Three Maclaurin Series to Recognize
The exponential, sine, and cosine functions have especially useful Maclaurin series. Their coefficients follow simple patterns: factorial denominators for all three, alternating signs for sine and cosine, and systematic gaps in the powers for the trigonometric functions.
A Maclaurin series is a Taylor series centered at \(x=0\). For a function with the required derivatives, its coefficient of \(x^n\) is \(f^{(n)}(0)/n!\). The three series in this tutorial are valid for every real \(x\).
Here \(n!\) means \(1\cdot2\cdot3\cdots n\) when \(n\) is a positive integer, and \(0!=1\). The index \(n\) starts at zero in each sigma expression. In the sine and cosine formulas, each increase of \(n\) advances the exponent by two.
Where the Coefficients Come From
The Maclaurin coefficient rule explains the patterns. For \(e^x\), every derivative is \(e^x\), so every derivative evaluated at zero is \(1\). Thus the coefficient of \(x^n\) is \(1/n!\).
For sine and cosine, the derivatives repeat in cycles. The cycle for sine is \(\sin x,\cos x,-\sin x,-\cos x\), then it repeats. Evaluating at zero gives \(0,1,0,-1,\ldots\). The cycle for cosine is \(\cos x,-\sin x,-\cos x,\sin x\), then it repeats; its values at zero are \(1,0,-1,0,\ldots\). Dividing each \(n\)th derivative value by \(n!\) gives the coefficient of \(x^n\).
For a Maclaurin series, the coefficient of \(x^n\) is \(f^{(n)}(0)/n!\).
The exponential cycle always gives \(1\); the trigonometric cycles give alternating nonzero values and zeros.
Sine has nonzero coefficients only for odd powers; cosine has nonzero coefficients only for even powers.
For example, the first few sine coefficients are \(f(0)=0\), \(f'(0)/1!=1\), \(f''(0)/2!=0\), \(f'''(0)/3!=-1/6\), and \(f^{(4)}(0)/4!=0\). These produce \(0+x+0x^2-x^3/3!+0x^4+\cdots\). The zero terms are often omitted, leaving the familiar sine series.
Why the Infinite Series Equal the Functions
A list of Maclaurin coefficients determines the Taylor polynomials, but a further fact is needed to conclude that the infinite series equals the function. Taylor's theorem gives a remainder after each finite polynomial. For these three functions, that remainder tends to zero for every fixed real \(x\).
For \(e^x\), every derivative is \(e^x\); at any point between \(0\) and \(x\), its absolute value is at most \(e^{|x|}\). The remainder after degree \(N\) is therefore bounded by \(e^{|x|}|x|^{N+1}/(N+1)!\). For sine and cosine, every derivative has absolute value at most \(1\), so the corresponding remainder is bounded by \(|x|^{N+1}/(N+1)!\). For fixed \(x\), these bounds tend to zero: the ratio of successive terms \(|x|^k/k!\) is \(|x|/(k+1)\), which eventually becomes less than any fixed number between zero and one. Consequently the Taylor polynomials converge to the functions, giving the stated series for every real input.
Reading the Patterns and the Missing Powers
In the exponential series, every nonnegative integer power appears, with coefficient \(1/n!\). In the sine series, the powers are \(1,3,5,7,\ldots\); in the cosine series, they are \(0,2,4,6,\ldots\). A power absent from an expanded sine or cosine series has coefficient zero, not an unlisted coefficient waiting to be inferred.
| Function | Nonzero powers | Coefficient pattern |
|---|---|---|
| \(e^x\) | \(0,1,2,3,\ldots\) | \(1/n!\) on \(x^n\) |
| \(\sin x\) | \(1,3,5,7,\ldots\) | \((-1)^n/(2n+1)!\) on \(x^{2n+1}\) |
| \(\cos x\) | \(0,2,4,6,\ldots\) | \((-1)^n/(2n)!\) on \(x^{2n}\) |
The alternating sign is indexed by the term number in the relevant series, not by the exponent alone. For sine, \(x^{2n+1}\) has sign \((-1)^n\); for cosine, \(x^{2n}\) has sign \((-1)^n\). This is why the constant term of cosine is positive, its \(x^2\) term is negative, and its \(x^4\) term is positive.
