July 07, 2026 - Calculus 2 - Power Series Multiplication, Calculus, and Taylor Series

Multiplication of Power Series

• General Principle: Multiplying power series is conceptually identical to multiplying polynomials, with the distinction that the operation continues infinitely. You must look at each individual degree term by multiplying every term in the first series by every term in the second series.

• Example Case: Find the power series representation for $\frac{1}{(1-x)(1-x^2)}$.

  • This can be viewed as the product of two separate power series: $f(x) = \frac{1}{1-x}$ and $g(x) = \frac{1}{1-x^2}$.

  • $f(x) = \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + x^4 + \dots$

  • $g(x) = \sum_{n=0}^\infty (x^2)^n = 1 + x^2 + x^4 + x^6 + \dots$

• Manual Multiplication Process:

  • Take the second polynomial and multiply it by each term of the first polynomial.

  • Term by Term Analysis:

    • Degree 0 (Constant): Obtained only by $1 \times 1$. Result: $1$.

    • Degree 1 ($x$): Obtained by $x \times 1$. There are no $x$ terms in the second series. Result: $1x$.

    • Degree 2 ($x^2$): Obtained by $1 \times x^2$ and $x^2 \times 1$. Result: $2x^2$.

    • Degree 3 ($x^3$): Obtained by $x \times x^2$ and $x^3 \times 1$. Result: $2x^3$.

    • Degree 4 ($x^4$): Obtained by $1 \times x^4$, $x^2 \times x^2$, and $x^4 \times 1$. Result: $3x^4$.

    • Degree 5 ($x^5$): Obtained by $x \times x^4$, $x^3 \times x^2$, and $x^5 \times 1$. Result: $3x^5$.

• Observed Pattern: The coefficients follow a repeating step pattern: $1, 1, 2, 2, 3, 3, 4, 4, \dots$.

• Mathematical Representation: Because there isn't a single simple formula for this repeating pattern, the series is best represented by breaking it into even and odd components:

  • Even terms: $\sum_{n=0}^\infty (n+1)x^{2n}$

  • Odd terms: $\sum_{n=0}^\infty (n+1)x^{2n+1}$

Calculus on Power Series

• Term-by-Term Differentiation: If a function behaves as a power series, you can find the derivative by applying the power rule to every individual term.

  • If $f(x) = \sum_{n=0}^\infty c_n x^n = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$

  • Then $f'(x) = c_1 + 2c_2x + 3c_3x^2 + \dots = \sum_{n=1}^\infty n c_n x^{n-1}$.

• Derivative of the Geometric Series:

  • Let $f(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$

  • $f'(x) = \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \dots$

• Integration of Power Series: Similarly, the anti-derivative is found by integrating each term individually and adding a constant of integration $C$.

  • $\int f(x)\,dx = C + \sum_{n=0}^\infty c_n \frac{x^{n+1}}{n+1}$.

• Example: Finding the Power Series for $\ln(1+x)$:

  • We know $\frac{d}{dx} \ln(1+x) = \frac{1}{1+x}$.

  • The series for $\frac{1}{1+x}$ is the alternating geometric series: $1 - x + x^2 - x^3 + \dots$

  • Integrating term-by-term: $\ln(1+x) = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$

  • To find $C$, plug in $x=0$: $\ln(1+0) = C + 0$. Since $\ln(1) = 0$, then $C=0$.

  • Final form: $\ln(1+x) = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1}$.

• Example: Finding the Power Series for $\arctan(x)$:

  • We know $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}$.

  • The series for $\frac{1}{1+x^2}$ is derived by replacing $x$ with $-x^2$ in the geometric series: $1 - x^2 + x^4 - x^6 + \dots$

  • Integrating term-by-term: $\arctan(x) = C + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$

  • Plugging in $x=0$: $\arctan(0) = 0$, so $C=0$.

  • Final form: $\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$.

General Taylor Series Coefficients

• Derivation of the Formula: We want to determine the coefficients $c_n$ for a general power series centered at $a$:

  • $f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + c_4(x-a)^4 + \dots$

  • Evaluated at $a$: $f(a) = c_0$.

