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Taylor Polynomials · Tutorial 815 of 998

Symmetry and Vanishing Taylor Coefficients

Use symmetry about zero to identify which Taylor coefficients must vanish and calculate only the derivatives that can contribute.

Beginner 9 min read

What You'll Learn

  • Define even and odd functions on intervals symmetric about zero.
  • Prove which Taylor coefficients vanish for each parity.
  • Use symmetry to decide which derivatives need to be calculated.
  • Construct Maclaurin polynomials for even, odd, and mixed functions.
  • Recognize when parity does not justify omitting a coefficient.

Why Symmetry Can Shorten a Taylor Calculation

A Taylor polynomial centered at zero is built from the derivatives of a function at zero. In principle, finding a polynomial of degree \(n\) may require calculating every derivative from order \(0\) through order \(n\). But if the function has symmetry about zero, some of those derivatives are already determined: they must be zero.

This means symmetry is more than a way to recognize the shape of a graph. It can tell us in advance which powers of \(x\) can appear in a Taylor polynomial. The key is to state the symmetry precisely and connect it to derivatives at the center.

Even and Odd Functions

For a function to be even or odd on an interval centered at zero, that interval must contain \(x\) and \(-x\) together. In particular, the function must be defined at both points whenever the condition is used.

Definition: A function \(f\) is even on an interval symmetric about zero if \(f(-x)=f(x)\) for every \(x\) in the interval. It is odd if \(f(-x)=-f(x)\) for every \(x\) in the interval.

For example, \(f(x)=x^2\) is even because \((-x)^2=x^2\), while \(g(x)=x^3\) is odd because \((-x)^3=-x^3\). These identities concern the function on the whole interval, not merely its value at zero.

For a function with derivatives through order \(n\) near zero, its degree-\(n\) Taylor polynomial centered at zero is

$$ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k. $$

Thus the coefficient of \(x^k\) is \(f^{(k)}(0)/k!\). Since \(k!\) is nonzero, that coefficient vanishes exactly when \(f^{(k)}(0)=0\). Symmetry will tell us which derivative values have to vanish.

Why Parity Forces Derivatives to Vanish

The result follows by differentiating the defining symmetry relation. Suppose \(f\) is \(n\) times differentiable on an interval symmetric about zero. Write \(s=1\) when \(f\) is even and \(s=-1\) when \(f\) is odd. In either case, the symmetry relation is \(f(-x)=s f(x)\).

Theorem: If \(f\) is even, then \(f^{(k)}(0)=0\) for every odd integer \(k\le n\). If \(f\) is odd, then \(f^{(k)}(0)=0\) for every even integer \(k\le n\).

Here is the justification. Differentiating \(f(-x)=s f(x)\) \(k\) times, the chain rule contributes a factor of \((-1)^k\) on the left. Therefore

$$ (-1)^k f^{(k)}(-x)=s f^{(k)}(x), \qquad\text{so}\qquad f^{(k)}(-x)=s(-1)^k f^{(k)}(x). $$

Set \(x=0\). If \(f\) is even, \(s=1\), so \(f^{(k)}(0)=(-1)^k f^{(k)}(0)\). For odd \(k\), this says \(f^{(k)}(0)=-f^{(k)}(0)\), which implies \(f^{(k)}(0)=0\). If \(f\) is odd, \(s=-1\), and \(f^{(k)}(0)=-(-1)^k f^{(k)}(0)\). For even \(k\), this again gives \(f^{(k)}(0)=-f^{(k)}(0)\), so that derivative is zero. This proves the result.

Because Taylor coefficients are those derivatives divided by factorials, the corresponding coefficients vanish as well. The conclusion applies only to the stated parity and derivative orders: an even function may have nonzero coefficients of even powers, and an odd function may have nonzero coefficients of odd powers.

Worked Example: An Even Function Through Degree Five

Worked Example: An Even Function Through Degree Five

Find the degree-five Taylor polynomial at zero for \(f(x)=e^{x^2}\), using its symmetry to reduce the work.

First, \(f(-x)=e^{(-x)^2}=e^{x^2}=f(x)\), so \(f\) is even. The theorem says the coefficients of \(x\), \(x^3\), and \(x^5\) vanish. We need only the values of \(f(0)\), \(f''(0)\), and \(f^{(4)}(0)\).

Differentiate to the required orders:

$$ \begin{aligned} f(x)&=e^{x^2},\\ f'(x)&=2xe^{x^2},\\ f''(x)&=(2+4x^2)e^{x^2},\\ f'''(x)&=(12x+8x^3)e^{x^2},\\ f^{(4)}(x)&=(12+48x^2+16x^4)e^{x^2}. \end{aligned} $$

At zero, \(f(0)=1\), \(f''(0)=2\), and \(f^{(4)}(0)=12\). The odd-order derivatives \(f'(0)\), \(f'''(0)\), and \(f^{(5)}(0)\) are zero by even symmetry. Hence

$$ P_5(x)=1+\frac{2}{2!}x^2+\frac{12}{4!}x^4 =1+x^2+\frac{x^4}{2}. $$

Check the coefficients by differentiating this polynomial: \(P_5(0)=1\), \(P_5''(0)=2\), and \(P_5^{(4)}(0)=12\). Its first, third, and fifth derivatives at zero are all zero, as required. These values match the derivatives of \(f\) through degree five.

