Error Bounds

TL;DR

When you use a Taylor polynomial to approximate a function, error bounds help you figure out the maximum possible difference between your approximation and the actual function value. There are specific formulas, like Taylor's Inequality (Lagrange Remainder) and the Alternating Series Estimation Theorem, to calculate these bounds. These bounds are crucial for understanding how good your approximation truly is.

1. The Mental Model

Think of error bounds as a "guarantee." If you use a Taylor polynomial to estimate a value, the error bound tells you the largest possible amount by which your estimate could be off from the true value. It's like knowing the maximum measurement error when using a tool.

2. The Core Material

When you approximate a function $f(x)$ with its $n$-th degree Taylor polynomial $P_n(x)$ centered at $a$, the difference between them is called the remainder or error, $R_n(x) = f(x) - P_n(x)$. Error bounds help us find an upper limit for the absolute value of this remainder, i.e., $|R_n(x)| \le \text{bound}$.

Taylor's Inequality (Lagrange Remainder)

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This is the general formula for the remainder of a Taylor series. If $f$ has $n+1$ derivatives on an interval $I$ containing $a$ and $x$, and $M$ is an upper bound for the absolute value of the $(n+1)$-th derivative on $I$ (i.e., $|f^{(n+1)}(t)| \le M$ for all $t$ between $a$ and $x$), then the remainder $R_n(x)$ satisfies:

$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$

To use this, you need to:
1. Find the $(n+1)$-th derivative of $f(x)$.
2. Find the maximum value, $M$, of $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$. This is often the trickiest part; you might need to check the derivative's behavior on that interval or simply evaluate it at the endpoints if it's monotonic.
3. Plug $M$, $n$, $a$, and $x$ into the formula.

Alternating Series Estimation Theorem

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This one is much simpler, but it only applies to alternating series that meet specific conditions:
1. The series terms are alternating in sign.
2. The absolute values of the terms are decreasing ($b_{k+1} \le b_k$).
3. The limit of the terms goes to zero ($\lim_{k \to \infty} b_k = 0$).

If these conditions are met, and you approximate the sum of the series by its first $n$ terms, the absolute value of the error ($|R_n|$) is less than or equal to the absolute value of the first neglected term ($b_{n+1}$).

$|R_n| \le b_{n+1}$

This means your error is smaller than the very next term you left out.

graph TD
    A["Need to estimate Taylor series error?"] --> B{Is it an alternating series?};
    B -- Yes --> C{Does it meet ASET conditions:
        1. Alternating signs
        2. Terms decreasing
        3. Limit of terms is 0?};
    C -- Yes --> D["Use Alternating Series Estimation Theorem:
        Error <= |first neglected term|"];
    C -- No --> E["Use Taylor's Inequality (Lagrange Remainder)"];
    B -- No --> E;
    E --> F["Steps for Taylor's Inequality:
        1. Find f^(n+1)(x)
        2. Find M = max |f^(n+1)(t)| on interval
        3. Calculate |R_n(x)| <= M/((n+1)!) * |x-a|^(n+1)"];

3. Worked Example

Let's approximate $e^{0.1}$ using a 2nd degree Maclaurin polynomial ($n=2$, so centered at $a=0$) and find the maximum error using Taylor's Inequality.

1. Function and desired approximation: $f(x) = e^x$, we want $e^{0.1}$. So $x=0.1$.

2. Maclaurin Polynomial: For $n=2$, $P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2$.

   • $f(x) = e^x \implies f(0) = 1$
   • $f'(x) = e^x \implies f'(0) = 1$
   • $f''(x) = e^x \implies f''(0) = 1$
     $P_2(x) = 1 + 1x + \frac{1}{2!}x^2 = 1 + x + \frac{x^2}{2}$.
     So, $P_2(0.1) = 1 + 0.1 + \frac{(0.1)^2}{2} = 1 + 0.1 + \frac{0.01}{2} = 1 + 0.1 + 0.005 = 1.105$.

