Tutorials › Calculus II › Recovering Coefficients from Derivatives

Home  /  Tutorials  /  Power Series Operations

Power Series Operations · Tutorial 760 of 998

Recovering Coefficients from Derivatives

Use derivatives at a power series’ center to recover its coefficients and compare candidate representations precisely.

Intermediate 9 min read

What You'll Learn

  • State the coefficient-recovery formula and the conditions under which it applies.
  • Explain why the \(n\)th derivative at the center isolates the \(n\)th coefficient.
  • Recover coefficients from a given power series.
  • Verify a geometric-series representation using derivatives at its center.
  • Use a derivative mismatch to rule out a proposed representation.

Why Derivatives Reveal Coefficients

A power series records a function through its coefficients. Differentiation offers a way to read those coefficients directly from the function: evaluate the appropriate derivative at the center, then divide by a factorial. This gives a precise test for a proposed power-series representation, not merely a way to check a few values of the function.

The center matters. For a series centered at \(c\), the coefficients describe powers of \(x-c\), so derivatives must be evaluated at \(x=c\). At that point, terms of higher degree still contain a positive power of \(x-c\) and vanish, while terms of lower degree have already been eliminated by differentiation.

Key idea: If a power series represents \(f\) near its center \(c\), then its \(n\)th coefficient is \(f^{(n)}(c)/n!\). The factorial accounts for differentiating the \(n\)th power \(n\) times.

The Coefficient-Recovery Theorem

Consider a real power series centered at \(c\):

$$ \sum_{k=0}^{\infty} a_k(x-c)^k. $$

Suppose its radius of convergence is \(R>0\), and let \(f\) be the function represented by this series for \(|x-c|<R\). The termwise differentiation theorem says that the series can be differentiated term by term at points strictly inside its interval of convergence, and the differentiated series represents \(f'\) there. This remains true after any finite number of differentiations. In particular, \(f^{(n)}(c)\) exists for every nonnegative integer \(n\).

Theorem: If \(f(x)=\sum_{k=0}^{\infty}a_k(x-c)^k\) for \(|x-c|<R\), where \(R>0\), then, for every \(n=0,1,2,\ldots\), $$ f^{(n)}(c)=n!a_n \qquad\text{and hence}\qquad a_n=\frac{f^{(n)}(c)}{n!}. $$

Here \(f^{(0)}\) means \(f\) itself, and \(0!=1\). Thus the case \(n=0\) says \(f(c)=a_0\). For \(n=1\), it says \(f'(c)=a_1\). For higher \(n\), the coefficient is not simply the derivative value: it is the derivative value divided by \(n!\).

Why the Formula Works

The key is to differentiate the series one step at a time, using termwise differentiation within the radius of convergence. Differentiating a term of degree \(k\) once multiplies its coefficient by \(k\) and reduces its power by one. Repeating this process makes each term of degree less than \(n\) disappear before the \(n\)th derivative is reached. The term of degree \(n\) becomes a constant, and every term of degree greater than \(n\) still contains a positive power of \(x-c\), so it vanishes when evaluated at \(c\).

1
Differentiate within the interval.
Because \(|x-c|<R\), termwise differentiation is valid. The same theorem can be applied again to the resulting power series.
2
Track the term of degree \(n\).
The term \(a_n(x-c)^n\), after \(n\) differentiations, is \(n!a_n\): its successive factors are \(n,n-1,\ldots,1\).
3
Evaluate at the center.
Terms of degree below \(n\) have vanished through differentiation. Terms of degree above \(n\) still contain \(x-c\), so they are zero at \(x=c\). Only \(n!a_n\) remains.

This proves \(f^{(n)}(c)=n!a_n\). Dividing by the nonzero number \(n!\) gives the coefficient formula. The argument depends on being inside the radius of convergence: it does not assert that the series can be differentiated term by term at an endpoint.

Worked Example: Reading Coefficients from a Series

Worked Example: Reading Coefficients from a Series

Suppose, for \(|x-2|<1\), a function is represented by

$$ f(x)=\sum_{k=0}^{\infty}\frac{(x-2)^k}{k+1}. $$

Here the center is \(c=2\), and the coefficient of \((x-2)^k\) is \(a_k=1/(k+1)\). The coefficient-recovery formula gives:

$$ f(2)=0!a_0=1\cdot 1=1, $$ $$ f'(2)=1!a_1=1\cdot\frac12=\frac12, $$ $$ f''(2)=2!a_2=2\cdot\frac13=\frac23, $$ $$ f'''(2)=3!a_3=6\cdot\frac14=\frac32. $$

For example, the second derivative at the center recovers the coefficient \(a_2\) as \(f''(2)/2!=(2/3)/2=1/3\), as required. The third derivative similarly gives \(f'''(2)/3!=(3/2)/6=1/4=a_3\). These checks confirm both the center and the factorial scaling.

