Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Textbook Question
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Chapter 4, Problem 1P

The following series can be used to approximate e x :

e x = 1 + x + x 2 2 + x 3 3 ! + + x n n !

(a) Prove that this Maclaurin series expansion is a special case of the Taylor series expansion [ Eq . ( 4.7 ) ]  with  x i = 0  and  h = x .

(b) Use the Taylor series to estimate f ( x ) = e x  at  x i + 1 = 1  for  x i = 0.2 . Employ the zero-, first-, second-, and third-order versions and compute the | ε t | for each case.

(a)

Expert Solution
Check Mark
To determine

To prove: The Maclaurin series ex=1+x+x22+x33!++xnn! is the special case of the Taylor series expansion f(xi+1)=f(xi)+f'(xi)h+f''(xi)2!h2+f(3)(xi)3!h3++f(n)(xi)n!hn+Rn where the remainder term Rn equals f(n+1)(ξ)(n+1)!hn+1 with xi=0 and h=x.

Explanation of Solution

Given Information:

The Maclaurin series of exponential function ex=1+x+x22+x33!++xnn!.

Formula Used:

The Taylor series expansion

f(xi+1)=f(xi)+f'(xi)h+f''(xi)2!h2+f(3)(xi)3!h3++f(n)(xi)n!hn+Rn.

Proof:

Let the function f(x) equal ex.

For this function,

f(n)(x)=f(x)=ex

For all integer n.

Now apply this in the Taylor series expansion to obtain:

f(xi+1)=f(xi)+f'(xi)h+f''(xi)2!h2+f(3)(xi)3!h3++f(n)(xi)n!hn+Rnexi+1=exi+hexi+h22exi+h33!exi+hnn!exi+hn+1(n+1)!eξ

Now replace xi with 0 and h with x to get,

ex=e0+xe0+x22e0+x33!e0+xnn!e0+xn+1(n+1)!eξ=1+x+x22+x33!++xnn!+xn+1(n+1)!eξ

Note that the remainder term xn+1(n+1)!eξ would tend to 0 as n tends to infinity.

This gives,

ex1+x+x22+x33!++xnn!

Hence, proved that the provided Maclaurin series is an approximation of ex and this is a generalisation of the Taylor series expansion.

(b)

Expert Solution
Check Mark
To determine

To calculate: The approximate value of f(x)=ex at the point xi+1=1 when xi=0.2 using the Taylor series with the simultaneous computation of |ξ| at the zero-, first-, second- and third order approximates.

Answer to Problem 1P

Solution:

The zero first, second and third order Taylor series approximations for the function f(x)=ex at the point xi+1=1 when xi=0.2 are 0.81873, 0.16375, 0.42574 and 0.35587 with the relative percentage error at the respective stages as 122.55%, 55.49%, 15.73% and 3.26% respectively.

Explanation of Solution

Given Information:

The function f(x)=ex with points xi+1=1 and xi=0.2.

Formula Used:

The Taylor series approximation of nth order is:

f(xi+1)=f(xi)+f'(xi)(xi+1xi)+f''(xi)2!(xi+1xi)2++f(n)(xi)n!(xi+1xi)n.

Calculation:

Consider the zero-order approximation for the provided function,

f(xi+1)=f(xi)

Now replace 1 for xi+1 and 0.2 for xi to obtain,

f(1)=f(0.2)=e0.2=0.81873

The zero order approximation gives 0.81873.

The exact value of the function at 1 would be:

f(1)=e1=0.36788

Thus,

|ξt|=|0.367880.818730.36788|×100%=122.55%

The relative percentage error at this stage is 122.55%.

The first-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1xi)

Now replace 1 for xi+1 and 0.2 for xi to obtain,

f(1)=f(0.2)+f'(0.2)(10.2)=e0.2e0.2(0.8)=0.16375

The first order approximation gives 0.16375.

Thus,

|ξt|=|0.367880.163750.36788|×100%=55.49%

The relative percentage error at this stage is 55.49%.

The second-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1xi)+f''(xi)2(xi+1xi)2

Now replace 1 for xi+1 and 0.2 for xi to obtain,

f(1)=f(0.2)+f'(0.2)(10.2)+f''(0.2)2(10.2)2=e0.2e0.2(0.8)+e0.2(0.32)=0.42574

The second order approximation gives 0.42574.

Thus,

|ξt|=|0.367880.425740.36788|×100%=15.73%

The relative percentage error at this stage is 15.73%.

The third-order Taylor series approximation would be:

f(xi+1)=f(xi)+f'(xi)(xi+1xi)+f''(xi)2(xi+1xi)2+f(3)(xi)3!(xi+1xi)3

Now replace 1 for xi+1 and 0.2 for xi to obtain,

f(1)=f(0.2)+f'(0.2)(10.2)+f''(0.2)2(10.2)2+f(3)(0.2)3!(10.2)3=e0.2e0.2(0.8)+e0.2(0.32)e0.2(0.08533)=0.35587

The third order approximation gives 0.35587.

Thus,

|ξt|=|0.367880.355870.36788|×100%=3.26%

The relative percentage error at this stage is 3.26%.

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