To find the degree-4 Taylor polynomial of f(x) = tan(x) about a = 0 we begin by determining the first four derivatives of f(x), as indicated below.

• In finding these derivatives we make use of the fact that the derivative of tan(x) is tan^2(x) + 1 so that

• (tan^n(x))' = n (tan(x))' tan^(n-1)(x). = n (tan^2(x) - 1) tan^(n-1)(x) = n tan^(n+1)(x) - tan^(n-1)(x).

Video Clip #1

To obtain the Taylor polynomial of degree three for the function f(x) = (x + 1) ^ p about x = 0 we proceed as in the figure below.

• We find the first three derivatives of the function.
• When evaluated at x = 0 these derivatives will have values p, p(p-1) and p(p-1)(p-2).
• The resulting Taylor polynomial is shown below.

Video Clip #2

We can apply the Taylor polynomial for (1+x)^p to the function 1 / (x+1)^2, which is (1+x)^p for p = -2.

• For p=-2, we have p-1=-3, p-2=-4, etc., resulting in the Taylor polynomial shown below.
• We note that p(p-1) = -2 * -3 = 3! = 3 * 2!, p(p-1)(p-2) = -2 * -3 * -4, = -4! = -4 * 3!, etc..

To determine the radius of convergence for the series, we note that the ratios of the absolute values of the coefficients in this expansion are 2/1, 3/2, 4/3, ..., (n+1)/n, ... .

• The reciprocals of these ratios are easily seen to approach the limit 1, which implies that the Taylor series has a radius of convergence of 1.
• Since the Taylor series is expanded about 0 we see that we have convergence for | x | < 1.

Video Clip #3

We now consider whether the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... converges.

• A series converges if the sequence of its partial sums Sn converges.
• As depicted in the graph below, we can represent the sum of the harmonic series by the area of an infinite number of rectangles lying beneath the curve y = 1/x.
• However, this is of no help since the integral diverges: While we can see that the sum is less than the divergent integral, this does not tell us whether the sum is also divergent or whether it is perhaps convergent.

Since the integral diverges, the only way a comparison can tell us anything is if the quantity we are comparing it with lies above the graph.

• We can make this happen by simply shifting all our rectangles 1 unit to the right, as depicted below.
• In the figure below we see that the sum of the first four members of the harmonic series is greater than the integral of 1/x from x = 1 to x = 5.

A similar construction for any n will show that Sn = 1 + 1/2 + 1/3 + ... + 1/n > int(1/x,1,n+1) = ln (n+1).

• Since the limit of ln (n+1) as n -> infinity diverges, so must the limit of Sn.

We can also see that ln (n+1) <  Sn provides a lower limit for a partial sum of a harmonic series.

We can place an upper limit on a partial sum of the harmonic series.

• The figure below shows the rectangles from the preceding figure shifted one unit to the left, so that they lie beneath the y = 1/x curve.
• The first rectangle, of area 1, has not been included; had it been included the rectangles would have started at x = 0 and the integral would have diverged at that point (note that int(1/x, x, 0, n) diverges at x = 0).
• Thus we see that 1/2 + 1/3 + 1/4 < int( 1/x, x, 0, 4 ) = ln(4).
• Adding 1 to both sides we see that 1 + 1/2 + 1/3 + 1/4 < 1 + nt( 1/x, x, 0, 4 ) = 1 + ln(4).

We thus obtain the inequality below in which the partial sum of the harmonic series is bounded below by ln(n+1) and above by 1 + ln(n).

• The actual sum tends for large x to lie within about .5 of both limits.

Video Clip #4

To find the Taylor series for e^(x^4), we could calculate the derivatives of e^(x^4), but this would be pretty tedious.

• Alternatively we could first find the Taylor series for e^x, then substitute x^4 for x in the series itself.
• The Taylor series for f(x) = e^x is shown below.
• In the last line we have found the Taylor series for f(x^4) = e^(x^4) by substituting x^4 for x.

To integrate 1 / `sqrt(1-x^2) we find a Taylor series for the function and integrate the series.

• We use the result we found before for the Taylor series of (1 + x) ^ p, into which we then substitute x = -z^2.
• The Taylor series for (1 + x) ^ p is given in the second line.
• In the third line we substitute p = -1/2 and x = z^2.

From the pattern of the coefficients we see that for the nth term

• the numerator will contain the odd numbers 3, 5, ..., (2n-1) as factors
• the denominator will contain 2^n and n! as factors and
• the power will be 2n.

To integrate the resulting series we integrate each term in turn.

• We place our constant of integration c at the beginning.
• The integral of 1 is clearly z.
• The integral of z^2 / 2 is z^3 / (3 * 2).
• The integral of 3z^4 / (2^2 * 2!) is 3 z^5 / (5 * 2^2 * 2!).
• The integral of the nth term will be as indicated.