Expand f (x) = ln (1 + x) into McLaurin series

Expand f (x) = ln (1 + x) into McLaurin series

ln(1+x)=x-1/2*x^2+1/3*x^3-1/4*x^4.+((-1)^n)/n+1)x^(n+1)

Finding the n-order McLaurin formula with piano remainder for f (x) = [ln (1 + x ^ 2)] / X

No need
ln(1+x)=∑[(-1)^n]x^(n+1)/n+1
ln(1+x^2)=∑[(-1)^n]x^2(n+1)/n+1
ln(1+x^2)/x=∑[-1)^n]x^(2n+1)/n+1

What is the McLaughlin formula of Ln (1-x)

ln(1-x)= -x+ x²/2 - x³/3 ...+(-1)^(n)x^(n)/n ...

McLaughlin expansion of function f (x) = (1 + x) ln (1 + x)
Thank you!
Pro claw machine can't see the picture and seek the text

f(x)=ln(1+x)+xln(1+x)=∑(-1)^(n-1)x^n /n +∑(-1)^(n-1)x^(n+1) /n
=x+∑(-1)^(n+1)x^(n+1) /[n(n+1)]

Prove X / 1 + x < ln (1 + x) < x (x is greater than 0) and verify the n-order McLaughlin formula of F (x) = ln (1 + x),

Look to the right first
Divide by two, then take e as the base index at the same time, and then expand e ^ X by McLaughlin
ln(1+x)/x=(1+x)/e^x=(1+x)/(1+x+x^2/2+x^3/6+.)

How to prove that sin (α + β) cos (α - β) = sin α cos α + sin β cos β

sinαcosα+sinβcosβ
=1/2（sin2α+sin2β）
=Sin (α + β) cos (α - β) and difference product formula
It can also be done on the left by using the sum difference formula

If z = cos θ - isin θ (I is an imaginary number), then the θ value of Z ^ 2 = - 1 may be

z=cosθ-isinθ
z^2=(cosθ-isinθ)^2
=(cosθ)^2-2icosθsinθ+(isinθ)^2
=(cosθ)^2-(sinθ)^2+2icosθsinθ
=cos2θ-isin2θ
=-1
So Cos2 θ = - 1, sin2 θ = 0
SO 2 θ = 2K π + π
So θ = k π + π / 2

Cos (15 π / 6) + isin (15 π / 6) = cos (5 π / 2) + isin (5 π / 2). How can 5 π / 2 be calculated

Divide 15 / 6 numerator and denominator by 3 to get
5/2

If z = cos (- π / 6) + isin (- π / 6), then argz ^ 3

z=cos(-π/6)+isin(-π/6)
z^3=cos(-π/2)+isin(-π/2)
=cos(3π/2)+isin(3π/2)
So argz ^ 3 = 3 π / 2

It is proved that the M + n power of a plus the M + n power of B is greater than the m power of a multiplied by the n power of B plus the n power of a multiplied by the m power of B

a^(m+n)+b^(m+n)-a^mb^n-a^nb^m
=a^m(a^n-b^n)+b^m(b^n-a^n)
=(a^m-b^m)(a^n-b^n)
Because a ≠ B
When a > b, a ^ m > b ^ m a ^ n > b ^ n, so (a ^ M-B ^ m) (a ^ N-B ^ n) > 0
When a