Nth Derivative of an Even Function

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Homework Help Overview

The problem involves the function f(x) = sin(x) / (b + cos(ax)) and asks to show that the nth derivative at zero is zero for even integers n. The subject area includes calculus, specifically derivatives and properties of even and odd functions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply Leibniz's product rule and considers the behavior of the derivatives of sin(x) and g(x) at x = 0. Some participants question the nature of the function in terms of evenness or oddness and suggest examining the Taylor series expansion of g(x). Others propose a simpler approach by asking fundamental questions about the properties of even and odd functions.

Discussion Status

The discussion is active with various approaches being explored. Some participants are considering the implications of the function's evenness or oddness, while others are focused on the application of the product rule. There is no explicit consensus yet, but multiple lines of reasoning are being investigated.

Contextual Notes

Participants are navigating the definitions of even and odd functions and their derivatives, which may influence the approach to the problem. The original poster's method involves assumptions about the derivatives at zero, which are being scrutinized.

FeDeX_LaTeX
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Homework Statement


Let ##f(x) = \frac{\sin x}{b + \cos(ax)}##. Show that the nth derivative ##f^{(n)}(0) = 0## if n is an even integer.


Homework Equations


Leibniz's generalised product rule:

##(f \cdot g)^{(n)} = \sum_{k = 0}^{n} \binom{n}{k} f^{(k)}g^{(n-k)}##


The Attempt at a Solution


I'm letting ##f(x) = \sin x## and ##g(x) = \frac{1}{b + \cos(ax)}## then applying Leibniz's rule. Clearly, the terms of the series k = 0, k = 2, ... (every even k) are all 0 when x = 0, since they all contain an even derivative of sin (which gives us sin again). But what do I do about the derivatives of g(x)? Is this the right approach?
 
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Odd functions of x have only odd powers of x in their Taylor-McLaurin series, so...
 
So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?
 
FeDeX_LaTeX said:
So you're saying I let ##\frac{1}{b + \cos(ax)} = a_0 + a_{1}x^2 + a_{2}x^4 + ... ##?

There is a much, much easier way. Just answer the following three questions.
1. The given function is (a) odd; (b) even; (c) neither.
2. The derivative of an even function is (a) even; (b) odd; (c) neither.
3. The derivative of an odd function is (a) even; (b) odd; (c) neither,
 

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