Let f⁽ᵏ⁾(0) exist for k = 0, 1, 2, ..., n. the the nth taylor polynomial of f about 0 is defined as the polynomial fo the degree with the most n given by
pₙ (x) = f (0) + f'(0)x + f" (0)/2!ˣ² + f⁽³⁾(0)/3!ˣ³ + ...+ f⁽ⁿ⁾(0)/n!ˣ⁽ⁿ⁾
show that Po (0) = f(0) and that p₁ has the same value at 0 and the same slope at 0 as f does