Why the Center Matters
A Taylor polynomial is local: it is built to match a function and several of its derivatives at one chosen input, called the center. The closer the target input is to that center, the more naturally the polynomial describes the function near the target. But closeness is not the only consideration. A center is useful when the needed function values and coefficients are also accessible, and when the expansion being used is valid for the displacement involved.
The practical task is to write a target \(x\) as \(x=a+h\), where \(a\) is a convenient center and \(h\) is a manageable displacement. Then use the Taylor polynomial centered at \(a\) in powers of \(h\), rather than automatically using a polynomial centered at zero.
Choosing a center does not change the target value. It changes the form of the calculation and which local information about the function is useful. A good choice usually makes \(h\) small, keeps the relevant expansion within its domain of validity, and makes the coefficients simple to obtain.
A Practical Center-Selection Method
Before calculating coefficients, identify the target and compare plausible centers. Ask three questions:
- How large is the displacement? A target close to \(a\) gives a small value of \(|x-a|\), which is generally favorable for a local approximation.
- What information is available at the center? A center where the function value and derivatives are simple can make the polynomial much easier to construct.
- Is the expansion valid at the displacement? If a known power series has a restricted interval of convergence, the displacement must lie in that interval. A short polynomial can still be calculated outside it, but the corresponding series no longer justifies using its terms as a convergent expansion there.
These considerations can compete. A center may be close to the target but have inconvenient derivatives. Another may give especially simple coefficients but leave a larger displacement. Choosing a center is therefore a mathematical decision, not a rule to always choose zero.
Write down the input \(x\) at which the function is to be approximated.
Look for a center with a known function value, manageable derivatives, or a useful known expansion.
For each candidate \(a\), calculate \(h=x-a\), so \(x=a+h\).
Check the size of \(h\), the domain of the function, and any validity condition on the expansion.
Why the Polynomial Is Built Around Its Center
The Taylor coefficients are chosen so that the polynomial and the function agree in value and in derivatives at the center. This explains why the powers are powers of \(x-a\), rather than powers of \(x\).
For \(P_{n,a}(x)=\sum_{k=0}^{n}c_k(x-a)^k\), differentiating \(j\) times and evaluating at \(x=a\) gives \(P_{n,a}^{(j)}(a)=j!c_j\). Terms with \(k<j\) have vanished after \(j\) differentiations, and terms with \(k>j\) still contain a positive power of \(x-a\), so they vanish at \(a\). To make \(P_{n,a}^{(j)}(a)=f^{(j)}(a)\), the coefficient must therefore be \(c_j=f^{(j)}(a)/j!\). This holds for each \(j=0,\ldots,n\), giving the Taylor polynomial formula.
Worked Example: Approximate an Exponential Near 1
Worked Example: Compare Centers for \(e^{1.1}\)
Approximate \(e^{1.1}\) with degree-2 Taylor polynomials, comparing centers \(0\) and \(1\). For \(f(x)=e^x\), every derivative is \(e^x\).
At the center \(a=0\), the displacement is \(h=1.1-0=1.1\). Since \(f(0)=f'(0)=f''(0)=1\),
Substitute \(x=1.1\):
Now use \(a=1\). The displacement is \(h=1.1-1=0.1\), and \(f(1)=f'(1)=f''(1)=e\). Therefore
At \(x=1.1\),
Here \(e\approx2.718282\), and the final approximation is rounded to six decimal places. Substitution into the displayed polynomial confirms the result: the linear contribution is \(0.1e\), and the quadratic contribution is \(0.005e\). The center \(1\) reduces the displacement from \(1.1\) to \(0.1\), while the center \(0\) gives simpler coefficients. For this target and degree, centering at \(1\) is a more local choice.
