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Proving Taylor-Series Identities With Differential Equations

Learn how differential equations and initial values establish that convergent power series are the Taylor series of the exponential, sine, and cosine functions.

Intermediate 10 min read

What You'll Learn

  • Derive the coefficient recurrence for the exponential series from its differential equation.
  • Derive the coupled coefficient recurrences for sine and cosine.
  • Check convergence and justify termwise differentiation for all real inputs.
  • Prove that matching differential equations and initial values identifies the represented functions.
  • Use the series differential equations to prove identities such as sine squared plus cosine squared equals one.

Why a Differential Equation Can Verify a Series

A proposed Taylor series is not established as a formula for a function merely because its coefficients resemble familiar derivatives. Two separate facts are needed: the series must converge where it is used, and its sum must actually be the function in question. Differential equations offer a systematic way to verify both the proposed coefficients and the identity of the sum.

The exponential, sine, and cosine functions are characterized by simple differential equations together with values at zero. We will construct power series that satisfy those equations and initial values. A uniqueness argument then shows that the sums of the series are the corresponding functions.

Key idea: If a convergent power series defines a differentiable function that satisfies the same differential equation and initial value as a known function, and that initial-value problem has a unique solution, then the two functions are equal.

The Candidate Series and Their Convergence

Consider the following three power series, defined for real \(x\):

$$ E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad S(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!},\qquad C(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}. $$

Here \(n!\) denotes the factorial, with \(0!=1\). Before differentiating, we check that these series converge for every real \(x\). For \(E\), the ratio of the absolute values of consecutive terms, when \(x\ne 0\), is

$$ \frac{|x|^{n+1}/(n+1)!}{|x|^n/n!} =\frac{|x|}{n+1}\longrightarrow 0 \qquad\text{as }n\longrightarrow\infty. $$

For the sine and cosine series, the corresponding ratios of consecutive absolute values are, respectively,

$$ \frac{|x|^{2}}{(2n+2)(2n+3)} \quad\text{and}\quad \frac{|x|^{2}}{(2n+1)(2n+2)}. $$

Each ratio tends to zero as \(n\) tends to infinity. The ratio test therefore gives absolute convergence for every real \(x\); all three power series have infinite radius of convergence. A power series can be differentiated term by term at every point strictly inside its radius of convergence, and the differentiated series has the same radius. Thus these series may be differentiated term by term for every real \(x\).

Worked Example: Deriving the Exponential Coefficients

Worked Example: Deriving the Exponential Coefficients

Suppose a power series \(Y(x)=\sum_{n=0}^{\infty}a_nx^n\) is to satisfy \(Y'=Y\) and \(Y(0)=1\). Differentiating term by term gives

$$ Y'(x)=\sum_{n=1}^{\infty}n a_nx^{n-1} =\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n. $$

The initial value gives \(a_0=1\). Matching the coefficient of \(x^n\) in \(Y'=Y\) gives \((n+1)a_{n+1}=a_n\), so \(a_{n+1}=a_n/(n+1)\). The first coefficients are

$$ a_0=1,\qquad a_1=\frac{a_0}{1}=1,\qquad a_2=\frac{a_1}{2}=\frac12,\qquad a_3=\frac{a_2}{3}=\frac16,\qquad a_4=\frac{a_3}{4}=\frac1{24}. $$

Continuing the recurrence gives \(a_n=1/n!\). Hence the candidate series is \(E(x)=\sum_{n=0}^{\infty}x^n/n!\). Its initial value is \(E(0)=1\). Differentiating and reindexing verifies the equation:

$$ E'(x)=\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n!} =\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!} =\sum_{k=0}^{\infty}\frac{x^k}{k!} =E(x). $$

For the final equality, the index change is \(k=n-1\). Thus the series satisfies both required conditions. The uniqueness argument below will identify its sum with \(e^x\).

Worked Example: Deriving the Sine and Cosine Coefficients

Worked Example: Deriving the Sine and Cosine Coefficients

Seek two power series \(U(x)=\sum_{n=0}^{\infty}a_nx^n\) and \(V(x)=\sum_{n=0}^{\infty}b_nx^n\) satisfying \(U'=V\), \(V'=-U\), \(U(0)=0\), and \(V(0)=1\). The initial values set \(a_0=0\) and \(b_0=1\). Differentiating and matching coefficients gives

$$ (n+1)a_{n+1}=b_n,\qquad (n+1)b_{n+1}=-a_n. $$

Applying these recurrences step by step yields

$$ a_1=b_0=1,\qquad b_1=-a_0=0, $$ $$ a_2=\frac{b_1}{2}=0,\qquad b_2=-\frac{a_1}{2}=-\frac12, $$ $$ a_3=\frac{b_2}{3}=-\frac16,\qquad b_3=-\frac{a_2}{3}=0, $$ $$ a_4=\frac{b_3}{4}=0,\qquad b_4=-\frac{a_3}{4}=\frac1{24}. $$

Continuing gives nonzero coefficients \(a_{2n+1}=(-1)^n/(2n+1)!\) and \(b_{2n}=(-1)^n/(2n)!\); the other coefficients vanish. These are exactly the coefficients in \(S\) and \(C\). Directly differentiating their convergent series checks the equations:

$$ S'(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=C(x), \qquad C'(x)=\sum_{n=1}^{\infty}(-1)^n\frac{x^{2n-1}}{(2n-1)!}=-S(x). $$

Also \(S(0)=0\) and \(C(0)=1\), since their constant terms are \(0\) and \(1\), respectively. The coupled series therefore satisfy the required equations and initial values.

