Methods for Finding and Using Taylor Series

Known Taylor Series for Common Functions

To find Taylor series efficiently, it is often more practical to utilize known series expansions rather than calculating derivatives manually, which can be laborious for many functions. The following standard Taylor series (Maclaurin series, centered at $x=0$) are fundamental:

• Exponential Function:     $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

• Sine Function:     $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$

• Cosine Function:     $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

• Natural Logarithm (Shifted):     $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$

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

• Binomial Series:     $(1+x)^p = 1 + px + \frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + \dots$

Finding Taylor Series via Substitution

New Taylor series can be derived by substituting a variable or expression into the standard series listed above.

• Example 1: $e^{-x}$

  • Start with $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

  • Replace $x$ with $-x$:

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

  • Result: $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$

• Example 2: $\ln(1-2y)$

  • Start with $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$

  • Substitute $x = -2y$:

  • $\ln(1-2y) = (-2y) - \frac{(-2y)^2}{2} + \frac{(-2y)^3}{3} - \frac{(-2y)^4}{4} + \dots$

  • Result: $\ln(1-2y) = -2y - 2y^2 - \frac{8}{3}y^3 - 4y^4 - \dots$

• Example 3: $z e^{-z^2}$

  • First, find the series for $e^{-z^2}$ by substituting $-z^2$ into the $e^x$ series:

  • $e^{-z^2} = 1 + (-z^2) + \frac{(-z^2)^2}{2!} + \frac{(-z^2)^3}{3!} + \dots = 1 - z^2 + \frac{z^4}{2!} - \frac{z^6}{3!} + \dots$

  • Multiply the entire series by $z$:

  • $z e^{-z^2} = z(1 - z^2 + \frac{z^4}{2} - \frac{z^6}{6} + \dots)$

  • Result: $z - z^3 + \frac{z^5}{2} - \frac{z^7}{6} + \dots$

• Example 4: $t \sin(3t)$

  • Substitute $x = 3t$ into $\sin(x)$.

  • $\sin(3t) = (3t) - \frac{(3t)^3}{3!} + \frac{(3t)^5}{5!} - \frac{(3t)^7}{7!} + \dots = 3t - \frac{27t^3}{6} + \frac{243t^5}{120} - \frac{2187t^7}{5040} + \dots$

  • Simplify and multiply by $t$:

  • $t \sin(3t) = 3t^2 - \frac{9}{2}t^4 + \frac{81}{40}t^6 - \frac{243}{560}t^8 + \dots$

• Example 5: $\sqrt{1-2x}$

  • This uses the Binomial series with $p = \frac{1}{2}$ and substituting $-2x$ for $x$.

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

  • $= 1 - x + \frac{(\frac{1}{2})(-\frac{1}{2})}{2}(4x^2) + \frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{6}(-8x^3) + \dots$

  • Result: $1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3 - \dots$

• Example 6: $\frac{t \sin(t^2) - t^3}{t^7}$

  • Substitute $x = t^2$ into $\sin(x)$: $\sin(t^2) = t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} - \dots$

  • Multiply by $t$: $t \sin(t^2) = t^3 - \frac{t^7}{3!} + \frac{t^{11}}{5!} - \dots$

  • Subtract $t^3$: $t \sin(t^2) - t^3 = -\frac{t^7}{3!} + \frac{t^{11}}{5!} - \dots$

  • Divide by $t^7$:

  • Result: $-\frac{1}{3!} + \frac{t^4}{5!} - \frac{t^{8}}{7!} + \dots = -\frac{1}{6} + \frac{1}{120}t^4 - \dots$

• Example 7: Hyperbolic Sine (Alternative Construction)

  • $\sinh(x) = \frac{e^x - e^{-x}}{2}$

  • Substitute expansions for $e^x$ and $e^{-x}$:

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

  • $\sinh(x) = \frac{2x + \frac{2x^3}{3!} + \frac{2x^5}{5!} + \dots}{2}$

  • Result: $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots$

Radius of Convergence Affected by Substitution

When substituting an expression into a known Taylor series, the radius of convergence ($R$) of the original series must be adjusted based on the substitute.

• Example: Series for $\frac{1}{1+x^2}$

  • Function: $\frac{1}{1 - (-x^2)}$. Using geometric series $\frac{1}{1-H}$ which converges if |H| < 1.

  • Substitute $H = -x^2$, so convergence is |-x^2| < 1 \implies |x|^2 < 1 \implies |x| < 1. Thus $R=1$.

  • Series: $1 - x^2 + x^4 - x^6 + \dots$

• Example: Series for $\frac{1}{1+4x^2}$

  • Substitute $-4x^2$ into the geometric series.

  • Convergence: |-4x^2| < 1 \implies 4|x|^2 < 1 \implies |x|^2 < \frac{1}{4} \implies |x| < \frac{1}{2}.

  • Thus, the new radius of convergence is $R = \frac{1}{2}$.

New Series via Differentiation and Integration

Just as we can use substitution, we can differentiate or integrate existing series term-by-term to find series for related functions.

• Finding the Taylor Series for $\text{arctan}(x)$:

  1. Observe that $\frac{d}{dx}(\text{arctan}(x)) = \frac{1}{1+x^2}$.

  2. Use the series for $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \dots$

  3. Integrate term-by-term:         $\text{arctan}(x) = \int (1 - x^2 + x^4 - x^6 + \dots) dx = C + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$

  4. Solve for the constant $C$: when $x=0$, $\text{arctan}(0) = 0$, so $0 = C + 0$, implying $C=0$.

  5. Final Series: $\text{arctan}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots$

• Finding the Taylor Series for $\ln(1-x)$ (Hint):

  • Use $\frac{d}{dx}(\ln(1-x)) = \frac{-1}{1-x}$.