Worked Example: Use a Known Logarithm Expansion
Worked Example: Approximate \(\ln(1.1)\)
To approximate \(\ln(1.1)\), use center \(a=1\), where \(\ln 1=0\), and set \(h=1.1-1=0.1\). The known Maclaurin expansion for \(\ln(1+u)\), valid for \(-1<u\leq 1\), gives the degree-4 polynomial in the displacement \(u\):
The target displacement \(u=0.1\) satisfies the stated condition. Substitute each power:
Rounded to six decimal places, the approximation is \(0.095308\). Checking the substitution directly, the four contributions are \(0.1\), \(-0.005\), \(0.000333333\ldots\), and \(-0.000025\), whose sum is the displayed value. The center \(1\) makes the logarithm’s input \(1+h\) with a small \(h\), exactly the form of the known expansion.
Worked Example: Respect a Geometric Series’ Domain
Worked Example: Approximate \(\frac{1}{1+1.1}\)
Let \(f(x)=1/(1+x)\) and approximate \(f(1.1)\). The geometric series \(1/(1+u)=1-u+u^2-u^3+\cdots\) converges when \(|u|<1\). Centering at zero would use \(u=x=1.1\), which is outside that interval. Instead, choose \(a=1\) and write \(x=1+h\). Then
At the target, \(h=1.1-1=0.1\), so \(|h/2|=0.05<1\). Applying the geometric expansion through degree 3 gives
Substitute \(h=0.1\), showing each term:
The finite polynomial calculation checks: the bracket is \(1-0.05=0.95\), then \(0.95+0.0025=0.9525\), then \(0.9525-0.000125=0.952375\). The exact function value is \(1/2.1=10/21\approx0.476190\), rounded to six decimal places, so this local approximation is close.
For comparison, the degree-3 polynomial centered at zero is \(1-x+x^2-x^3\). At \(x=1.1\), it gives \(1-1.1+1.21-1.331=-0.221\). The associated geometric series is not convergent there, and this polynomial is far from the positive function value. This illustrates why checking the expansion’s domain can matter as much as making the displacement small.
What Makes a Center Practical?
The examples show that a good center is not simply the nearest familiar number. A useful choice balances several features:
| Feature | Why it helps |
|---|---|
| Small displacement \(h=x-a\) | The target is close to the point where the polynomial matches the function’s derivatives. |
| Simple values or derivatives at \(a\) | The polynomial is easier to construct and evaluate. |
| A known expansion in \(h\) | Existing formulas can be used directly after rewriting the input. |
| Valid domain for the expansion | The series representation being used applies at the target displacement. |
For some functions, the target is naturally close to a center other than zero. For others, rewriting the expression reveals a known series even when the original input looks unfamiliar. A target such as \(1.1\) may be better expressed as \(1+0.1\); a rational function at a large input might become a geometric expansion in a smaller displacement after shifting its center.
Common Mistake: Choosing Zero Automatically
A common error is to use a Maclaurin polynomial simply because it is familiar, without checking the target displacement or the expansion’s validity. For the rational example, the geometric series centered at zero would tempt us to substitute \(x=1.1\) into \(1-x+x^2-x^3+\cdots\). But its convergence condition is \(|x|<1\), and \(1.1\) does not satisfy it.
Another mistake is to assume that a small displacement alone guarantees a highly accurate answer. The degree of the polynomial, the behavior of the function, and the validity range still matter. This tutorial’s calculations show how to choose and use centers; they do not supply a general numerical error bound. A particular approximation should not be assigned a guaranteed error without an applicable error result.
Check Your Understanding
For each question, identify the center and displacement before deciding how to proceed.
- To approximate \(e^{2.05}\), compare the displacements from centers \(0\) and \(2\). Which one is more local?
- For \(\ln(0.9)\), what center writes the input in the form \(1+h\), and what is \(h\)?
- The expansion \(1/(1-u)=1+u+u^2+\cdots\) requires \(|u|<1\). If a substitution gives \(u=1.2\), can that series be used as a convergent expansion there?
- In the rational-function example, why does centering at \(1\) produce the small quantity \(h/2\) rather than \(x\)?
- Name two practical features, besides distance to the target, that can make a Taylor center useful.
