Informative line

### 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  $$|x| <1$$

#### 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

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

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

#### 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

# 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.

#### 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  $$(1+x)^k$$ .

#### 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

# 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 $$x =a$$).

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

#### 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