Taylor Series for MATH 141
Exam Relevance for MATH 141
Taylor series are heavily weighted on the MATH 141 final. This late-course topic gets significant exam coverage. Know e^x, sin x, cos x, ln(1+x) by heart.
What is a Taylor Series?
In the previous lesson, we saw that power series can represent functions like $e^x$ and $\frac{1}{1-x}$. But where do those power series come from? How do we know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$?
Taylor series answer this question. Given any "nice" function $f(x)$, the Taylor series gives us a systematic way to build a power series that represents it. The idea is beautifully simple: match the function's value and all its derivatives at a single point.
The Key Insight: Matching Derivatives
Suppose we want a power series $\sum c_n (x-a)^n$ that equals $f(x)$ near $x = a$.
If the series equals the function, then at $x = a$:
- The series value should equal $f(a)$
- The series' first derivative should equal $f'(a)$
- The series' second derivative should equal $f''(a)$
- And so on for all derivatives...
Working this out (plug in $x = a$ and solve for each $c_n$), we discover:
$$c_n = \frac{f^{(n)}(a)}{n!}$$
where $f^{(n)}(a)$ means the $n$th derivative of $f$ evaluated at $a$.
The remarkable implication: Knowing all the derivatives at a single point means you know the function's complete "local DNA" — its value, its slope, its curvature, the rate of change of its curvature, and so on. This complete local information at ONE point is enough to reconstruct the entire function (within the radius of convergence). Perfect knowledge at a single location lets you predict the function's behavior even at points far away from where you started.
The Taylor Series Formula
The Taylor series of $f(x)$ centered at $x = a$ is:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
Expanded out:
$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$
Special case — Maclaurin series: When $a = 0$, we call it a Maclaurin series:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$$
What Does "Centering" Mean?
The center $a$ is the point where you're gathering all your derivative information. Different choices of $a$ give different (but equally valid) series representations of the same function.
Key insight: Accuracy depends on distance from the center.
Taylor polynomials (finite approximations) are most accurate near the center and become less accurate as you move away. This makes intuitive sense — you're using information gathered at $x = a$ to predict values elsewhere, so predictions close to $a$ should be better than predictions far from $a$.
Practical example: Suppose you want to approximate $\ln(1.1)$.
- Using the series for $\ln x$ centered at $a = 1$: You're approximating at $x = 1.1$, which is only $0.1$ away from center. Very accurate!
- Using the series for $\ln x$ centered at $a = 2$: You're approximating at $x = 1.1$, which is $0.9$ away from center. Less accurate with the same number of terms.
Same function, different representations: The Taylor series for $\ln x$ centered at $a = 1$ and centered at $a = 2$ both represent $\ln x$, but each is "optimized" for its own neighborhood. Choose your center based on where you need accuracy.
Why Taylor Series Matter
1. They turn complicated functions into polynomials
Want to compute $e^{0.1}$ without a calculator? Use the Taylor series: $$e^{0.1} \approx 1 + 0.1 + \frac{0.01}{2} + \frac{0.001}{6} + \cdots \approx 1.1052$$
2. They let us integrate "impossible" functions
The function $e^{-x^2}$ has no elementary antiderivative. But its Taylor series can be integrated term by term!
3. They reveal hidden structure
Why is $e^{ix} = \cos x + i\sin x$? Compare their Taylor series — it falls right out.
How to Find a Taylor Series
Step 1: Compute derivatives of $f(x)$: $f(x), f'(x), f''(x), f'''(x), \ldots$
Step 2: Evaluate each derivative at $x = a$: $f(a), f'(a), f''(a), f'''(a), \ldots$
Step 3: Plug into the formula: $$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
Step 4: Look for a pattern and write the general term.
Find the Maclaurin series for $f(x) = e^x$.
Step 1: Compute derivatives
$$f(x) = e^x, \quad f'(x) = e^x, \quad f''(x) = e^x, \quad f'''(x) = e^x, \ldots$$
Every derivative of $e^x$ is just $e^x$!
Step 2: Evaluate at $a = 0$
$$f(0) = 1, \quad f'(0) = 1, \quad f''(0) = 1, \quad f'''(0) = 1, \ldots$$
Step 3: Write the series
$$e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
Radius of convergence: $R = \infty$ (converges for all $x$).
Find the Maclaurin series for $f(x) = \sin x$.
