Why Symmetry Predicts Missing Powers
A function may have a Taylor polynomial with many zero coefficients. Those zeros are not always accidental: symmetry can force them. If a function has the same value at \(x\) and \(-x\), its Maclaurin polynomial can contain only even powers. If its values have opposite signs at those inputs, only odd powers can appear.
This gives a useful prediction before any derivatives are calculated. It also provides a check on a derivative table or an algebraic expansion: a coefficient that conflicts with the function’s symmetry signals an error.
Even and Odd Functions
A function’s parity describes how its values compare at opposite inputs. The comparison requires that both \(x\) and \(-x\) are in the function’s domain. For conclusions about Maclaurin coefficients, it is enough that the relevant symmetry holds on an interval around \(0\).
For example, \(x^2\) is even because \((-x)^2=x^2\), while \(x^3\) is odd because \((-x)^3=-x^3\). A function need not be even or odd: \(x^2+x\) has both an even-power term and an odd-power term, and it satisfies neither symmetry condition.
In the Maclaurin polynomial of degree \(n\), the coefficient of \(x^k\) is \(f^{(k)}(0)/k!\), provided the derivatives through order \(n\) exist. Thus, predicting which coefficients vanish amounts to determining which derivatives at \(0\) must be zero.
Why Parity Forces Derivatives to Vanish
Suppose \(f(-x)=s f(x)\) near \(0\), where \(s=1\) for an even function and \(s=-1\) for an odd function. Assume \(f\) has derivatives through order \(n\) on an interval around \(0\). Differentiating the symmetry relation \(n\) times gives a factor of \((-1)^n\) on the left, by the chain rule:
Setting \(x=0\) yields \(\bigl((-1)^n-s\bigr)f^{(n)}(0)=0\). If \((-1)^n\ne s\), the factor multiplying \(f^{(n)}(0)\) is nonzero, so that derivative must be zero. Since the Maclaurin coefficient is \(f^{(n)}(0)/n!\), the corresponding coefficient vanishes too.
In particular, an odd function defined at \(0\) satisfies \(f(0)=-f(0)\), so \(f(0)=0\). The theorem does not say that the remaining derivatives are nonzero; symmetry only guarantees the indicated zeros. Other features of the function may make additional coefficients vanish.
Worked Example: An Odd Rational Function
Worked Example: An Odd Rational Function
Find the Maclaurin polynomial through degree \(8\) for \(f(x)=\frac{x}{1+x^2}\), and identify the missing powers.
First check the symmetry. For every real \(x\), the denominator is nonzero and
Thus \(f\) is odd, so its even-indexed Maclaurin coefficients vanish. To calculate the terms through degree \(8\), start with the exact identity
Indeed, multiplying the first four terms by \(1+u\) gives \(1-u^4\); adding \(u^4\) gives \(1\). Now set \(u=x^2\) and multiply by \(x\):
The remainder has a factor \(x^9\), so this gives the polynomial through degree \(8\):
There are no constant, \(x^2\), \(x^4\), \(x^6\), or \(x^8\) terms, as odd symmetry predicts. Check the identity at \(x=\frac12\). The function value is \(\frac{1/2}{1+1/4}=\frac25\). The polynomial value is \(\frac12-\frac18+\frac1{32}-\frac1{128}=\frac{51}{128}\), and the remainder is \(\frac{(1/2)^9}{1+1/4}=\frac1{640}\). Since \(\frac{51}{128}+\frac1{640}=\frac{255}{640}+\frac1{640}=\frac25\), the exact identity checks.
Worked Example: An Even Rational Function
Worked Example: An Even Rational Function
Find the Maclaurin polynomial through degree \(7\) for \(g(x)=\frac{x^2}{1-x^2}\), and explain which coefficients must be missing.
The function is defined near \(0\), and wherever both sides are defined,
So \(g\) is even near \(0\), and all odd-indexed coefficients vanish. For the polynomial, use the identity
Multiplication verifies it: \((1-u)(1+u+u^2)=1-u^3\), and adding \(u^3\) gives \(1\). Set \(u=x^2\) and multiply by \(x^2\):
The remainder begins with a factor \(x^8\), so the terms through degree \(7\) give
The constant term and every odd power through \(x^7\) are absent. To check the arithmetic at \(x=\frac12\), the function value is \(\frac{1/4}{3/4}=\frac13\). The polynomial value is \(\frac14+\frac1{16}+\frac1{64}=\frac{21}{64}\), and the remainder is \(\frac{(1/2)^8}{1-1/4}=\frac1{192}\). Their sum is \(\frac{63}{192}+\frac1{192}=\frac13\), as required.
