geometric series, Fourier polynomials

Calculus II

Class Notes, 3/22/99

The nth term of a geometric series is of the form s_n = a rⁿ.

We thus see that for any n, s_n+1 / s_n = r.
It follows that s_n / s_n+1 = 1 / r is the radius of convergence of the geometric series a + ax + ax² + . . . + a xⁿ + ... .

For the series in the figure below, we see that since s₂ / s₁ = -1/5, the series will be geometric provided every term is of the form a rⁿ with r = -1/5.

Since the first term is 1, we have a = 1.
This checks out for the third and fourth terms. As indicated below the third term is a r² = 1 (-1/5)² = 1/25 and the fourth is a r³ = 1 (-1/5)³ = -1/125.

To see if the series in the figure below is geometric, we see that s₂ / s₁ = 1/2, so if it is geometric r = 1/2.

However, s₃ / s₂ is not 1/2, so the series cannot be geometric.

Video File #2

To find the sum of the series in the figure below, we separate it into two series.

Since (2ⁿ + 5) / 3ⁿ = (2ⁿ / 3ⁿ) + 5 / 3ⁿ, we can separate the series as shown.
The first series is geometric, with terms of the form a rⁿ = 1 * (2/3)ⁿ; we see that a = 1 and r = 2/3 so the sum of the series is a / (1-r) = 1 / (1-2/3).
The second series is geometric with a rⁿ = 5 * (1/3)ⁿ, so a = 5 and r = 1/3 and the sum is 5 / (1 - 1/3).
The resulting quantities are easily evaluated and simplified.

Video File #3

We wish to find how much of a drug is left in the bloodstream if 6.3 hours is required to reduce the amount of drug to half its original amount.

The amount of drug is given by an exponential function, as indicated in the figure below.
The halflife condition is expressed in the second equation.
Substituting the form of the function in the left-hand side we obtain the equation in the third line, which we easily solve for k.
We obtain an approximate value k = .11.
Thus our function is as indicated in the last line.

Using this function we easily see, as shown below, that after 24 hours only 7.14% of the drug is left.

If originally there is no drug in the bloodstream, then the amount Q1 we have right after taking a dose Q0 at the time 24 hours after the first dose (also of size Q0) will be the .0714 Q0 remaining from the first dose plus the new dose Q0, as indicated in the first line.

The amount Q2 right after taking a dose after two 24-hour periods have elapsed will be .0714 of the Q1 present 24 hours previously, plus the dose Q0, as indicated in the second line.
We can substitute the expression for Q1, as indicated in the third line, to obtain the expression for Q2 shown in the fourth line.
We can interpret the .0714² Q0 as the amount left from the first dose after two 24-hour periods, the .0714 Q0 as the amount left from the second dose after one 24-hour period, and Q0 as the dose just taken.

This process continues, as indicated in the last two lines.

It should be clear that after the dose is taken following n twenty-four-hour periods, the amount will be Qn = .0714ⁿ Q0 + .0714^(n-1) Q0 + ... + .0714² Q0 + .0714 Q0 + Q0.

This is just the nth partial sum of the geometric series a rⁿ = Q0 * .0714 ⁿ with a = Q0 and r = .0714.

Video File #4

The Fourier polynomial of degree n for the function f(x) is defined as shown below.

To calculate the Fourier coefficients a1, b1, a2 and b2 of the function f(x) = x we proceed as in the figure below.

a0 is given by the integral in the second line, which is easily seen to be zero.
This integral is easily evaluated, but since x is an odd function evaluated between -`pi and `pi, the integral from -`pi to 0 will be exactly the negative of the integral from zero to `pi, which shows us without evaluating the integral that the result will be zero.

a1 is found by the integral in the third line, which can easily be seen by either evaluating the integral or by considering the symmetry of the graph to be zero.

b1 is found using the integral in the fourth line. The integrand is a product of two odd functions and is hence even, so we do not expect the integral to be 0.

Using either integration by parts or tables we easily find that b1 = -2 (note in evaluating the integral that `pi cos(`pi) - -`pi cos(-`pi) = -2`pi).

Using the symmetry argument again we see that a2 = 0. In fact all the functions x cos(kx) will be odd so their integrals will be zero.

The integral for b2 has sin(2x) where the integral for b1 had sin(x). Integration by parts shows why this integral will therefore give half the absolute value of the integral for b1.

Since `pi cos(2`pi) - -`pi cos(-2`pi) = 2`pi, we will obtain a positive value for the integral.

We obtain b2 = 1.

The pattern established above continues with b3 = -1/3 * b1 = 2/3, b4 = 1/4 * b1 = 1/2, and in general bn = (-1)ⁿ * 1/n * b1 = (-1)ⁿ * 2 / n.

We can easily check the result for bn, using integration by parts as shown below.

We thus see that x is given approximately by the Fourier expansion in the figure below.

Video File #5

The graph below shows how the even cosine function and the odd y = x function yield an odd function whose integral, corresponding to the shaded region of the graph of x cos(x) in the figure below, must be zero due to the equal regions above and below the x axis.

On the other hand the graph of x sin(x) is even, with congruent regions on the same side of the x axis, so the integral is nonzero.

The graph of the y = x and y = sin(2x) functions and their product is depicted below.

The area under the graph of the product function x sin(2x) is seen to consist of to congruent regions above the x axis and two congruent regions below the x axis, with the regions below the axis having the greater area.
We see that the positive and negative areas do not cancel, but in opposing one another they do make the absolute value of the integral less than that of x sin(x) (as seen before the integral of x sin(2x) is -1/2 that of x sin(x)).

For larger values of k, the areas above and below the x axis come closer and closer to cancelling one another, and the integrals become smaller and smaller, consistent with our result that the integral of x sin(nx) is 1/n times that of x sin(x).

Video File #6

Video File #7