Step 1: Compute derivatives
$$f(x) = \sin x \implies f(0) = 0$$ $$f'(x) = \cos x \implies f'(0) = 1$$ $$f''(x) = -\sin x \implies f''(0) = 0$$ $$f'''(x) = -\cos x \implies f'''(0) = -1$$ $$f^{(4)}(x) = \sin x \implies f^{(4)}(0) = 0$$
The pattern repeats: $0, 1, 0, -1, 0, 1, 0, -1, \ldots$
Step 2: Write the series
Only odd powers survive (even derivatives are 0 at $x = 0$):
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
Radius of convergence: $R = \infty$.
Find the Maclaurin series for $f(x) = \cos x$.
Step 1: Compute derivatives at 0
$$f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -1, \quad f'''(0) = 0, \quad f^{(4)}(0) = 1, \ldots$$
The pattern: $1, 0, -1, 0, 1, 0, -1, 0, \ldots$
Step 2: Write the series
Only even powers survive:
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$
Radius of convergence: $R = \infty$.
Find the Taylor series for $f(x) = \ln x$ centered at $a = 1$.
Step 1: Compute derivatives
$$f(x) = \ln x \implies f(1) = 0$$ $$f'(x) = \frac{1}{x} \implies f'(1) = 1$$ $$f''(x) = -\frac{1}{x^2} \implies f''(1) = -1$$ $$f'''(x) = \frac{2}{x^3} \implies f'''(1) = 2$$ $$f^{(4)}(x) = -\frac{6}{x^4} \implies f^{(4)}(1) = -6$$
Step 2: Identify the pattern
$$f^{(n)}(1) = (-1)^{n+1}(n-1)! \text{ for } n \geq 1$$
Step 3: Write the series
$$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n$$
$$= (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$$
Radius of convergence: $R = 1$ (interval: $(0, 2]$).
Note: We can't center $\ln x$ at 0 because $\ln 0$ is undefined!
Find the Maclaurin series for $f(x) = \frac{1}{1-x}$.
Step 1: Compute derivatives
$$f(x) = (1-x)^{-1} \implies f(0) = 1$$ $$f'(x) = (1-x)^{-2} \implies f'(0) = 1$$ $$f''(x) = 2(1-x)^{-3} \implies f''(0) = 2$$ $$f'''(x) = 6(1-x)^{-4} \implies f'''(0) = 6$$
Pattern: $f^{(n)}(0) = n!$
Step 2: Write the series
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} \frac{n!}{n!}x^n = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$$
This is the geometric series! Taylor series confirms what we already knew.
Radius of convergence: $R = 1$.
Important Maclaurin Series to Know
These come up frequently and are worth memorizing:
| Function | Maclaurin Series | Radius |
|---|---|---|
| $e^x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ | $\infty$ |
| $\sin x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ | $\infty$ |
| $\cos x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ | $\infty$ |
| $\frac{1}{1-x}$ | $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ | $1$ |
| $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ | $1$ |
Taylor Polynomials: Finite Approximations
The $n$th-degree Taylor polynomial $T_n(x)$ is the partial sum of the Taylor series:
$$T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$$
This gives a polynomial approximation to $f(x)$ near $x = a$.
Example: The 3rd-degree Taylor polynomial for $e^x$ at $a = 0$:
$$T_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$
At $x = 0.5$: $T_3(0.5) = 1 + 0.5 + 0.125 + 0.0208 = 1.6458$
Actual value: $e^{0.5} \approx 1.6487$ — pretty close!
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting the factorial in the denominator
Wrong: Writing the Taylor series as $f(a) + f'(a)(x-a) + f''(a)(x-a)^2 + \cdots$
Why it's wrong: Missing the $n!$ in the denominator. The formula is $\frac{f^{(n)}(a)}{n!}$, not just $f^{(n)}(a)$.
Correct: $f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$
❌ Mistake: Centering at the wrong point
Wrong: Finding the Maclaurin series for $\ln x$ by plugging in $a = 0$.
Why it's wrong: $\ln 0$ is undefined! You can't center a Taylor series at a point where the function doesn't exist.
Correct: Center $\ln x$ at $a = 1$ (or any positive number), giving a series in powers of $(x - 1)$.
❌ Mistake: Confusing Taylor series with Taylor polynomial
Wrong: "The Taylor series for $e^x$ is $1 + x + \frac{x^2}{2}$."
Why it's wrong: That's the 2nd-degree Taylor polynomial $T_2(x)$, not the full series.
Correct: The Taylor series is the infinite sum. Taylor polynomials are finite truncations used for approximations.
❌ Mistake: Assuming Taylor series always equals the function
Wrong: "The Taylor series for $f(x)$ IS $f(x)$ everywhere."
Why it's wrong: The Taylor series only equals $f(x)$ within the radius of convergence. Outside that interval, the series either diverges or converges to something else.
Correct: The Taylor series represents $f(x)$ within its interval of convergence.
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