choose to approximate the same value as the series is centered about, why doesn't (x-a) become zero?

Answer 8

I believe I see what you meant now. You meant why doesn't the approximation become 0 when we choose to approximate at x = a? (because (x-a) will become 0 and 0 multiply any constant is still 0). But instead, it's not 0

right?

Answer 9

Exactly

Answer 10

if you look back at how the Taylor series is derived, you'll see that it's true that all terms should be 0 for n GREATER than 0. But when n = 0, the convention is that (x-a)^0 = 1 and f^(0)(a) = f(a)

Answer 11

and 0! = 1

Answer 12

Yes, I totally get that, but then I don't understand how the approximation value would ever change beyond n = 0; my question still stands. The first term would be one, and then everything else would be zero, so how do we still get accurate approximations that aren't just equal to 1? If you had a maclauren series and plugged in x and a in (x-a) where x = a, you would get zero beyond the first term, and every concievable maclauren series would just be equal to 1.

Answer 13

no, it's not 1. It's the value of the function at x = a, name f(a) when n = 0

Answer 14

Is there any way you could write out an example of this in the eqn editor and I could switch browsers to see the pics and then wsitch back to post here? I want to say thanks so much in the first place for stcicking around and helping despite the tech issues and the time, and I would much appreciate knocking out this last confusion.

Answer 15

also, wouldn't (x-a)^0 in this situation equal 0^0? I could undersand if it was a constant, but isn't there a big issue with 0^0 = 1?

Answer 16

not in Taylor series. This is because of the assumption that a function CAN be approximated as.. (hold on picture coming)

Answer 17

it might take a while so hold on...

Answer 18

no problem

Answer 19

\[f(x) = C _{0} + C_{1}(x-a) + C_{2}(x-a)^2 + C_{3}(x-a)^3+...\]

Answer 20

when x = a, f(a) = C_0, and C_0 is the value of the function

Answer 21

the constant is determined by (pic coming)

Answer 22

\[c_n =\frac{f ^{(n)}(a) }{ n! }\]

Answer 23

so when n = 0, (picture coming)

Answer 24

\[C_0 = \frac{ f^0(x) }{ 0! }\]

Answer 25

when x = a (picture coming again lol)

Answer 26

\[C_0 = \frac{ f^0(a) }{ 0! }\]

Answer 27

when x = a, f(a) = C_0, that is to say C_0 *is* the value of the function at x = a, that's why when we evaluate (x-a)^n, (x=a, n=0) (x-a)^0 has to be DEFINED for C_0 to exist. and we define (0^0) in Taylor series to be 1

Answer 28

so when we evaluate at x = a (pic coming)

Answer 29

\[C_0 = \frac{ f^0(x) }{ 0! } (a - a)^0 = \frac{ f(a) }{ 1 } (1) = f(a)\]

Answer 30

which is precisely the value of the function. Not any other values. I hope this helps

Answer 31

I think I just figured out my confusion.

Answer 32

did my answer clear that confusion? lol

Answer 33

Partly, I'm having a hell of a time with my tablet trying to explain why but give me a moment, but yeah, it did, one sec, retyping answer.

Answer 34

Say we have a Taylor Series centered about a = 0 (Maclauren, derp) and we evaluate it at x = 0. In every instance of whatever the Taylor Series is, all terms but the first will always cancel out, e.g.

Answer 35

\[f(x) = e^{x} = 1 + x + \frac{x^{2}}{2!}+ \frac{x^{3}}{3!} ... + \frac{x^{n}}{n!} = \]
(My tablet can'trender the siga, but I did find a workaround to using the LaTeX editor)

Answer 36

1 = f(0) = e^0

Answer 37

If evaluated at x = 0,

\[f(0) = e^{0} = 1 + (0) + \frac{(0)^{2}}{2!} + \frac{(0)^{3}}{3!} ... + \frac{(0)^{n}}{n!} = 1\]

Answer 38

Similarly, any Taylor Series evaluated at x = a will lead to all terms cancelling except for the first. Correct?

I think my confusion was just a matter of thinking about how taylor series are supposed to be made; I think that I thought (lol) that setting x = a for the sole purpose of finding the respective coefficients of each term was something more than that that later the series *had* to be evaluated at x = a but all it really was was conjecture to get that coefficient.

After the general formula of a Taylor or Taylor/Maclauren Series has been found, it can be evaluated at x = a, but it does not *have* to be evaluated at x = a.

Answer 39

that's correct

Answer 40

yes, the part where we set x = a to find the constants is merely just a hypothesis.
That is, if x = a, then this is how you find the constants (see the formula)

If we were to set x to be a different value, then i'm not even sure if we can even find a general formula for the constants. Even if we did, I would suppose the formula is very complicated.

Answer 41

Thanks so much! This has helped so, so, so much, hah. Taylor Series has been something I've struggled with for a while.

Answer 42

Gladly! :)

Answer 43

r u in Cal2 btw?