How did RN (x) of Lagrange remainder of Taylor formula come from? I mean how to find RN (x), that is to say, how to use its expansion

If the function f (x) has a derivative up to order n + 1 in the open interval (a, b), then when the function is in this interval, it can be expanded as the sum of a polynomial about (X-X.) and a remainder: F (x) = f (X.) + F '(X.) (X-X.) + F' '(X.) / 2! & # 8226; (X-X.) ^ 2, + F' '(X.) / 3! & # 8226; (X-X.) ^ 3 + +f(n)(x...

How to determine the remainder of Taylor's formula, RN (x), I want to write the detailed process of each step,

RN (x) can be divided into piano remainder and Lagrange remainder
There is nothing to say about the piyano remainder (infinitesimal of higher order of the nth power of x-x0)
Lagrange remainder will first write the general term of degree n + 1, and replace x with a (Kesi sign) between X and x0

Finding the n-order derivative of arctanx without Taylor formula

What do you think?
y'=1/(1+x^2)
(1 + x ^ 2) * y & # 39; = 1, then find the derivative of order n:

Why should Taylor's formula be written as the sum of n-order derivatives?

In fact, this problem can also be understood as the proof of Taylor's formula, which is how Taylor thought of this formula. The following is the proof process: F (x) = f (X.) + F '(X.) (X-X.) + α (the finite increment theorem derived from Lagrange's mean value theorem has Lim Δ x → 0 f (X. + Δ x) - f (X.) = f' (X.) Δ x), where the error α

How to use Taylor's formula to calculate the n-order derivative (if you say to expand the original formula, it's better to directly calculate the n-order derivative)?

The derivation of Taylor's formula is to expand. First, the derivative of the order is expanded abstractly. Then, the function is expanded concretely to the order. If the coefficients of the two are equal, it is the higher-order derivative. Indeed, for some problems, it is more convenient to directly calculate the n-order derivative. But some problems must be solved by Taylor's expansion, and then the coefficients of the two are compared, The f (x) in the title is quite complicated, but it's easy to do some Taylor expansion

Finding the local Taylor formula at x = 0
ln((1－x)/(1＋x))

ln((1－x)/(1＋x))
=ln(1－x)-ln(1＋x)
=-2(1+x+x^2/2+x^3/3+…… +x^n/n+……）

Find the limit of [x-x & sup2; ㏑ (1 + 1 / x)] when x tends to positive infinity (Taylor formula can be used)

The 1 / 2 prior 0 / 0 ln formula is expanded by Taylor's formula of order 2

When x tends to infinity, the limit of Ln (1 + 1 / x) is calculated by Taylor formula

Taylor formula to find the limit x
X tends to zero to find the limit of x-sinx / x + SiNx. I use Taylor formula to get - 1 and important limit to get 1. What's wrong

This problem does not need Taylor's formula. It can be simply changed
　　　lim(x→0)(x-sinx)/(x+sinx)
　　= lim(x→0)(1-sinx/x)/(1+sinx/x)
　　= (1-1)/(1+1)
　　= 0.

Primitive function with derivative (SiNx + cosx) ^ 2

∫(sinx+cosx)^2 dx
=∫(1+2sinxcosx)dx
=∫dx+∫sin2xdx
=x+∫sin2xdx
=x-(1/2)∫d(cos2x)
=x-(1/2)*cos2x+c.

How did RN (x) of Lagrange remainder of Taylor formula come from? I mean how to find RN (x), that is to say, how to use its expansion