Why a Shift Makes a New Center Useful
A Taylor polynomial describes a function near a chosen center. A power series we already know can often give that polynomial without calculating every derivative: rewrite the function so the input to the known series is zero at the desired center, substitute that shifted input, and keep the terms through the requested degree.
The key is choosing a representation that is actually valid near the new center. Substituting into a series is not the same as taking a Maclaurin polynomial and shifting that finite polynomial. If the original series is written in powers of \(x\), then near a nonzero center, contributions from arbitrarily high powers of \(x\) can affect the low-degree powers of \(x-a\). The method works cleanly when the known series is expressed in a variable that vanishes at the center.
The Shifted-Series Rule
Suppose a known power series represents a function \(F\) when its input \(u\) is within its radius of convergence \(R\):
If the function of interest can be written as \(f(x)=F(\lambda(x-a))\), then \(u=\lambda(x-a)\) is zero when \(x=a\). Substitution gives a series directly in powers of \(x-a\). To find a polynomial through degree \(n\), retain terms \(k=0\) through \(k=n\). The substitution is valid wherever \(|\lambda(x-a)|<R\).
Here is why the truncation gives the Taylor polynomial, rather than just a plausible approximation. The omitted terms have at least \(n+1\) factors of \(u\). Near \(u=0\), they are therefore of order at least \(u^{n+1}\). More explicitly, choose a positive number \(\rho<R\). For \(|u|\leq \rho\), absolute convergence gives a finite sum \(C=\sum_{k=n+1}^{\infty}|a_k|\rho^k\). Then
Thus the omitted part is bounded by a constant times \(|u|^{n+1}\). With \(u=\lambda(x-a)\), it is bounded by a constant times \(|x-a|^{n+1}\). The retained polynomial therefore agrees with the function through degree \(n\), as required of its Taylor polynomial. The argument also explains why the series must be valid near the chosen center.
Worked Examples
Worked Example: Recenter the Exponential Function at \(x=1\)
Find the degree-3 Taylor polynomial for \(f(x)=e^x\) centered at \(x=1\), using the known series \(e^u=\sum_{k=0}^{\infty}u^k/k!\).
Set \(h=x-1\), so \(x=1+h\). The exponential identity gives \(e^x=e^{1+h}=e\,e^h\). Substitute \(h\) into the known series:
Keep only terms through degree 3 in \(h\), and then write \(h=x-1\):
Check the coefficients against derivatives. Since every derivative of \(e^x\) is \(e^x\), the value and first three derivatives at \(x=1\) are all \(e\). The polynomial has constant-in-\(h\) term \(e\), linear coefficient \(e\), quadratic coefficient \(e/2=e/2!\), and cubic coefficient \(e/6=e/3!\), as required. The exponential series converges for every real input, so this recentered series is valid for every real \(x\).
Worked Example: A Trigonometric Polynomial About \(x=\frac{\pi}{2}\)
Find the degree-4 Taylor polynomial for \(f(x)=\sin(2x-\pi)\) centered at \(x=\frac{\pi}{2}\), using the known Maclaurin series for sine.
Let \(h=x-\frac{\pi}{2}\). Then \(2x-\pi=2h\), so the function is already written as sine of a quantity that is zero at the center. Use \(\sin u=u-u^3/3!+u^5/5!-\cdots\) with \(u=2h\):
Through degree 4, retain the terms of degrees 1 and 3. The degree-5 term and all later terms are omitted:
Check at the center. The function value is \(\sin(0)=0\). Its first derivative is \(2\cos(2x-\pi)\), which equals \(2\) at \(x=\frac{\pi}{2}\). Its second derivative is \(-4\sin(2x-\pi)\), which equals \(0\) there. Its third derivative is \(-8\cos(2x-\pi)\), which equals \(-8\), giving cubic Taylor coefficient \(-8/3!=-4/3\). Its fourth derivative is \(16\sin(2x-\pi)\), also \(0\) there. These are exactly the coefficients represented by the polynomial. The sine series converges for every real input.