Worked Example: Verifying a Geometric-Series Representation

Worked Example: Verifying a Geometric-Series Representation

For \(|x|<1\), the finite geometric-sum identity is \((1-x)\sum_{k=0}^{N}x^k=1-x^{N+1}\). Since \(x^{N+1}\) tends to zero when \(|x|<1\), taking the limit gives

$$ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k. $$

To check what the coefficient-recovery theorem says here, let \(f(x)=1/(1-x)\), with center \(c=0\). We calculate its derivatives. The first derivative is

$$ f'(x)=\frac{1}{(1-x)^2}. $$

More generally, \(f^{(n)}(x)=n!/(1-x)^{n+1}\). This follows by induction: it holds for \(n=0\), and differentiating \(n!/(1-x)^{n+1}\) gives \((n+1)!/(1-x)^{n+2}\). Therefore, at the center,

$$ f^{(n)}(0)=\frac{n!}{(1-0)^{n+1}}=n!, \qquad a_n=\frac{f^{(n)}(0)}{n!}=\frac{n!}{n!}=1. $$

Every coefficient is \(1\), matching the geometric series. As a check at \(n=2\), \(f''(0)=2\), and \(f''(0)/2!=2/2=1\), which is the coefficient of \(x^2\).

Worked Example: Rejecting a Proposed Representation

Worked Example: Rejecting a Proposed Representation

Consider \(g(x)=1/(1-x^2)\) for \(|x|<1\). A proposed representation is

$$ g(x)=\sum_{k=0}^{\infty}x^{2k}+x. $$

At \(x=0\), both sides equal \(1\), so checking only the function value does not reveal a problem. Differentiate the function directly:

$$ g'(x)=\frac{2x}{(1-x^2)^2}, \qquad g'(0)=\frac{2\cdot 0}{(1-0^2)^2}=0. $$

Now check the proposed series. Its terms \(x^{2k}\) have zero derivative at \(x=0\) for every \(k\geq 1\), and the constant term has derivative zero. The additional term \(x\) has derivative \(1\). Thus the proposed series has derivative \(1\) at the center, not \(0\).

$$ \left.\frac{d}{dx}\left(\sum_{k=0}^{\infty}x^{2k}+x\right)\right|_{x=0}=1 \ne g'(0)=0. $$

The representation is therefore false on any interval around \(0\). The value at the center agreed, but the first derivative exposed the incorrect coefficient of \(x\).

Using Derivatives to Compare Series

The recovery formula has a useful uniqueness consequence. Suppose two power series, both centered at \(c\), converge on a neighborhood of \(c\) and represent the same function there. Their \(n\)th derivatives at \(c\) are then equal, since they are derivatives of the same function. Dividing that common derivative by \(n!\) shows that their \(n\)th coefficients are equal. This holds for every \(n\), so two power-series representations of the same function at the same center cannot have different coefficients.

Conversely, if two proposed representations give different coefficients at any index, they cannot represent the same function on a neighborhood of their common center. A practical comparison often starts with the first few derivatives: a mismatch at even one order is enough to disprove equality. Agreement for only finitely many derivatives is not enough to establish that two series represent the same function; it confirms only that their coefficients agree through the corresponding finite order.

Common Pitfall: Forgetting the Factorial

A frequent error is to identify \(a_n\) with \(f^{(n)}(c)\) rather than with \(f^{(n)}(c)/n!\). For instance, in the geometric-series example, \(f''(0)=2\). Writing the coefficient of \(x^2\) as \(2\) is incorrect. The correct calculation is

$$ a_2=\frac{f''(0)}{2!}=\frac{2}{2}=1. $$

Another common error is evaluating derivatives at the wrong point. For a series centered at \(c\), its coefficients are recovered from derivatives at \(c\), not automatically at \(0\). A series in powers of \(x-2\), for example, must be checked using derivatives at \(x=2\). Also keep the scope of the theorem in view: the formula applies to a power series representing the function in its open interval of convergence, and it does not by itself justify termwise differentiation at a boundary point.

Why Coefficient Recovery Matters

Power-series operations can produce new coefficient sequences, but a coefficient calculation alone does not always settle what function a series represents. Derivatives provide an independent check: if the series converges near its center and represents a specified function, its coefficients are forced by the function’s derivatives there. This is especially useful when comparing two candidate expansions, checking a result after algebraic operations, or identifying precisely where a proposed series first disagrees with the function.

The central procedure is simple: identify the center, compute the derivative of the matching order at that center, and divide by the factorial. The theorem explains why this works and why a single coefficient mismatch is decisive.

Check Your Understanding

Use the coefficient-recovery theorem and show the key calculation in each case.

  1. If \(f(x)=\sum_{n=0}^{\infty}a_n(x-3)^n\) near \(x=3\), which derivative recovers \(a_4\), and what factor must be used?
  2. A series centered at \(0\) represents \(f\), and \(f'''(0)=12\). What is the coefficient of \(x^3\)?
  3. Why do terms of degree greater than \(n\) contribute zero to \(f^{(n)}(c)\)?
  4. Two power series centered at the same point represent the same function near that point. Explain why their coefficients at every index must agree.
  5. A proposed series agrees with a function’s value at its center but has a different first derivative there. What does this tell you about the proposed representation?

Previous

Next

Back to Tutorials