From Derivatives to a Series
Taylor polynomials give polynomial models of a function near a chosen center. For the exponential function, the derivatives have an especially simple pattern: differentiating does not change the function. This makes its Maclaurin coefficients easy to calculate. The important further question is whether the infinite series suggested by those coefficients actually equals the function, and for which inputs.
Here a Maclaurin polynomial means a Taylor polynomial centered at zero. We will first compute its coefficients for the exponential function, then use Taylor’s theorem to prove that the sequence of these polynomials converges to the function for every real input.
Constructing the Maclaurin Polynomials
Let \(f(x)=e^x\). Since every derivative of \(f\) is again \(e^x\), we have \(f^{(k)}(x)=e^x\) for every nonnegative integer \(k\). At the center \(x=0\), this gives \(f^{(k)}(0)=e^0=1\). The degree-\(n\) Maclaurin polynomial uses the \(k\)th derivative value divided by \(k!\) as the coefficient of \(x^k\).
The first few polynomials display the pattern:
The factorials are essential. For instance, the coefficient of \(x^3\) is \(f^{(3)}(0)/3!=1/6\), not \(1\). A derivative value determines the coefficient only after it is divided by the appropriate factorial.
Worked Example: Constructing the Degree-Five Polynomial
For \(f(x)=e^x\), calculate the derivative values at zero through order five. The function and all its derivatives are \(e^x\), so \(f(0)=1\), \(f'(0)=1\), \(f''(0)=1\), \(f^{(3)}(0)=1\), \(f^{(4)}(0)=1\), and \(f^{(5)}(0)=1\).
Insert these values into the Maclaurin formula: \(P_5(x)=1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\frac{1}{5!}x^5\). Since \(2!=2\), \(3!=6\), \(4!=24\), and \(5!=120\), this is \(P_5(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\).
Check the coefficients against the defining derivative data: the \(k\)th derivative at zero of \(x^k/k!\) is \(1\), while terms of higher degree contribute zero to that derivative at zero. Thus the polynomial’s derivatives at zero, through order five, are all \(1\), as required.
The Candidate Infinite Series
As the degree increases, each polynomial adds one more term. This suggests the infinite series
At this stage, the formula is a candidate representation, not yet a proof of equality. Knowing every Taylor coefficient does not by itself guarantee that the Taylor polynomials approach the function. To establish the equality, we need to control the error left after any finite number of terms.
Why the Series Equals the Function
Taylor’s theorem with the Lagrange form of the remainder states that if a function has continuous derivatives through order \(n+1\) on the interval between \(0\) and \(x\), then \(f(x)=P_n(x)+R_n(x)\), where \(R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}\) for some \(\xi\) between \(0\) and \(x\). The exponential function has continuous derivatives of every order, so the theorem applies for every real \(x\).
For \(f(x)=e^x\), its \((n+1)\)st derivative is \(e^x\). Therefore the remainder has the form \(R_n(x)=e^\xi\frac{x^{n+1}}{(n+1)!}\), where \(\xi\) lies between \(0\) and \(x\). We now show that this remainder tends to zero as \(n\) increases.
Because \(\xi\) lies between \(0\) and \(x\), we have \(\xi\leq |x|\). The exponential function is increasing, so \(e^\xi\leq e^{|x|}\).
Taking absolute values gives \(\lvert R_n(x)\rvert\leq e^{|x|}\frac{|x|^{n+1}}{(n+1)!}\).
For fixed \(x\neq 0\), let \(b_n=|x|^{n+1}/(n+1)!\). Then \(b_{n+1}/b_n=|x|/(n+2)\), which is at most \(1/2\) for all sufficiently large \(n\). From that point onward, the terms are bounded by a geometric sequence with ratio \(1/2\), and hence tend to zero. For \(x=0\), the remainder is zero for every \(n\).
For any fixed real \(x\), \(e^{|x|}\) is a constant independent of \(n\). The bound in Step 2 therefore tends to zero, so \(R_n(x)\to 0\). Since \(e^x=P_n(x)+R_n(x)\), the polynomials \(P_n(x)\) approach \(e^x\). This proves the theorem for every real input, including negative inputs.
