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Composite Maclaurin Series for Exponentials and Trigonometric Functions

Learn how polynomial substitution produces Maclaurin expansions for exponential and trigonometric functions, and why these expansions converge for every real input.

Intermediate 9 min read

What You'll Learn

  • Substitute polynomial arguments into the standard Maclaurin series for the exponential, sine, and cosine.
  • Write the resulting terms and general summation formulas accurately.
  • Use term ratios to verify convergence after substitution.
  • Determine the interval of validity for these composite expansions.
  • Recognize and correct common errors with exponents and factorials.

Substituting a Polynomial into a Maclaurin Series

A familiar Maclaurin series expresses a function in powers of its input. When that input is itself a polynomial, substitution produces a series whose terms involve powers of the polynomial. For simple arguments such as \(x^2\) or \(x^3\), this immediately gives a useful expansion in powers of \(x\).

The essential point is that substitution preserves the identity wherever the original series converges. The exponential, sine, and cosine Maclaurin series converge for every real input, so replacing their input by a real-valued polynomial gives expansions valid for every real \(x\).

Definition: A composite Maclaurin expansion here means the series obtained by substituting a polynomial \(p(x)\) for the input \(u\) in a known Maclaurin series for a function of \(u\). Its validity follows from the convergence of that original series at \(u=p(x)\).

The Three Series to Substitute Into

The standard Maclaurin series for the exponential, sine, and cosine are valid for every real number \(u\):

$$ e^u=\sum_{n=0}^{\infty}\frac{u^n}{n!}, \qquad \sin u=\sum_{n=0}^{\infty}(-1)^n\frac{u^{2n+1}}{(2n+1)!}, \qquad \cos u=\sum_{n=0}^{\infty}(-1)^n\frac{u^{2n}}{(2n)!}. $$

Here \(n\) ranges over the nonnegative integers, and \(0!=1\). The exponential series uses every nonnegative power of its input. The sine series uses odd powers, while the cosine series uses even powers. Those patterns remain visible after substitution, although the resulting powers of \(x\) depend on the polynomial being substituted.

Worked Example: Expanding \(e^{x^2}\)

Begin with the exponential series and replace \(u\) by \(x^2\):

$$ e^{x^2} =\sum_{n=0}^{\infty}\frac{(x^2)^n}{n!} =\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. $$

For \(n=0,1,2,3,4\), the terms are \(1\), \(x^2\), \(x^4/2!\), \(x^6/3!\), and \(x^8/4!\). Thus:

$$ e^{x^2} =1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+\cdots. $$

To check the substitution, set \(u=x^2\) in \(e^u=\sum_{n=0}^{\infty}u^n/n!\). Since \(u^n=(x^2)^n=x^{2n}\), this gives exactly the displayed expansion. The original exponential series converges for every real \(u\), including every value \(u=x^2\); therefore the expansion is valid for all real \(x\).

Why the Substitution Works

For each fixed real \(x\), a polynomial \(p(x)\) is just a real number. The known series identity can therefore be evaluated at that number. Convergence can also be checked directly from the terms after substitution; the following ratios tend to zero for every fixed real \(x\).

Theorem: Let \(p\) be a polynomial with real coefficients. For every real \(x\), $$ e^{p(x)}=\sum_{n=0}^{\infty}\frac{p(x)^n}{n!},\qquad \sin(p(x))=\sum_{n=0}^{\infty}(-1)^n\frac{p(x)^{2n+1}}{(2n+1)!}, $$ and $$ \cos(p(x))=\sum_{n=0}^{\infty}(-1)^n\frac{p(x)^{2n}}{(2n)!}. $$ Each series converges absolutely for every real \(x\). In particular, the interval of validity on the real line is \((-\infty,\infty)\).

Here is a direct convergence justification. If \(p(x)\ne 0\), the ratio of successive absolute terms in the exponential series is \(\lvert p(x)\rvert/(n+1)\), which tends to zero. In the sine series, that ratio is \(\lvert p(x)\rvert^2/((2n+3)(2n+2))\). In the cosine series, it is \(\lvert p(x)\rvert^2/((2n+2)(2n+1))\). Both also tend to zero. The ratio test therefore gives absolute convergence in each case. If \(p(x)=0\), the substituted series reduce directly to their values at zero and converge as well.

To justify the identities, not just convergence, use the known Maclaurin series at the real input \(u=p(x)\). Those series equal \(e^u\), \(\sin u\), and \(\cos u\) for every real \(u\). Substituting \(u=p(x)\) gives the stated functions and sums. The convergence check confirms that the resulting expressions are legitimate series at every real \(x\).

1
Start with a known identity.
Choose the exponential, sine, or cosine series in its input \(u\).
2
Replace the input consistently.
In every power, factorial, and sign pattern, substitute \(u=p(x)\).
3
Write several terms.
Evaluate the first few indices to expose the powers and coefficients.
4
State the validity.
For these three functions and a real polynomial input, the expansion converges absolutely for every real \(x\).

Trigonometric Composites

For sine and cosine, the alternating signs and factorials belong to the powers of the original input. Substitution changes the powers themselves but does not change which factorial or sign accompanies a given index.