Worked Examples
Worked Example: Use the Exponential Series to Approximate \(e^{-1}\)
Use terms through \(n=5\) in the exponential series. Substituting \(x=-1\) gives
Compute using denominator \(120\):
Thus this partial sum is \(11/30\), or approximately \(0.3667\) to four decimal places. The check is that the six terms correspond exactly to \(n=0,1,2,3,4,5\), and each has the required \((-1)^n/n!\) value. The partial sum is an approximation, not an equality with \(e^{-1}\); the omitted terms have not been included.
Worked Example: Build a Partial Sum for \(\sin 1\)
The sine series contains odd powers, with signs alternating from positive to negative. Keeping the terms through \(x^7\) and substituting \(x=1\) gives
Using denominator \(5040\), the numerator is \(5040-840+42-1=4241\). Therefore
The check is that the exponents are \(1,3,5,7\), and the corresponding signs are \(+,-,+,-\), as required by \((-1)^n\). The result is a partial-sum approximation to \(\sin 1\); no claim about its error is being made here.
Worked Example: Identify Coefficients and Missing Powers
Find the coefficients of \(x^6\) in \(e^x\), of \(x^7\) in \(\sin x\), and of \(x^7\) in \(\cos x\).
For \(e^x\), the coefficient of \(x^n\) is \(1/n!\). Setting \(n=6\) gives \(1/6!=1/720\).
For sine, the power \(x^7\) occurs when \(2n+1=7\), so \(n=3\). Its coefficient is \((-1)^3/7!=-1/5040\).
For cosine, the powers have the form \(2n\). There is no integer \(n\) satisfying \(2n=7\), so \(x^7\) is missing and its coefficient is zero.
Here \([x^k]f(x)\) denotes the coefficient of \(x^k\) in the series for \(f\). The check is that \(6\) is a nonnegative integer power allowed in \(e^x\), \(7\) is an odd power allowed in sine, and \(7\) is not an even power allowed in cosine.
A Common Mistake: Losing the Factorial or the Sign
A frequent error is to list the sine powers correctly but use the wrong factorial or sign. For example, one might write
This is incorrect: the denominators are factorials, and the sign pattern applies to the terms. The correct beginning is
The wrong expression gives coefficient \(-1/3\) for \(x^3\), but the Maclaurin coefficient rule gives \(\sin'''(0)/3!=-1/6\). Comparing a disputed term with its derivative coefficient is a direct way to catch this error. Also do not put an \(x^2\) term into sine or an \(x^3\) term into cosine: those coefficients are zero.
Using the Series Carefully
These series are useful for expressing function values as limits of finite sums, building polynomial approximations, and spotting coefficients without differentiating repeatedly. When a problem asks for a specified degree, include only powers up to that degree; when it asks for a specified number of terms, count the nonzero terms rather than counting all powers.
Keep the distinction between a series and a partial sum clear. An infinite series represents the function as the limit of its partial sums. A finite truncation is an approximation unless the omitted terms are known to vanish. The coefficient patterns make it straightforward to construct a truncation, but a separate error analysis is needed to certify how accurate that approximation is.
Check Your Understanding
Use the coefficient patterns and show the key steps in each response.
- Write the first five nonzero terms of the Maclaurin series for \(e^x\).
- What is the coefficient of \(x^9\) in \(\sin x\)?
- What is the coefficient of \(x^8\) in \(\cos x\)?
- Using terms through \(x^6\), write the partial-sum approximation to \(\cos 1\) and simplify it as a fraction.
- Why does the fact that a power is absent from the sine series imply its coefficient is zero?
