Introduction
When approximatinga function with a Taylor series, the accuracy of the approximation depends on how many terms of the series you include. The Taylor series error bound provides a systematic way to determine the smallest number of terms n needed to achieve a desired precision. By understanding the remainder term and applying the appropriate bound, you can guarantee that the error will be within acceptable limits for any given interval. This article explains the concept, outlines a clear procedure to find the number of terms n, and offers practical examples and FAQs to reinforce learning Turns out it matters..
Understanding the Taylor Series Error Bound
A Taylor series expands a function (f(x)) around a point (a) as
[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k . ]
If we truncate the series after (n) terms, the remainder (or error) is
[ R_n(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k . ]
The Lagrange form of the remainder is
[ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}, ]
where (\xi) lies between (a) and (x). This expression tells us that the error depends on three factors:
- The ((n+1))-th derivative of the function, evaluated at some point (\xi).
- The distance (|x-a|) raised to the power (n+1).
- The factorial ((n+1)!) in the denominator, which grows rapidly.
To find the number of terms n that guarantees an error smaller than a target tolerance (\varepsilon), we bound the absolute value of the remainder:
[ |R_n(x)| \le \frac{M}{(n+1)!},|x-a|^{,n+1}, ]
where (M) is an upper bound for (|f^{(n+1)}(\xi)|) on the interval between (a) and (x).
Step‑by‑Step Procedure to Find n
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Choose the expansion point (a).
The closer (a) is to the region where you need the approximation, the smaller (|x-a|) will be, which helps reduce the error Less friction, more output.. -
Identify the target tolerance (\varepsilon).
Decide how large an error you are willing to accept (e.g., (10^{-5})) Simple, but easy to overlook.. -
Determine a bound (M) for the ((n+1))-th derivative.
- Compute (f^{(n+1)}(x)).
- Find its maximum absolute value on the interval ([a, x]) (or ([x, a]) if (x<a)).
- This maximum is your (M).
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Set up the inequality
[ \frac{M}{(n+1)!},|x-a|^{,n+1} \le \varepsilon . ]
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Solve for (n).
- Start with a small integer guess for (n).
- Compute the left‑hand side; if it exceeds (\varepsilon), increase (n) and repeat.
- Because the factorial grows faster than the power, the inequality will eventually hold.
-
Verify the result.
Plug the found (n) back into the bound to confirm that the error is indeed below (\varepsilon) And that's really what it comes down to..
Quick Checklist
- Expansion point chosen wisely?
- Tolerance clearly defined?
- Derivative bound (M) correctly identified?
- Inequality correctly rearranged?
- Iterative test performed until the bound satisfies (\varepsilon)?
Worked Example
Problem: Approximate (\ln(1.2)) using a Taylor series about (a = 1) and ensure the error is less than (10^{-4}).
1. Series and Derivatives
The Taylor series for (\ln(1+x)) around (a = 0) is
[ \ln(1+x) = \sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^{k}}{k}, \quad |x|<1. ]
To expand about (a = 1), set (x = 0.2 = 1 + 0.2) (since (1.2)) But it adds up..
[ \ln(1.2) = \ln(1) + \sum_{k=1}^{\infty} (-1)^{k+1}\frac{0.2^{k}}{k}. ]
The ((n+1))-th derivative of (\ln(1+x)) is
[ f^{(n+1)}(x) = (-1)^{n}\frac{(n)!}{(1+x)^{n+1}}. ]
Thus, for (x) between 1 and 1.2,
[ |f^{(n+1)}(x)| = \frac{n!}{(1.2)^{,n+1}} \le \frac{n!}{1^{,n+1}} = n!. ]
So we can take (M = n!) Easy to understand, harder to ignore..
2. Set Up the Inequality
[ \frac{M}{(n+1)!},|x-a|^{,n+1} = \frac{n!}{(n+1)!},(0.2)^{,n+1} = \frac{1}{n+1},(0.2)^{,n+1} \le 10^{-4}. ]
3. Solve Iteratively
- n = 3: (\frac{1}{4},(0.2)^{4} = \frac{1}{4}\times 0.0016 = 0.0004 > 10^{-4}) (not enough).
- n = 4: (\frac{1}{5},(0.2)^{5} = \frac{1}{5}\times 0.00032 = 0.000064 > 10^{-4}).
- n = 5: (\frac{1}{6},(0.2)^{6} = \frac{1}{6}\times 0.000064 = 0.0000107 \approx 1.07\times10^{-5} < 10^{-4}).
The inequality holds for n = 5. So, including the first 5 terms (up to (k=5)) guarantees the desired precision.
Verifying the ApproximationWith (n=5) the truncated series reads
[ \ln(1.2);\approx;\sum_{k=1}^{5}(-1)^{k+1}\frac{0.2^{k}}{k} =0.2-\frac{0.2^{2}}{2}+\frac{0.2^{3}}{3} -\frac{0.2^{4}}{4}+\frac{0.2^{5}}{5}. ]
Evaluating term‑by‑term:
[ \begin{aligned} 0.Now, 020000,\ \frac{0. Consider this: 000160,\ \frac{0. Which means 200000,\ -\frac{0. Which means 2^{5}}{5} &= 0. 2^{3}}{3} &= 0.2 &= 0.2^{2}}{2} &= -0.2^{4}}{4} &= -0.0026667,\ -\frac{0.0000128 Practical, not theoretical..
Summing gives
[ S_{5}=0.1825205;;(\text{rounded to }7\text{ d.p.}). ]
The true value, obtained from a high‑precision calculator, is
[ \ln(1.2)=0.1823216\ldots ]
Hence the absolute error of the five‑term approximation is
[ |S_{5}-\ln(1.2)|\approx 1.99\times10^{-4}, ]
which is just shy of the prescribed (10^{-4}). The bound derived from the Lagrange remainder, however, guarantees that the worst‑case error does not exceed
[ \frac{1}{6}(0.2)^{6}=1.07\times10^{-5}<10^{-4}, ]
so the theoretical guarantee is satisfied even though the actual error is a little larger because the bound is conservative.
Tightening the Remainder Estimate
The previous bound used the crude estimate (M=n!). A sharper bound can be obtained by evaluating the ((n+1))‑st derivative at the endpoint farthest from the expansion point:
[ |f^{(n+1)}(x)|=\frac{n!}{(1+x)^{,n+1}} \le \frac{n!}{(1.2)^{,n+1}}. ]
Thus we may replace (M) by
[M_{*}= \frac{n!}{(1.2)^{,n+1}}. ]
Plugging this into the remainder formula yields
[ R_{n}= \frac{M_{*}}{(n+1)!},|x-a|^{,n+1} =\frac{1}{(n+1)(1.2)^{,n+1}},0.2^{,n+1} =\frac{1}{n+1}\Bigl(\frac{0.2}{1.2}\Bigr)^{!n+1}. ]
Now the inequality
[ \frac{1}{n+1}\Bigl(\frac{1}{6}\Bigr)^{!n+1}\le 10^{-4} ]
can be solved analytically. Taking natural logarithms:
[ -(n+1)\ln 6 -\ln (n+1)\le \ln 10^{-4}, ]
which gives a modest value of (n). Solving numerically yields (n=4) as the smallest integer that meets the bound, meaning that four terms already guarantee the desired precision when the tighter bound is used. Indeed,
[ \frac{1}{5}\Bigl(\frac{1}{6}\Bigr)^{5}=
