Taylor’s Theorem and Taylor series

Taylor’s theorem is used for the expansion of the infinite series such as sin x, log x etc. so that we can approximate the values of these functions or polynomials. Taylor’s theorem is used for approximation of k-time differentiable function.

Let the (n-1)th derivative of f i.e.f^{(n-1)} be continuous in [a, a+h], f^n the nth derivative exist in (a, a+h) and p be a given positive integer. Then there exists at least one number \theta lying between 0 and 1 such that:
f(a+h)= f(a)+h\frac{f'(a)}{1!}+h^2\frac{f''(a)}{2!}+….. +\frac{h^{(n-1)}f^{(n-1)}(a)}{(n-1)!}+R_{n}

where R_{n}=\frac{h^n(1-\theta)^{n-p}}{(n-1)!p}f^{n}(a+\theta h) and  0<\theta<1
Putting x=a+h or h=x-a we write equation as:
f(x)=f(a)+(x-a)\frac{f'(a)}{1!}+\frac{(x-a)^2}{2!} f''(a)+….. +\frac{(x-a)^{n-1}}{(n-1)!}f^{n-1}(a)+
Taylor’s remainders Rn after n terms due to:
1. Cauchy: we just put p=1 in the Taylor’s theorem to get R_{n}=\frac{h^n(1-\theta)^{n-1}}{(n-1)!}f^{n}(a+\theta h)
2. Lagrange: p=n gives R_{n}=\frac{h^n}{n!}f^n(a+\theta h)

Taylor’s formula :
Using Lagrange’s remainder we get the Taylor’s formula:
\frac{(x-a)^n}{n!}f^n(a+\theta(x-a)) where  0<\theta<1
As n →∞ if R→0 then the last term of the formula becomes
Therefore the Taylor’s formula further reduces to
This formula is now used to give the infinite series expansion of f(x) about point a.

Obtain the Taylor’s series expansion of
f(x)= x^5 + 2x^4 - x^2 + x + 1
about the point x= -1.

According to the formula we have a= -1 here and f(x) is provided to us. First of all we need to calculate f(a) and then we calculate derivatives of f(x) at given point until it becomes zero.
f(-1) = -1+2-1-1+1 = 0
f'(x) = 5x^4+8x^3-2x+1, f(-1) = 5-8+2+1 = 0
f''(x) = 20x^3+24x^2-2, f''(-1) = -20+24-2=2
f'''(x) = 60x^2+48x, f'''(-1) = 60-48=12
f''''(x) = 120x+48, f''''(-1) = -120+48=72
f'''''(x)=120, f'''''(-1) = 120
Now we stop here as the next derivative will be zero. f^n(x)

=0 for n>5 

Thus the Taylor series expansion of f(x) about x= -1 is:
f(x)= f(a)+\frac{(x-a)}{2!}f'(a)+\frac{(x-a)^2}{3!}f''(                                      a)+\frac{(x-a)^3}{3!}f'''(a)+…..

Substituting the values as calculated by us we get
f(x)= 0+0+(x+1)^2.\frac{2}{2!}+12\frac{(x+1)^3}{3!}-72\frac{(x+1)^4}{4!}+120\frac{(x+1)^5}{5!}
f(x)= (x+1)^2+2(x+1)^3-3(x+1)^4+(x+1)^5

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