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Taylor Maclaurin Binomial Power Series

Find the power and taylor series representation for the function, evaluate the limits using maclaurin series & differentiation and integration of power series calculus.

Representation of Function as Power Series

  • Certain types of function can be expressed as sum of power series by manipulating geometric series or by differentiation or integration of such a series.

\(\dfrac{1}{1-x} = 1+x+x^2+x^3 + .......= \sum\limits_{n=0}^\infty x^n\)    if  \(|x| <1\)

 

Illustration Questions

Find power series representation of the function  \(f(x) = \dfrac{x}{1+x^2}\) .

A \(f(x) = \sum (-1)^n \,x^{2n}\)

B \(f(x) = \sum\limits_{n=0}^\infty \,x^{2n}\)

C \(f(x) = \sum\limits_{n=0}^\infty \,x^{2n+1}\)

D \(f(x) = \sum\limits_{n=0}^\infty (-1)^n \,x^{2n+1}\)

×

\(\dfrac{1}{1-x} = 1+x+x^2+x^3 + .......= \sum\limits_{n=0}^\infty x^n\)    if  \(|x| <1\) 

 

\( f(x) = \dfrac{x}{1+x^2}\\ = x\left[\dfrac{1}{1-(-x^2)}\right]\)

\(= x\,[1-x^2+x^4-x^6+x^8.....]\)

\( = x\sum\limits_{n=0}^\infty (-1)^n \,x^{2n}\)

\(= \sum\limits_{n=0}^\infty x^{2n+1} × (-1)^n\)

\(\therefore f(x) = \sum\limits_{n=0}^\infty (-1)^n \,x^{2n+1}\)

Find power series representation of the function  \(f(x) = \dfrac{x}{1+x^2}\) .

A

\(f(x) = \sum (-1)^n \,x^{2n}\)

.

B

\(f(x) = \sum\limits_{n=0}^\infty \,x^{2n}\)

C

\(f(x) = \sum\limits_{n=0}^\infty \,x^{2n+1}\)

D

\(f(x) = \sum\limits_{n=0}^\infty (-1)^n \,x^{2n+1}\)

Option D is Correct

Differentiation and Integration of Power Series  

\(f(x) = \sum\limits_{n=0}^\infty C_n(x-a)^n = C_0 + C_1(x-a) + C_2 (x-a)^2 +.....\)

is differentiable on the interval \((a-R, a+R)\) if power series  \(\sum\limits_{n=0}^\infty C_n(x-a)^n\) has radius of convergence R>0;  We say that 

  • \(f'(x) = C_1 +2C_2 (x-a) + 3C_3(x-a)^2 + .....= \sum\limits_{n=1}^\infty n \,C_n (x-a)^{n-1} \)  .........(1) 

       and 

  • \(\int f(x) \,dx = C+ C_0 (x-a) +C_1 \dfrac{(x-a)^2}{2}+ C_2 \dfrac{(x-a)^3}{3}+.....\)

            \(= C+ \sum\limits_{n=0}^{\infty} C_n \dfrac{(x-a)^{n+1}}{n+1}\)    .........(2)

  • The radius of convergence of power series of both (1) and (2)  is also R.
  • We can also say that 

\(\Rightarrow \dfrac{d}{dx} \left(\sum\limits_{n=0}^{\infty}C_n (x-a)^n\right) = \sum\limits_{n=0}^{\infty}\dfrac{d}{dx} (C_n (x-a)^n)\)

\(\Rightarrow \displaystyle\int\sum\limits_{n=0}^{\infty}C_n (x-a)^n \,dx= \sum\limits_{n=0}^{\infty}\int C_n (x-a)^n\,dx\)

Illustration Questions

Find a power series representation of  \(f(x) = \ell n (x^2+4)\) .

A \(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^n}{n!}\)

B \(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^{2n}}{2^{2n+1}{(2n+1)}}\)

C \(\sum\limits_{n=0}^\infty\dfrac{x^{2n+2}}{2^{2n+1}{(2n+2)}}\)

D \(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^{2n+2}}{2^{2n+1}{(2n+2)}}\)

×

\(\displaystyle\int \sum\limits_{n=0}^\infty C_0 (x-a)^n dx = \sum\limits_{n=0}^\infty \displaystyle\int C_n \,(x-a)^n\,dx\)

\(\dfrac{d}{dx} (\ell n (x^2+n)) = \dfrac{2x}{x^2+n}\)        (Chain Rule)

\(\therefore \ell n (x^2+4) =\displaystyle \int \dfrac{2x}{x^2+4} \,dx\)

\(\therefore \ell n (x^2+4) = 2 \displaystyle\int\dfrac{x}{4}\left( \dfrac{1}{1-\left(\dfrac{-x^2}{4}\right)}\right)dx\)

