FP1 2.1 - Taylor series

FP1 2.1 - Taylor series

Build a power series around a point of your choice, rather than always around zero. You will derive the Taylor coefficients, expand explicit functions and obtain a local series from a differential equation without first solving it.

Moving the expansion point

How can you approximate ln⁡2.1\ln2.1 with a short polynomial when ln⁡x\ln x has no Maclaurin series at zero? Move the centre to x=2x=2. The small quantity becomes x−2=0.1x-2=0.1, and derivatives at 22 replace the unavailable derivatives at zero.

You already know

A Maclaurin series has coefficient f(n)(0)/n!f^{(n)}(0)/n! on xnx^n. A finite truncation generally gives a local approximation. The full series must actually represent the function on the interval where you use it; infinitely many derivatives alone do not establish that representation.

Set h=x−ah=x-a. A power series about aa has the form c0+c1h+c2h2+⋯c_0+c_1h+c_2h^2+\cdots. Since dh/dx=1\mathrm{d}h/\mathrm{d}x=1, evaluating at x=ax=a, where h=0h=0, gives

f(a)=c0set h=0f′(a)=c1differentiate oncef′′(a)=2!c2differentiate twicef(n)(a)=n!cnisolate the coefficient\begin{aligned} f(a)&=c_0 && \text{set }h=0 \\ f'(a)&=c_1 && \text{differentiate once} \\ f''(a)&=2!c_2 && \text{differentiate twice} \\ f^{(n)}(a)&=n!c_n && \text{isolate the coefficient} \end{aligned}

Thus moving the centre changes where the derivatives are evaluated, but not the factorial rule.

Taylor expansion about a

Where the Taylor series represents ff,

f(x)=f(a)+(x−a)f′(a)+(x−a)22!f′′(a)+(x−a)33!f′′′(a)+⋯f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a) +\frac{(x-a)^3}{3!}f'''(a)+\cdots

The general term is f(n)(a)(x−a)n/n!f^{(n)}(a)(x-a)^n/n!. The expansion variable is x−ax-a, and every derivative coefficient is evaluated at aa. The booklet prints the zero-centre Maclaurin version; this shifted formula follows from it.

Not in the formula booklet: learn it.

The polynomial through degree mm matches the function's value and its first mm derivatives at the centre. It need not match the function far away. Use ≈\approx for a numerical truncation and retain an ellipsis when writing the start of an infinite series.

[DIAGRAM: asset_name: A Taylor polynomial centred at x equals 2; asset_slug: 9fm0-fp163-taylor-centre; caption: Derivatives at 2 produce a polynomial that follows the logarithm locally; description: A Cartesian graph shows ln x, the tangent ln 2 plus (x minus 2) over 2 and the cubic Taylor polynomial about x equals 2. Their common point (2,ln 2) is marked. The cubic follows the logarithm more closely near 2; the legend is outside the curves.]
Derivatives at 2 produce a polynomial that follows the logarithm locally

Key idea

The centre aa determines both the powers (x−a)n(x-a)^n and the derivative values f(n)(a)f^{(n)}(a). Move both together.

The next check separates the new expansion variable from the old one.

Expanding a trigonometric function about a non-zero point

Differentiate first, substitute the centre second and assemble the series last. A zero derivative value removes that power; it does not move the expansion point or shift the other factorials.

Worked example 1: sine about pi

Question. Derive the Taylor expansion of sin⁡x\sin x about x=πx=\pi, up to and including the term in (x−π)3(x-\pi)^3.

Plan. Evaluate the function and its first three derivatives at π\pi, then use powers of h=x−πh=x-\pi.

n0123f(n)(π)0−101\begin{array}{c|rrrr} n&0&1&2&3\\\hline f^{(n)}(\pi)&0&-1&0&1 \end{array}

Therefore

sin⁡x=0−h+02!h2+13!h3+⋯Taylor coefficients=−(x−π)+(x−π)36+⋯restore the requested variable\begin{aligned} \sin x&=0-h+\frac0{2!}h^2+\frac1{3!}h^3+\cdots && \text{Taylor coefficients} \\ &=-(x-\pi)+\frac{(x-\pi)^3}{6}+\cdots && \text{restore the requested variable} \end{aligned}

Why this step? The negative linear term comes from cos⁡π=−1\cos\pi=-1. Copying the Maclaurin sine coefficients unchanged would reverse the signs.

Answer. sin⁡x=−(x−π)+(x−π)3/6+⋯\sin x=-(x-\pi)+(x-\pi)^3/6+\cdots.

Check. sin⁡(π+h)=−sin⁡h\sin(\pi+h)=-\sin h, whose standard series is −h+h3/6−⋯-h+h^3/6-\cdots.

Change only the function from sine to cosine, keeping the centre π\pi and the highest permitted power.

Worked example 2: cosine about pi

Question. Derive the Taylor expansion of cos⁡x\cos x about π\pi through the term in (x−π)3(x-\pi)^3.

Plan. Evaluate the cosine derivative cycle at the same centre.

n0123f(n)(π)−1010\begin{array}{c|rrrr} n&0&1&2&3\\\hline f^{(n)}(\pi)&-1&0&1&0 \end{array}

Thus

cos⁡x=−1+0h+h22!+03!h3+⋯insert derivative values=−1+(x−π)22+⋯zero odd coefficients\begin{aligned} \cos x&=-1+0h+\frac{h^2}{2!}+\frac0{3!}h^3+\cdots && \text{insert derivative values} \\ &=-1+\frac{(x-\pi)^2}{2}+\cdots && \text{zero odd coefficients} \end{aligned}

Why this step? “Through the cubic term” asks you to account for that power, even when its coefficient is zero. It does not require inventing a non-zero cubic term.

Answer. The requested polynomial is −1+(x−π)2/2-1+(x-\pi)^2/2; the cubic coefficient is zero.

Check. cos⁡(π+h)=−cos⁡h=−1+h2/2−⋯\cos(\pi+h)=-\cos h=-1+h^2/2-\cdots, agreeing with the expansion.

The standard sine and cosine series are printed in the booklet. Their shifted expansions above are not separately printed; either derivative values or the angle-shift identities establish them.

Watch out

Wrong: write f(x−a)=f(a)+⋯f(x-a)=f(a)+\cdots when expanding f(x)f(x) about aa. The function remains f(x)f(x); only the powers use the displacement x−ax-a.

Change the inner scale of cosine and keep the same centre.

Shifting a standard series and making an estimate

You can also introduce h=x−ah=x-a inside a familiar function and reuse its standard Maclaurin series. For a logarithm, extracting the centre as a factor makes the small quantity visible.

Continue this lesson.

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