find a formula for the truncation error if we use P6(x) to approximate 1/(1-2x) on (-1/2,1/2)
I would have to get my old numerical analysis book out of the attic.
well i appreciate the acknowledgement
I think you find that most questions are answered fairly quickly and almost all answerable questions are answered within 12 hours are so. When a question isn't answered, it's often because several people have looked at it and haven't been able to answer it. However, they hope that someone else may have that knowledge.
To find the formula for the truncation error when using a polynomial approximation, we need to consider the Taylor series expansion of the function being approximated. In this case, we want to approximate the function f(x) = 1/(1-2x) using the polynomial P6(x).
The Taylor series expansion of f(x) about a point a is given by:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
To use P6(x) as an approximation for f(x), we can truncate the Taylor series expansion at the term involving the (x-a)^6 power, since P6(x) is a polynomial of degree 6.
Let's find the Taylor series expansion of f(x) at a=0:
f(0) = 1/(1-2*0) = 1
f'(x) = 2/(1-2x)^2
f''(x) = 8/(1-2x)^3
f'''(x) = 48/(1-2x)^4
f''''(x) = 384/(1-2x)^5
f'''''(x) = 3840/(1-2x)^6
Using the above derivatives, we can write the Taylor series expansion at a=0 as:
f(x) = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + 32x^5 + ...
Since we're approximating f(x) with P6(x), we only consider terms up to the x^6 power:
f(x) ≈ 1 + 2x + 4x^2 + 8x^3 + 16x^4 + 32x^5
The truncation error is the difference between the actual function and the approximation:
Truncation error = f(x) - P6(x)
Substituting the expressions for f(x) and P6(x), we get:
Truncation error = (1 + 2x + 4x^2 + 8x^3 + 16x^4 + 32x^5) - P6(x)
To find the formula for the truncation error, we can simplify this expression further and write it in a more compact form.