Question

Proof of the Taylor series?

Answer 1

Why if a point shares all the derivatives with a function does it equate to that function?

Answer 2

taylor series assumes that a polynomial function exists for another

equal function have equal derivatives

Answer 3

Do I take it as a given, then?

Answer 4

the derivatives of a function tell us about how the function moves at a given point
if 2 functions move the same at a given point, then at that point they are equal in all aspects

Answer 5

I'm feeling a little pedantic today, but is there no proof of this (despite it being seemingly evident)? What about non-continuous functions?

Answer 6

taylor series have radiuses and intervals of convergence; such that not every taylor expansion is equal at all points of the functions

Answer 7

im not sure what you mean about non continuous functions in the regard

Answer 8

I meant that if a function jumps weirdly, it may muck up its Taylor (as you said that not every TE matches all points: incidentally, what sort of functions does this happen to?)

Answer 9

*series
*TS

Answer 10

\[f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+...+c_nx^n\]
h(x) = f(x) when all their derivatives are equal
\[h'(x) = c_1+2c_2x+3c_3x^2+4c_4x^3+...+nc_nx^{(n-1)}\]
\[h''(x) = 2c_2+3(2)c_3x+4(3)c_4x^2+...+n(n-1)c_nx^{(n-2)}\]
\[h^{(3)}(x) = 3(2)c_3+4(3)(2)c_4x+...+n(n-1)(n-2)c_nx^{(n-3)}\]
etc...
and from there it goes into solving for the coeffs using methods appropriate for systems of equations

Answer 11

mostly we are concerned with solving for an interval that can be determined

Answer 12

http://www-math.mit.edu/~djk/18_01/chapter06/contents.html
try to find more higher order approximations

Answer 13

knowing where the function and its polynomial converge allows us to use the poly in place of the function to simplify computations

Answer 14

i think the logarithmic and their TS polys have an interval of convergence that is not "at all points"

Answer 15

Excellent, thanks for that- I just wanted to know more about TS without having a meaningful question

Answer 16

:) youre welcome