From Taylor Polynomials to a Coefficient Formula
A Taylor polynomial records the value of a function and its derivatives at a chosen center \(a\). This tutorial focuses on how those derivative values determine the coefficients, and on an important distinction: a Taylor series can be formed from derivatives even when it does not converge to the function.
Write a polynomial in powers of \(x-a\), rather than powers of \(x\). Its coefficients are then tied directly to its derivatives at \(a\). The same formula defines the coefficients of the Taylor series when derivatives of every order are available.
“Formal” matters. It means that the coefficients have been specified by the derivative formula; it does not assert that the infinite sum converges, or that its sum equals \(f(x)\). If only derivatives through order \(N\) are used, the corresponding finite Taylor polynomial is \(\sum_{n=0}^{N}\frac{f^{(n)}(a)}{n!}(x-a)^n\).
Why the Derivative Formula Gives the Coefficients
Begin with a polynomial written in centered powers: \(P(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_N(x-a)^N\). At \(x=a\), every positive power of \(x-a\) is zero, so \(P(a)=c_0\). Differentiating once and setting \(x=a\) leaves \(P'(a)=c_1\). Differentiating twice leaves \(P''(a)=2!c_2\). In general, the \(n\)th derivative of the term \(c_n(x-a)^n\) is \(n!c_n\); the lower-degree terms have become zero, and higher-degree terms still contain a positive power of \(x-a\) at \(a\).
This proves the coefficient formula for polynomials. For an infinite power series, the corresponding result uses a standard power-series fact: inside the radius of convergence, a power series can be differentiated term by term, and the differentiated series has the same radius of convergence. Repeating this operation and evaluating at the center gives the same formula.
The value \(f(a)\) gives the constant coefficient \(c_0\).
Find \(f^{(n)}(a)\) for the desired orders.
Set \(c_n=f^{(n)}(a)/n!\), then multiply by \((x-a)^n\).
Coefficient calculation constructs the series; convergence and equality to \(f\) require further analysis.
Worked Coefficients at a Nonzero Center
Worked Example: A Cubic Centered at \(a=2\)
Find the Taylor series of \(f(x)=x^3-2x+1\) centered at \(a=2\).
First evaluate the function: \(f(2)=2^3-2(2)+1=8-4+1=5\). Thus \(c_0=5\).
The first derivative is \(f'(x)=3x^2-2\), so \(f'(2)=3(2^2)-2=12-2=10\), and \(c_1=10/1!=10\).
The second derivative is \(f''(x)=6x\), so \(f''(2)=12\), and \(c_2=12/2!=12/2=6\).
The third derivative is \(f'''(x)=6\), so \(f'''(2)=6\), and \(c_3=6/3!=6/6=1\). All higher derivatives are zero.
Therefore the Taylor series is finite: \((x-2)^3+6(x-2)^2+10(x-2)+5\).
Check: Set \(h=x-2\), so \(x=2+h\). Direct substitution gives \((2+h)^3-2(2+h)+1=8+12h+6h^2+h^3-4-2h+1 =5+10h+6h^2+h^3\), exactly the polynomial found from the coefficients. Differentiating that polynomial and evaluating at \(x=2\) gives \(5,10,12,6\) for orders \(0,1,2,3\), as required.
Worked Coefficients from Successive Derivatives
Worked Example: Four Taylor Coefficients for \(\sqrt{x}\) at \(a=4\)
Find the Taylor polynomial through degree four for \(f(x)=\sqrt{x}\), centered at \(a=4\). The function is defined for \(x>0\), so all derivatives used here exist near \(4\).
Evaluate the function and derivatives: \(f(4)=2\); \(f'(x)=\frac12x^{-1/2}\), so \(f'(4)=\frac14\); \(f''(x)=-\frac14x^{-3/2}\), so \(f''(4)=-\frac1{32}\); \(f'''(x)=\frac38x^{-5/2}\), so \(f'''(4)=\frac3{256}\); and \(f''''(x)=-\frac{15}{16}x^{-7/2}\), so \(f''''(4)=-\frac{15}{2048}\).
Divide each derivative value by the corresponding factorial: \(c_0=2\), \(c_1=\frac14\), \(c_2=\frac{-1/32}{2}=-\frac1{64}\), \(c_3=\frac{3/256}{6}=\frac1{512}\), and \(c_4=\frac{-15/2048}{24}=-\frac5{16384}\).
With \(h=x-4\), the degree-four Taylor polynomial is \(P_4(x)=2+\frac14h-\frac1{64}h^2+\frac1{512}h^3-\frac5{16384}h^4\).
Check: At \(x=4\), \(P_4(4)=2\). Its first four derivative values there are \(P_4'(4)=\frac14\), \(P_4''(4)=2(-\frac1{64})=-\frac1{32}\), \(P_4'''(4)=6(\frac1{512})=\frac3{256}\), and \(P_4''''(4)=24(-\frac5{16384})=-\frac{15}{2048}\). These agree with the function and derivative values calculated above.
