Question

I'm failing to understand the premises of Taylor Series and how they function. (Screenshot, explanation of question below).

Answer 1

If this is the Taylor Series generated at
\[x = a\]How does (x-a) in the formula not equal zero? If x = a, shouldn't the term always cancel out? I don't get it.

Answer 2

your x is your variable ... i could be anything. if x=a, then your f(x) = f(a)

Answer 3

*it

Answer 4

But it says, right there, that x = a.

Answer 5

Is it true or not true that x = a? If it is true, then x - a must equal zero.

Answer 6

Or could someone better explain to me what the last line means, this "by f at x = a" business?

Answer 7

@John_ES

Answer 8

(Regardless of x being variable, x should always cancel out with a the way I understand it, because as it says prior, x = a. Obviously my understanding is wrong, but could somebody actually *point out* how it's wrong, please.)

Answer 9

Obviously, it isn't, because math doesn't just fall apart in Calc II, but clearly my interpretation is wrong. What is incorrect with me saying that since x = a, x - a = 0. Clearly that can't be the case.

Answer 10

No problem, I'm just really frustrated with trying to get this at the moment, if x = a, then x - a must equal zero, and the nth derivative, whatever that may be, divided by n factorial, times that last term (which would always end up being zero to the nth power) would end up being zero.

Answer 11

hmm, we are just beginning to learn taylor series now. sorry i'm no help but from what i know so far, taylor series are just composed of taylor polynomials...so i'm guessing that yes x-a must equal zero.

Answer 12

The idea of a Taylor series of a function f(x) is
to give a series that "re-produces" the values of f(x) as you vary x.

See https://en.wikipedia.org/wiki/Taylor_series
for some examples matching sin(x) and e^x

Generally, f(x) is defined over all x. However, the Taylor series picks a specific x as its "starting point", i.e. x=a, and approximates f(x) *in the vicinity* of x=a
As x gets further away from a, the taylor series approximation needs more terms to "match" f(x).

notice that the series is defined by
\[\sum_{k=0}^{\infty}\frac{ 1 }{ k! }f ^{(k)}(a) \ (x-a)^k\]
the first term, where k=0, we have
\[ \frac{ 1 }{ 0! }f ^{(0)}(a) \ (a-a)^0\]
1/0! (0! is *defined* to be 1) = 1
f^(0)(a) meaning the 0th derivative of f(x)... i.e. the original function
evaluated at x=a
times (a-a)^0 (0^0 is indeterminate, but = 1 in this case)
we see the first term is f(a)

that means with one term of the taylor series we can match f(x) *only at x=a*, at f(b) b≠a, we expect this one term series to not match.. however, as we add more terms to the series, it will do a better job of matching f(x), even as x is far from a.

Answer 13

the x^0 term creates some debate since 0^0 is undefined. I have seen other texts try to clarify the issue by pulling out the first term as a constant so that we are dealing with an affine function.

Answer 14

in essense, a polynomial is the simplest form that can be worked. If we can match a poly that moves and bends and acts just like a more persnickety function then we can develop simpler tools to assess the poly.

\[f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+...\]
\[f'(x) = c_1+2!c_2x+3c_3x^2+4c_4x^3+5c_5x^4+6c_6x^5+...\]
\[f''(x) = 2!c_2+3!c_3x+12c_4x^2+20c_5x^3+30c_6x^4+...\]
\[f'''(x) = 3!c_3+4!x+60c_5x^2+120c_6x^3+...\]
\[f^{(4)}(x) = 4!c_4+5!c_5x+360c_6x^2+...\]
\[f^{(5)}(x) = 5!c_5+6!c_6x+...\]
\[f^{(6)}(x) = 6!c_6+...\]
in order to zero out the non constant terms, for some x=a, we can then shift the poly left or right by (x-a) and solve the system for the coefficients