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

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

RN (x) is divided into piano remainder and Lagrange remainder
There is nothing to say about the piano remainder (the n-th power infinitesimal of x-x0)
The Lagrangian remainder term first writes out the N + 1 general term and replaces x with a (Kesi symbol) between X and x0

Find the n-order derivative of arctanx without Taylor formula

What do you think?
y'=1/(1+x^2)
(1 + x ^ 2) * y '= 1, and then find the n-order derivative:

Why should Taylor's formula be written in the form of n-order derivative as the sum of coefficients?

In fact, this problem can also be understood as the proof of Taylor's formula, that 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 mean value theorem is Lim Δ x→0 f(x.+ Δ x)-f(x.)=f'(x.) Δ x) , where error α...

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

The derivation of Taylor's formula is to be expanded. First, it is expanded abstractly to the derivative of the order, and the function is expanded concretely to the order. If the coefficients of the two are equal, it is the higher-order derivation. Indeed, it is more convenient to directly calculate the n-order derivation for some problems. However, some problems must be expanded by Taylor and then compared with the coefficients of the two. I have seen a postgraduate entrance examination problem in which F (x) is quite complex, But it's easy to expand a part of Taylor

It is known that a = (COSH, Sinh), B = (cosx, SiNx) (0

∵a=(cosh,sinh),b=(cosx,sinx).∴a+b=(cosh+cosx,sinh+sinx).a-b=(cosh-cosx,sinh-sinx).∴(a+b)*(a-b)=(cos ² h-cos ² x)+(sin ² h-sin ² x)=(cos ² h+sin ² h)-(cos ² x+sin ² x)=1-1=0....

Let the function f (x) = 2x (the x power of e minus the negative x power of AE) (x belongs to R) be an even function, then the real number a =?

f(-1)=f(1)
f(1)=2(e-a/e),f(-1)=-2(1/e-ae)
2(e-a/e)=-2(1/e-ae)
E-A / E = - 1 / E + AE
I.e. E ²- a=-1+ae ²
Namely: (A-1) e ²- (a-1)=0
Namely: (A-1) (E) ²- 1)=0
So: a = 1