\(= \dfrac{1}{2} \displaystyle\int x\left(1-\dfrac{x^2}{4}+\dfrac{x^4}{16}-\dfrac{x^6}{64}.....\right)dx\)

\(= \dfrac{1}{2} \displaystyle\int (x-\dfrac{x^3}{4}+\dfrac{x^5}{16}-\dfrac{x^7}{64}.....)dx\)

\(= \dfrac{1}{2} \left[\dfrac{x^2}{2}-\dfrac{x^4}{16}+\dfrac{x^6}{16× 6}+\dfrac{x^8}{64× 8}.....\right]\)

\(\therefore \ell n (x^2+4) = \dfrac{1}{2} \sum \dfrac{(-1)^n\,x^{2n+2}}{4^n (2n+2)}\)

 \(\Rightarrow \sum\limits_{n=0}^{\infty} \,\,\dfrac{(-1)^n\,x^{2n+2}}{2^{2n+1}(2n+2)} \)

Find a power series representation of  \(f(x) = \ell n (x^2+4)\) .

A

\(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^n}{n!}\)

.

B

\(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^{2n}}{2^{2n+1}{(2n+1)}}\)

C

\(\sum\limits_{n=0}^\infty\dfrac{x^{2n+2}}{2^{2n+1}{(2n+2)}}\)

D

\(\sum\limits_{n=0}^\infty(-1)^n\dfrac{x^{2n+2}}{2^{2n+1}{(2n+2)}}\)

Option D is Correct

Taylor Series

Suppose \(f\) is a function which we know and we are interested in the power series of this function .

Let \(f(x) = C_0 + C_1 (x-a) +C_2 (x-a)^2 +C_3 (x-a)^3 +.......|x-a|<R\) 

Where  \(C_0 , C_1, C_2,.....\)  are unknown.

  • Put  \(x=0\) on both sides

           \(\Rightarrow f(0) =C_0 \)

          \(\Rightarrow C_0 = f(0)\)

  • Differentiate both sides and put  \(x =a\) 
  • \(C_1=f'(a)\)
  • Similarly  \(C_2 = \dfrac{f''(a)}{2!} , \,\,C_3 = \dfrac{f'''(a)}{3!} .........C_n = \dfrac{f^n(a)}{n!}\) 

              \(\therefore \) If \(f\) has a power series representation at a , that is if

                   \(f(x) = \sum\limits_{n=0}^\infty C_n (x-a)^n \,\,\,\,\,\,\,(x-a)<R\)

                   then its coefficients are given by the formula

\(\Rightarrow C_n= \dfrac{f^n (a)}{n!}\)

  • We can say that ,

         \(\Rightarrow f(x) = \sum\limits_{n=0}^\infty \dfrac{f^n (a)}{n!} (x-a)^n\)

        \(\Rightarrow f(a) +\dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(a)}{2!}(x-a)^2+......\)

  • The above series is called the Taylor series  of the function at \(x =a\)  (or about 'a' or centered at \(x =a\)).

Illustration Questions

Find the Taylor series of  \(f(x) = e^{2x}\) centered at a=3.

A \(f(x) = \sum\limits_{n=0}^\infty \dfrac{2^n}{n!} (x-3)^n\)

B \(f(x) = \sum\limits_{n=0}^\infty \dfrac{2^n×e^6}{n!} (x-3)^n\)

C \(f(x) = \sum\limits_{n=0}^\infty \dfrac{3^n}{n!} (x-3)^n\)

D \(f(x) = \sum\limits_{n=0}^\infty e^6 (x-3)^n\)

×

The Taylor series for any function \(f\) is  given by 

\(f(x) = f(a) + \dfrac{f'(a)}{1!} (x-a) + \dfrac{f''(a)}{2!} (x-a)^2....... \)

\(\Rightarrow \sum\limits_{n=0}^\infty \dfrac{f^n(a)}{n!} (x-a)^n\)  

In the case  \(f(x) = e^{2x},\,\,\,a=3\)

\(\therefore f(3) =e^6,\,\,\,f'(3) = 2\,e^6,\,\,\,\,\,f''(3) = 2^2\,e^6,......f^n(3) = 2^n\,e^6\)  

 

\(\therefore e^{3x } = \sum\limits_{n=0}^\infty \dfrac{2^n \,e^6 (x-3)^n}{n!}\)

\(= e^6 \left[1+2(x-3) +\dfrac{4}{2! }(x-3)^2 +\dfrac{8}{3!}(x-3)^3+....\right]\)

Find the Taylor series of  \(f(x) = e^{2x}\) centered at a=3.

A

\(f(x) = \sum\limits_{n=0}^\infty \dfrac{2^n}{n!} (x-3)^n\)

.

