Hello,
For the first one, calculate the successive derivatives of p(x) at x=2.
p(x)=2x + 4x^3 - 5x^4
p(2)=-44
p'(x)=2+12x^2-20x^3
p'(2)=-110
p''(x)=24x-60x^2
p''(2)=...
Hi I got problems regarding Taylor and need your help on this please
1.
p(x) = 2x + 4x^3 - 5x^4
Find the polynomial i.e
and prove that
Using the fact that two polynomials of degree k are equal if and only if they
have the same coefficients.
AND
2.
Find the following:
Whichever help I can get would be gr8
Thx for tip Moo I did the following based on your idea...
p(x)=2x + 4x^3 - 5x^4
p(2)=-44
p'(x)=2+12x^2-20x^3
p'(2)=-110
p''(x)=24x-60x^2
p''(2)=-456
p'''(x)=24 - 120x
p'''(2)= -216
p''''(x)=-120
Taylor Polynomial = [(P(2)/0!)(x-2)^0] + [(P'(2)/1!)(x-2)^1] + [(P''(2)/2!)(x-2)^2] + [(P'''(2)/3!)(x-2)^3] + [(P''''(2)/4!)(x-2)^4]
After all substitutions and simplifying I got the following:
= -5x^4 + 4x^3 - 168x^2n+ 530x -528
which does look slightly similar to P(x).
Did I go wrong somewhere or is there a few steps to perform?
Any hints on part 2?
Following the method revamped I ended up with the following:
P(4)p,2(x)=-44-110x+220-96x^2+384x-384-36x^3+216x^2-432x+288-5x^4+4-x^3-120x^2+160x-80
which cancelled down to....-5x^4+4x^3+2x which confirms the proof.
As for part 2 I got the following:
f(x) = arctan
f'(x)= 1/(1+x^2)
f''(x)= -2x/(1+x^2)2
f'''(x)= [8x^2/(1+x^2)^3] + [2/(1+x^2)^2]
f(x) = 0
f'(x)= 1
f''(x)= 0
f'''(x)= -2
0+x+ (0x^2) (-2/3!)x^3
Any miscalculations?