In Taylor series for any function do we have to look for the coefficient C_n and then take it's nth derivatives ?

C_n is the derivative of the initial function at the intended point I believe

divided by n!

yeah, the n! just happens to be a sideeffect of the longhanded version:)

\[f(x) = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4\] \[f'(x) = c_1+2c_2x+3c_3x^2+4c_4x^3\] \[f''(x) =2.3c_2+3.2c_3x+4.3c_4x^2\] \[f'''(x) =3.2.1c_3+4.3.2c_4x \] \[f^{(4)}(x) =4.3.2.1c_4 \]

I know the formula which is f(x) = Summation from n=0 to inf (f^n (a) / n!) *(x-a)^n (where f^n means the nth derivative) but for example if we have to find the taylor series for f(x)= x^4 e^(-3x^2) then we take e^(-3x^2) as as f(a) why not the whole x^4 e^(-3x^2) is taken as f(a) ok because x^4 e^(-3x^2) is equal to f(x) so it means that f(a) is just equals to the coefficient of the power series and therefore in Taylor series we just first go after that coefficient & then we go after the whole formula? I hope that I have explained well actually I am trying to read it since last night ( http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx )

if you have the power series expansion for \[e^{-3x^2}\] then you have it for \[x^4e^{-3x^2}\] by multiplying each term by \[x^4\] i.e. kicking up the exponent by 4 in each term

in the above link I got stuck at example number 3 that why not we have taken the derivative of x^4 and kept it outside and just took the derivative of the exponential, is it because exponential is common in the all sequesnce terms of the series as everytime somehow the exoponential repeats itself therefore it's a coefficient of the power series from whom the Taylor series has been derived?

by expansion do you mean the representation of the power series?

ok in example 4 he is taking the power series around x = -4, i.e. writing the taylor series in powers of (x+4)

since they already have a viable solution to e^x they adapt is as shown and multiply in the x^4 for brevity

I haven't reached at example 4 letme check it please

maybe you were referring to example 3

oh sorry ignore me i misread what you wrote

yes example 3 :|

what amistre said find taylor series, multiply each term by x^4

?????????

from example 1 they determined \[e^x=\sum \frac{x^n }{n!}\] -3x^2 is just inserted in the place of x to adapt it it appears

exactlly

is your question as to why they can adapt it like that?

Why we have not taken the derivative of x^4 why we have kept it outside the Taylor formula ?

because they are taking a shortcut; it can still be done the long way but it takes up alot of realestate

isn't because in taylor series we just take the derivative of that part of the given function which appears as coefficient of power series?

no, it because they are simply avoiding a long and drawn out attempt at finding the results.

thats the stated reason as to why; if there is some other explanation tho I would not know of it :)

OKKKKK so if we take the derivative of the x^4 along exponential then the result will be same as when we take the derivative of alone exponential? as by taking a term out of the series and then putting it back in the series doesn't harm it, isn't?

sounds about right. But what they are doing is more akin to distributing the x^4 they the taylor for e^(-3x^2) say T(x) = taylor expansion for e^{-3x^2} then x^4 T(x) = x^4 (taylor expansion of e^{-3x^2})

bravo

gotcha

thanksssssssssssssssss a tonne

youre welcome :)

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