The second order McLaughlin formula for f (x) = e ^ SiNx I want to know how to work out the rest... It's a very complicated thing

The second order McLaughlin formula for f (x) = e ^ SiNx I want to know how to work out the rest... It's a very complicated thing

f(x)＝e^sinx,f(0)＝1
f'(x)＝e^sinx×cosx,f'(0)＝1
f''(x)＝e^sinx×cosx×cosx－e^sinx×sinx,f''(0)＝1
So, e ^ SiNx = 1 + X + 1 / 2 × x ^ 2 + O (x ^ 2)
－－－－－－－－
There are two forms of remainder, O (x ^ 2) is Peano type remainder, Lagrange type remainder needs third derivative
f'''(x)＝－e^sinx×cosx×(sinx)^2－3e^sinx×sinxcosx
So the remainder of Lagrange type is 1 / 6 × f '' '(ξ) × x ^ 3
e^sinx＝1＋x＋1/2×x^2－1/6×[e^sinξ×cosξ×(sinξ)^2＋3e^sinξ×sinξcosξ]×x^3

The McLaurin series expansion of F (x) = arctanx is________ ?

Σ (- 1) ^ n * x ^ (2n + 1) / (2n + 1) (n from 0 to ∞)
|x|

The McLaurin series of function f (x) = ∫ (1-cos √ x) / X DX (upper limit x, lower limit 0) is
∑（-1）^(n-1) x^n/[(2n)!n]

cosx=1-x^2/2!+x^4/4!-.+(-1)^nx^(2n)/(2n)!+.
cos√x=1-x/2!+x^2/4!-.+(-1)^nx^(n)/(2n)!+.
1-cos√x=x/2!-x^2/4!+.+(-1)^(n-1)x^(n)/(2n)!+.
(1-cos√x)/x=1/2!-x/4!+.+(-1)^(n-1)x^(n-1)/(2n)!+.
So: F (x) = ∫ (1-cos √ x) / X DX
=∫(1/2!-x/4!+.+(-1)^(n-1)x^(n-1)/(2n)!+.）dx
=x/2!-x^2/4!2+.+(-1)^(n-1)x^(n)/(2n)!n+.
=∑(1,+∞)(-1)^(n-1)x^(n)/[(2n)!n]

The second derivative of COS series
We know that f (x) is a series of COS, n = 0 - to - infinite [(- 1) ^ n * x ^ 2n] / (2n)!
Please prove f '' (x) + F (x) = 0 by series
Don't seek the second derivative cosx directly

f(x)=cosx
The sum of cosx = n from 0 to infinity [(- 1) ^ n * x ^ 2n] / (2n)!, X belongs to R
So f '(x) = {n sum from 0 to infinity [(- 1) ^ n * x ^ 2n] / (2n)!}'
={n sum from 0 to infinity [(- 1) ^ n * x ^ 2n] / (2n)!] '}
={n sum from 0 to infinity [(- 1) ^ n * 2n * x ^ (2n-1)] / (2n)!] '}
={n sum from 1 to infinity [(- 1) ^ n * x ^ (2n-1)] / (2n-1)!]}
therefore
f''(x)=[f'(x)]'
={n sum from 1 to infinity [(- 1) ^ n * x ^ (2n-1)] / (2n-1)!]} '
={n sum from 1 to infinity [(- 1) ^ n * x ^ (2n-1)] / (2n-1)!] '}
={n sum from 1 to infinity [(- 1) ^ n * (2n-1) * x ^ (2n-2)] / (2n-1)!]}
={n sum from 1 to infinity [(- 1) ^ n * x ^ (2n-2)] / (2n-2)!]}
={n sum from 0 to infinity [(- 1) ^ (n + 1) * x ^ 2n] / (2n)!}
=-{n sum from 0 to infinity [(- 1) ^ n * x ^ 2n] / (2n)!}
=-f(x)
So f '' (x) + F (x) = 0

What is f (x) = 2 ^ x expanded into McLaughlin series?

f '(x)=ln2 * 2^x
f ''(x)= (ln2)^2 * 2^x
……
The Nth derivative of F (x) = (LN2) ^ n * 2 ^ x,
So when x = 0, 2 ^ x = 1,
So "(LNF) = (2, LNF) = 0 (LNF) = (2, LNF) = (0, LNF) = (2, LNF) = (0, LNF) = (0, LNF) = (2, LNF) = (0, LNF) = (2, LNF) = (0, LNF) f (n) (0)=(ln2)^n,
So f (x) = 2 ^ x develops into McLaurin series,
f(x)=f(0) +f '(0) x +f "(0) x^2 /2!+ f "'(0) x^3 /3!+…… + f(n) (0) x^n /n!+Rn(x)
=1 +ln2 *x +(ln2)^2 * x^2/2!+ (ln2)^3 * x^3 /3!+…… + (ln2)^n * x^n /n!+Rn(x)
Where RN (x) is the remainder

Given f (x) = [(1-x) ^ n] cos (π x), find the n-th derivative of F (1)

It is easy to understand that f (x) = [(1-x) ^ n] cos (π x) is equivalent to ∑ {[(1-x) ^ n] m-order derivative * cos (π x) n-m-order derivative}, where m is a natural number less than or equal to N. obviously, when m is not equal to N, when x = 1, the value of m-order derivative of [(1-x) ^ n] is 0

How to find the McLaurin series of 1 / x?

Taylor series of function f (x) at x = 0 is called McLaughlin series
Taylor series requires f (x) to be differentiable of any order in some field of x 0
But f (x) = 1 / X has no definition at x = 0, let alone differentiable
Therefore, there is no McLaughlin series with F (x) = 1 / X

The - T ^ 2 power DT of F (x) = ∫ (0 to x) e is expanded into Maclaurin series
Write out the process or idea

The second derivative: - 2xe ^ (- x ^ 2) the third derivative: - 2E ^ (- x ^ 2) + 4x ^ 2E ^ (- x ^ 2) the fourth derivative: - 4xe ^ (- x ^ 2) + 8XE ^ (- x ^ 2) - 8XE ^ 3E ^ (- x ^ 2). Obviously, f (0) = 0f '(0) = 1F' '(0) = 0f' '(0) = - 2

The sin and COS expansion terms of McLaurin's formula with Lagrange remainder
McLaurin expansion of sin (x)
sin(x) = x - x^3 / + x^5 / + ...+ ((-1)^(m-1))*((x^(2m-1)) / (2m - 1)!) + ((-1)^(m))*(cos(θx)*(x^(2m+1)) / (2m + 1)!)
cos(x) = 1 - x^2 / + x^4 / + ...+ ((-1)^(m))*((x^(2m)) / (2m)!) + ((-1)^(m+1))*(cos(θx)*(x^(2m+2)) / (2m + 2)!)
According to the definition of Taylor formula
If f has a continuous derivative of order n on [a, b] and a derivative of order n + 1 in (a, b), then for any given x, x0 ∈ [a, b], there is at least one point ξ ∈ [a, b] that makes the formula hold
So the definition mentioned that f has n + 1 order derivative, but did not mention the existence of N + 2 order derivative
Then the N + 1 derivative of COS in the remainder of COS should be 0, which is written in the above formula
Why can n + 2 derivative of COS be written like this

Because COS is differentiable of any order

SiNx expands the McLaughlin series and the result is sin (x + n π / 2)

Let y = SiNx
y '=cosx=sin(x+π/2)
y ''=(sin(x+π/2))'=cos(x+π/2)=sin(x+π)
y'''=(sin(x+π))'=cos(x+π)=sin(x+3π/2)
and so on
The n-order derivative of Y is sin (x + n π / 2)