Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 6

To determine the power series centered at c, we may apply the formula for Taylor series:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) =f(c)+f'(c)(x-c) +(f''(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f'^4(c))/(4!)(x-c)^4 +...
To list the f^n(x) for the given function f(x)=2/(6-x) centered at c=-2 , we may apply Law of Exponent: 1/x^n = x^-n and Power rule for derivative: d/(dx) x^n= n *x^(n-1) .
f(x) =2/(6-x)
=2(6-x)^(-1)
Let u =6-x then (du)/(dx) = -1
d/(dx) c*(6-x)^n = c *d/(dx) (6-x)^n
= c *(n* (6-x)^(n-1)*(-1)
= -cn(6-x)^(n-1)
f'(x) =d/(dx)2(6-x)^(-1)
=-2*(-1)(6-x)^(-1-1)
=2(6-x)^(-2) or 2/(6-x)^2
f^2(x) =d/(dx) 2(6-x)^(-2)
=-2(-2)(6-x)^(-2-1)
=4(6-x)^(-3) or 4/(6-x)^3
f^3(x) =d/(dx)4(6-x)^(-3)
=-4(-3)(6-x)^(-3-1)
=12(6-x)^(-4) or 12/(6-x)^4
f^4(x) =d/(dx)12(6-x)^(-4)
=-12(-4)(6-x)^(-4-1)
=48(6-x)^(-5) or 48/(6-x)^5
Plug-in x=-2 for each f^n(x) , we get:
f(-2)=2/(6-(-2))
=2/ 8
=1/4
f'(-2)=2/(6-(-2))^2
=2/8^2
= 1/32
f^2(-2)=4/(6-(-2))^3
=4/8^3
=1/128
f^3(-2)=12/(6-(-2))^4
=12/8^4
= 3/1024
f^4(-2)=48/(6-(-2))^5
=48/8^5
= 3/2048
Plug-in the values on the formula for Taylor series, we get:
2/(6-x) = sum_(n=0)^oo (f^n(-2))/(n!) (x-(-2))^n
= sum_(n=0)^oo (f^n(-2))/(n!) (x+2)^n
=1/4+1/32(x+2) +(1/128)/(2!)(x+2)^2 +(3/1024)/(3!)(x+2)^3 +(3/2048)/(4!)(x+2)^4 +...
=1/4+1/32(x+2) +(1/128)/2(x+2)^2 +(3/1024)/6(x+2)^3 +(3/2048)/24(x+2)^4 +...
=1/4+1/32(x+2) + 1/256(x+2)^2 +1/2048(x+2)^3 + 1/16384(x+2)^4 +...
=1/2^2+1/2^5(x+2) + 1/2^8(x+2)^2 +1/2^11(x+2)^3 + 1/2^14(x+2)^4 +...
=sum_(n=1)^oo (x+2)^(n-1)/2^(3n-1)
=sum_(n=1)^oo( (x+2)^n*(x+2)^(-1))/(2^(3n)2^(-1))
=sum_(n=1)^oo (2(x+2)^n)/(2^(3n)(x+2))
=sum_(n=1)^oo (2/(x+2))((x+2)/2^3)^n
=sum_(n=1)^oo (2/(x+2))((x+2)/8)^n
Note: Exponents of 2 as 2,5,8,11,14,... follows arithmetic sequence a_n=a_0+(n-1)d.
a_n = 2 +(n-1)3
=2+3n-3
=3n-1
To determine the interval of convergence, we may apply geometric series test wherein the series sum_(n=0)^oo a*r^n is convergent if |r|lt1 or -1 ltrlt 1 . If |r|gt=1 then the geometric series diverges.
By comparing sum_(n=1)^oo (2/(x+2))((x+2)/8)^n with sum_(n=0)^oo a*r^n , we determine: r = (x+2)/8.
Apply the condition for convergence of geometric series: |r|lt1 .
|((x+2)/8)|lt1
-1lt(x+2)/8lt1
Multiply each sides by 8:
-1*8lt(x+2)/8*8lt1*8
-8ltx+2lt8
Subtract 2 from each sides:
-8-2ltx+2-2lt8-2
-10ltxlt6
Thus, the power series of the function f(x) = 2/(6-x) centered at c=-2 is sum_(n=1)^oo (x+2)^(n-1)/2^(3n-1) with an interval of convergence: -10ltxlt6 .