B

\(f(x) = \sum\limits_{n=0}^\infty \dfrac{2^n×e^6}{n!} (x-3)^n\)

C

\(f(x) = \sum\limits_{n=0}^\infty \dfrac{3^n}{n!} (x-3)^n\)

D

\(f(x) = \sum\limits_{n=0}^\infty e^6 (x-3)^n\)

Option B is Correct

Maclaurin Series

Taylor Series

Suppose \(f\) is a function which we know and we are interested in the power series of this function .

Let \(f(x) = C_0 + C_1 (x-a) +C_2 (x-a)^2 +C_3 (x-a)^3 +.......|x-a|<R\) 

Where  \(C_0 , C_1, C_2,.....\)  are unknown.

  • Put  \(x =0\) on both sides

           \(f(0) =C_0 \)

          \(\Rightarrow C_0 = f(0)\)

  • Differentiate both sides and put  \(x =a\) 
  • \(C_1 = f'(a)\)
  • Similarly  \(C_2 = \dfrac{f''(a)}{2!} , \,\,C_3 = \dfrac{f'''(a)}{3!} .........C_n = \dfrac{f^n(a)}{n!}\) 

              \(\therefore \) If \(f\) has a power series representation at a , that is if

                   \(f(x) = \sum\limits_{n=0}^\infty C_n (x-a)^n \,\,\,\,\,\,\,(x-a)<R\)

there its coefficients are given by the formula

\(\Rightarrow C_n= \dfrac{f^n (a)}{n!}\)

  • We can say that ,

         \(\Rightarrow f(x) = \sum\limits_{n=0}^\infty \dfrac{f^n (a)}{n!} (x-a)^n\)

        \(\Rightarrow f(a) +\dfrac{f'(a)}{1!}(x-a) + \dfrac{f''(a)}{2!}(x-a)^2+......\)

  • The above series is called the Taylor series  of the function at \(x =a\)  (or about 'a' or centered at \(x =a\)).
  • If we put \(a=0\)  in the Taylor series we get

           \(f(x) = \sum\limits_{n=0}^\infty \dfrac{f^n (0)}{n!} x^n = f(0) +\dfrac{f'(0)}{1!}x+\dfrac{f''(0)}{2!}x^2 +....\)

  •  which is called Maclaurin series of \(f\) .
  • This Maclaurin series is a speed Taylor series centered about \(x=0\) .
  • Maclaurin series is defined separately  as it is easy to calculate.

Given below is the Maclaurin series for some common function which  can be derived easily.

(1)  \(\dfrac{1}{1-x} = \sum\limits_{n=0}^\infty x^n = 1+ x+x^2+x^3+......R=1 \)

(2) \(sin \,x = \sum\limits_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{(2n+1)!} = x-\dfrac{x^3}{3!} +\dfrac{x^5}{5!} .......R= \infty\)

 

(3) \(cos\,x = \sum\limits_{n=0}^\infty (-1)^n \dfrac{x^{2n}}{(2n)!} = 1-\dfrac{x^2}{2!} +\dfrac{x^4}{4!} .......R= \infty\)

(4) \(e^x \, = \sum\limits_{n=0}^\infty \dfrac{x^{n}}{n!} =1+ x+\dfrac{x^2}{2!} +\dfrac{x^3}{3!} +.......R= \infty\)

(5)  \(\ell n(1+x) \, = \sum\limits_{n=1}^\infty (-1)^{n-1}\dfrac{x^{n}}{n} = x-\dfrac{x^2}{2} +\dfrac{x^3}{3} +.......R= 1\)

 

Illustration Questions

Using the table of Maclaurin series, find series for  \(f(x) = x^2 \,\ell n\, (1+x^3)\) .

A \(f(x) = \sum\limits_{n=1}^\infty (-1)^n \dfrac{x^{3n+2}}{n}\)

B \(f(x) = \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{x^{3n+2}}{n}\)

C \(f(x) = \sum\limits_{n=1}^\infty \dfrac{x^{n}}{n}\)

D \(f(x) = \sum\limits_{n=1}^\infty \dfrac{x^{3n}}{n}\)

×

\(\ell n (1+x) = \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{x^n}{n}\)

\(=x-\dfrac{x^2}{2}-\dfrac{x^3}{3} ......\)

\(\therefore f(x) = x^2 \ell n(1+x^3) = x^2 \,\ell n (1+x^3)\)

\(= x^2 \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{(x^3)^n}{n}\)

\(\Rightarrow \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{x^{3n+2}}{n}\)

\(\Rightarrow \dfrac{x^5}{1} - \dfrac{x^8}{2} + \dfrac{x^{11}}{3}.....\)

Using the table of Maclaurin series, find series for  \(f(x) = x^2 \,\ell n\, (1+x^3)\) .

A

\(f(x) = \sum\limits_{n=1}^\infty (-1)^n \dfrac{x^{3n+2}}{n}\)

.

