Problem 4 & 5 Consider f(x) = Squareroot 1+2x Find the Taylor polynomial of degr

Question

Problem 4 & 5

Consider f(x) = Squareroot 1+2x Find the Taylor polynomial of degree three, p3(x), for f(x) centered at x_0 = 0 Give a maximum bound for the absolute value of error for 0 lessthanorequalto x lessthanorequalto 0.1 Calculate the absolute error at x = 0.1. How is the error at x = 0.1 compared to part (b)? Let f(x) = e^x - x, 1 lessthanorequalto x lessthanorequalto 2. Find the Lagrange interpolating polynomial for the data x_1 = 1 and x_2 = 2 Find the maximum error bound for the Lagrange Polynomial approximation for 1 lessthanorequalto x lessthanorequalto 2.

Explanation / Answer

4 )

f(x)=(1+2x)^0.5; f(0)= 1

f'(x) = 1/(2*x + 1)^(1/2) f'(0)= 1

f''(x) = -1/(2*x + 1)^(3/2) f''(0) =-1;

f'''(x) = 3/(2*x + 1)^(5/2) f'''(0) = 3

by taylor series

f(x) = f(0) + x.f'(0) + x^2/2! f''(0) + x^3/3! . f'''(0)

= 1 + x - x^2/2 + x^3/2

b ) Upper bound on error is given by

err = f^(n+1)(a) / (n+1) ! . (x-a) ^n+1

since we calculated taylor expansion upto degree 3

err = f^4(a)/4! . (x-a)^4

f^4 = f '''' = -15/(2*x + 1)^(7/2)

3 )

fx=(1+2x)^0.5 = 1.095445

by taylor series expansion

f(x) = 1.0955

abs (err)= 1.0955-1.095445=0.050055

by max error bound

err =f^4(a)/4! . (x-a)^4 = -7.924226815757/24*0.1^4=0000330176

2)

a)

f(x) =e^x-x

f(1) = e-1=1.7183

f(2)= e^2-2= 5.3891

f(x) =(x-2)/(1-2) f(1) + (x-1)/(2-1).f(2) =

= -1.7183*(x-2) + (x-1) 5.3891

=  3.6708*x -1.9525

for polynomial of order n

max err = f^(n+1) / (n+1)! . (x-1)(x-2)

f''=e^x= e^2= 7.3891

max ( x-1)(x-2) = max ( x^2- 3x + 2 ) = 2

maxx err = 7.3891/24*2= 0.6158

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