1. Sec:4.1
THE TAYLOR SERIES

2. Sec:4.1 THE TAYLOR SERIES Taylor’s theorem
states that any smooth function can be approximated as a polynomial.
Taylor series ( center is a )
If the function f and its first n + 1 derivatives are continuous on an interval containing a and x,
there exists a point ξ between a and x such that the reminder
Maclaurin series ( center is 0 )
ξ between 0 and x

3. Sec:4.1 THE TAYLOR SERIES Example: Taylor series ( center is a )
Example:
Let f (x) = cos x and a = 0. Determine
(a) the second order approximation for f about a; and
(b) Use (a) to approximate cos(0.01)
f(x) = cos x,
f'(x) = −sin x,
f’’(x) = −cos x,
f’’’(x) = sin x,
f’’’’(x) = cos x,
f(0) = 1,
f'(0) = 0,
f’’(0) = −1,
f’’’(0) = 0,
f’’’’(0) = 1,
𝑓 𝑥 =𝑐𝑜𝑠𝑥
syms x
taylor(cos(x),x,0,'order',3)
1 - x^2/2
p 𝒙 =𝟏− 𝟏 𝟐 𝒙 𝟐
ezplot('cos(x)',[-pi,pi]); hold on
ezplot('1-x.^2/2',[-pi,pi]); hold off
grid on

4. Sec:4.1 THE TAYLOR SERIES Example:
Let f (x) = cos x and a = 0. Determine
(a) the second order approximation for f about a; and
(b) Use (a) to approximate cos(0.01)
How accurate this approximation:
exact
approximation
Hence the approximation matches at least the first five digits of the exact

5. Sec:4.1 THE TAYLOR SERIES Example:
Let f (x) = cos x and a = 0. Determine
(a) the second order approximation for f about a; and
(b) Use (a) to approximate cos(0.01)
This example illustrates the two objectives of numerical analysis:
Find an approximation to the solution of a given problem.
(ii) Determine a bound for the accuracy of the approximation

6. Sec:4.1 THE TAYLOR SERIES Example:
Why: we want to approximate a function by a polynomial.
Example:
Let f (x) = cos x and a = 0. Determine
(a) the third order approximation for f about a; and
(b) Use (a) to approximate
integrate both sides
exact
approximation

7. Sec:4.1 THE TAYLOR SERIES Taylor’s theorem
states that any smooth function can be approximated as a polynomial.
Taylor series ( center is a )
there exists a point ξ between a and x such that
Another expression for the Reminder

8. Sec:4.1 THE TAYLOR SERIES Taylor series ( center is a )
It is often convenient to simplify the Taylor series by defining a step size h = xi+1 − xi
and expressing above as
Taylor series ( center is a )

9. Sec:4.1 THE TAYLOR SERIES 𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊
𝑓 𝑥 𝑖+1 =𝑓 𝑥 𝑖 + 𝑓′( 𝑥 𝑖 ) 1! ℎ+ 𝑓 2 ( 𝑥 𝑖 ) 2! ℎ 2 +…+ 𝑓 𝑛 𝑥 𝑖 𝑛! ℎ 𝑛 + 𝑹 𝒏
Taylor series
𝑓 𝑥 𝑖+1 =𝑓 𝑥 𝑖 + 𝑓 ′ 𝑥 𝑖 1! ( 𝑥 𝑖+1 − 𝑥 𝑖 )+ 𝑓 2 ( 𝑥 𝑖 ) 2! ( 𝑥 𝑖+1 − 𝑥 𝑖 ) 2 +…+ 𝑓 𝑛 𝑥 𝑖 𝑛! ( 𝑥 𝑖+1 − 𝑥 𝑖 ) 𝑛 + 𝑹 𝒏
First-order approximation
forward finite difference
𝑓 𝑥 𝑖+1 =𝑓 𝑥 𝑖 + 𝑓 ′ 𝑥 𝑖 1! ( 𝑥 𝑖+1 − 𝑥 𝑖 )+ 𝑓 2 (𝜉) 2! ( 𝑥 𝑖+1 − 𝑥 𝑖 ) 2
𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊
Move the first term to left side and divid by h
𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊 − 𝒇 ′ 𝒙 𝒊 = 𝒇 𝟐 (𝝃) 𝟐! 𝒉
Truncation error
Approximation to the derivative