  • Integrate the negative of the geometric series to find the result.

Function Expansions in Specific Variables

Sometimes functions must be expanded in terms of specific variables by algebraic manipulation to fit standard forms.

• Example A: Expand $\frac{1}{5-x}$ in terms of $x$:

  • Factor out the constant: $\frac{1}{5(1 - \frac{x}{5})} = \frac{1}{5} \cdot \frac{1}{1 - (\frac{x}{5})}$.

  • Apply geometric series for $H = \frac{x}{5}$:

  • Result: $\frac{1}{5} [1 + \frac{x}{5} + (\frac{x}{5})^2 + (\frac{x}{5})^3 + \dots]$

• Example B: Expand $\frac{x}{2+x^2}$ in terms of $x$:

  • Manipulate: $x \cdot \frac{1}{2(1 - (-\frac{x^2}{2}))} = \frac{x}{2} \cdot \frac{1}{1 - (-\frac{x^2}{2})}$.

  • Apply geometric series: $\frac{x}{2} [1 + (-\frac{x^2}{2}) + (-\frac{x^2}{2})^2 + \dots] = \frac{x}{2} [1 - \frac{x^2}{2} + \frac{x^4}{4} - \dots]$

  • Result: $\frac{x}{2} - \frac{x^3}{4} + \frac{x^5}{8} - \dots$

• Example C: Expand $(2+x)^3$ in terms of $x$ using the Binomial formula:

  • Manipulate: $[2(1 + \frac{x}{2})]^3 = 8(1 + \frac{x}{2})^3$.

  • Substitute $p=3$ and $H=\frac{x}{2}$ into $(1+H)^p = 1 + pH + \frac{p(p-1)}{2!}H^2 + \frac{p(p-1)(p-2)}{3!}H^3$.

  • $8 [1 + 3(\frac{x}{2}) + \frac{3(2)}{2}(\frac{x}{2})^2 + \frac{3(2)(1)}{6}(\frac{x}{2})^3] = 8 [1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}]$.

  • Result: $8 + 12x + 6x^2 + x^3$.

Ordering and Comparison Near Zero

Taylor series allow us to compare the relative magnitudes of different functions for values of $y$ near $0$ by looking at the leading terms.

Functions to compare:

1. $a = 1 + \sin(y) \approx 1 + y - \frac{y^3}{6}$

2. $b = e^y \approx 1 + y + \frac{y^2}{2}$

3. $c = \frac{1}{\sqrt{1-2y}} = (1-2y)^{-1/2}$. Using Binomial Series with $p=-1/2$ and substitution $-2y$:     $1 + (-\frac{1}{2})(-2y) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-2y)^2 = 1 + y + \frac{3}{2}y^2$

Comparison Analysis:

• All three series match up to the linear term (all are $1+y$).

• Function (a) has no quadratic term ($0y^2$), while (b) has $\frac{1}{2}y^2$ and (c) has $\frac{3}{2}y^2$.

• Since \frac{3}{2} > \frac{1}{2} > 0, for small positive $y$:     Order: 1 + \sin(y) < e^y < \frac{1}{\sqrt{1-2y}}

Applications: Approximations and Graphing

• Expansion for $e^x - e^{-x}$:

  • Subtracting the two series (as seen in the $\sinh(x)$ derivation above) leaves only odd powers.

  • $e^x - e^{-x} \approx 2x + \frac{x^3}{3} + \frac{x^5}{60} + \dots$ (Adjusting for coefficients: $2\frac{x^3}{3!} = \frac{x^3}{3}$; $2\frac{x^5}{120} = \frac{x^5}{60}$).

  • Graphing note: Near $x=0$, the graph of $e^x - e^{-x}$ resembles a cubic polynomial, specifically $P_3(x) = 2x + \frac{x^3}{3}$.

• Approximating Integrals:

  • To find $\int_0^1 e^{x^2} dx$, we first expand $f(x) = e^{x^2}$.

  • $e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$

  • Integrate: $\int_0^1 (1 + x^2 + \frac{x^4}{2}) dx = [x + \frac{x^3}{3} + \frac{x^5}{10}]_0^1$

  • Evaluate: $1 + \frac{1}{3} + \frac{1}{10} = 1 + 0.333 + 0.1 = 1.433$.

Exercise Solutions

• Problem 1b: First four terms for $e^{x^2}$ are $1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6}$.

• Problem 1e: First four terms for $e^t \cos(t)$.

  • $(1 + t + \frac{t^2}{2} + \frac{t^3}{6} + \dots)(1 - \frac{t^2}{2} + \frac{t^4}{24} - \dots)$

  • Multiplying and keeping terms up to $t^4$: $1 + t - \frac{t^3}{3} - \frac{t^4}{6}$.

• Problem 2b: Expand $\frac{r}{(a+r)^2}$ in terms of $r$.

  • $\frac{r}{a^2(1+\frac{r}{a})^2} = \frac{r}{a^2}(1+\frac{r}{a})^{-2}$.

  • Using Binomial series for $p=-2$: $\frac{r}{a^2} [1 - 2(\frac{r}{a}) + 3(\frac{r}{a})^2 - 4(\frac{r}{a})^3 + \dots]$.

• Problem 3: For $g(x) = \frac{1}{x^2} f(\sqrt{x})$ with $f(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n+1)!}$.

  • Value for $c_3$ in the expansion of $g(x)$.

  • $c_3 = -\frac{1}{11!}$.