Question

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

Short answer

If \(f\) is an even function, its nth Maclaurin polynomial has only even powers of \(x\) because the coefficients for the odd power terms \(f^n(0)\) turn out to be zero due to the property of even functions.

Step-by-step solution

Defining the even function

An even function is mathematically defined such that \(f(x) = f(-x)\) for all \(x\) in the domain of \(f\). This means that the function's value is the same at \(x\) and \(-x\). This property is the key to this proof.

Definition of Maclaurin series

The Maclaurin series is a Taylor series expansion of a function about 0. The general formula for a Taylor series around \(a\) is: \(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots \). The Maclaurin series is a special case of the Taylor series where \(a=0\). Hence, the nth term in the Maclaurin expansion is: \(\frac{f^n(0)x^n}{n!}\).

Proving the statement

In the general Maclaurin's series expansion, the nth derivative of the function \(f(x)\) at 0, i.e. \(f^n(0)\), has a power of n. So if \(n\) is odd, then the nth derivative of \(f(-x)\) at 0, i.e. \(f^n(-0)\) would be \(-f^n(0)\) due to the property of the even function. But \(f(-x)\) is the same as \(f(x)\), hence \(f^n(-0)\) is the same as \(f^n(0)\), yielding a contradiction unless \(f^n(0)\) is zero. Therefore, for an even function, all coefficients of odd powers in its Maclaurin expansion, i.e. \(f^n(0)\) for odd \(n\), vanish. Therefore, the Maclaurin polynomial of an even function only has terms of even powers of \(x\).

Key concepts

Even Function Properties

When we talk about even functions, the most defining feature is their symmetry with respect to the y-axis. This naturally leads to a simple but powerful property: for every even function, you'll find that applying the function to a positive input or its negative counterpart yields the same result. In mathematical terms, this is written as \(f(x) = f(-x)\) for every value that lies within the function's domain.

One of the key repercussions of this property is the shape of the graph of an even function. Try plotting a few points by substituting positive and negative values for \(x\), and you'll observe that the points reflect over the y-axis, creating a mirror image. This symmetry isn't just aesthetically pleasing - it has profound effects on the function's analytical properties, particularly the terms in its polynomial expansion, which leads us to aspects of calculus concerning power series.

Maclaurin Series Expansion

The Maclaurin series expansion is like a personal ID for functions, expressing them in a unique series of powers of \(x\). But it's not just any representation; it's one centered around the point zero. This stems from the general recipe for a Taylor series, which unfolds a function into an infinite sum of terms calculated from the function's derivatives at a single point. The Maclaurin series is this special type of Taylor series where that point is zero.

The nth term of the Maclaurin series for a function \(f\) is given by \(\frac{f^n(0)x^n}{n!}\), with \(n!\) representing the factorial of \(n\). This expansion includes all powers of \(x\), from the zeroth up to infinity. But when we tie this to even functions, there's a twist. Due to the symmetry of even functions, only even powers of \(x\) survive in the expansion; odd powers become zero because of the pattern of derivatives at zero, as we will explore further in the application of these concepts.

Taylor Series

If the Maclaurin series is the expansion around zero, then the Taylor series is its adventurous sibling, allowing for expansions around any point \(a\). This general form is written as \(f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots\). It's a powerful tool because it lets us approximate functions near the chosen point \(a\) with impressive accuracy.

Important to note, though, is that not all functions have a Taylor series that converges to the function for all \(x\). When it does converge, however, it allows us to boil down complex functions to simpler polynomial forms, which are much easier to work with for both theoretical analysis and practical calculations. Now, when we speak about even functions through the lens of Taylor series, and in particular Maclaurin series, we see a synchronization between the function's inherent symmetry and the resulting series: only even powers remain, reinforcing the initial property we spoke about in even functions.