A Taylor Series Can Have a Limited Interval of Convergence
The coefficients determine a formal series, but not the set of \(x\)-values where its sum exists. The interval of convergence is the interval of real numbers for which the power series converges; an endpoint may or may not be included. Here is a direct example in which both the coefficients and the interval can be determined.
Worked Example: The Taylor Series of \(\frac{1}{1-x}\) at \(a=0\)
Let \(f(x)=\frac{1}{1-x}\), defined for \(x\ne1\). Its \(n\)th derivative is \(f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}}\). This follows by induction: it holds for \(n=0\), and differentiating the displayed expression gives \((n+1)!/(1-x)^{n+2}\). At \(x=0\), \(f^{(n)}(0)=n!\), so every Taylor coefficient is \(c_n=n!/n!=1\). The formal Taylor series is therefore \(\sum_{n=0}^{\infty}x^n\).
For \(x\ne1\), the finite partial sum satisfies \((1-x)(1+x+\cdots+x^N)=1-x^{N+1}\), hence \(1+x+\cdots+x^N=(1-x^{N+1})/(1-x)\). If \(|x|<1\), then \(x^{N+1}\) tends to zero, so these partial sums tend to \(1/(1-x)=f(x)\).
At \(x=1\), every term equals \(1\), so the terms do not tend to zero and the series diverges. At \(x=-1\), its partial sums alternate between \(1\) and \(0\), so they do not approach a limit. If \(|x|>1\), the terms \(x^n\) do not tend to zero, so the series diverges there as well.
Check: The series converges exactly for \(-1<x<1\), so its interval of convergence is \((-1,1)\). On that interval its sum agrees with the function. The endpoint tests are essential: neither endpoint is included.
Convergence Does Not Guarantee Equality to the Function
Even when a Taylor series converges, its sum need not be the function from which its coefficients were calculated. A function that has derivatives of every order is called smooth; smoothness alone does not ensure that the Taylor series represents the function near its center.
For a concrete example, define \(g(0)=0\) and \(g(x)=e^{-1/x^2}\) for \(x\ne0\). Every derivative of \(g\) at zero is zero. To see why, away from zero each derivative has the form \(P(1/x)e^{-1/x^2}\), where \(P\) is a polynomial: differentiating such an expression again gives another polynomial in \(1/x\) multiplied by \(e^{-1/x^2}\). For every fixed nonnegative integer \(m\), \(|x|^{-m}e^{-1/x^2}\) tends to zero as \(x\) tends to zero. Indeed, putting \(t=1/x^2\), the expression is \(t^{m/2}e^{-t}\), which tends to zero because exponential growth dominates every fixed power of \(t\).
This limit shows inductively that each derivative extends to zero with value zero, and that its derivative at zero is also zero: the difference quotient is a polynomial in \(1/x\) times \(e^{-1/x^2}\), which still tends to zero. Consequently every Taylor coefficient at zero is zero. The Taylor series is the zero series, which converges for every real \(x\). But \(g(x)=e^{-1/x^2}>0\) whenever \(x\ne0\). Thus the series converges everywhere without representing \(g\) except at its center.
Common Pitfall: Forgetting the Factorial
A frequent error is to identify \(f^{(n)}(a)\) itself as the coefficient of \((x-a)^n\). The derivative value must be divided by \(n!\). For example, in the cubic calculation above, the third derivative at \(2\) is \(6\), but the coefficient of \((x-2)^3\) is \(6/3!=1\), not \(6\).
The incorrect expression \(5+10(x-2)+12(x-2)^2+6(x-2)^3\) fails a direct check: its second derivative at \(2\) is \(2(12)=24\), not \(f''(2)=12\), and its third derivative at \(2\) is \(6(6)=36\), not \(f'''(2)=6\). Dividing the order-\(n\) derivative by \(n!\) corrects both errors.
Why the Distinction Matters
Knowing how coefficients are determined makes derivative information useful: a table of derivatives at one point specifies the Taylor polynomial to any available order. But the polynomial and series answer different questions. A finite Taylor polynomial is a local approximation, while an infinite Taylor series additionally raises questions about convergence and whether the sum reproduces the function.
When working with Taylor series, keep the reasoning in this order: calculate coefficients from derivatives, determine where the series converges, and then establish whether its sum equals the function on the region of interest. Do not infer the last two conclusions from the coefficient formula alone.
Check Your Understanding
Use the coefficient formula and the distinction between formal series and convergence to answer these questions.
- If the coefficient of \((x-a)^4\) is \(c_4\), what is \(f^{(4)}(a)\), assuming the Taylor series represents \(f\) near \(a\)?
- For \(f(x)=x^2+1\) centered at \(a=1\), calculate the coefficients through degree two.
- Why does a Taylor series that converges at a point not automatically equal the function at that point?
- What are the endpoint behaviors of \(\sum_{n=0}^{\infty}x^n\) at \(x=1\) and \(x=-1\)?
- What specific error occurs if one uses \(f^{(n)}(a)\), rather than \(f^{(n)}(a)/n!\), as the coefficient of \((x-a)^n\)?
