find a formula for the truncation error if we use P6(x) to approximate 1/(1-2x) on (-1/2,1/2)
To find the formula for the truncation error when using P6(x) to approximate 1/(1-2x) on the interval (-1/2, 1/2), we need to understand a few concepts.
First, let's define P6(x) as the polynomial of degree 6, which represents the Taylor series expansion of 1/(1-2x) around x = 0. The Taylor series expansion of a function f(x) around a point x = a is given by:
f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
In our case, we have f(x) = 1/(1-2x), and we want to approximate it using P6(x). So, we need to determine the coefficients of P6(x):
P6(x) = a0 + a1x + a2x^2 + a3x^3 + a4x^4 + a5x^5 + a6x^6
To find these coefficients, we can differentiate the function f(x) and evaluate it at x = 0:
f(x) = 1/(1-2x)
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
f''''''(x) = 46080/(1-2x)^7
Now, let's evaluate these derivatives at x = 0:
f(0) = 1/(1-2*0) = 1
f'(0) = 2/(1-2*0)^2 = 2
f''(0) = 8/(1-2*0)^3 = 8
f'''(0) = 48/(1-2*0)^4 = 48
f''''(0) = 384/(1-2*0)^5 = 384
f'''''(0) = 3840/(1-2*0)^6 = 3840
f''''''(0) = 46080/(1-2*0)^7 = 46080
We can use these values to determine the coefficients of P6(x):
P6(x) = 1 + 2x + 8x^2 + 48x^3 + 384x^4 + 3840x^5 + 46080x^6
The truncation error, denoted by E(x), is defined as the difference between the exact value of the function and its approximation using the polynomial:
E(x) = f(x) - P6(x)
In our case, E(x) = 1/(1-2x) - (1 + 2x + 8x^2 + 48x^3 + 384x^4 + 3840x^5 + 46080x^6)
Hence, the formula for the truncation error when using P6(x) to approximate 1/(1-2x) on (-1/2, 1/2) is:
E(x) = 1/(1-2x) - (1 + 2x + 8x^2 + 48x^3 + 384x^4 + 3840x^5 + 46080x^6)