10. Sec:4.1 THE TAYLOR SERIES
forward finite difference
𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊 − 𝒇 ′ 𝒙 𝒊 = 𝒇 𝟐 (𝝃) 𝟐! 𝒉
𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊
SIMILAR TO PROBLEM 4.5/p105:
𝒙 𝒊 =2
𝒙 𝒊+𝟏 =2.2
Use forward difference approximations to estimate the first derivative of the function
𝒇 (𝒙) = 𝟐𝟓 𝒙 𝟑 − 𝟔 𝒙 𝟐 + 𝟕𝒙
Evaluate the derivative at x = 2 using a step size of h = 0.2, 0.1, Compare your results with the true value of the derivative. Interpret your results on the basis of the remainder term of the Taylor series expansion.
𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊
h = 0.2
𝒇 ′ 𝟐 ≈ 𝒇 𝟐.𝟐 −𝒇 𝟐 𝟎.𝟐 = 𝟐𝟓𝟐.𝟓𝟔−𝟏𝟗𝟎 𝟎.𝟐 =𝟑𝟏𝟐.𝟖
h = 0.1
𝒇 ′ 𝟐 ≈ 𝒇 𝟐.𝟏 −𝒇 𝟐 𝟎.𝟏 =𝟐𝟗𝟕.𝟔𝟓
h = 0.001
𝒇 ′ 𝟐 ≈ 𝒇 𝟐.𝟎𝟎𝟏 −𝒇 𝟐 𝟎.𝟎𝟎𝟏 =𝟐𝟖𝟑.𝟏𝟒𝟒𝟎𝟐𝟓
𝒇 ′ 𝟐 =𝟐𝟖𝟑

11. Sec:4.1 THE TAYLOR SERIES
first finite divided difference
𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊 − 𝒇 ′ 𝒙 𝒊 = 𝒇 𝟐 (𝝃) 𝟐! 𝒉
𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊
SIMILAR TO PROBLEM 4.5/p105:
𝒙 𝒊 =2
𝒙 𝒊+𝟏 =2.2
Use forward difference approximations to estimate the first derivative of the function
𝒇 (𝒙) = 𝟐𝟓 𝒙 𝟑 − 𝟔 𝒙 𝟐 + 𝟕𝒙
Evaluate the derivative at x = 2 using a step size of h = 0.2, 0.1, Compare your results with the true value of the derivative. Interpret your results on the basis of the remainder term of the Taylor series expansion.
h = 0.001
𝒇 ′ 𝟐 ≈ 𝒇 𝟐.𝟎𝟎𝟏 −𝒇 𝟐 𝟎.𝟎𝟎𝟏 =𝟐𝟖𝟑.𝟏𝟒𝟒𝟎𝟐𝟓
𝒇 ′ 𝟐 =𝟐𝟖𝟑
Error = 0.𝟏𝟒𝟒𝟎𝟐𝟓
𝒇 𝟐 (𝝃) 𝟐! 𝒉
𝒇 (𝟐) (𝒙) =𝟏𝟓𝟎𝒙−𝟏𝟐
𝒇 𝟐 (𝝃) 𝟐! 𝒉= 75ξ−6 ∗(0.001)
≤ 75∗2.001−6 ∗(0.001) ≤

12. Sec:4.1 THE TAYLOR SERIES 𝒂 𝒙 𝒙 𝒊+𝟏 𝒙 𝒊−𝟏 𝒙 𝒊
Taylor series ( center is a ) 𝒙 𝒂
𝒙 𝒊−𝟏
𝒙 𝒊+𝟏
𝒙 𝒊
Taylor series
𝑹 𝒏 = (−1) 𝑛+1 𝑓 𝑛+1 𝑥 𝑖 (𝑛+1)! ℎ 𝑛+1
𝑓 𝑥 𝑖−1 =𝑓 𝑥 𝑖 − 𝑓 ′ 𝑥 𝑖 1! ℎ+ 𝑓 𝑥 𝑖 2! ℎ 2 +…+ (−1) 𝑛 𝑓 𝑛 𝑥 𝑖 𝑛! ℎ 𝑛 + 𝑹 𝒏

13. Sec:4.1 THE TAYLOR SERIES 𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊 −𝒇 𝒙 𝒊−𝟏 𝒙 𝒊 − 𝒙 𝒊−𝟏
𝑹 𝒏 = (−1) 𝑛+1 𝑓 𝑛+1 𝑥 𝑖 (𝑛+1)! ℎ 𝑛+1
𝑓 𝑥 𝑖−1 =𝑓 𝑥 𝑖 − 𝑓 ′ 𝑥 𝑖 1! ℎ+ 𝑓 𝑥 𝑖 2! ℎ 2 +…+ (−1) 𝑛 𝑓 𝑛 𝑥 𝑖 𝑛! ℎ 𝑛 + 𝑹 𝒏
First-order approximation
backward finite difference
𝑓 𝑥 𝑖−1 =𝑓 𝑥 𝑖 − 𝑓 ′ 𝑥 𝑖 1! ( 𝑥 𝑖 − 𝑥 𝑖−1 )+ 𝑓 2 (𝜉) 2! ℎ 2
𝒇 ′ 𝒙 𝒊 ≈ 𝒇 𝒙 𝒊 −𝒇 𝒙 𝒊−𝟏 𝒙 𝒊 − 𝒙 𝒊−𝟏
Move the first term to left side and divid by h
𝒇 𝒙 𝒊 −𝒇 𝒙 𝒊−𝟏 𝒙 𝒊 − 𝒙 𝒊−𝟏 − 𝒇 ′ 𝒙 𝒊 =− 𝒇 𝟐 (𝝃) 𝟐! 𝒉
Truncation error
Approximation to the derivative