Using the Polynomials and Bounding the Error
The theorem justifies using a finite Maclaurin polynomial as an approximation: its error is the remainder \(R_n(x)\). The same remainder formula gives a bound. For a particular \(x\), if \(e^\xi\leq M\) for every \(\xi\) between zero and \(x\), then \(\lvert e^x-P_n(x)\rvert\leq M\frac{|x|^{n+1}}{(n+1)!}\). The bound depends on the interval between the center and the input, so the value of \(M\) should be chosen for that interval.
Worked Example: Approximating \(e^{-1}\) with Degree Four
The degree-four polynomial is \(P_4(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\). At \(x=-1\), substitute each power: \(P_4(-1)=1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\). Using denominator \(24\), this becomes \(P_4(-1)=\frac{24}{24}-\frac{24}{24}+\frac{12}{24}-\frac{4}{24}+\frac{1}{24}=\frac{9}{24}=\frac{3}{8}\). Thus \(e^{-1}\) is approximated by \(3/8\).
To bound the error, the remainder is \(R_4(-1)=e^\xi\frac{(-1)^5}{5!}\) for some \(\xi\in[-1,0]\). On this interval, \(e^\xi\leq 1\), so \(\lvert e^{-1}-3/8\rvert=\lvert R_4(-1)\rvert\leq 1/120\).
Check: the signed remainder is negative because \(e^\xi>0\) and \((-1)^5<0\), so \(e^{-1}<3/8\). Its magnitude is at most \(1/120\), consistent with the stated approximation and error bound.
Worked Example: Evaluating the Degree-Five Polynomial at \(x=2\)
From the polynomial already constructed, \(P_5(2)=1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+\frac{2^5}{5!}\). Evaluate each term: \(2^2/2!=4/2=2\), \(2^3/3!=8/6=4/3\), \(2^4/4!=16/24=2/3\), and \(2^5/5!=32/120=4/15\). Therefore \(P_5(2)=1+2+2+\frac{4}{3}+\frac{2}{3}+\frac{4}{15}\).
The first three terms sum to \(5\), and \(4/3+2/3=2\), giving \(7+4/15=105/15+4/15=109/15\). Thus \(P_5(2)=109/15\), approximately \(7.2667\).
For a check on the approximation, the remainder theorem gives \(e^2-P_5(2)=e^\xi\frac{2^6}{6!}\) for some \(\xi\in[0,2]\). Since \(e^\xi\leq e^2\), its magnitude is at most \(e^2\frac{64}{720}=\frac{4e^2}{45}\). The exact value is therefore \(e^2=109/15+R_5(2)\), with the displayed bound on the remainder.
A Common Coefficient Error
A frequent mistake is to record the derivative values correctly but use them directly as the coefficients. Since \(f^{(3)}(0)=1\), one might incorrectly write the cubic term as \(x^3\). That would give \(1+x+x^2+x^3\) through degree three. The Taylor coefficient of \(x^3\), however, is \(f^{(3)}(0)/3!=1/6\), so the correct cubic term is \(x^3/6\).
Why the Result Matters
The Maclaurin series provides polynomial approximations to \(e^x\) whose accuracy can be assessed using the remainder. More fundamentally, the proof establishes that the Taylor polynomials do not merely match the exponential function’s derivatives at zero: their limit is the function itself, at every real input. The convergence argument uses the factorial in the denominator to control the remainder even when \(x\) is negative or far from zero.
For any fixed input, higher-degree polynomials include more terms of the series. The theorem guarantees that the resulting values approach \(e^x\); the remainder estimate tells how large the error can be for a particular finite degree. Both statements are needed: the series gives the proposed representation, while the vanishing remainder proves that it represents the function.
Check Your Understanding
Use the derivative formula and the remainder argument to answer these questions.
- What is the coefficient of \(x^4\) in the Maclaurin polynomial for \(e^x\), and how does the fourth derivative at zero determine it?
- Write the degree-three Maclaurin polynomial for \(e^x\).
- State the Lagrange remainder for the degree-\(n\) Maclaurin polynomial of \(e^x\).
- Why does the bound \(e^{|x|}|x|^{n+1}/(n+1)!\) tend to zero for a fixed nonzero real \(x\)?
- What is the error bound when \(P_4(-1)\) is used to approximate \(e^{-1}\)?