Question

Prove that if \(f\) is an odd function, then its \(n\) th Maclaurin polynomial contains only terms with odd powers of \(x\)

Short answer

The nth Maclaurin polynomial of an odd function contains only terms with odd powers of \( x \) as odd functions yield zero when evaluated at x=0 except for terms with odd powers.

Step-by-step solution

Understand the properties of odd function

A function \( f(x) \) is said to be odd if, for all \( x \) in the function's domain, \( f(-x) = -f(x) \). This property essentially mirrors the function about the origin.

Define the nth Maclaurin polynomial

The nth term in a Taylor (or Maclaurin) series, which is centered at \( x=0 \), is defined as \( f^n(0)/n! *x^n \). Where \( f^n (0) \) denotes the nth derivative of the function evaluated at 0.

Apply odd function rule to Maclaurin polynomial

Using the property of odd function, the nth derivative of an odd function will be odd if \( n \) is odd, but will be even if \( n \) is even. Odd functions yield zero when evaluated at x=0. This is because by definition, an odd function is symmetric about the origin.

Conclusion

Therefore, the nth Maclaurin polynomial of an odd function has only terms with odd powers of \( x \), since for these terms, the nth derivative of the odd function, evaluated at \( x=0 \), is not zero.