Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 12

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)=4/(3x+2) 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) =4/(3x+2)
=4(3x+2)^(-1)
Let u =3x+2 then (du)/(dx) = 3 .
d/(dx) c*(3x+2)^n = c *d/(dx) (3x+2)^n
= c *(n* (3x+2)^(n-1)*3
= 3cn(3x+2)^(n-1)
f'(x) =d/(dx) 4(3x+2)^(-1)
=3*4*(-1) *(3x+2)^(-1-1)
=-12(3x+2)^(-2) or 2/(3x+2)^2
f^2(x) =d/(dx) -12(3x+2)^(-2)
=3*(-12)(-2)(3x+2)^(-2-1)
=72(3x+2)^(-3) or 72/(3x+2)^3
f^3(x) =d/(dx) 72(3x+2)^(-3)
=3*(72)(-3)(3x+2)^(-3-1)
=-648(3x+2)^(-4) or -648/(3x+2)^4
Plug-in x=3 for each f^n(x) , we get:
f(3)=4/(3(3)+2)
=4/ 11
f'(3)=-12/(3(3)+2)^2
=-12/11^2
= -12/121
f^2(3)=72/(3(3)+2)^3
=72/11^3
=72/1331
f^3(3)=-648/(3(3)+2)^4
=-648/11^4
= -648/14641
Plug-in the values on the formula for Taylor series, we get:
4/(3x+2)= sum_(n=0)^oo (f^n(3))/(n!) (x-3)^n
= sum_(n=0)^oo (f^n(3))/(n!) (x-3)^n
=4/11+(-12/121)(x-3) +(72/1331)/(2!)(x-3)^2 +(-648/14641)/(3!)(x-3)^3 +...
=4/11-(12/121)(x-3) +(72/1331)/2(x-3)^2 - (648/14641)/6(x-3)^3 +...
=4/11-12/121(x-3) +36/1331(x-3)^2 -108/14641(x-3)^3 +...
= sum_(n=1)^oo 4(-3(x-3))^(n-1)/11^n
= sum_(n=1)^oo 4(-3(x-3))^(-1)(-3(x-3))^n/11^n
= sum_(n=1)^oo 4/(-3(x-3))((-3(x-3))/11)^n
=sum_(n=1)^oo 4/(-3x+9)((-3(x-3))/11)^n
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 4/(-3x+9)((-3(x-3))/11)^n with sum_(n=0)^oo a*r^n , we determine: r = (-3(x-3))/11 .
Apply the condition for convergence of geometric series: |r|lt1 .
|(-3(x-3))/11|lt1
|-1|*|(3(x-3))/11|lt1
1*|(3(x-3))/11|lt1
|(3(x-3))/11|lt1
|(3x-9)/11|lt1
-1lt(3x-9)/11lt1
Multiply each sides by 11 :
-1*11lt(3x-9)/11*11lt1*11
-11lt3x-9lt11
Add 9 on each sides:
-11+9lt3x-9+9lt11+9
-2lt3xlt20
Divide each side by 3 :
-2/3lt(3x)/3lt20/3

-2/3ltxlt20/3
Thus, the power series of the function f(x) =4/(3x+2) centered at c=3 is sum_(n=1)^oo 4/(-3x+9)((-3(x-3))/11)^n
with an interval of convergence: -2/3ltxlt20/3 .

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