f(x) = 1/x = x^-1f(x)=1x=x−1
so:
(df)/dx = (-1)x^(-2)dfdx=(−1)x−2
(d^2f)/dx^2 = (-1)(-2)x^(-3)d2fdx2=(−1)(−2)x−3
and clearly in general:
(d^nf)/dx^n = (-1)^n(n!)x^(-n-1) = (-1)^n(n!)/x^(n+1)dnfdxn=(−1)n(n!)x−n−1=(−1)nn!xn+1
The coefficients of the Taylor series around x=2x=2 are then:
c_n = f^((n))(2)/(n!) = (-1)^n/2^(n+1)cn=f(n)(2)n!=(−1)n2n+1
so that:
1/x = sum_(n=0)^oo (-1)^n (x-2)^n/2^(n+1)1x=∞∑n=0(−1)n(x−2)n2n+1
or:
1/x = 1/2sum_(n=0)^oo (-1)^n (x-2)^n/2^n = 1/2sum_(n=0)^oo (-1)^n ((x-2)/2)^n = 1/2sum_(n=0)^oo (-(x-2)/2)^n = 1/2sum_(n=0)^oo (1-x/2)^n1x=12∞∑n=0(−1)n(x−2)n2n=12∞∑n=0(−1)n(x−22)n=12∞∑n=0(−x−22)n=12∞∑n=0(1−x2)n
We can also find the radius of convergence of this series by noting that:
1/x = 1/(2-2+x) = 1/2 1/(1-1+x/2) = 1/2 1/(1-(1-x/2))1x=12−2+x=1211−1+x2=1211−(1−x2)
We know that the sum of a geometric series is:
sum_(n=0)^oo alpha^n = 1/(1-alpha)∞∑n=0αn=11−α
and is convergent in the interval alpha in (-1,1)α∈(−1,1)
So we have:
1/x = 1/2 1/(1-(1-x/2)) = 1/2 sum_(n=0)^oo (1-x/2)^n1x=1211−(1−x2)=12∞∑n=0(1−x2)n
converging for:
-1 < 1-x/2 < 1−1<1−x2<1
-2 < -x/2 < 0−2<−x2<0
0 < x/2 <= 20<x2≤2
0 < x < 40<x<4
