In this recitation, we covered Taylor series and some problems from the mock series midterm.

Approximations

One way of framing calculus is that we want to study nice enough functions which “locally” look like lines. This leads us to the concept of derivatives and tangent lines; the derivative is the slope of the tangent line, and the tangent line is the best linear approximation at a point. However, a really good question to ask is how good of an approximation this is? For example, if we’re given a line, then the tangent line is equal to the line, and so the approximation isn’t even an approximation; it is an equality. On other hand, if we look at the function \(f(x) = x^2\), then the tangent line at \(x=0\) is

This is a bad approximation the further we get from \(x=0\), so we should probably avoid using it. The question now boils down to the following: 1) Is there a way to get a better approximation than a line? 2) Is there a way to tell how bad of an approximation we have?

The answer to both questions can be found in Taylor polynomials. Before going into Taylor polynomials, let’s review the tangent line. Remember that the goal of constructing a tangent line at \(x=a\) is to find a line \(L_a(x)\) which is the “best approximation” of \(f(x)\) at the point \(x=a\). Since it is a line, we know

for some \(m\) and \(b\). If we plug in \(x=a\), then we should have

So we can rearrange our original equation to be in terms of one variable:

To figure out what \(m\) is, we want the derivatives of \(f(x)\) and \(L_a(x)\) to be the same at \(x=a\) (i.e. the slope of the infinitesmal line should be the same). Taking the derivative, we have

and this eliminates the last variable. We can write

What if we try to do the same thing with a quadratic instead? Let’s write

where we need to figure out \(t_0, t_1,\) and \(t_2\). Again, we want them to be the same at \(x=a\), so we should have

and we can solve for \(t_0\) and substitute this in to get

Next, we want the derivatives to be the same, so we have

If we solve for \(t_1\) and plug it in to the original equation, we have

Notice

so we can factor the last part as

If we want to eliminate this last variable, we will need \(f(x)\) to have a second derivative. If we assume that, then we want the second derivatives to agree, and we get

Solving for \(t_2\), we get

Taylor Polynomials

Let’s call \(p_n^a(x)\) the \(n\)th order Taylor polynomial at \(x=a\). We found \(p_1\) (which is the tangent line) and \(p_2\) (which is this \(Q_a\) I wrote a second ago). In general, one can show that

where \(f^{(k)}(x)\) is the \(k\)th derivative of \(f(x)\). Note that we have a nice recursive definition for finding the next Taylor polynomial:

This can be useful if we are given partial information about the Taylor polynomial. Intuitively, it might be clear that a higher order Taylor polynomial gives a better approximation, and this is indeed the case. One might also wonder what happens when we keep taking derivatives – maybe a function is infinitely differentiable. Is this an alternative way of expressing the same function?

Example: Recall that derivatives of \(\sin(x)\) follow a nice pattern. For an integer \(n\), let’s denote \(R(n)\) to be the remainder of the integer when we divide by \(4\). For example, \(R(1) = 1, R(2) = 2, R(3) = 3, R(4) = 0, R(5) = 1, R(6) = 2, \ldots\). We have the following pattern:

If we want to find the \(n\)th order Taylor polynomial for \(\sin(x)\) at \(x=0\), we would have

In the next section, we’ll explore whether or not

but if you want to play around with this, try plugging in some values into Wolfram and see if they agree.

Notice there’s a big question I’m skipping over. Does the series on the right even converge? Sadly, this is outside of the scope of the course, but if you are interested, look into the alternating series test.

Convergence

This section is optional, and we probably will not get to this in class. But I think it’s important to think about this at least once in your life.

Remember one of our favorite theorems from calculus: the mean value theorem. One way of interpreting the theorem is as follows: Suppose \(f(x)\) is a function that is differentiable on the interval \((a,b)\). Then for every \(s\) and \(t\) such that \(a <s < t < b\) we have that there is a \(c\) such that \(s < c < t\) and such that

Notice you’ll get the original mean value theorem if you rearrange this formula and use the fact that differentiable implies continuous. It turns out that this is generalizable as well – if \(f(x)\) is a function such that the \(n\)th derivative exists on the interval \((a,b)\), then for every \(s\) and \(t\) such that \(a < s < t < b\) we have that there is a \(c\) such that \(s < c < t\) and such that

If we look carefully at this, we might notice that the sum is a familiar friend. Using the previous section, we can write

In particular, this gives us a way of calculating the error between a function and its \(n\)th Taylor polynomial. Suppose there is an \(M\) such that \(-M \leq f^{(n)}(x) \leq M\) for all \(x\) in the interval \((a,b)\). Then we have

For example, let’s go back to \(\sin(x)\). Since \(\sin(x)\) and \(\cos(x)\) are bounded by \(1\), we get

so as long as \(t\) and \(s\) are fixed, we can use growth rates to deduce that the error goes to zero as \(n\) tends to infinity (remember factorials grow faster than exponentials).

In general, there’s not a way of uniformly choosing an \(n\) so that the error is small. As the distance between \(t\) and \(s\) gets larger, we have to take more derivatives in order for the approximation to be accurate.

Suppose we’re designing a calculator. Calculating \(\sin(t)\) is generally bad news – outside of some particular values of \(t\), we are pretty much stuck. However, Taylor polynomials are great. A calculator only needs to be so accurate (say within four decimal places), so we can just use our formula for the Taylor polynomial to approximate \(\sin(t)\). Polynomials are easy to program, so we’re back in business. Except, bad news, if \(t\) and \(s\) are far, we don’t have a good formula for the polynomial. Is there something we can do to salvage this?

