1. ## Taylor Probs

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

2. 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)=...

\displaystyle \begin{aligned} T_{p,2}^{(4)} (x) & =\frac{p(2)}{0!} (x-2)^0+\frac{p'(2)}{1!} (x-2)^1 +\frac{p''(2)}{2!} (x-2)^2+\dots \\ & =\frac{-44}{1} \cdot 1+\frac{-110}{1} \cdot (x-2)+\dots \end{aligned}

3. 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?

4. 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?