Worked Example: Recenter a Geometric Series
Find the degree-3 Taylor polynomial for \(f(x)=\frac{1}{3-x}\) centered at \(x=1\), using the geometric series.
Write \(h=x-1\). Then \(3-x=2-h=2(1-h/2)\), so
The geometric series \(1/(1-u)=\sum_{k=0}^{\infty}u^k\) is valid for \(|u|<1\). Here \(u=h/2\), so the series converges when \(|h|<2\), or \(|x-1|<2\). Substitute and retain degrees 0 through 3:
Check by multiplying the polynomial by the original denominator. In terms of \(h\), \(T_3=\frac12+\frac h4+\frac{h^2}{8}+\frac{h^3}{16}\), so
The product equals 1 through degree 3; the remaining error begins in degree 4, as expected for a degree-3 Taylor polynomial. The calculation also verifies the coefficients and signs.
How to Choose and Truncate the Shift
The practical work is often finding the right input \(u\). Start by locating the desired center \(a\), then rewrite the function so the known series receives an input equal to zero at \(x=a\). If the input is \(u=\lambda(x-a)\), the series powers are already powers of \(x-a\); the requested degree tells you exactly where to stop.
Set \(h=x-a\), so the center corresponds to \(h=0\).
Rewrite the function in the form \(F(u)\), where the known power series represents \(F\) and \(u=0\) when \(x=a\).
Translate the known condition \(|u|<R\) into a condition on \(x\); this ensures the substitution is valid near the center.
Keep only terms whose powers contribute through the requested degree, and express the result in powers of \(x-a\).
Compare coefficients with derivatives at the center, or use the original function to verify the polynomial through the requested degree.
For an affine shift \(u=\lambda(x-a)\), the \(k\)th series term has degree \(k\), so the rule is especially direct. If the shifted input is more complicated than a constant multiple of \(x-a\), a term’s degree in \(x-a\) may differ from its series index. In that situation, expand the retained terms and discard only powers above the requested degree; do not assume the index alone gives the final degree.
A Common Recentering Mistake
A tempting shortcut is to truncate a Maclaurin series first and then substitute \(x=1+h\). That generally fails: high powers of \(x\), after the shift, contain low powers of \(h\). For example, truncating \(e^x\) at degree 2 gives \(1+x+x^2/2\). Shifting this polynomial yields
But the degree-2 Taylor polynomial for \(e^x\) at \(x=1\) is \(e(1+h+h^2/2)\). Even the constant term is wrong: the truncated shifted polynomial has constant \(5/2\), not \(e\). The missing terms of the original infinite series contribute to that constant and to the other coefficients after shifting. The correction is to express \(e^x=e\,e^h\) first and then truncate the series in \(h\). More generally, arrange for the known series’ input to vanish at the new center before deciding which terms to keep.
Why This Method Matters
Known power series let you build local polynomial models without repeatedly calculating higher derivatives. The shift makes the center visible in the powers: the constant term describes the function at \(a\), the coefficient of \(x-a\) describes its slope there, and higher powers encode higher-order behavior. The convergence condition also provides a check that the representation genuinely applies around the center. On an AP problem, these checks help distinguish a valid recentering from a polynomial that merely looks plausible.
Check Your Understanding
For each question, identify the shifted input, the degree to retain, and any convergence restriction needed.
- Use \(e^u=\sum_{k=0}^{\infty}u^k/k!\) to find the degree-2 Taylor polynomial for \(e^{3x}\) centered at \(x=1\).
- For \(f(x)=\frac{1}{5-x}\) centered at \(x=1\), rewrite the function using a geometric-series input that is zero at the center. State the interval condition for convergence.
- Using the sine Maclaurin series, find the degree-3 Taylor polynomial for \(\sin(4(x-2))\) centered at \(x=2\).
- Explain why substituting \(x=a+h\) into a finite Maclaurin polynomial for a function may not produce its Taylor polynomial at \(a\).
- For \(F(u)=\sum_{k=0}^{\infty}a_k u^k\) with radius \(R\), state the condition on \(x\) that makes \(F(2(x-3))\) valid near \(x=3\).
