Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 8

A power series centered at c=0 is follows the formula:
sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...
The given function h(x)= 1/(1-5x) resembles the power series centered at c=0 :
(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n
or
(1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...
To evaluate the given function h(x) =1/(1-5x) centered at c=0 , we may apply Law of exponents: 1/x^n = x^(-n) .
h(x)= (1-5x) ^(-1)
Apply the aforementioned formula for power series on (1-5x) ^(-1) or (1+(-5x))^(-1) , we may replace "x " with "-5x " and "k " with "-1 ". We let:
(1+(-5x))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-5x) ^n
=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-5)^nx ^n
=1+(-1)(-5)^1x +(-1(-2))/(2!)(-5)^2x ^2+(-1(-2)(-3))/(3!)(-5)^3x ^3+(-1(-2)(-3)(-4))/(4!)(-5)^4x ^4+...
=1+5x +2/2*25*x ^2+(-6)/6(-125)x ^3+24/24*625*x ^4+...
=1+5x +25x ^2+125x ^3+625x ^4+...
= sum_(n=0)^oo (5x)^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=0)^oo (5x)^n with sum_(n=0)^oo a*r^n , we determine: r = 5x .
Apply the condition for convergence of geometric series: |r|lt1 .
|5x|lt1
-1 lt5xlt1
Divide each part by 5 :
(-1)/5 lt(5x)/5lt1/5
-1/5ltxlt1/5
Thus, the power series of the function h(x)=1/(1-5x) centered at c=0 is sum_(n=0)^oo(5x)^n with an interval of convergence: -1/5 ltxlt1/5 .

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