We found out that \(p_n^0(t)\) is easy to calculate earlier. We have

Since \(\sin(t)\) and \(\cos(t)\) are \(2\pi\) periodic, it turns out we actually have a formula for infinitely many choices of centers: if \(m\) is an integer, then

So what’s the strategy for our calculator? If someone types \(t\) into our calculator, then we find an \(m\) so that \(2\pi m \leq t < 2\pi (m+1)\) (which is unique), and this will guarantee that \(0 \leq (t- 2\pi m)^{2k+1} < (2\pi)^{2k+1}.\) Now we can find \(k\) so that \(\frac{(2\pi)^{2k+1}}{(2k+1)!}\) is as small as our calculator needs, and then we just plug \(t\) into the corresponding Taylor polynomial. We’re back in business, and this is not hard to program: try doing it with your favorite programming language. Here is an example in Python.

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import math

# Recursive function for factorial
def factorial(n):
  if n == 0:
    return 1
  else:
    return n * factorial(n-1)

# t is the value we want to plug into the polynomial,
# n is how far out in the series we go,
# and m determines the center
def taylor_approximation_sine(t,n,m):
  sum = 0
  for k in range(0,n+1):
    sum += (-1)**k/factorial(2*k+1) * (t - (2 * math.pi*m))**(2*k+1)
  return sum

# Determine the correct m value and return the approximation.
def approximate_sine(t,n):
  m = math.floor(t/(2*math.pi))
  return taylor_approximation_sine(t,n,m)

# Compare sin(14) and the approximate value
d = abs( approximate_sine(14, 4) - math.sin(14))
print(d)

Notice that the error between our estimate for \(\sin(14)\) and the actual value is approximately \(1.3000097202064964 \cdot 10^{-6}\). In general, if we take \(n = 4\), then the distance between this approximation of \(\sin(x)\) and the actual value will be bounded above by \((2\pi)^5/(11 !) \approx 0.000245.\)

Mock Series Midterm

Here are some variations of the harder problems from the mock series midterm.

Problem: Taylor is asked to determine if

converges. They give the following argument: Since \(\ln(k+4) - \ln(k+3) = \ln \left(\frac{k+4}{k+3} \right)\), we see that

so the series converges. We are supposed to evaluate Taylor’s argument, and if it is wrong state a correct argument. The idea is that Taylor is trying to use the divergence test to prove convergence, which as we discussed is not useful. What they have shown is that the divergence test fails, but this does not tell us that the series converges. We need to do more in order to prove this.

Let \(a_k = \ln(k+3)\), then \(a_{k+1} = \ln((k+1) + 3) = \ln(k+4)\) and

The series is telescoping! If we look at the partial sums, we have

Plugging in numbers,

Taking the limit, we see that

so the series diverges.

Problem: Brad needs to determine if the series

converges and is confused how to start. Observe that growth rates do not apply because it is a degree two polynomial in the exponent instead of a linear term. To determine if it converges, let’s try a ratio test. Define

so

Rearranging, we get

An exponential grows faster than any polynomial, so we get \(\lim_{k \rightarrow \infty} a_{k+1}/a_k = 0.\) The ratio test tells us the series converges. Since the series converges, notice that the divergence test tells us that

Problem: We wish to calculate

Let

For \(n\) large we have that \(a_n > 0,\) so

Next, notice

This is indeterminate of the form \(0/0\), so we can apply L’Hopital’s rule:

Here we use growth rates. By continuity of the exponential:

Remark: The definition of \(e\) comes from compound interest. If we invest \(P\) dollars into an account, then the idea behind interest is that we should earn a percentage back for each time interval (the number of times it is “compounded”). So if the interest rate is \(r\) and it is compounded four times a year, then in two years we should have

If it is compounded \(n\) times a year, then after \(t\) years we should have

How much would we earn if it was continuously compounded? If we compound it “infinitely” often, then we would earn

Dividing by \(P\) and setting \(t=1\) tells us the percentage we would earn after one year:

Just like in the previous problem, we can solve this using logarithms. Let

We can do the L’Hopital trick again:

So

So the percentage of money we’ve made after one year is \(e^r\). If \(r=1\), then we’ve rederived Euler’s constant following in the footsteps of Jacob Bernoulli; see also here.

Problem: Consider

Suppose we wished to calculate \(\lim_{n \rightarrow \infty} a_n\). Recall that logarithms grow slower than exponentials, so

Since polynomials grow slower than exponentials:

Problem: Consider

Suppose we wished to calculate \(\lim_{n \rightarrow \infty} a_{n+1}/a_n.\) Notice that using growth rates, we don’t need to do anything – \(a_{n+1}\) grows at a rate of \(5n^2\) and \(a_n\) grows at a rate of \(5n^2\), so this limit is one.

Problem: Consider

We wish to determine whether this series converges or diverges. The ratio test is probably the right move here. Let

so

and

Taking the ratio:

Let

so we can write

Limit laws tell us that as long as \(\lim_{k \rightarrow \infty} b_k\) and \(\lim_{k \rightarrow \infty} c_k\) exist and are finite, then we have

We need to treat these separately. By growth rates, the dominant term for \(b_k\) in the numerator and denominator is \(k\), so

The limit for \(c_k\) is more complicated. Observe that it is indeterminate of the form \(1^\infty\). We can write

Notice \(\lim_{k \rightarrow \infty} \ln(c_k)\) is also indeterminate of the form $$ \infty \cdot 0$, where we use growth rates for the limit inside of the natural log. This can be rearranged to apply L’Hopital:

Again, we use growth rates to determine this limit is \(-1\), since the dominant term on top and bottom is \(k^2\). Thus

and we have

By the ratio test, this series converges.