Why is there a factorial in the Taylor series and the expression in parentheses has a degree?

Question

Mathematics

It’s just that the formula is given in the descriptions, but it’s not clear where it came from. Can anyone explain how this formula works? Each term of the series is a gradual refinement, this is understandable, but the details are not clear.
When I wrote the question there was clearly more than 15 characters, why it was not sent is not clear.

Answer 1

The funniest thing about Taylor's series is why he has such a residual member.

Lagrange and Cauchy
have very good limbs.
And Schlomilch and Rosh
have the very best, they say.

Let's solve the problem in a simpler way: let's estimate on the fingers the form of the power series Sum{a _i x ⁱ }, which approximates the function in the vicinity of x=0.
0th approximation: f(x) ≈ f(0).
1st approximation: f(x) ≈ f(0) + f'(0) x.
So far, no complaints. Let's think about the second approximation.
f(x) ≈ f(0) + f'(0) x + ax².
I would like this polynomial to have the same derivatives up to the second as the function f. (x²)| _x=0 =(x²)'| _x=0 =0, no problem with that. Because (x²)''| _x=0 =2, it turns out that a=f''(0)/2.
And immediately the nth approximation.
f(x) ≈ f(0) + f'(0) x + f''(0) x²/2 + … + bx ⁿ .
And this polynomial must have the same nth derivative as the function f. What is (x ⁿ ) ⁽ⁿ⁾ | _x=0 ? Of course, n!. Hence the coefficient f ⁽ⁿ⁾ (0)/n!.