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+...`
No comments:
Post a Comment