Taylor series and it’s derivation

Question

Hello, could someone help me solve this problem? Show that the Taylor polynomial at x0 has the same derivatives as the underlying function, i.e. 0≤k≤N

I already tried to differentiate the series but I can’t solve the problem

d^k/dx^k TN​[f(x0​)](x)

∣x=x0​​

=d^k/dx^k ​f(x) ∣x=x0​​

Answer 1

the (x-x₀)⁰ has multiplier f(x₀) in Taylor Series of f(x)

the (x-x₀)¹ has multiplier f'(x₀)/1! in Taylor Series of f(x)

the (x-x₀)² has multiplier f''(x₀)/2! in Taylor Series of f(x)

the (x-x₀)³ has multiplier f'''(x₀)/3! in Taylor Series of f(x)

etc.

Have f(x)=taylor series for x in radius of convergence around x₀

f'(x)=0+[1*f'(x₀)/1!](x-x₀)⁰+[2f''(x₀)/2!](x-x₀)¹+[3f'''(x₀)/3!](x-x₀)²+....

f'(x₀)=0+f'(x₀)+0+0+...

f''(x)=0+0+[2*1*f''(x₀)/2!](x-x₀)⁰+[3*2f'''(x₀)/3!](x-x₀)¹+....

f''(x₀)=0+0+f''(x₀)+0+0...

f'''(x)=0+0+0+[3*2*1*f'''(x₀)/3!](x-x₀)⁰+....

f'''(x₀)=0+0+0+f'''(x₀)+0+0...

etc.