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Taylor’s Theorem and Taylor series
  • Difficulty Level : Hard
  • Last Updated : 03 May, 2020

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.

Statement:
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)+
\frac{(x-a)^n(1-\theta)^{n-p}}{(n-1)!p}f(a+\theta(x-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:
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)+
\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
\frac{(x-a)^n}{n!}f^n(a)
Therefore the Taylor’s formula further reduces to
f(x)=f(a)+\sum\frac{f^n(a)}{n!}(x-a)^n
This formula is now used to give the infinite series expansion of f(x) about point a.

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



Explanation:
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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