Strain Correction Series for a Composite Rod
A materials laboratory calibrates a strain gauge bonded to a carbon-fibre test rod. The gauge responds nonlinearly, and the manufacturer's correction factor is C(x) = (1 + 3x)^(1/2), where x is the strain recorded by the logger, a small dimensionless number normally kept below 0.1. The logger's firmware can evaluate polynomials only, so C(x) must be replaced by a truncated binomial series before it can be uploaded. A technician proposes keeping terms up to and including x³, then applying the result at x = 0.08, where the raw output is 1.1000 units. The quality procedure demands that the corrected value be within 1 part in 10⁴ of the exact value; anything outside that band must be flagged and recalculated. A second technician objects that the plan is unsafe because a series has a limited range of validity, and the pair must decide whether to accept the four-term version, add a further term, or abandon the approach for the higher-strain tests planned next month at x = 0.5.
1. Expand (1 + 3x)^(1/2) in ascending powers of x, up to and including the term in x³, and state precisely the values of x for which the expansion is valid.
Read the worked answer
Use the standard binomial expansion with n = 1/2 and the substitution y = 3x:
(1 + y)^(1/2) = 1 + (1/2)y + [(1/2)(−1/2)/2!]y² + [(1/2)(−1/2)(−3/2)/3!]y³ + …
Coefficients:
- y term: 1/2, so with y = 3x the term is (1/2)(3x) = (3/2)x = 1.5x
- y² term: (1/2)(−1/2)/2 = −1/8, so with y² = 9x² the term is −9x²/8 = −1.125x²
- y³ term: (1/2)(−1/2)(−3/2)/6 = 3/48 = 1/16, so with y³ = 27x³ the term is 27x³/16 = 1.6875x³
So C(x) = 1 + 1.5x − 1.125x² + 1.6875x³ + …
Validity: the binomial series converges only for |y| < 1, that is |3x| < 1, giving |x| < 1/3. At |x| = 1/3 the terms do not tend to zero fast enough to be truncated safely, so the strict inequality matters.
2. Use the four-term expansion to estimate the correction at x = 0.08 (that is, an estimate of √1.24), compare it with the calculator value, and use the first omitted term to judge whether the 1 part in 10⁴ standard is met.
Read the worked answer
Substituting x = 0.08, with x² = 0.0064 and x³ = 0.000512:
C ≈ 1 + 1.5(0.08) − 1.125(0.0064) + 1.6875(0.000512) = 1 + 0.12 − 0.0072 + 0.000864 = 1.113664
Calculator value: √1.24 = 1.113553 (6 d.p.).
Actual error = 1.113664 − 1.113553 = 0.000111, so the relative error is 0.000111/1.113553 ≈ 9.97 × 10⁻⁵, which is (just) inside 1 × 10⁻⁴.
Error bound from the first omitted term: the x⁴ coefficient is (1/2)(−1/2)(−3/2)(−5/2)/4! = (−15/16)/24 = −5/128, and multiplying by 3⁴ = 81 gives −405/128 = −3.1640625. At x = 0.08 this term equals −3.1640625 × 4.096 × 10⁻⁵ ≈ −1.30 × 10⁻⁴.
Since the term magnitudes decrease and alternate from the x term onward, the truncation error after x³ is at most about 1.30 × 10⁻⁴, i.e. a relative bound near 1.17 × 10⁻⁴. That worst-case bound exceeds the standard even though the actual computed value passes. Judgement: the four-term version is not safely compliant; add the x⁴ term (giving 1.113534, error ≈ 1.9 × 10⁻⁵) or compute exact values in the flagged band.
3. Explain, using the validity condition, why the same polynomial cannot be used for the planned tests at x = 0.5, and suggest how the correction could still be handled there.
Read the worked answer
The series is an expansion in powers of 3x and converges only when |3x| < 1. At x = 0.5 we have 3x = 1.5, so the expansion would be a series in powers of 1.5, whose terms (1.5)ⁿ grow without limit; truncating it gives a number with no connection to √2.5 = 1.5811. In fact the four-term polynomial gives 1 + 0.75 − 0.28125 + 0.2109375 = 1.6797, an error of about 6 per cent, and adding terms makes it worse, not better.
As x approaches 1/3 from below, 3x approaches 1 and convergence becomes extremely slow, so even values just inside the interval need many terms for 10⁻⁴ accuracy.
Remedy: factor out the large factor before expanding, e.g. (1 + 3x)^(1/2) = (3x)^(1/2)(1 + 1/(3x))^(1/2), which is a convergent binomial series when |1/(3x)| < 1, that is x > 1/3. For x = 0.5 this gives √1.5 × (1 + 1/1.5)^(1/2) with argument 2/3 < 1, which converges quickly. Alternatively, for the few high-strain tests, evaluate the square root directly rather than with a polynomial.