10-20-2025 Power Series

f(x)=11+xf(x) = \frac 1{1+x}

Reminder:

  • This is a geometric series:

∑n=0∞arn=a1−r\sum_{n=0}^\infty ar^n = \frac a{1-r}

A power series contains a variable, xx raised to the nn power

∑n=0∞xn=11−x\sum_{n=0}^\infty x^n = \frac 1{1-x}

  • This formula looks very similar to the one at the top

  • This center is centered at x = 0

    • We can rewrite this as ∑n=0∞(x−c)n\sum_{n=0}^\infty (x-c)^n

Example:

f(x)=11+xf(x) = \frac 1{1+x}

We need to rewrite this in the form of a1−r\frac a{1-r}

  • We need to convert 1+x1 + x to 1−r1 - r

11−x=11−(−x)=a1−r\frac 1{1-x} = \frac 1{1-(-x)} = \frac a{1-r}

  • 1 = a

  • -x = r

∑n=0∞arn=∑n=0∞1(−x)n=∑n=0∞(−1)n(x)n\sum_{n=0}^\infty a r^n = \sum_{n=0}^\infty 1 (-x)^n =\sum_{n=0}^\infty(-1)^n (x)^n

Finding the interval of convergence

  • |r| < 1 \rightarrow Converge

  • |-x| < 1

    • |x-c| < R

    • R = 1

|x| < 1 \rightarrow -1 < x < 1

c=0c = 0

R=1R=1

f(x)=11+x=∑n=0∞(−1)nxnf(x) = \frac 1{1+x} = \sum_{n=0}^\infty (-1)^n x^n

  • Sn=1−x1+x2−x3+x4−x5+…S_n = 1-x^1+x²-x³+x^4-x^5+…

f(0.5)=11+0.5=0.667f(0.5) = \frac 1{1 + 0.5} = 0.667

Using the power series to estimate the answer:

1−0.5+0.52−0.53+0.54−0.55+0.56−0.57=0.66411 - 0.5 + 0.5² - 0.5³ + 0.5^4 - 0.5^5 + 0.5^6 - 0.5^7 = 0.6641

Practice 2:

f(x)=1xf(x) = \frac 1x

a1−r\frac a{1-r}

We need to change xx to 1−r1-r

1x=1x+1−1=11+x−1\frac 1x = \frac 1{x+1-1} = \frac 1{1 + x - 1}

→11−(−x+1)=a1−r\rightarrow \frac 1{1-(-x+1)} = \frac a{1-r}

a=1a = 1

r=(−x+1)r = (-x+ 1)

∑n=0∞arn=∑n=0∞(−x+1)n\sum_{n=0}^\infty ar^n = \sum_{n=0}^\infty (-x+1)^n

∑n=0∞(x−c)n\sum_{n=0}^\infty (x-c)^n

→∑n=0∞[−1(x−1)]n=∑n=0∞(−1)n(x−1)n\rightarrow \sum_{n=0}^\infty [-1(x-1)]^n = \sum_{n=0}^\infty (-1)^n(x-1)^n

  • =1x=\frac 1x

C=1C = 1

r=−1(x−1)r = -1(x-1)

|r| < 1

|-1(x-1)| < 1

|x-1| < 1 \rightarrow -1<x-1<1

0<x<2

Interval of convergence is (0,2)(0,2)

Practice 3:

f(x)=11−x3f(x) = \frac 1{1-x³}

a1−r\frac a{1-r}

We need to change 1−x31-x³ to 1−r1-r

a=1a=1

r=x3r = x³

∑n=0∞arn=∑n=0∞(x3)n=∑n=0∞x3n\sum_{n=0}^\infty ar^n = \sum_{n=0}^\infty (x³)^n = \sum_{n=0}^\infty x^{3n}

Sn=1+x3+x6+x9=11−x3S_n = 1 + x³ + x^6 + x^9 = \frac 1{1-x³}

Interval of convergence:

|r| < 1 \rightarrow |x³| < 1

-1 < x³ < 1 \rightarrow -1 < x < 1

Interval of convergence is (−1,1)(-1, 1)

Practice 3:

f(x)=13−xf(x) = \frac 1{3-x}

Changing 13−x⇒a1−r\frac 1{3-x} \Rightarrow \frac a{1-r}

c=0;c=1c = 0; c = 1

When centered at 0:

  • ∑n=0∞xn\sum_{n=0}^\infty x^n

When centered at another number:

  • ∑n=0∞(x−c)n\sum_{n=0}^\infty(x-c)^n

c = 0

  • a1−r\frac a{1-r}

f(x)=13−x=13−x⋅1313=131−(x3)=a1−rf(x) = \frac 1{3-x} = \frac 1{3-x} \cdot \frac {\frac 13}{\frac 13} = \frac {\frac 13}{1-(\frac x3)} = \frac a{1-r}

a=13a = \frac 13

r=x3r = \frac x3

∑n=0∞arn=∑n=0∞13(x3)n\sum_{n=0}^\infty ar^n = \sum_{n=0}^\infty \frac 13(\frac x3)^n

∑n=0∞131⋅xn3n=∑n=0∞xn3n+1\sum_{n=0}^{\infty}\frac1{3^1}\cdot\frac{x^n}{3^n}=\sum_{n=0}^{\infty}\frac{x^{n}}{3^{n+1}}

Interval of convergence:

|r| < 1

r=x3r = \frac x3

|\frac x3| < 1

-1 < \frac 13 x < 1

-3 < x < 3