14. Sec:4.1 THE TAYLOR SERIES
forward finite difference
backward finite difference
𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊 𝒙 𝒊+𝟏 − 𝒙 𝒊 − 𝒇 ′ 𝒙 𝒊 = 𝒇 𝟐 (𝝃) 𝟐! 𝒉
𝒇 𝒙 𝒊 −𝒇 𝒙 𝒊−𝟏 𝒙 𝒊 − 𝒙 𝒊−𝟏 − 𝒇 ′ 𝒙 𝒊 =− 𝒇 𝟐 (𝝃) 𝟐! 𝒉
Centered finite difference
𝒇 𝒙 𝒊+𝟏 −𝒇 𝒙 𝒊−𝟏 𝟐𝒉 − 𝒇 ′ 𝒙 𝒊 =𝑶( 𝒉 𝟐 )
Notice that the truncation error is of the order of ℎ 2 in contrast to the forward and backward approximations that were of the order of h. Consequently, the Taylor series analysis yields the practical information that the centered difference is a more accurate representation of the derivative

15. Sec:4.1 THE TAYLOR SERIES
SIMILAR TO PROBLEM 4.5/p105:
Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2)to estimate the first derivative of the function
𝒇 (𝒙) = 𝟐𝟓 𝒙 𝟑 − 𝟔 𝒙 𝟐 + 𝟕𝒙
Evaluate the derivative at x = 2 using a step size of h = 0.1, 0.01, , , Compare your results with the true value of the derivative.
1.0e+02 * h
backward difference
forward difference
centered difference
0.1
0.01
0.001
0.0001
back diff error
forward diff error
centered diff error
0.1
0.01
0.001
0.0001
Error Table

16. Sec:4.1 THE TAYLOR SERIES format short fun = @(x) 25*x^3 - 6*x^2+7*x;
dfun 75*x^2 - 12*x + 7;
x=2;
exact = dfun(2);
h_vec=[ ];
for i=1:5
h = h_vec(i);
[bd,fd,cd] = num_diff(fun,x,h);
bd_vec(i) = bd;
fd_vec(i) = fd;
cd_vec(i) = cd;
end
format long
res1 = [ bd_vec' fd_vec' cd_vec']
res2 = [h_vec' bd_vec' fd_vec' cd_vec']
err = [bd_vec' fd_vec' cd_vec']-exact;
err_res = [h_vec' err]
function [bd,fd,cd] = num_diff(f,x,h)
xi = x;
xip1 = xi+h;
xim1 = xi-h;
bd = (f(xi) - f(xim1))/h;
fd = (f(xip1) - f(xi) )/h;
cd = (f(xip1) - f(xim1))/(2*h);
end
fun 25*x^3 - 6*x^2+7*x;
dfun 75*x^2 - 12*x + 7;
h=0.2;
x=2;
[bd,fd,cd] = num_diff(fun,x,h)
dfun(2)

17. Sec:4.1 THE TAYLOR SERIES 𝒇 𝟐 (𝒙) Second Derivative Approximation
𝒇 𝟐 (𝒙)
Second Derivative Approximation
forward finite difference
𝒙 𝒊−𝟏
𝒙 𝒊
𝒙 𝒊+𝟏
𝒙 𝒊+𝟐
𝒙 𝒊−𝟐
𝒇 𝒙 𝒊+𝟐 −𝟐𝒇 𝒙 𝒊+𝟏 +𝒇 𝒙 𝒊 𝒉 𝟐 − 𝒇 ′′ 𝒙 𝒊 =𝑶(𝒉)
backward finite difference
𝒙 𝒊−𝟏
𝒙 𝒊
𝒙 𝒊+𝟏
𝒙 𝒊+𝟐
𝒙 𝒊−𝟐
𝒇 𝒙 𝒊 −𝟐𝒇 𝒙 𝒊−𝟏 +𝒇 𝒙 𝒊−𝟐 𝒉 𝟐 − 𝒇 ′′ 𝒙 𝒊 =𝑶(𝒉)
Centered finite difference
𝒙 𝒊−𝟏
𝒙 𝒊
𝒙 𝒊+𝟏
𝒙 𝒊+𝟐
𝒙 𝒊−𝟐
𝒇 𝒙 𝒊+𝟏 −𝟐𝒇 𝒙 𝒊 +𝒇 𝒙 𝒊−𝟏 𝒉 𝟐 − 𝒇 ′′ 𝒙 𝒊 =𝑶( 𝒉 𝟐 )