Taylor formula can be expressed as: F (a + H) = f (a) + F '(a) H + F' '(a) H & # 178; / 2! + O (H & # 178;)

First of all, f (x) is expanded at a point: F (x) = f (a) + F '(a) (x - a) + F' '(a) (x - a) & # 178; / 2! + O (x - a) & # 178; x = a + H is replaced by: F (a + H) = f (a) + F' (a) H + F '' (a) H & # 178; / 2

Taylor's formula of order n with Lagrange complement for function f (x) = 1 / X expanded by power of (x + 1)

f(x)=1-(x-1)+(x-1)^2-(x-1)^3+...+(-1)^(n-1)(x-1)^n+R
R = (- 1) ^ n (x-1) ^ (n + 1) / ξ ^ (n + 2) ξ is a value between 1 and X
If f '(x) f "(x)... Is solved, just bring in 1. Expand by X-1, that is, the Taylor expansion at x = 1

F (x) = 1 / x, which is expanded into Taylor formula with Lagrange remainder at x = - 1
The - x power of F (x) = e develops Taylor's formula at x = a. the expansions of these two formulas

f(x)=1/x
=-1/[1-(x+1)]
=-[1+(x+1)+(x+1)²+...+(x+1)^n]+[f(ζ)^(n+1)×(x+1)^(n+1)]/(n+1)!
f(x)=e^(-x)
=e^[-(x-a)-a]
=e^(-a)×e^[-(x-a)]
=e^(-a)×[1-(x-a)+(x-a)²/2!+...+(-1)^n(x-a)^n/n!]+o((x-a)^n)

Finding the denominator of RN (x) in the answer of n-order Taylor formula with Lagrange remainder expanded by the power of F (x) = 1 / X (x + 1)
Finding Taylor's formula of order n with Lagrange remainder of F (x) = 1 / X expanded by power of (x + 1)
How to find [- 1 + θ (x + 1)] in the denominator of RN (x) in the answer?

Taylor formula:

Lagrangian remainder:

To expand by the power of (x + 1) is to make a = - 1 in the formula

In Lagrange remainder, let a = - 1, the independent variable in N + 1 derivative = - 1 + θ (x + 1)

The expansion of Taylor formula at 0 is not f (x) = f (x0) + F '(x0) (x-x0) +... FN (x0) / N! (x-x0) n power
Why do I expand ln (1 + x) four times
Expand from 0, 0 equals 0
The expansion of order 1 equals x-x0 / 1 + X x0 = 0, so it equals X
If the second-order expansion is equal to the second-order expansion, we don't know how to expand it. We shouldn't take ln (1 + x) as the second derivative and then multiply it by (x-x0) x0 = 0
The second order expansion is not the square of F '' (x0) (x-x0). Why do I calculate that f '' (x0) is equal to - 1? Why is it equal to - 2?

The expansion of Ln (1 + x) at x = 0 is ln (1 + x) = x-x ^ 2 / 2 + x ^ 3 / 3-x ^ 4 / 4 +. + (- 1) ^ (n + 1) * x ^ n / N +. (- 1)

The relationship between X and x0 in Taylor formula
Does x in Taylor's formula tend to x0?
Let's see an example: finding the approximate value of 30 ^ (1 / 3) with the third order Taylor formula
Why decompose 30 into 3 (1 + 1 / 9) instead of 1 + 29

It does not mean that it must tend to x0, but that the closer X and x0 are, the closer the calculated value is to the exact value,
The example you gave uses McLaughlin's formula, x0 = 0, so x must be close to 0, so 30 is decomposed into 3 (1 + 1 / 9), and 1 / 9 is close to 0, so it can be decomposed in this way. If it is decomposed into (1 + 29), there is a big difference between 29 and 0, and the calculated value is far from the accurate value, so it is not called approximate value

Finding Taylor's (F-2) expansion with power LNX

What is the LNX Taylor expansion? The LNX reciprocal 1 / X obtained by applying McLaughlin formula directly has no definition on a = 0?

There is no definition at x = 0, because ln 0 is not defined. How can we expand it~
Taylor's expansion is OK, but it's boring. It's usually to expand ln (x + 1)
ln(x+1) = x - x^2/2 + x^3/3 ...+(-1)^(n-1)x^n/n+...

On the peyano remainder in Taylor formula
I read in the book that SiNx = x - X3 / 6 + O (x3), and SiNx = x - X3 / 6 + O (x4) also holds. Why can you do both
Can the X2 = 1 + x2 + X4 / 2 + O (x5) of E be written as 1 + x2 + X4 / 2 + O (x4)?
What I want to find out is how to find the peyano remainder in Taylor expansion?
I want to know if a function f (x) is expanded to x ^ n, then the peyano remainder must be o (x ^ n)?

SiNx = x-x 3 / 6 + O (x 3) and SiNx = x-x 3 / 6 + O (x 4) are OK
Because the next term of Taylor's formula of SiNx is X5 / 5! Which is higher order than X3 and x4, it is OK to write o (x3) or O (x4) in this place
However, if the topic is to ask you to write the Taylor formula of SiNx, this place is still decided to use o (x3) according to the last item of the previous expansion - X3 / 6. If the Taylor formula is used to calculate the limit, then whether to use o (x3) or O (x4) depends on the topic
Similarly, X2 = 1 + x2 + X4 / 2 + O (x5) and 1 + x2 + X4 / 2 + O (x4) of E are OK, because the next term of Taylor formula of x2 of E is X6 / 6, which is higher order than X4 and X5
Generally, if a function f (x) is expanded to x ^ n, the peyano remainder is written as O (x ^ n)

Finding the extreme value of function: z = - x square + xy-y square + 2x-y

dz/dx = -2x +y +2=0
dz/dy = x -2y -1 =0
We get x = 2Y + 1
-2(2y+1)+y+2=0
We get y = 0, x = 1
The minimum value is Z = - 1 + 2 = 1

Taylor formula can be expressed as: F (a + H) = f (a) + F '(a) H + F' '(a) H & # 178; / 2! + O (H & # 178;)