The function f (x) = 1 / (4-3x + x ^ 2) is expanded into a power series about X

The function f (x) = 1 / (4-3x + x ^ 2) is expanded into a power series about X

It can be expanded into a power series of (x + 3 / 2)
f(x)=1/(4-3x+x^2)=1/[(X-3/2)^2+7/4]=4/7*﹛1/[1+(x2/√7-3/√7) ²] Take (x2 / √ 7-3 / √ 7) as a whole and get f (x) = (4 / 7)* Ε (n = 0 ~ ∞) (- 4 / 7) ^ n * (x-3 / 2)) ^ 2n where (3 - √ 7) / 2

The function f (x) = 1 / (2x ^ 2-3x + 1) is expanded into a power series of X RT

First decompose into partial fractions and then expand
f(x)=1/[(2x-1)(x-1)]
=1/(x-1)-2/(2x-1)
=-1/(1-x)+2/(1-2x)
=-[1+x+x^2+x^3+.+x^n+..]+2[1+2x+4x^2+8x^3+...+2^nx^n+..]
=1+3x+7x^2+15x^3+...+(2^(n+1)-1)x^n+..

Find the power series that expands the function f (x) = 1 / (2-3x + x) into x? Prawns help answer

F (x) = 1 / (X-2) (x-1) = 1 / (X-2) - 1 / (x-1) = 1 / 2 (1-x / 2) + 1 / (1-x) = 1 / 2 Σ (x / 2) n + ∑ x n Σ above is infinity, below is n = 0, X range is (- 1,1)

The function f (x) = 1 / (x-3) is expanded into a power series of X

f(x)=-1/3*1/(1-x/3)
=-1/3*[1+x/3+x^2/9+x^3/27+x^4/81+.]
=-1/3-x/9-x^2/27-x^3/81-...
The convergence region is | x | < 3

F (x) = x Λ 2 SiNx is expanded into a power series of X, which contains the coefficients of X Λ 7

g(x) = sinxg'(x) = cosxg''(x) = -sinxg'''(x)= -cosxg''''(x) = sinxg'''''(x) = cosxg(x) = g(0) +(g'(0)/1!)x +(g''(0)/2!)x^2+.f(x) = x^2.sinx= g(0) x^2+(g'(0)/1!)x^3 +(g''(0)/2!)x^4+.coef.of x^7=g^(5)(0...

Series expansion of sine function, such as series expansion of SiNx, cosx

sin x = x-x^3/3!+ x^5/5!-... (-1)^(k-1)*x^(2k-1)/(2k-1)!+ Rn(x)(-∞