Find the Values of the Radius Where the N = 2, L = 0 Radial Probability

January 14, 2022 Post a Comment

The Radius of Convergence

The partial sums increase, because all the 's were assumed to be positive.

If on the other hand a series satisfies--for some q>1-- the condition

then the series will diverge; it will go "off to infinity''.

Draw a picture similar to the one on the previous page to convince yourself that this is what will happen!

In fact, we can generalize the idea behind these two results slightly to obtain the classical ratio test for series. Here we do not insist anymore that all the

's are positive and we only compare the limit of the ratios of the absolute values of the

's to 1:

Ratio Test

If , then the series

is convergent.
If , then the series

is divergent.
The test is inconclusive when the limit equals 1! (This will not bother us much when we consider power series!)

An Example.

It's time to exploit this for power series. Consider the series

We want to find out for what values of x the series converges. If we view this power series as a series of the form

then , , and so forth. The general term will have the form

(Plug in to see that this formula works!) Consequently the ratios are given by

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Since

we obtain

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What's next? Do you remember the question we are trying to answer? For what values of x does the power series converge! The ratio test tells us now that the series will converge as long as |x|<1. It also tells us that the series will diverge for |x|>1. That gives us a pretty complete picture about what's going on:

The biggest interval (it is always an interval!) where a power series is converging is called interval of convergence of the power series. The interval of convergence is always centered at the center of the power series. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end points of the interval), its length is 2, so the radius of convergence equals 1.

Another example.

Consider the power series

Step 1. Find the general term of the power series. In our case

will do the job for . Since we will be taking the limit as n goes to infinity, the odd ``5'' at the beginning is of no consequence!
Step 2. Compute the ratios . Don't forget the absolute values!
Step 3. Compute the limit of the ratios. Since , in our case

for all x.
Step 4. Apply the ratio test. Since 0<1 (in this example the limit does not depend on the value of x), the series converges for all x.
Thus the interval of convergence is the interval . The radius of convergence in this case is said to be .

One more example.

Consider the power series

Step 1. Find the general term of the power series. This is not as easy as in the last examples! The exponent of the (x + 2)'s jumps by 2 each time, up front we have a power of 2. Let's try to rewrite the absolute values of the first terms slowly: | A ₀| = 2^{0 .}| x + 2|¹, | A ₁| = 2^{1 .}| x + 2|^{(2^.1 + 1)}, | A ₂| = 2^{2 .}| x + 2|^{(2^.2 + 1)}, | A ₃| = 2^{3 .}| x + 2|^{(2^.3 + 1)}, and so forth. Thus the general term is given by
| A _n| = 2^{n .}| x + 2|^{(2^.n + 1)}.
Step 2. Compute the ratios .
Step 3. Compute the limit of the ratios. Since in this case the ratios do not depend on n, there's nothing to do!
Step 4. Apply the ratio test. The series will converge, when the ratio in Step 3 is less than 1 (diverge when the ratio exceeds 1):

Bingo! The radius of convergence in this case is . The interval of convergence is the interval from to .