B

\(f(x) = \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{x^{3n+2}}{n}\)

C

\(f(x) = \sum\limits_{n=1}^\infty \dfrac{x^{n}}{n}\)

D

\(f(x) = \sum\limits_{n=1}^\infty \dfrac{x^{3n}}{n}\)

Option B is Correct

Evaluation of Limits Using Maclaurin Series 

  • We can use Maclaurin series to evaluate certain limits  which are otherwise very difficult to evaluate .
  • We use the standard Maclaurin series of the table. 

Illustration Questions

Evaluate   \(\lim\limits_{x\to0} \dfrac{sin\,x-x+\dfrac{x^3}{6}}{x^5}\) .

A \(\dfrac{1}{60}\)

B \(\dfrac{1}{120}\)

C \(\dfrac{5}{17}\)

D \(\dfrac{1}{20}\)

×

Use Maclaurin Series,

\(\ell n (1+x) = \sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{x^n}{n}\)

\(=x-\dfrac{x^2}{2}-\dfrac{x^3}{3} ......\)

 

\(\Rightarrow sin\,x = x- \dfrac{x^3}{3! } +\dfrac{x^5}{5!} ......\)

\(\therefore \lim\limits_{x\to0} \dfrac{\sin\,x-x+\dfrac{x^3}{6}}{x^5}\)

\(\Rightarrow \lim\limits_{x\to0} \dfrac{\left(x-\dfrac{x^3}{6}+\dfrac{x^5}{120}....\right)-x+\dfrac{x^3}{6}}{x^5}\)

\(\Rightarrow \dfrac{1}{120}\)

Evaluate   \(\lim\limits_{x\to0} \dfrac{sin\,x-x+\dfrac{x^3}{6}}{x^5}\) .

A

\(\dfrac{1}{60}\)

.

B

\(\dfrac{1}{120}\)

C

\(\dfrac{5}{17}\)

D

\(\dfrac{1}{20}\)

Option B is Correct

Binomial Series 

  • If k is any real number and \(|x| <1\) then  

    \(\Rightarrow(1+x)^k = 1+k\,x +k(k-1)\dfrac{x^2}{2!} +k(k-1)(k-2)\dfrac{x^3}{3!}+......\)

  • The above result can be derived form the Maclaurin series for  \((1+x)^k\) .

Illustration Questions

Use Binomial series to expand  \(\dfrac{1}{(2+x)^3}\) as power of \(x \) .

A \(\dfrac{1}{(2+x)^3} = 1+2x+4\,x^2 +8\,x^3 .....\)

B \(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} - \dfrac{3x}{4}+\dfrac{3x^2}{6}....\)

C \(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} +\dfrac{3x}{16}-\dfrac{3x^2}{16}+\dfrac{5x^3}{32}.....\)

D \(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} -\dfrac{3x}{16}+\dfrac{3x^2}{16}-\dfrac{5x^3}{32}.....\)

×

Binomial Series 

\(\Rightarrow(1+x)^k = 1+k\,x +k(k-1)\dfrac{x^2}{2!} +k(k-1)(k-2)\dfrac{x^3}{3!} +......\)

Where  \(|x| <1\)  

\(\Rightarrow \dfrac{1}{(2+x)^3} = (2+x)^{-3} = 2^{-3}\left(1+\dfrac{x}{2}\right)^{-3}\)  (to express in standard form)

\(= \dfrac{1}{8} \left(1+\dfrac{x}{2}\right)^{-3}\)

\(\Rightarrow \left(1+\dfrac{x}{2}\right)^{-3} = 1+ \left(-3×\dfrac{x}{2}\right)+\dfrac{(-3)(-4)}{2!} \left(\dfrac{x}{2}\right)^2 +\dfrac{(-3)(-4)(-5)}{3!} \left(\dfrac{x}{2}\right)^3 +........\)

\(\Rightarrow 1-\dfrac{3x}{2} +\dfrac{3x^2}{2} - \dfrac{5x^3}{4} ......\)

\(\Rightarrow \dfrac{1}{8} \left(1+\dfrac{x}{2}\right)^{-3} = \dfrac{1}{8} -\dfrac{3x}{16}+\dfrac{3x^2}{16}-\dfrac{5x^3}{32}.......\)

Use Binomial series to expand  \(\dfrac{1}{(2+x)^3}\) as power of \(x \) .

A

\(\dfrac{1}{(2+x)^3} = 1+2x+4\,x^2 +8\,x^3 .....\)

.

B

\(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} - \dfrac{3x}{4}+\dfrac{3x^2}{6}....\)

C

\(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} +\dfrac{3x}{16}-\dfrac{3x^2}{16}+\dfrac{5x^3}{32}.....\)

D

\(\dfrac{1}{(2+x)^3} = \dfrac{1}{8} -\dfrac{3x}{16}+\dfrac{3x^2}{16}-\dfrac{5x^3}{32}.....\)

Option D is Correct

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