convergence; Taylor polynomial error

Calculus II

Class Notes, 3/26/99

The sum in the figure below consist of the i = 0 to i = 10 terms of the geometric sequence defined by a r^n = 4(.91)^n, with a = 4 and r = .91.

The sum of the first i=1 to i =10 terms is easily calculated from the formula a (1 - r^N) / (1-r).

To calculate the sum in the first line of the figure below, we note that every term is the product of 11 and a power of 1.1.

The problem is that some of the powers are positive and some are negative.
We can remedy this by multiplying the expression by 1.1 ^ 10; if we also divide the result by 1.1^10, as in the second line below, we will not have changed the value of the sum.
The numerator in the third line below consists of the i = 0 to i = 12 terms of the geometric sequence with a = 11 and r = 1.1.
Using the formula a (1 - r^N)/(1-r) we obtain the expression in the fourth line, which is easily evaluated.

Video File #1

At first the expression the first line below doesn't look exactly like a geometric series because many powers are missing.

However, a closer look shows that the exponents are equally spaced, which allows us to express each power as a power of .8^3 = .512, as shown in the second and third lines below.
The result is then easily found and evaluated.

The sum of a geometric series converges if | r | < 1 and diverges if | r | > = 1.

We see this by taking the limit of the partial sum Sn = a (1-r^n) / (1-r).
If r > 1 the r^n in the numerator will grow exponentially and hence diverge.
If r = 1, Sn will consist of n+1 a's, so Sn = a * (n+1), which clearly diverges.
If r = -1, Sn will consist of n+1 alternating a's and -a's, and will hence equal a when n is odd and 0 when n is even, so the sequence again fails to converge.
When | r | < 1, r^n -> 0 for large n and Sn approaches a (1 / (1-r)).

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Our definition of convergence is just that used implicitly above. A sequence converges provided the large-n limit of its partial sums exists.

From the geometric series we see that, at least for this series, convergence is associated with individual terms a r^n that approach 0.

Convergence is associated with | r | < 1, so that a r^n -> 0 for large n.

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We recapitulate the argument shows that the sum of the harmonic series, for which an = 1/n, diverges.

Recall that the partial sums can be represented by small rectangles either beneath or above the curve y = 1/x.
Recall that the integral of y = 1/x is divergent as x -> infinity.
If the rectangles are located beneath the curve, we will conclude that the sum is less than the area under a divergent integral, which tells us nothing.

If however we position the rectangles as indicated in the figure below, we see that the area in the rectangles, which corresponds to the sum of the harmonic series, is greater than the area under the curve.

We therefore conclude that the partial sums of harmonic series approach infinity as n approaches infinity.

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We conclude that, though the terms 1/n of the harmonic series certainly approach 0 as n approaches infinity, they don't approach 0 fast enough to prevent the series from diverging.

A measure of how fast the terms of the series approach zero is the behavior of the subsequent ratios of terms a(n+1) / a(n).
For a geometric series we have seen that this ratio is always r, and that for | r | < 1 the series converges and for | r | > 1 the series diverges.

For the harmonic series the limiting value of the ratio is the limit of a(n+1) / an = 1 / (n+1) / [ 1 / n ]. This expression simplifies to n / (n+1), which is n approaches infinity approaches a limit of 1.

Since for a geometric series we have divergence when r = 1, we might suspect that if the limit of a(n+1) / an is 1 the series must diverge.

However, the series 1 / n^2 can be compared with the integral of 1 / x^2, which converges for large x.

Since the limit of a(n+1) / an is also 1 for this sequence, as seen below, we see that we can have convergence when the limit of the ratios is 1.

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We see that though the limits of 1/n and 1/n^2 are both 0 for increasing n, one sequence converges while the other diverges.

We see also that though the limits of the ratios a(n+1) / an of the sequences are both 1, one converges well the other diverges.
One of the sequences is getting smaller, in fact approaching zero, but not fast enough while the other is approaching 0 fast enough.

`The figure below summarizes a fundamental theorem about the convergence of series in terms of the limits of the ratios a(n+1) / an.

As we have seen, if the limit of the ratios is 1 the series might converge or diverge.
As with a geometric sequence, if the limit of the ratio is < 1 the series will converge.
As with a geometric sequence, if the limit of the ratio is > 1 the series will diverge.

Again we reemphasize the fact that for a geometric series the limit of the ratio is r, and that the sequence converges when an only when r < 1.

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For a Taylor polynomial of degree 1 (i.e., a tangent-line approximation) about a = 0, we define the error to be the difference f(x) - P1(x) between the function f(x) and the Taylor polynomial P1(x).

As we will show next time, if the second derivative f''(x) for x within some distance d of the origin lies between the lower limit L and and upper limit M, then the error will lie between L/2 x^2 and M/2 x^2.

This result is related to the behavior of the functions L x^2 and M x^2 on this interval.

If we let E1(x) stand for the error of the first-degree polynomial at x, E2(x) for the error of the second-degree polynomial at x, and in general En(x) for the error of the nth-degree polynomial at x, we have the sequence of error limits depicted below.

Note that the denominators in the left- and right-hand terms of the error limits for an nth-degree approximation are always equal to (n+1)!, while the power of x is n+1.

Video File #7

If the Taylor polynomial is expanded about x = a, then the maximum errors are given by the expressions below, which follow from the previous expressions by replacing x by (x - a).

The figure below depicts a function f(x) and a possible Taylor polynomial Pn(x), expanded about x = a.

At x = a, the function and its polynomial approximation both pass-through the point (a, f(a)).
The interval | x - a | < d consists of all x values which lie within distance d of x = a; this interval is indicated in the figure.

The Error Theorem states that if the n+1 derivative of f has an absolute value bounded by M for all x in the interval, then for any xi interval the error bounded is valid.

As an example we find and error bound for the degree-5 approximation to the cosine function, expanded about x = 0, for values of x that lie between -.5 and .5.

We must first find an upper bound for the n+1 derivative of our function.
Since n = 5, we must find the 6th derivative of the cosine function.

We easily see that the 6th derivative of cos(x) is -cos(x).

We therefore must find an M such that | cos(x) | < M for -.5 < x < .5.

Since | cos(x) | < 1 for all values of x, we would be safe letting M = 1.

Since 0 lies between -.5 and .5, and since cos(0) = 1, we cannot let M be < 1, so we let M = 1 and understand that this is the best possible upper limit we can place on the 6th derivative.

Note that it is not necessary to always use the best possible upper limit; in many cases any reasonable limit will allow less to make reasonable estimates of our possible errors.

We now write down the error bound for the error E6(x), using the formula for En(x).

We obtain the result in the second line.
Noting that 7! = 5040 and that for | x - 0 | < .5, | x - 0 | ^ 7 < .01, we see that E6(x) must be less that .000002 for | x | < .5.

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