10-24-2025 Power Series

Basic geometric series

∑n=0∞xn=1+x+x2+x3+…+xn+…=11−x\sum_{n=0}^\infty x^n = 1 + x + x² + x³ + … + x^n + … = \frac 1{1-x}for |x| < 1

Theorem 22. Properties of Functions Defined by Power Series

  • If a power series ∑n=0∞cn(x−a)n\sum_{n=0}^\infty c_n(x-a)^n has a non-zero convergence R, then the function defined by f(x)=∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+…f(x) = \sum_{n=0}^\infty c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)²+… is continuous and differentiable on the interval (a−R,a+R)(a-R,a+R) and 

    • f’(x)=ddx∑n=1∞cn(x−a)n→∑n=0∞d(cn(x−a)n)dx⇒∑n=1∞ncn(x−a)n−1=c1+2c2(x−a)+3c3(x−a)2+…f’(x) = \frac d{dx} \sum_{n=1}^\infty c_n(x-a)^n \rightarrow \sum_{n=0}^\infty \frac{d(c_n(x-a)^n)}{dx} \Rightarrow \sum_{n=1}^\infty nc_n(x-a)^{n-1}=c_1+2c_2(x-a)+3c_3(x-a)²+…

    • ∫f(x)dx=∫∑n=0∞cn(x−a)ndx\int f(x)dx = \int \sum_{n=0}^\infty c_n(x-a)^n dx

      • =∑n=0∞∫cn(x−a)ndx=\sum_{n=0}^\infty \int c_n(x-a)^n dx

      • =C+∑n=0∞cn(x−a)n+1n+1=C+\sum_{n=0}^\infty c_n \frac {(x-a)^{n+1}}{n+1}

      • =C+c0(x−a)+c1(x−1)22+c2(x−a)33+…=C+c_0(x-a)+c_1 \frac {(x-1)²}2 + c_2 \frac {(x-a)³} 3 + …

Practice: Find a power series representation for f(x)=1(1+x)2f(x) = \frac 1{(1+x)²} and find its radius of convergence

For g(x)=11+x,g’(x)=−1(1+x)2=−f(x)g(x) = \frac 1{1+x}, g’(x) = - \frac 1{(1+x)²} = -f(x)

11+x=11−(−x)=1−x+x2−x3+x4−…=∑(−1)n⋅xn\frac 1{1+x} = \frac 1{1-(-x)} = 1 - x + x² - x³ + x^4 - … = \sum (-1)^n \cdot x^n for |-x| < 1

d(11+x)dx=−1(1+x)2=−1+2x−3x2+…=∑n=0∞d((−1)n⋅xn)dx=∑n=1∞(−1)n⋅nxn−1\frac {d(\frac 1{1+x})}{dx} = \frac {-1}{(1+x)²} = -1 + 2x - 3x² + … = \sum_{n=0}^\infty \frac {d((-1)^n \cdot x^n)}{dx} = \sum_{n=1}^\infty (-1)^n \cdot nx^{n-1}

Therefore:

  • f(x)=1(1+x)2=1−2x+3x2−…=∑n=1∞(−1)n+1⋅nxn−1f(x) = \frac 1{(1+x)²} = 1 - 2x + 3x² - … = \sum _{n=1}^\infty (-1)^{n+1} \cdot nx^{n-1}

The radius of convergence is the same as that of the original series, R = 1. We leave it as an exercise to show that (-1, 1) is the interval of convergence for both series.

Find a power series representation for ln(1+x)ln(1+x) and find its radius of convergence.

Recall that           ln⁡(1+x)+C=∫11+xdx\ln(1+x) + C = \int \frac 1{1+x} dx

Using equation 2:

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

   =1+(−x)+(−x)2+(−x)3+(−x)4=…= 1 + (-x) + (-x)² + (-x)³ + (-x)^4 = …

=1−x+x2−x3+x4−…= 1 - x + x² - x³ + x^4 - …

Therefore, the following families of functions are equal.

ln⁡(1+x)+C=∫11+xdx=∫(1−x+x2−x3+x4−…)dx\ln(1+x) + C = \int \frac 1{1+x} dx = \int (1-x+x²-x³+x^4-…)dx

=∫∑n=0∞(−1)nxndx=∑n=0∞∫(−1)nxndx=\int \sum_{n=0}^\infty (-1)^nx^n dx = \sum_{n=0}^\infty \int (-1)^n x^n dx

=x−x22+x33−x44+…+C‾= x - \frac {x²} 2 + \frac {x³} 3 - \frac {x^4}4 + … + \overline C

=∑n=0∞(−1)nxn+1n+1+C‾=\sum_{n=0}^\infty (-1)^n \frac {x^{n+1}}{n+1} + \overline C

Therefore there is a constant C‾\overline C such that 

ln⁡(1+x)=∑n=0∞(−1)nxn+1n+1+C‾\ln (1+x) = \sum_{n=0}^\infty (-1)^n \frac {x^{n+1}}{n+1} + \overline C

To fine the C‾\overline C that gives ln⁡(1+x)\ln (1+x), we put x=0x=0 into this equation which gives us

ln⁡(1+x)=∑n=0∞(−1)nxn+1n+1\ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac {x^{n+1}}{n+1}


By Theorem 22 the radius of convergence is R = 1, since this is the radius of convergence for 11+x=∑n=0∞(−1)nxn\frac 1{1+x} = \sum_{n=0}^\infty (-1)^nx^n. Notice that (-1, 1) is the largest open interval of values of x centered at 0 on which ln⁡(1+x)\ln(1+x) is defined. At the endpoints −1-1 and 11 we have the negative of the harmonic series and the alternating harmonic series, respectively. Thus, the interval of convergence is (-1, 1].