Deducing Maclaurin series converges from Leibniz formula

[sr3056]

Given f(x) = xe^-x2 I can differentiate once and use Leibniz to show that for n greater than 1

f⁽ⁿ⁾ = -2nf^(n-2) - 2xf^(n-1)

I want to show that the Maclaurin series for f(x) converges for all x.

At x = 0, the above Leibniz formula becomes f⁽ⁿ⁾ = -2nf^(n-2)

I know that f(0) = zero so this implies that even terms of Maclaurin series are zero, whilst
f⁽¹⁾(0) = 1, f⁽³⁾(0) = -6*1 = -6, f⁽⁵⁾(0) = -10*-6*1 = 60 and so on.

I assume that to show convergence, I need to find a formula for the nth term of the Maclaurin series, then use the ratio test to show that terms are decreasing. I can see the pattern (the derivative increases by a factor of 2(2n + 1) each time) but am unsure how to express this in a formula. Perhaps using factorials?

Thanks for any help.

 

[chiro]

Hey sr3056 and welcome to the forums.

One suggestion is to use the Mclaurin series expansion for e^(-x^2) and then multiply all terms by x. So you start by expanding e^(-x) and then replace every x by an x^2, and multiply each term by x.

The above should give you a series expansion which should allow you to do a further test for convergence.

 

[sr3056]

Thanks. Is there no way of proving convergence from the Leibniz formula though?

 

[chiro]

If you get a series that is the correct definition, then you should be able to do what you said (i.e. the ratio test), so as long as you get the correct series expansion, it will be OK.

Both series expansions should be equal though and testing this will test whether your above approach is the same as the one I discussed above.

 

[CellCoree]

To show that the Maclaurin series for f(x) converges for all x, we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of successive terms of a series is less than 1, then the series converges. In this case, we can express the nth term of the Maclaurin series as:

f(n)(0) = (-1)n (2n)! / (2n + 1)!

Using the ratio test, we have:

|f(n+1)(0) / f(n)(0)| = |(-1)n+1 (2n+2)! / (2n+3)!| * |(2n)! / ((2n + 1)!)|

= |(-1)n+1 (2n+2) / (2n+3)|

= (2n+2) / (2n+3)

As n approaches infinity, this ratio approaches 1, which means that the series converges for all x.

Additionally, we can also use the Maclaurin series representation of f(x) to show that it converges for all x. The Maclaurin series for f(x) is given by:

f(x) = ∑ (-1)n (2n)! / (2n + 1)! * x2n

This series converges for all x because it is a power series with a radius of convergence of infinity, meaning it converges for all values of x. Therefore, we can conclude that the Maclaurin series for f(x) converges for all x.