Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 2

Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of f^n(x) centered at x=c . The general formula for Taylor series is:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...
To apply the definition of Taylor series for the given function f(x) = e^(-4x) centered at x=0 , we list f^n(x) using the derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx) .
Let u =-4x then (du)/(dx)= -4 .
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(-4x) = e^(-4x) *(-4)
= -4e^(-4x)
Applying d/(dx) e^(-4x)= -4e^(-4x) and d/(dx) c*f(x) = c d/(dx) f(x) for each f^n(x) , we get:
f'(x) = d/(dx) e^(-4x)
= -4e^(-4x)
f^2(x) = -4 *d/(dx) e^(-4x)
= -4*(-4e^(-4x))
=16e^(-4x)
f^3(x) = 16*d/(dx) e^(-4x)
= 16*(-4e^(-4x))
=-64e^(-4x)
f^4(x) =- 64*d/(dx) e^(-4x)
= -64*(-4e^(-4x))
=256e^(-4x)
Plug-in x=0 , we get:
f(0) =e^(-4*0) =1
f'(0) =-4e^(-4*0)=-4
f^2(0) =16e^(-4*0)=16
f^3(0) =-64e^(-4*0)=-64
f^4(0) =2564e^(-4*0)=256
Note: e^(-4*0)=e^0 =1 .
Plug-in the values on the formula for Taylor series, we get:
e^(-4x) =sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n
=sum_(n=0)^oo (f^n(0))/(n!) x^n
= 1+(-4)/(1!)x+16/(2!)x^2+(-64)/(3!)x^3+256/(4!)x^4+...
=1- 4/1x +16/(1*2)x^2 - 64/(1*2*3)x^3 +256/(1*2*3*4)x^4 +...
=1- 4x + 16/2x^2 - 64/6x^3 +256/24x^4 +...
= 1-4x+ 8x^2 - 32/3x^3 + 32/3x^4+...
The Taylor series for the given function f(x)=e^(-4x) centered at c=0 will be:
e^(-4x) =1-4x+ 8x^2 - 32/3x^3 + 32/3x^4+...

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