Worked Example: An Odd Function Through Degree Six

Worked Example: An Odd Function Through Degree Six

Find the degree-six Taylor polynomial at zero for \(g(x)=\sin x+x^3\).

Both terms are odd: \(\sin(-x)=-\sin x\) and \((-x)^3=-x^3\). Their sum is therefore odd. Its even-order derivatives at zero, including the sixth derivative, vanish. Calculate the remaining derivative values:

$$ \begin{aligned} g(0)&=0,\\ g'(x)&=\cos x+3x^2, & g'(0)&=1,\\ g'''(x)&=-\cos x+6, & g'''(0)&=5,\\ g^{(5)}(x)&=\cos x, & g^{(5)}(0)&=1. \end{aligned} $$

Thus the only potentially nonzero terms through degree six are \(x\), \(x^3\), and \(x^5\):

$$ P_6(x)=x+\frac{5}{3!}x^3+\frac{1}{5!}x^5 =x+\frac{5}{6}x^3+\frac{1}{120}x^5. $$

For a check, the first, third, and fifth derivatives of this polynomial at zero are \(1\), \(5\), and \(1\), respectively; its value and all its even-order derivatives at zero through order six are zero. These agree with the required derivative data for \(g\).

Using Both Parities in One Calculation

A function need not be entirely even or entirely odd to benefit from symmetry. Any function defined at both \(x\) and \(-x\) can be split into an even part and an odd part:

$$ f_{\mathrm{even}}(x)=\frac{f(x)+f(-x)}{2}, \qquad f_{\mathrm{odd}}(x)=\frac{f(x)-f(-x)}{2}. $$

The first expression is even, and the second is odd. Adding them gives \(f(x)\). Their Taylor polynomials can be handled according to their separate parity: the even part contributes only even powers, and the odd part contributes only odd powers.

Worked Example: A Function with Even and Odd Parts

Find the degree-four Taylor polynomial at zero for \(h(x)=(1+x)e^{x^2}\).

Write \(h(x)=e^{x^2}+xe^{x^2}\). The first term is even and the second is odd. The derivatives of \(e^{x^2}\) at zero through order four are \(1,0,2,0,12\). For \(xe^{x^2}\), its value at zero is zero, and its derivatives at zero through order four are \(1,0,6,0\). (For orders \(k\ge1\), the derivative at zero of \(xg(x)\) is \(k g^{(k-1)}(0)\), by the product rule.) Adding the two sets of derivative values gives

$$ h(0)=1,\qquad h'(0)=1,\qquad h''(0)=2,\qquad h'''(0)=6,\qquad h^{(4)}(0)=12. $$

Therefore

$$ P_4(x)=1+x+\frac{2}{2!}x^2+\frac{6}{3!}x^3+\frac{12}{4!}x^4 =1+x+x^2+x^3+\frac{x^4}{2}. $$

Differentiate the polynomial to check: its value and first four derivatives at zero are \(1,1,2,6,12\). Each agrees with the corresponding derivative value of \(h\).

A Common Symmetry Mistake

A frequent error is to reverse the conclusion and assume that an even function has only odd powers, or that an odd function has only even powers. The theorem gives the opposite: even functions have zero coefficients at odd powers; odd functions have zero coefficients at even powers.

Worked Example: Correcting a Parity Error

For \(f(x)=\cos x\), evenness does not imply that the \(x^2\) coefficient vanishes. In fact, \(f''(x)=-\cos x\), so \(f''(0)=-1\). The \(x^2\) coefficient is therefore \(f''(0)/2!=-1/2\), not zero. What evenness forces to vanish are the odd-order derivatives at zero, and therefore the coefficients of odd powers.

Symmetry must also hold on an interval around zero for this reasoning to apply to derivatives. A function that happens to satisfy \(f(-x)=f(x)\) at a few selected points is not thereby even on an interval. And if the function does not have the required derivatives, the corresponding Taylor polynomial is not defined to that order.

Why the Result Matters

Before differentiating, check whether the function is even or odd. That simple step can eliminate nearly half the derivative values needed for a Taylor polynomial. If a function has neither parity, its even and odd parts can still reveal which contributions belong to which powers.

Use symmetry as a justified shortcut, not as a guess based on the appearance of a formula. Verify the identity \(f(-x)=f(x)\) or \(f(-x)=-f(x)\), identify exactly which derivative orders are forced to zero, and calculate the remaining derivatives needed for the stated degree.

Check Your Understanding

Use the parity of each function to answer the questions. When a coefficient is requested, state whether symmetry forces it to be zero.

  1. If \(f\) is even and differentiable through order seven near zero, which of its Taylor coefficients through degree seven must vanish?
  2. If \(g\) is odd, what does symmetry imply about \(g(0)\) and \(g''(0)\)?
  3. Is \(x^4-3x^2\) even, odd, or neither? Which powers can occur in its Taylor polynomial at zero?
  4. For \(h(x)=x^2+x^3\), explain why neither whole-function parity rule applies, and identify its even and odd parts.
  5. Why does checking parity at only a few points not establish the theorem’s hypotheses?

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