Taylor's Theorem For Power Series

Taylor's Theorem For Power Series  :

  Statement :  

 

      Let         ∞                                        

                      Σ aₙ xⁿ be a power series with

                    n= 0

 radius of convergence R , and let 

                     ∞

           f(x) = Σ aₙ xⁿ  , |x| < R 

                    n=0

Then for any a∈ ]-R ,R[ , prove that f can be expanded in a power series about 'a' which converges for  |x-a| < R- |a| , and 

              ∞

    f(x) = Σ  f⁽ⁿ⁾(a) (x-a)ⁿ /n! , |x-a|<R-|a|

             n=0

Proof :

        Suppose |x-a|<R-|a|.

Then |x|≤|x-a|+|a|<R and thus Σaₙxⁿ converges .
                         ∞            ∞
   Now , f(x) = Σ aₙxⁿ = Σ aₙ(x-a + a)ⁿ
                        n=0        n=0
                        ∞      n
                     = Σ aₙ Σ   ⁿCₘ aⁿ⁻ᵐ(x-a)ᵐ ......(1)
                       n=0  m=0

We wish to change the order of summation in this expression . To prove its validity we notice that it is the summation by rows of double series , which if convergent absolutely , will converge by columns as well to the same sum .

   Replacing all quantities by their moduli and taking all terms with positive sign , expression on the right of (1) gives
       ∞          n
       Σ |aₙ| Σ ⁿCₘ |aₙ|ⁿ⁻ᵐ|x-a|ᵐ
     n=0      m=0
                       ∞
                    = Σ |aₙ|(|x-a|+|a|)ⁿ
                      n=0

  which is a power series and converges , since |x-a|+|a|<R .

   Hence , for |x-a|<R-|a| , change in the order of summation is justified and so
             ∞       ∞
    f(x)= Σ  {   Σ  ⁿCₘ aₙaⁿ⁻ᵐ} (x-a)ᵐ ......(2)
            m=0 n=m

   This is the required result , but for the coefficients which we shall now express in terms of the values of f and its derivatives at a point .

    We know

     f(x) =a₀+a₁x+a₂x²+...+aₘxᵐ+...+aₙxⁿ+..

∴ f⁽¹⁾(x)=a₁+2a₂x+3a₃x²+..+maₘxᵐ⁻¹+

                                                 ..+naₙxⁿ⁻¹+....

 f⁽²⁾(x)= 2!a₂+3.2a₃x+..+m(m-1)aₘxᵐ⁻²+..

                                      +n(n-1)aₙxⁿ⁻²+...

    ..............................................................

f⁽ᵐ⁾(x)= m!aₘ+ (m+1)m(m-1)..3.2aₘ₊₁x

                        +..+n(n-1)..(n-m+1)aₙxⁿ⁻ᵐ+...

        = m![aₘ+ ᵐ⁺¹Cₘaₘ₊₁x+ᵐ⁺²Cₘaₘ₊₂x²+...]
                 ∞

        = m! Σ  ⁿCₘ aₙxⁿ⁻ᵐ
               n=m
                         ∞
 ∴  f⁽ᵐ⁾(a) = m Σ ⁿCₘaₙaⁿ⁻ᵐ .
                       n=m

Hence from equation(2) ,
                   ∞
         f(x) = Σ  f⁽ᵐ⁾(a) (x-a)ᵐ /m!
                  m=0

                    for |x-a|<R-|a| .(Proved)