Separating the Even and Odd Parts
A function that is neither even nor odd can still be split into two parts with definite parity. Define its even part and odd part by
Adding these expressions gives \(f_{\mathrm{even}}(x)+f_{\mathrm{odd}}(x)=f(x)\). Replacing \(x\) by \(-x\) in the first expression leaves it unchanged, while doing so in the second changes its sign. Hence the first part is even and the second is odd. This decomposition can expose the missing powers even when the original function has no single parity.
Worked Example: Recovering Parity Parts from Opposite Inputs
Let \(h(x)=(1+x)^4\). Find its even and odd parts by comparing \(h(x)\) and \(h(-x)\), then check the result at opposite inputs.
Expand each expression:
The even part is their average:
The odd part is half their difference:
Adding the parts recovers \(h(x)=1+4x+6x^2+4x^3+x^4\). For a direct numerical check, \(h(1)=16\) and \(h(-1)=0\). The even part at \(1\) is \(1+6+1=8\), and the odd part at \(1\) is \(4+4=8\). Their sum is \(16\); at \(-1\), the even part remains \(8\) and the odd part becomes \(-8\), giving \(0\). The opposite-input values therefore agree with the parity of the two parts.
Checking the Pattern with Derivatives
The symmetry test uses function values, while the derivative test examines the coefficients directly. Consider \(F(x)=e^x+e^{-x}\). Opposite inputs give \(F(-x)=e^{-x}+e^x=F(x)\), so \(F\) is even.
Repeated differentiation gives \(F^{(n)}(x)=e^x+(-1)^n e^{-x}\). This follows because differentiating the second term once changes its sign, and each further differentiation alternates that sign. At \(x=0\),
For example, the derivative values through order \(6\) are \(F(0)=2\), \(F'(0)=0\), \(F''(0)=2\), \(F^{(3)}(0)=0\), \(F^{(4)}(0)=2\), \(F^{(5)}(0)=0\), and \(F^{(6)}(0)=2\). Dividing each by its factorial gives the Maclaurin polynomial through degree \(6\):
For a coefficient check, the \(x^4\) coefficient is \(F^{(4)}(0)/4!=2/24=1/12\), and the \(x^6\) coefficient is \(F^{(6)}(0)/6!=2/720=1/360\). Every odd-power coefficient through degree \(6\) is zero because the corresponding derivative is zero. This is the derivative version of the same even-symmetry pattern.
Common Pitfalls
Confusing even powers with even-numbered coefficient values. “Only even powers” means that terms such as \(x^0,x^2,x^4\) may appear; it does not mean their numerical coefficients must be even integers. For instance, the even function \(1+x^2\) has coefficient \(1\) on \(x^2\), and the even function \(e^x+e^{-x}\) has coefficient \(1/12\) on \(x^4\). Both are consistent with even symmetry.
Using symmetry without checking the domain. The comparison \(f(-x)=f(x)\) requires that both inputs are allowed. For example, \(x^2/(1-x^2)\) is even near \(0\), but it is undefined at \(x=1\) and \(x=-1\). The parity argument applies on an interval around \(0\) that avoids those points.
Assuming symmetry determines every coefficient. Parity identifies coefficients that must be zero; it does not supply the values of the remaining coefficients. Those values still come from derivatives or a valid algebraic expansion. Also, the vanishing pattern for Taylor coefficients does not by itself prove that an infinite Taylor series converges to the function.
Check Your Understanding
Use the definitions and results above to answer these questions.
- If \(p\) is even and has derivatives through order \(7\) near \(0\), which of \(p'(0),p^{(2)}(0),\ldots,p^{(7)}(0)\) must be zero?
- If \(q\) is odd near \(0\), what must be true about \(q(0)\) and its even-indexed Maclaurin coefficients?
- For a function with neither parity, how do the formulas for its even and odd parts use \(f(x)\) and \(f(-x)\)?
- Why does knowing that a function is even not determine the numerical values of its even-power coefficients?
- What domain condition is needed before using a comparison between \(f(x)\) and \(f(-x)\)?
