Questions about expansion of function into power series In the part of learning Taylor's formula, we know that if the function f (x) has a derivative up to (n + 1) order in a neighborhood of x0, then the n-order Taylor's formula of F (x) in that neighborhood is a polynomial + RN (x) remainder term. This formula should be constant as long as the function f (x) has a derivative up to (n + 1) order in a neighborhood of x0, But if f (x) has all order derivatives in a neighborhood of x0, we can write down the polynomial infinitely, that is, f (x) = polynomial (infinite term), then the corresponding formula should also be tenable, as long as f (x) has all order derivatives in a neighborhood of x0, and the function is in this neighborhood? If I'm right, Then look at the following But why does the expansion of a function into a power series only hold in the common part of the convergence domain of the series and the definition domain of the function? For example, 1 / 1-x = 1 + X + x ^ 2 +. + x ^ n +... Only in (- 1

Questions about expansion of function into power series In the part of learning Taylor's formula, we know that if the function f (x) has a derivative up to (n + 1) order in a neighborhood of x0, then the n-order Taylor's formula of F (x) in that neighborhood is a polynomial + RN (x) remainder term. This formula should be constant as long as the function f (x) has a derivative up to (n + 1) order in a neighborhood of x0, But if f (x) has all order derivatives in a neighborhood of x0, we can write down the polynomial infinitely, that is, f (x) = polynomial (infinite term), then the corresponding formula should also be tenable, as long as f (x) has all order derivatives in a neighborhood of x0, and the function is in this neighborhood? If I'm right, Then look at the following But why does the expansion of a function into a power series only hold in the common part of the convergence domain of the series and the definition domain of the function? For example, 1 / 1-x = 1 + X + x ^ 2 +. + x ^ n +... Only in (- 1

If the function f (x) has a derivative up to order (n + 1) in a neighborhood of x0, then the Taylor formula of order n of F (x) in this neighborhood is a polynomial + RN (x) remainder term. This formula should be constant as long as the function f (x) has a derivative up to order (n + 1) in a neighborhood of x0 and the function is in this neighborhood

How to understand that a divided by B equals a of B
It's not clear

A divided by B = a △ B = a * 1 / b = A / b divided by a number equals to the reciprocal of the number

If the domain is an odd function of R, y = (x) period is t, (T > 0), then why is f (T / 2) equal to 0?
F (- t / 2) = f (T - (T / 2)) = f (T / 2), so f (T / 2) = 0, why?

f(-T/2)+f(T/2)=0
And f (- t / 2 + T) = f (- t / 2) = f (T / 2)
So 2F (T / 2) = 0
Get proof

What is the difference between the reciprocal of 2 and the quotient of 2.5 divided by 4.2, minus the quotient of 0.5 divided by 1 / 4

（½+2.5）÷4.2-0.5÷¼
=3÷4.2-2
=5/7-2
=﹣9/7

The product of algebraic formula (AX ^ 2 + BX + 1) and (3x ^ 2-2x + 1) does not contain x ^ 3 and X term 1. Find the value of a and B 2. Find the value of these two polynomials

X ^ 3 is generated because x ^ 2 is multiplied by X
a * （-2） + b * 3 = 0
Similarly, the sum of the factors that produce the X term is also 0:
b + （-2） = 0
The results are as follows
a = 3,b = 2；
The product is:
9X^4 + 2X^2 + 1

Li Bai bought wine: "no matter walking on the street, pick up a pot to buy wine. If you meet a shop, you can double it. If you meet a flower, you can drink a bucket. If you meet a shop and a flower three times, you can drink all the wine in the pot."
How many buckets are there in the pot

Because after drinking the wine, if you meet the shop later, you can't have no wine
If there is x Dou wine in the pot, we can get the following relations
Once in the shop and after the flower, the wine in the pot is 4x-3;
For eryudian and Huahou, Huzhong liquor was 4 (4x-3) - 3;
The results showed that: 4 [4 (4x-3) - 3] - 3; 3;
Therefore, there is a relation: 4 [4 (4x-3) - 3] - 3 = 0 one
Solution 1: x = 63 / 64 (bucket)
So the original pot 63 / 64 Doujiu
checking calculation:
First store: 63 * 4 / 64 = 252 / 64
First flower: (252-64 * 3) / 64 = 60 / 64
Second store: 60 * 4 / 64 = 240 / 64
Second flower: (240-64 * 3) / 64 = 48 / 64
The third store: 48 * 4 / 64 = 192 / 64
The third flower: (192-64 * 3) / 64 = 0

Given (M2 + N2) (M2 + n2-9) - 10 = 0, find the value of the algebraic formula M2 + N2

Let y = M2 + N2, then the original equation becomes: Y (Y-9) - 10 = 0, sorted out: y2-9y-10 = 0, (Y-10) (y + 1) = 0, the solution is: y = 10 or y = - 1, then M2 + N2 = 10, or M2 + N2 = - 1 (rounding off), so M2 + N2 = 10

Simple calculation 5 / 6 minus bracket 1 / 6 minus 3 / 4 reverse bracket

The original formula = 5 / 6 minus 1 / 6 plus 3 / 4
=Four sixths plus three fourths
=17 out of 12

A moving point is inscribed with the fixed circle X & sup2; + Y & sup2; + 4y-32 = 0 and passes through the fixed point a (0,2), and the trajectory equation of the center P of the moving circle is obtained
The focus is on the x-axis and the y-axis. What do you do next
Wrong. The focus is on the y-axis

If the radius of the moving circle is r, then the distance between the centers of the moving circle and the fixed circle is equal to the difference of the radius. That is: | PM | = 6-r. if the moving circle passes through the fixed point a (0,2), then | PA | = R, then | PM | = 6 - | PA |, that is, | PC | + | PA | = 6, and | PC | + |

Zero - bracket 2 and 2 / 3 - bracket negative 4 has 1 / 4 + bracket 2 and 2 / 3
Please answer quickly now

- 4 and 4 branches 1