Power Series & Taylor Expansions

Endpoints.

\(x = 1\): \(\sum 4^n/(n^2 4^n) = \sum 1/n^2\), a p-series with \(p = 2\). Converges.

\(x = -7\): \(\sum (-4)^n/(n^2 4^n) = \sum (-1)^n/n^2\). Converges absolutely.

Note the pattern: an \(n^2\) in the denominator makes both endpoints converge, so the interval is closed. An \(n^1\) would have made it half-open (compare Unit 4, worked example 1). The power of \(n\) decides the endpoint behavior.

Substitute \(-x\) into the exponential series:

Multiply through by \(x^2\) — multiplication by a power just shifts the exponents:

\(R = \infty\), unchanged: the exponential series converges everywhere, and multiplying by \(x^2\) cannot break that.

Why this beats the definition: computing five derivatives of \(x^2e^{-x}\) by the product rule is a long, error-prone slog. Substitute-and-multiply takes fifteen seconds.

Do not differentiate repeatedly. Force the expression into geometric form by writing \(x = 1 + (x-1)\):

That is \(1/(1-u)\) with \(u = -(x-1)\):

Convergence needs \(|u| < 1\), i.e. \(|x-1| < 1\):

Endpoints: at \(x = 0\) and \(x = 2\) the terms are \(\pm 1\) and do not go to zero, so both diverge by the \(n\)th-term test. The interval is open.

Makes sense: \(1/x\) blows up at \(x = 0\), which is distance 1 from the center — so no series centered at 1 could possibly reach past it. The radius is always the distance to the nearest singularity.

Substitute \(x^2\) into the sine series:

Integrate term by term:

Accuracy justification: the series alternates with decreasing terms, so the error is smaller than the first omitted term. After the third term that bound is well under \(10^{-7}\) — far tighter than the four decimals requested. Even truncating after two terms would have sufficed.

Note: \(\sin(x^2)\) has no elementary antiderivative. Series is not a shortcut here — it is the method.

Substitute \(2x\) into the exponential series:

Subtract the 1 and the \(2x\) — that is exactly what the numerator does:

Watch the \((2x)^2 = 4x^2\): the substitution must go into every occurrence of \(x\), including inside the powers. Writing \(2x^2/2\) instead of \(4x^2/2\) gives 1 and is the standard slip on this problem.

Error bound. All derivatives of \(e^x\) are \(e^x\), increasing, so on \([0,\ 0.3]\) we can take \(M = e^{0.3} < 1.35\):

Check against truth: \(e^{0.3} = 1.349859\), so the actual error is

The bound holds and is not far off — a healthy sign. A Taylor bound is always conservative (it assumes the worst-case derivative across the whole interval), so the actual error should come in below it. If yours ever comes in above, you have made an error somewhere.

This is \((1+x)^k\) with \(k = -1/2\):

Compute the coefficients one at a time:

• \(x\):  \(k = -\tfrac{1}{2}\)
• \(x^2\):  \(\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)/2 = \left(\tfrac{3}{4}\right)/2 = \tfrac{3}{8}\)
• \(x^3\):  \(\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)\left(-\tfrac{5}{2}\right)/6 = \left(-\tfrac{15}{8}\right)/6 = -\tfrac{5}{16}\)

Quick check at \(x = 0.1\): the series gives \(1 - 0.05 + 0.00375 - 0.0003125 = 0.953438\), and \(1/\sqrt{1.1} = 0.953463\). Agreement to five digits with three correction terms.

(a) Put \(x = i\theta\) into the exponential series and use \(i^2 = -1\):

Separate real and imaginary parts and recognize the two series:

(b) Write \(\cos\theta = (e^{i\theta} + e^{-i\theta})/2\) and cube it. The binomial expansion gives four terms:

Pair the conjugates back into cosines:

Check at \(\theta = 0\): left side 1; right side 3/4 + 1/4 = 1. ✓

(c) The engineering content. Feed a pure tone \(\cos\omega t\) into a device with a cubic term in its transfer characteristic, and the output contains both the original frequency and a component at \(3\omega\). A cubic nonlinearity generates third-harmonic distortion — and, with two input tones, third-order intermodulation, the IP3 spec on every RF datasheet.

Note also that the fundamental picks up a 3/4 factor: the nonlinearity does not only add a harmonic, it also changes the gain at the original frequency. That is gain compression, the mechanism behind the 1 dB compression point. Both effects fall out of one trigonometric identity that you derived from a power series.

(a) Bias current: \(I_Q = 0.01(2)^2 =\) 0.040 A = 40 mA.

(b) Small-signal parameters — the first-order Taylor coefficient at the bias point:

(c) Distortion. Put \(v = 2 + 0.2\cos\omega t\) into the exact law and expand:

Now use \(\cos^2\omega t = (1 + \cos 2\omega t)/2\):

Read off the three pieces: a slightly shifted DC term, the fundamental, and a second harmonic.

General result worth keeping: for a pure square-law device, \(\mathrm{HD}_2 = \hat{V}/(4V_Q)\). Check: \(0.2/(4\times 2) = 0.025\) ✓. Distortion scales linearly with signal amplitude and falls as you bias harder — which is exactly why amplifiers trade quiescent power for linearity.

Note the DC term also shifted, from 0.0400 to 0.0402 A. A nonlinear device rectifies: applying an AC signal changes the average current. That is how a square-law detector works.

Factor out the big term — this is the move that makes every approximation of this type work:

Now the quantity inside is 1 + (small), so apply \(\sqrt{1+u} \approx 1 + u/2\) with \(u = (X/R)^2\):

Numbers. With \(R = 100\), \(X = 10\):

Error 0.0012 Ω, a relative error of about \(1.2\times10^{-5}\) — and that is with \(X/R = 0.1\), which is not especially small.

Why engineers like this form: it shows the reactance contributes only at second order. A 10% reactive component changes the impedance magnitude by 0.5%. That is why a small series reactance can often be ignored in magnitude calculations — while still mattering enormously for phase, which shifts by \(\arctan(0.1) \approx 5.7^\circ\).