Find the taylor series generated by f(x) = 7x^3 + 2x^2 + 8x - 3 at x = 2

Question

I got $ 77 + 100(x-2) + 44(x-2)^2 + 7(x-2)^3 + 0 $ but I can't find the pattern to simplify in sigma notation.

Answer 1

Given:

f(x) = 7x^3 + 2x^2 + 8x - 3 at x = 2

Taylor's polynomial is:

f(c) + f'(c)(x-c) + f''(c)(x-c)²/2 + f'''(c)(x-c)³/6 + ... +f⁽ⁿ⁾(c)(x-c)ⁿ/n!

f(c) = f(2) = 7(2)³+2(2)² +8(2) -3 = 56+8+16-3=77

f'(x)=21x²+4x+8, f'(c)=f'(2)=21(2)²+4(2)+8=84+8+8=100

f''(x)=42x+4, f''(c)=f''(2)=42(2)+4=84+4=88

f'''(x)=42,f'''(2)=42

f''''(x)=f''''(2)=0

Now plugin to the Taylor's polynomial:

77+100(x-2)+88(x-2)²/2+42(x-2)³/6

=77+100(x-2)+44(x-2)²+7(x-2)³

=∑³_n=0 f⁽ⁿ⁾(2)(x-2)ⁿ/n!

To compare with original f(x):

=77+(100x-200)+44(x²-4x+4)+7(x³-6x²+12x-8)

= 77+100x-200+44x²-176x+176+7x³-42x²+84x-56

=-3+8x+2x²+7x³