Power Series & Taylor Expansions

Endpoints.

x = 1: Σ4ⁿ/(n²4ⁿ) = Σ1/n², a p-series with p = 2. Converges.

x = −7: Σ(−4)ⁿ/(n²4ⁿ) = Σ(−1)ⁿ/n². Converges absolutely.

Note the pattern: an n² in the denominator makes both endpoints converge, so the interval is closed. An n¹ 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 x² — multiplication by a power just shifts the exponents:

R = ∞, unchanged: the exponential series converges everywhere, and multiplying by x² cannot break that.

Why this beats the definition: computing five derivatives of x²e^−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 ±1 and do not go to zero, so both diverge by the nth-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 x² 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(x²) 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 = 4x²: the substitution must go into every occurrence of x, including inside the powers. Writing 2x²/2 instead of 4x²/2 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 = −¹⁄₂
• x²:  (−¹⁄₂)(−³⁄₂)/2 = (³⁄₄)/2 = ³⁄₈
• x³:  (−¹⁄₂)(−³⁄₂)(−⁵⁄₂)/6 = (−¹⁵⁄₈)/6 = −⁵⁄₁₆

Quick check at x = 0.1: the series gives 1 − 0.05 + 0.00375 − 0.0003125 = 0.953438, and 1/√1.1 = 0.953463. Agreement to five digits with three correction terms.

(a) Put x = iθ into the exponential series and use i² = −1:

Separate real and imaginary parts and recognize the two series:

(b) Write cos θ = (e^iθ + e^−iθ)/2 and cube it. The binomial expansion gives four terms:

Pair the conjugates back into cosines:

Check at θ = 0: left side 1; right side 3/4 + 1/4 = 1. ✓

(c) The engineering content. Feed a pure tone cos ωt into a device with a cubic term in its transfer characteristic, and the output contains both the original frequency and a component at 3ω. 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 ωt into the exact law and expand:

Now use cos²ωt = (1 + cos 2ωt)/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, HD₂ = V̂/(4V_Q). Check: 0.2/(4×2) = 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 √(1+u) ≈ 1 + u/2 with u = (X/R)²:

Numbers. With R = 100, X = 10:

Error 0.0012 Ω, a relative error of about 1.2×10⁻⁵ — 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) ≈ 5.7°.