Why Taylor Polynomials Are Useful for Estimation
A function value may be difficult to calculate directly, but it can sometimes be estimated by using a polynomial whose value and derivatives match the function at a nearby point. Polynomials are convenient to evaluate, and a nearby center often makes the calculation simpler than expanding around zero.
The central choice is the center \(a\). To estimate \(f(x)\), write the displacement as \(h=x-a\), then evaluate the Taylor polynomial using powers of \(h\). The polynomial uses information about the function at \(a\), while the displacement tells us how far we are applying that local information.
The Taylor Polynomial as an Estimation Tool
If \(f\) has derivatives through order \(n\) at the center \(a\), its degree-\(n\) Taylor polynomial centered at \(a\) is
Here \(f^{(0)}(a)\) means \(f(a)\), and \(k!\) is the factorial of \(k\), with \(0!=1\). The estimate is \(f(x)\approx P_n(x)\). A Taylor polynomial is constructed to match the function and its first \(n\) derivatives at \(a\); near \(a\), its values can therefore provide useful estimates. The notation \(\approx\) indicates an estimate, not an equality guaranteed for every \(x\).
For practical work, choose a center where the function’s value and derivatives are known or easy to compute. Then calculate the displacement, evaluate the terms in order, and combine them. For trigonometric functions, angles are measured in radians.
Why the Formula Uses the Displacement
The reason powers of \(x-a\) appear is that the polynomial is designed to match derivatives at the center. For example, a polynomial centered at \(a\) begins with a constant term \(f(a)\), followed by a linear term proportional to \(x-a\). At \(x=a\), the displacement is zero, so the polynomial has value \(f(a)\). Its derivative’s value at \(a\) is the coefficient of the linear term, namely \(f'(a)\).
The same pattern continues for higher derivatives: differentiating the \(k\)th-power term \(k\) times and then setting \(x=a\) leaves \(k!\) times its coefficient. Choosing that coefficient to be \(f^{(k)}(a)/k!\) makes the \(k\)th derivative agree at the center. Thus the Taylor polynomial matches the function’s derivative information through order \(n\) there. This explains the construction, though it does not by itself provide a numerical error bound at a different point.
Worked Example: Estimate an Exponential Value
Worked Example: Estimate \(e^{1.05}\)
Use the degree-three Taylor polynomial for \(f(x)=e^x\) centered at \(a=1\). The derivatives of \(e^x\) are all \(e^x\), so each derivative at \(1\) equals \(e\). The displacement is
Therefore, the degree-three polynomial is
Substitute \(h=0.05\) into the parenthesis:
Using \(e\approx2.718281828\), the estimate is
Check: Substituting \(h=0.05\) again gives the four contributions \(1\), \(0.05\), \(0.00125\), and \(0.0000208333\); their sum is \(1.0512708333\). Multiplication by \(2.718281828\) gives approximately \(2.8576504\), as stated. This checks the polynomial arithmetic, not the approximation’s error.
Worked Example: Estimate a Trigonometric Value
Worked Example: Estimate \(\sin(\frac{\pi}{6}+0.1)\)
Use the degree-three Taylor polynomial for \(f(x)=\sin x\) centered at \(a=\frac{\pi}{6}\). The angle and displacement are in radians. At the center,
The Taylor polynomial, with \(h=x-a\), is
For \(x=\frac{\pi}{6}+0.1\), \(h=0.1\). Substituting gives
Evaluate each term:
Thus \(\sin(\frac{\pi}{6}+0.1)\approx0.58396\). Check: The quadratic contribution is \(-0.25(0.01)=-0.0025\), and the cubic contribution is \(-(\sqrt3/12)(0.001)\approx-0.00014434\). Adding these to \(0.5+0.08660254\) gives \(0.58395820\), confirming the evaluation.
Worked Example: Estimate a Logarithmic Value
Worked Example: Estimate \(\ln(1.04)\)
Use the degree-four Taylor polynomial for \(f(x)=\ln x\) centered at \(a=1\). Since the derivatives at \(1\) give alternating coefficients, the polynomial in \(h=x-1\) is
For \(x=1.04\), the displacement is \(h=1.04-1=0.04\). Evaluate each term:
Therefore, \(\ln(1.04)\approx0.03922069\). Check: The powers are \(h^2=0.0016\), \(h^3=0.000064\), and \(h^4=0.00000256\); dividing by their indicated denominators and applying the alternating signs reproduces the displayed sum.
Worked Example: Estimate a Rational Function Value
Worked Example: Estimate \(\frac{1}{0.98}\)
Let \(f(x)=1/x\) and use its degree-three Taylor polynomial centered at \(a=1\). Its derivatives at \(1\) give the polynomial \(1-h+h^2-h^3\), where \(h=x-1\). At \(x=0.98\),
Substitute this displacement carefully, including its negative sign:
Thus \(1/0.98\approx1.020408\). Check: The exact reciprocal satisfies \(0.98(1.020408)=0.99999984\), which is very close to \(1\) at the displayed precision. The small discrepancy comes from using a truncated polynomial estimate.
A Reliable Estimation Routine
Use the same sequence of decisions each time. Separating the choice of center from the arithmetic helps prevent common errors.
Write the requested input \(x\) and the chosen center \(a\). The center should be a point where the needed function values and derivatives are manageable.
Set \(h=x-a\). Keep its sign: a target below the center has negative displacement.
Use powers of \(x-a\), or replace \(x-a\) by \(h\) throughout. The coefficients must come from derivatives evaluated at \(a\).
Compute powers and terms separately, combine them, and check the signs and arithmetic. Round only the final estimate to the requested precision.
Common Mistake: Substituting the Input Instead of the Displacement
A Taylor polynomial centered at \(1\) is written in powers of \(x-1\), not powers of \(x\). For the logarithm example, the relevant displacement at \(x=1.04\) is \(h=0.04\). A common error is to substitute \(1.04\) for \(h\) in the polynomial \(h-h^2/2+h^3/3-h^4/4\). That produces
which is not the degree-four Taylor polynomial centered at \(1\) evaluated at \(1.04\). It treats \(1.04\) as the distance from the center, even though the distance is \(1.04-1=0.04\). The correction is to calculate the displacement first and use it consistently:
This distinction is essential whenever the center is nonzero. The input and the displacement coincide only when the center is zero.
Interpreting the Estimate
The examples show how the calculation works, but a careful interpretation matters too. Taylor polynomials are local approximations: they are built from function information at one center. A small displacement often makes the successive powers small, which can make omitted higher-order terms less influential. That observation is useful for judging whether an estimate seems plausible, but it is not a proof of a particular number of correct digits.
For a rigorous statement about the error, one needs an error theorem and suitable information about derivatives on an interval containing the center and target. Those error bounds are a separate tool. Until such a bound is established, report the polynomial value as an estimate rather than claiming a guaranteed accuracy.
Check Your Understanding
For each question, write the displacement before evaluating or describing the Taylor polynomial.
- For a Taylor polynomial centered at \(2\), what displacement should be used to estimate \(f(1.9)\)?
- Using \(P_2(x)=e\left(1+h+\frac{h^2}{2}\right)\) centered at \(1\), evaluate the estimate for \(e^{1.02}\).
- For a Taylor polynomial of \(\cos x\) centered at \(0\), what is the displacement when estimating \(\cos(0.08)\)?
- Why is a polynomial value an estimate rather than automatically an exact value of the function?
- When would a rigorous error bound be needed in addition to evaluating the Taylor polynomial?
