Question

Suppose you know the Maclaurin series for \(f\) and that it converges to \(f(x)\) for \(|x|<1 .\) How do you find the Maclaurin series for \(f\left(x^{2}\right)\) and where does it converge?

Short answer

Short Answer: The Maclaurin series for \(f(x^2)\) is given by: $$f(x^2) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^{2n}$$ Its interval of convergence is: $$0 < x^2 < 1$$

Step-by-step solution

Recall Maclaurin series for a function f(x)

A Maclaurin series is a special case of a Taylor series centered at 0. It can be defined as a power series representation of a function, \(f(x)\), around the point \(x = 0\). The Maclaurin series for \(f(x)\) is given by: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$ where \(f^{(n)}(0)\) is the nth derivative of the function, \(f(x)\), at the point x = 0, and \(n!\) is the factorial of n.

Find the Maclaurin series for \(f(x^2)\)

In order to find the Maclaurin series for \(f\left(x^2\right)\), we will replace \(x\) by \(x^2\) in the series for \(f(x)\) as follows: $$f(x^2) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x^2)^n$$ This can be simplified further: $$f(x^2) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^{2n}$$ So, the Maclaurin series for \(f(x^2)\) is given by: $$f(x^2) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^{2n}$$

Find the interval of convergence for \(f(x^2)\)

We are given that the Maclaurin series for \(f(x)\) converges for \(|x| < 1\). Now, since we've replaced \(x\) by \(x^2\) in the Maclaurin series of \(f\left(x^2\right)\), we need to find the interval of convergence for the new series. Let \(x^2 = u\), so \(x = \sqrt{u}\). The original interval of convergence is \(|x| < 1\), or \(-1 < x < 1\). In terms of \(u\), we can rewrite this inequality as follows: $$ -1 < \sqrt{u} < 1 $$ Squaring the inequality, we get: $$0 < u < 1$$ Therefore, the interval of convergence for the Maclaurin series of \(f\left(x^2\right)\) is: $$0 < x^2 < 1$$

Key concepts

Power Series Representation

When we talk about the power series representation of a function, we are referring to the expression of that function as an infinite sum of terms, each term being a multiple of a power of the variable. Here, the variable is represented usually by 'x'. This can seem quite abstract at first glance, but it is a very powerful tool in calculus, especially for functions that cannot be expressed in a simple algebraic form.

In the context of the exercise, the power series of the function 'f' at 'x = 0', also known as its Maclaurin series, looks like this:
\[f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\]
Here, the notation \( f^{(n)}(0) \) represents the nth derivative of the function evaluated at zero, and \( n! \) stands for n factorial. This simple yet profound representation allows us to analyze and compute the values of functions even when 'x' is not a simple number, since a power series can often be evaluated when other methods fail.

Interval of Convergence

The interval of convergence is a crucial concept when dealing with power series. It refers to the set of all 'x' values for which the series converges, meaning the infinite sum of the terms in the series will approach a finite value. Outside of this interval, the series may diverge, or head towards infinity, which is not useful from a computational perspective.

To visualize this, imagine that each value of 'x' is like a customer trying to enter a club. The interval of convergence is like the club's door policy; it lets some customers ('x' values) in where they can contribute to the party (the finite sum), while it excludes others for whom the party never stops (infinite sum). For the Maclaurin series of 'f(x)', we're given that the series converges for \( |x| < 1 \).
But what happens when we transform 'x' into 'x^2' in the exercise? The interval has to be squared, leading to a new interval of \( 0 < x^2 < 1 \) for the Maclaurin series of 'f(x^2)'. This interval of convergence helps us determine for which values of 'x' we can confidently use the power series to find the function's value.

Taylor Series

Now, let us not forget about the Taylor series, of which the Maclaurin series is essentially a special case. A Taylor series is the power series representation of a function about any point 'a', not just zero. It’s like having the power series club at every location on the number line, not just the origin!

The general form of a Taylor series for a function 'g(x)' around a point 'a' is given by:
\[g(x) = \sum_{n=0}^\infty \frac{g^{(n)}(a)}{n!}(x-a)^n\]
What does this mean? It tells us that any function that is Taylor series-representable (which in practice, a lot of functions are) can be expressed as an infinite sum of polynomial terms based on its derivatives at a single point. This single point is 'a' in the general formula, and for a Maclaurin series, this 'a' is simply 0. It is a fascinating and mind-bending leap to think that even complex functions have these polynomial ‘doppelgängers’ lurking around any corner of the number line, waiting to reveal their sameness within a certain range!