3. Find the error bound using Taylor's Inequality: We need $|R_2(0.1)|$.

   • We need the $(n+1)$-th derivative, which is $f^{(2+1)}(x) = f^{(3)}(x)$.
   • $f^{(3)}(x) = e^x$.
   • We need to find $M$, the maximum value of $|f^{(3)}(t)| = |e^t|$ on the interval between $a=0$ and $x=0.1$.
   • Since $e^t$ is an increasing function, its maximum value on $[0, 0.1]$ will be at $t=0.1$.
   • So, $M = e^{0.1}$. We don't know this exactly, but we know $e < 3$, so $e^{0.1} < 3^{0.1}$. Also, $e^{0.1} = \sqrt[10]{e} < \sqrt[10]{3}$. A safe upper bound for $e^{0.1}$ (since $e^0=1$ and $e^1 \approx 2.718$) is just $e^{0.1} < e^1 \approx 2.718$, or even simpler, since $0.1 < 1$, we can just say $e^{0.1} < e^1 = e$. Let's use $M=e^{0.1}$ for now, and estimate it. Since $0.1$ is small, $e^{0.1}$ is just a bit over 1. For a practical bound, it's often acceptable to use a slightly larger, easily calculable number. For instance, $e^{0.1} < e^{0.5} = \sqrt{e} < \sqrt{3} \approx 1.732$. Or, since $0.1 < 1$, we know $e^{0.1} < e^1 < 3$. Let's use $M=e^{0.1} \approx 1.105$. (Sometimes you'll be told to use an easier bound like $M=3$ for $e^x$ on $[0,1]$). Let's use $M = e^{0.1}$ for precision, but acknowledge a numerical bound would be needed. A common simplification is to use $M=e^1$ for $x \in [0,1]$ which gives $M \approx 2.718$. For $[0, 0.1]$, we can be tighter: $M = e^{0.1}$. Using a calculator for this specific $M$: $e^{0.1} \approx 1.10517$.
   • Plug into the formula: $|R_2(0.1)| \le \frac{M}{(2+1)!}|0.1-0|^{2+1} = \frac{e^{0.1}}{3!}(0.1)^3$
   • $|R_2(0.1)| \le \frac{1.10517}{6}(0.001) \approx 0.184195 \times 0.001 = 0.000184195$.

So, our approximation $P_2(0.1) = 1.105$ is within approximately $0.000184$ of the true value of $e^{0.1}$.
(Actual value of $e^{0.1} \approx 1.1051709$, so the error is $|1.1051709 - 1.105| = 0.0001709$, which is indeed less than our bound of $0.000184$.)

4. Key Takeaways

• Error bounds quantify the maximum possible difference between a function's actual value and its Taylor polynomial approximation.
• Taylor's Inequality (Lagrange Remainder) is a general method for bounding the error, requiring you to find an upper bound ($M$) for the absolute value of the next derivative.
• The Alternating Series Estimation Theorem offers a simpler error bound for specific alternating series: the error is less than or equal to the absolute value of the first neglected term.
• Choosing the correct $M$ for Taylor's Inequality is crucial; it must be an upper bound for the derivative on the entire interval between the center and the point of approximation.
• Error bounds allow you to determine how many terms are needed in a series to achieve a desired level of accuracy.

Common Mistakes to Avoid

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• Using the wrong error bound formula: Don't use the Alternating Series Estimation Theorem if the series isn't strictly alternating and decreasing to zero.
• Incorrectly finding $M$ for Taylor's Inequality: $M$ must be the maximum of $|f^{(n+1)}(t)|$ over the relevant interval, not just $|f^{(n+1)}(a)|$ or $|f^{(n+1)}(x)|$.
• Forgetting the factorial or power in Taylor's Inequality: The formula is $\frac{M}{(n+1)!}|x-a|^{n+1}$.
• Misinterpreting the interval for $M$: The interval for $t$ is between $a$ and $x$. For instance, if $a=0$ and $x=0.5$, check the maximum derivative on $[0, 0.5]$.

5. Now Try It

Use a 3rd degree Maclaurin polynomial for