What is the Taylor formula for expanding a power series of a function into a Taylor series? If so, please give an example!

What is the Taylor formula for expanding a power series of a function into a Taylor series? If so, please give an example!

F (x) = f (x0) + F '(x0) (x-x0) + [f' '(x0) / 2!] / (x-x0) ∧ 2 +. + [FN (x0) / N!] (x-x0) ∧ n +... The right side is the Taylor expansion of F at x = 0
In practical application, the expansion at x0 = 0 is mainly discussed
For example, find the expansion of F (x) = e Λ X
Because FN (x) = e Λ x, FN (0) = 1, (n = 1,2,3.)
So the Lagrangian remainder of F is RN (x) = [e Λ (θ x) / (n + 1)!] · x Λ (n + 1), (0 ≤ θ ≤ 1)
It is obvious that | RN (x) | ≤ [(E ∧ x |) / (n + 1)!] ·| x | (n + 1),
It has Lim [(E Λ|x|) / (n + 1)!] ·|x| (n + 1) = 0 for any real number X
So limrn (x) = 0
So f (x) = e Λ x = 1 + (1 / 1!) x + (1 / 2!) x Λ 2 +. + (1 / N!) x Λ n +,

Expand f (x) = 1 / radical (1-x ^ 2) into a power series of X
Serious answer, score is not enough to apply for additional!

We can refer to the generalized binomial theorem

Open the function f (x) = 1 / radical (1-x ^ 2) into a power series of X

Use McLaughlin's formula
f(x)=1-x^2/2+x^4/4!+.

Expansion of x ^ 2 cosx into power series of X

The function f (x) = 1 / (4-x) is expanded into a power series of (X-2)

1 / (4-x) = 1 / 2 [1 / (1-1 / 2 (X-2))] and then the sum formula A1 / (1-Q) = ∑ (n from 1 to infinity) an has [1 / (1-1 / 2 (X-2))] = ∑ (n from 0 to infinity). Note that n here is the index of common ratio, so it starts from 0 [1 / 2 (X-2)] ^ n so 1 / (4-x) = 1 / 2 [1 / (1-1 / 2 (X-2))] = 1 / 2 ∑ (n

A sequence of natural numbers, the sum of the first n odd numbers equal to

The odd number columns composed of natural numbers are 1, 3, 5, 7
The first term is an arithmetic sequence with 1 tolerance of 2, according to the formula
Sn=1*n+n（n-1）*2/2
=n+ n²-n
=n²

The first term of an infinite equal ratio sequence is a natural number, and the common ratio is the reciprocal of a natural number. If the sum of the items of the sequence is 3, then the sum of the first two items of the sequence is 3
A 8/3 b 2 c 2/3 d 1/3
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d

The first term of an infinite equal ratio sequence {an} is a natural number, the common ratio is Q, and the reciprocal of Q is a natural number greater than 1
If the first term of infinite equal ratio sequence {an} is a natural number, the common ratio is Q, and the reciprocal of Q is a natural number greater than 1, then the sum of the first two terms of this sequence is 3____

The reciprocal of Q is greater than 1
So 00
So 3Q = 3-a1 > 0
a1

It is known that the common ratio of the equal ratio sequence {an} is Q > 1, n belongs to the natural number, and the square of the 17th term is equal to the 24th term
Find a1 + A2 + a3 + +an>1/a1 + 1/a2 + 1/a3 +…… +If 1 / an holds, the value range of n
(A)n>9 (B)n19 (D) n>16
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A:
If the square of item 17 is equal to item 24, A1 * q ^ 9 = 1
From a1 + A2 + a3 + +an>1/a1 + 1/a2 + 1/a3 +…… +The result is ^ an: 1

a1*q^9=1 ==> a1=1/(q^9)=q^(-9)
a1^2*q^(n-1)>1
(q^(-9))^2*q^(n-1)>1
q^(-18)*q^(n-1)>1
q^(n-19)>1
q>1
Then n-19 > 0
==> n > 19

When A1 = 1 and N is greater than 1 in the known sequence {an}, the square of 2Sn = 2ansn an to find an

2sn^2=2ansn-an2sn^2=2[sn-s(n-1)]sn-[sn-s(n-1)]2sn^2=2sn^2-2sns(n-1)-sn+s(n-1)2sns(n-1)+sn-s(n-1)=01/sn-1/s(n-1)=2 1/sn=1/s1+2(n-1)=2n-1sn=1/(2n-1)an=sn-s(n-1)=1/(2n-1)-1/(2n-3),a1=1,an=1/(2n-1)-1/(2n-...