Why Shift a Maclaurin Series?
A Maclaurin series describes a function using powers of \(x\), which makes it most directly useful near \(x=0\). A Taylor series centered at a different number \(a\) instead uses powers of \(x-a\). This shift lets us describe the same function in a form organized around the point of interest.
For the exponential, sine, and cosine, the shift can be obtained directly from their Maclaurin series. The resulting expansions are valid for every real \(x\), regardless of the chosen real center.
The Shifted Variable
Set \(h=x-a\), so that \(x=a+h\). The earlier Maclaurin expansions are
Replacing \(h\) by \(x-a\) gives a series in the desired powers. For sine and cosine, we must also relate the value at \(a+h\) to the functions of \(h\). The angle-addition identities do this, while the exponential identity \(e^{a+h}=e^a e^h\) does the same for \(e^x\).
Deriving the Three Centered Series
For the exponential, use the product identity and substitute the Maclaurin expansion of \(e^h\):
For sine, the angle-addition identity is \(\sin(a+h)=\sin a\cos h+\cos a\sin h\). Substituting the two Maclaurin series gives
Similarly, \(\cos(a+h)=\cos a\cos h-\sin a\sin h\), so
Why These Are Taylor Coefficients
The shift derivation shows that the displayed series equal the functions. We can also check that their coefficients have the defining Taylor form. The derivatives of \(e^x\) are all \(e^x\), so its \(n\)th derivative at \(a\) is \(e^a\). Dividing by \(n!\) gives the coefficient \(e^a/n!\).
For sine and cosine, derivatives cycle through the four functions \(\sin x\), \(\cos x\), \(-\sin x\), and \(-\cos x\). At \(a\), the even-order derivatives of sine alternate between \(\sin a\) and \(-\sin a\), while its odd-order derivatives alternate between \(\cos a\) and \(-\cos a\). For cosine, the even-order derivatives alternate between \(\cos a\) and \(-\cos a\), and the odd-order derivatives between \(-\sin a\) and \(\sin a\). Dividing each \(n\)th derivative by \(n!\) gives exactly the coefficients in the shifted formulas.
Write \(x=a+h\), so the powers in the new series will be powers of \(h=x-a\).
For the exponential use \(e^{a+h}=e^a e^h\); for sine and cosine use their angle-addition identities.
Replace each function of \(h\) by its established Maclaurin expansion.
Every power is a power of \(x-a\), and the coefficient of that power agrees with the corresponding derivative at \(a\) divided by its factorial.
Worked Examples
Worked Example: Expand \(e^x\) About \(a=2\)
Set \(h=x-2\), so \(x=2+h\). The exponential identity gives \(e^x=e^2e^h\). Substituting the Maclaurin series for \(e^h\) yields
Thus the coefficient of \((x-2)^n\) is \(e^2/n!\). To check the first coefficients, the function value and first three derivatives at \(2\) are all \(e^2\). Their Taylor coefficients are \(e^2/0!=e^2\), \(e^2/1!=e^2\), \(e^2/2!=e^2/2\), and \(e^2/3!=e^2/6\), matching the displayed expansion. Substituting \(x=2\) also leaves only the constant term \(e^2\), as required.
Worked Example: Expand \(\sin x\) About \(a=\pi/6\)
Let \(h=x-\pi/6\). Since \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\), the shifted sine formula gives
Equivalently, replacing \(h\) by \(x-\pi/6\),
Check the coefficients using derivatives at the center. The value is \(\sin(\pi/6)=1/2\); the first derivative is \(\cos(\pi/6)=\sqrt{3}/2\); the second derivative is \(-\sin(\pi/6)=-1/2\); the third is \(-\cos(\pi/6)=-\sqrt{3}/2\); and the fourth is \(\sin(\pi/6)=1/2\). Dividing these by \(0!,1!,2!,3!,4!\), respectively, gives \(1/2,\sqrt{3}/2,-1/4,-\sqrt{3}/12,1/48\), as in the expansion. At \(x=\pi/6\), all positive powers vanish and the series gives \(1/2\), the correct function value.
Worked Example: Expand \(\cos x\) About \(a=\pi/3\)
Set \(h=x-\pi/3\). Here \(\cos(\pi/3)=1/2\) and \(\sin(\pi/3)=\sqrt{3}/2\). The cosine addition identity gives
The first terms, written in \(x-\pi/3\), are
The derivative values at \(\pi/3\) in orders zero through four are \(1/2,-\sqrt{3}/2,-1/2,\sqrt{3}/2,1/2\). Dividing by the corresponding factorials gives \(1/2,-\sqrt{3}/2,-1/4,\sqrt{3}/12,1/48\), which verifies each displayed coefficient. Setting \(x=\pi/3\) makes \(h=0\), so the series gives \(1/2=\cos(\pi/3)\), also confirming its constant term.
Convergence for Every Real Input
The centered formulas come from Maclaurin series evaluated at \(h=x-a\). Each of the three Maclaurin series converges for every real value of its input. For any real \(x\) and any real center \(a\), the shifted input \(h=x-a\) is still a real number. Therefore the substitutions converge at every such \(x\).
The convergence can also be seen from the terms. For the exponential series, when \(h\ne 0\), the ratio of the absolute values of consecutive terms is
which tends to zero as \(n\) increases. For the sine series, the ratio between consecutive nonzero absolute terms is \(|h|^2/((2k+2)(2k+3))\), which tends to zero; for cosine it is \(|h|^2/((2k+1)(2k+2))\), which also tends to zero. At \(h=0\), the series reduce to their constant terms. Thus these series converge absolutely for every real \(h\), and hence for every real \(x\). There are no finite endpoints at which a separate endpoint test is needed.
A Common Translation Error
A frequent mistake is to replace \(x\) by \(x-a\) inside the original function and then treat that as the desired expansion of the original function. For example, at \(a=\pi/6\), using \(\sin(x-\pi/6)\) as the series for \(\sin x\) would give zero when \(x=\pi/6\). But the original function has value \(\sin(\pi/6)=1/2\), so that cannot be right.
The shifted variable is \(h=x-a\), but the function being expanded remains \(\sin x=\sin(a+h)\). The angle-addition identity supplies the necessary center-dependent factors:
At \(x=a\), this expression is \(\sin a\cdot 1+\cos a\cdot 0=\sin a\), as required. The same care applies to cosine, whose shifted identity includes a minus sign before \(\sin a\sin(x-a)\).
Why the Center Matters Even When Convergence Does Not
For these three functions, changing the center does not change the convergence interval: every centered series converges for all real inputs. It does change the coefficients and the powers used to represent the function. Near a chosen center \(a\), the terms are organized in powers of the distance \(x-a\); at the center itself, every term of positive degree vanishes. The coefficients encode the function's value and successive derivatives at that particular point.
This distinction is useful when constructing local descriptions at points other than zero. The Maclaurin expansion is the special case \(a=0\); translating it does not create a new convergence restriction for the exponential, sine, or cosine.
Check Your Understanding
Use the shifted variable \(h=x-a\) and the centered identities to answer the following.
- Write the Taylor series for \(e^x\) about an arbitrary real center \(a\).
- In the expansion of \(\sin x\) about \(a\), which factor multiplies the odd powers of \(x-a\)?
- Write the first four nonzero terms of the Taylor series for \(\cos x\) about \(a=0\), and compare them with the centered cosine formula.
- What is the convergence interval for the Taylor series of \(\sin x\) about any real center?
- Explain why \(\sin(x-a)\) alone cannot be the Taylor expansion of \(\sin x\) about \(a\), unless the relevant values happen to agree.