Worked Example: Expanding \(\sin(x^3)\)

Use \(\sin u=\sum_{n=0}^{\infty}(-1)^n u^{2n+1}/(2n+1)!\), then replace \(u\) by \(x^3\):

$$ \sin(x^3) =\sum_{n=0}^{\infty}(-1)^n\frac{(x^3)^{2n+1}}{(2n+1)!} =\sum_{n=0}^{\infty}(-1)^n\frac{x^{6n+3}}{(2n+1)!}. $$

The first four terms come from \(n=0,1,2,3\):

$$ \sin(x^3) =x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}-\frac{x^{21}}{7!}+\cdots =x^3-\frac{x^9}{6}+\frac{x^{15}}{120}-\frac{x^{21}}{5040}+\cdots. $$

For a check, substituting \(u=x^3\) into the original sine series gives the displayed powers because \((x^3)^{2n+1}=x^{6n+3}\); its denominator remains \((2n+1)!\). For fixed \(x\), the ratio of successive absolute terms tends to zero, so the series converges. Since the sine series is valid for every real \(u\), this expansion equals \(\sin(x^3)\) for every real \(x\).

Worked Example: Expanding \(\cos(2x^2)\)

Substitute \(u=2x^2\) into the cosine series:

$$ \cos(2x^2) =\sum_{n=0}^{\infty}(-1)^n\frac{(2x^2)^{2n}}{(2n)!} =\sum_{n=0}^{\infty}(-1)^n\frac{2^{2n}x^{4n}}{(2n)!}. $$

The first four terms are obtained as follows:

$$ \begin{aligned} n=0 &: 1,\\ n=1 &: -\frac{(2x^2)^2}{2!}=-\frac{4x^4}{2}=-2x^4,\\ n=2 &: \frac{(2x^2)^4}{4!}=\frac{16x^8}{24}=\frac{2}{3}x^8,\\ n=3 &: -\frac{(2x^2)^6}{6!}=-\frac{64x^{12}}{720}=-\frac{4}{45}x^{12}. \end{aligned} $$

Therefore:

$$ \cos(2x^2)=1-2x^4+\frac{2}{3}x^8-\frac{4}{45}x^{12}+\cdots. $$

To check, the general term comes from the cosine term \((-1)^n u^{2n}/(2n)!\) with \(u=2x^2\); hence the power is \(x^{4n}\), the factor from the input is \(2^{2n}\), and the factorial is \((2n)!\). This series converges absolutely for every real \(x\), and substituting \(u=2x^2\) into the cosine identity verifies that its sum is \(\cos(2x^2)\).

A Common Substitution Error

A frequent mistake is to change the power of \(x\) while leaving the substituted exponent incomplete. For example, substituting \(u=x^2\) into the sine term \(u^{2n+1}\) does not give \(x^{2n+1}\). The entire expression \(x^2\) must be raised to the power \(2n+1\).

Common mistake and correction: The incorrect expression $$ \sin(x^2)\ne \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!} $$ uses the powers from \(\sin x\) without applying the substitution to them. The correct calculation is $$ \sin(x^2) =\sum_{n=0}^{\infty}(-1)^n\frac{(x^2)^{2n+1}}{(2n+1)!} =\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+2}}{(2n+1)!} =x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\cdots. $$

A reliable check is to compute the first term from the original function: \(\sin(x^2)\) begins with \(x^2\), since \(\sin u\) begins with \(u\). An expansion beginning with \(x\) cannot be correct. Also retain the original factorial: in the sine series it is \((2n+1)!\), regardless of the polynomial substituted for \(u\).

Validity and Interpretation

For the exponential, sine, and cosine series, polynomial substitution creates no finite restriction on real \(x\). A polynomial \(p(x)\) is finite for every real \(x\), and the original series converge for every real input. Consequently, each composite expansion above has the entire real line as its interval of convergence and its interval of validity.

The powers that appear can still be sparse. For example, \(e^{x^2}\) contains only even powers, \(\sin(x^3)\) contains powers \(x^{6n+3}\), and \(\cos(2x^2)\) contains powers \(x^{4n}\). This comes directly from the exponent pattern in the original series combined with the power in the argument. Recognizing that pattern helps both with writing terms and with checking them.

These expansions are useful for identifying the leading behavior of a function near zero and for constructing approximations by keeping a finite number of terms. A finite truncation is an approximation, whereas the infinite series equals the composite function throughout its interval of validity. The convergence statement applies to the full infinite series; it does not mean that a short truncation is equally accurate for every \(x\).

Key takeaway: Substitute the complete polynomial argument into each power of the known series. For \(e^u\), \(\sin u\), and \(\cos u\), the resulting series converge absolutely and equal their composite functions for every real \(x\).

Check Your Understanding

Use the original exponential, sine, and cosine series to answer these questions.

  1. Write the general-term expansion for \(e^{x^3}\), and state its interval of validity on the real line.
  2. Find the first four nonzero terms of \(\sin(2x^2)\).
  3. Write the general-term expansion for \(\cos(x^3)\). Which powers of \(x\) occur?
  4. For a fixed real \(x\), what happens to the ratio of successive absolute terms in the series for \(e^{p(x)}\) as the index increases?
  5. Explain why substituting a real polynomial argument does not impose a finite real convergence interval for these three functions.

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