Why the Differential Equations Identify the Functions

Having verified the equations, we still need to show that their solutions are unique. For the exponential, the standard function \(e^x\) satisfies \((e^x)'=e^x\) and \(e^0=1\). Suppose \(Y\) is any differentiable function satisfying \(Y'=Y\) and \(Y(0)=1\). The product rule gives

$$ \frac{d}{dx}\bigl(e^{-x}Y(x)\bigr) =-e^{-x}Y(x)+e^{-x}Y'(x) =e^{-x}\bigl(Y'(x)-Y(x)\bigr)=0. $$

A differentiable function whose derivative is zero on an interval is constant there. Since \(e^0Y(0)=1\), this constant is \(1\), so \(e^{-x}Y(x)=1\) and \(Y(x)=e^x\). In particular, the series sum \(E\) equals \(e^x\).

For sine and cosine, use the standard derivative relations \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\), with \(\sin 0=0\) and \(\cos 0=1\). Suppose \(U'=V\), \(V'=-U\), \(U(0)=0\), and \(V(0)=1\). Define the differences \(p=U-\sin\) and \(q=V-\cos\). Then \(p'=q\), \(q'=-p\), and \(p(0)=q(0)=0\). Consequently,

$$ \frac{d}{dx}\bigl(p(x)^2+q(x)^2\bigr) =2p(x)p'(x)+2q(x)q'(x) =2p(x)q(x)-2q(x)p(x)=0. $$

The sum \(p^2+q^2\) is constant and has value zero at \(x=0\). Since squares are nonnegative, \(p(x)^2+q(x)^2=0\) implies \(p(x)=q(x)=0\). Thus \(U=\sin\) and \(V=\cos\). Applying this to the series sums \(S\) and \(C\) proves their identities with sine and cosine.

Theorem: The three convergent power series above represent \(e^x\), \(\sin x\), and \(\cos x\) for every real \(x\). The justification requires both termwise differentiation inside the radius of convergence and uniqueness of the corresponding initial-value problems.

Worked Example: Proving a Trigonometric Identity from the Series

Worked Example: Proving a Trigonometric Identity from the Series

Since \(S=\sin\) and \(C=\cos\), consider \(F(x)=S(x)^2+C(x)^2\). Using the verified differential equations \(S'=C\) and \(C'=-S\), the product rule gives

$$ F'(x)=2S(x)S'(x)+2C(x)C'(x) =2S(x)C(x)-2C(x)S(x)=0. $$

Therefore \(F\) is constant. Its value at zero is

$$ F(0)=S(0)^2+C(0)^2=0^2+1^2=1. $$

Hence \(S(x)^2+C(x)^2=1\) for every real \(x\), and the established identities \(S=\sin\), \(C=\cos\) give \(\sin^2x+\cos^2x=1\). This check uses the differential equations and initial values; it does not require multiplying two infinite series.

A Common Indexing Pitfall

When differentiating the exponential series, a common error is to forget the factor produced by the power rule. The incorrect calculation

$$ \frac{d}{dx}\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right) \overset{\text{wrong}}{=} \sum_{n=1}^{\infty}\frac{x^{n-1}}{n!} $$

omits the factor \(n\). For example, the coefficient of \(x\) in the incorrect derivative is \(1/2\), whereas differentiating the \(x^2/2!\) term correctly gives \(x\), with coefficient \(1\). The correct calculation includes \(n\) and then reindexes:

$$ \frac{d}{dx}\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right) =\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n!} =\sum_{n=1}^{\infty}\frac{x^{n-1}}{(n-1)!} =\sum_{k=0}^{\infty}\frac{x^k}{k!}. $$

There is also a broader logical pitfall: matching a few coefficients, or observing that a function has specified derivatives at zero, is not by itself a proof that a series equals that function. Here convergence licenses differentiation, and uniqueness of the differential equations supplies the equality.

Why This Method Matters

Power-series coefficients can be difficult to guess by repeatedly calculating derivatives. Differential equations turn the problem into coefficient recurrences, and initial values fix the starting coefficients. This is especially effective when the function has a simple derivative relation, as \(e^x\), \(\sin x\), and \(\cos x\) do.

The method also clarifies the roles of the different steps. A recurrence proposes and determines the coefficients. Convergence ensures the series defines a function and permits termwise differentiation. The differential equation and initial values then identify that function. Leaving out any of these steps weakens the argument: a formal coefficient pattern alone does not establish a represented function.

Check Your Understanding

Use the differential equations, coefficient recurrences, and convergence facts developed above to answer these questions.

  1. What recurrence for \(a_n\) follows from \(Y'=Y\) when \(Y(x)=\sum_{n=0}^{\infty}a_nx^n\)?
  2. Why is termwise differentiation valid for each of the three candidate series at every real \(x\)?
  3. How do the initial values determine the first coefficients in the coupled sine-and-cosine recurrences?
  4. Why does the identity \(\frac{d}{dx}(e^{-x}Y(x))=0\), together with \(Y(0)=1\), imply \(Y=e^x\)?
  5. How do the differential equations and initial values prove \(S(x)^2+C(x)^2=1\)?

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