  • First derivative: $f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + 4c_4(x-a)^3 + \dots$ Evaluated at $a$: $f'(a) = c_1$.

  • Second derivative: $f''(x) = 2c_2 + 3 \times 2c_3(x-a) + 4 \times 3c_4(x-a)^2 + \dots$ Evaluated at $a$: $f''(a) = 2c_2$, so $c_2 = \frac{f''(a)}{2}$.

  • Third derivative: $f'''(x) = 3 \times 2 \times 1 c_3 + 4 \times 3 \times 2 c_4(x-a) + \dots$ Evaluated at $a$: $f'''(a) = 3! c_3$, so $c_3 = \frac{f'''(a)}{3!}$.

• General Taylor Coefficient Formula: $c_n = \frac{f^{(n)}(a)}{n!}$.

• Taylor Series Definition: A representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

  • $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$

Power Series for Exponential and Trigonometric Functions

• Exponential Function ($e^x$):

  • Centered at $a=0$ (Maclaurin series).

  • Since $\frac{d}{dx} e^x = e^x$, all derivatives evaluated at $0$ are $e^0 = 1$.

  • Series: $\sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

  • Radius of Convergence ($R$): Using the Ratio Test, the limit of $\frac{x^{n+1}}{(n+1)!} \times \frac{n!}{x^n} = \frac{x}{n+1}$ as $n \to \infty$ is $0$. Thus, the series converges everywhere ($R = \infty$).

• Sine Function ($\sin(x)$):

  • Derivatives at $x=0$: $\sin(0)=0$, $\cos(0)=1$, $-\sin(0)=0$, $-\cos(0)=-1$. The sequence of coefficients repeats $0, 1, 0, -1$.

  • Since all even derivatives are $0$, the series contains only odd powers.

  • Series: $\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

  • Radius of Convergence: Similar to $e^x$, the factorial in the denominator ensures convergence everywhere ($R = \infty$).

• Cosine Function ($\cos(x)$):

  • Derivatives at $x=0$: $\cos(0)=1$, $-\sin(0)=0$, $-\cos(0)=-1$, $\sin(0)=0$. The sequence repeats $1, 0, -1, 0$.

  • Since all odd derivatives are $0$, the series contains only even powers.

  • Series: $\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

  • Alternatively, this is the derivative of the sine series.

Euler's Formula

• Connection and Manipulation: Euler discovered a profound connection between exponential and trigonometric functions by substituting an imaginary value ($ix$) into the power series for $e^x$.

• Derivation:

  • $e^{ix} = 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \dots$

  • Simplify powers of $i$ ($i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i$):

  • $e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \dots$

• Separation of Terms:

  • Real parts (even powers): $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots = \cos(x)$.

  • Imaginary parts (odd powers): $i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots) = i\sin(x)$.

• Final Identity: $e^{ix} = \cos(x) + i\sin(x)$.

• The "Most Beautiful Formula": Evaluating Euler's identity at $x = \pi$:

  • $e^{i\pi} = \cos(\pi) + i\sin(\pi)$

  • $e^{i\pi} = -1 + i(0)$

  • $e^{i\pi} + 1 = 0$, which relates the five fundamental mathematical constants ($e, i, \pi, 1, 0$).

Questions & Discussion

• Question (Triangle Pattern): When discussing the multiplication of series, a student asked if the result follows a triangle pattern like Pascal's triangle.

• Response: Not quite. The coefficients follow a pair-based counting pattern ($1, 1, 2, 2, 3, 3 \dots$). For example, $x^2$ has a coefficient of $2$, $x^3$ has $2$, $x^4$ has $3$, and $x^5$ has $3$. This occurs because as you multiply by higher degrees, the lower degree terms eventually stop being produced.

• Question (Shifting Center): How does shifting the center to a value other than zero (e.g., $a=2$) change the calculus?

• Response: Shifting the center (e.g., using $x-2$ instead of $x$) is fundamentally just a left or right shift of the function. The derivative and integral rules remain the same, provided you maintain the shift ($x-a$) inside the terms. For instance, finding arcsin or arctan centered at $2$ would involve plugging $x-2$ into the series and solving for the constant $C$ by plugging in $2$.