What You Need to Know To Understand TAYLOR SERIES
One Calculus topic that seems to challenge my tutoring clients every year is the topic of Taylor series. It usually comes up near the tail end of their Calculus class, and it is touted as a method to come up with an “approximation” to a function. Most Calculus textbooks introduce Taylor series with a scary-looking formula like
[math]\sum\limits_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!} \cdot (x-a)^n[/math]
Often, clients tell me that they can plug in the right numbers into the formula more or less adequately, but they have no idea what the formula means or why this would even generate an approximation of the original function. If you feel this way, then you have come to the right place! I will show you a way of thinking about Taylor series so that the formula to generate them won’t be such a mystery anymore.
To start, let’s imagine that we have a function [math]f(x)[/math], which is continuous and differentiable in a region around [math]x = a[/math]:
What I would like to do is to come up with a power series [math]p(x)[/math] that exactly mimics the original function [math]f(x)[/math] at [math]a[/math]. I don’t care if the power series mimics the original function anywhere else, but it has to be a perfect match for the function at [math]a[/math]:
If your Calculus class has not gone over power series, please be assured that a power series is just a polynomial, except that it has an infinite number of terms. This is not as crazy as it sounds! You may have learned that an infinite series can in fact add up to a finite number even if it has an infinite number of terms; this is what we call a convergent series.
Since the number of terms is infinite, I can’t write the power series starting with the highest power of [math]x[/math] term as the leading term like I normally do. Instead, I write the constant term first, then the [math]x[/math] term, then the [math]x^2[/math] term, as follows:
[math]c_0 + c_1 \cdot x + c_2 \cdot x^2 + c_3 \cdot x^3 + \dots[/math]
In this case, I want to make sure that this power series [math]p(x)[/math] is defined at [math]a[/math], so I will write it in powers of [math](x-a)[/math] instead of powers of just [math]x[/math]:
[math]p(x) = c_0 + c_1 \cdot (x-a) + c_2 \cdot (x-a)^2 + c_3 \cdot (x-a)^3 + \dots[/math]
If you have learned about function transformations, you know that this just amounts to shifting the polynomial over [math]a[/math] units to the right and doesn’t make the polynomial any more or less valid. My task is now to set the [math]c_0[/math], [math]c_1[/math], etc. so that [math]p(x)[/math] is a perfect replica of [math]f(x)[/math] at [math]x = a[/math].
Well, if [math]p(x)[/math] is supposed to be a duplicate of [math]f(x)[/math] at [math]x = a[/math], then the least I could do is to make sure that both functions have the same value at [math]x = a[/math]:
[math]p(a) = f(a)[/math]
If I plug [math]a[/math] into [math]p(x)[/math], I get
[math]\begin{align} p(a) &= c_0 + c_1 \cdot (a-a) + c_2 \cdot (a-a)^2 + c_3 \cdot (a-a)^3 + \dots \\ &= c_0 + c_1 \cdot (0) + c_2 \cdot (0)^2 + c_3 \cdot (0)^3 + \dots \\ &= c_0 \\ \end{align}[/math]
that, is, all of the terms become zero except for the constant term. So if I want my [math]p(x)[/math] to have the same value as [math]f(x)[/math] at [math]x = a[/math], then I have to set [math]c_0[/math] to the value of [math]f(a)[/math]:
[math]c_0 = f(a)[/math]
Remember that [math]f(a)[/math] is just a number — the value of [math]f(x)[/math] when [math]x[/math] is equal to [math]a[/math].
OK, that seems good as far as it goes. But I think I can do better. For [math]p(x)[/math] to be a good duplicate of [math]f(x)[/math] at [math]x = a[/math], I also want the slopes of the two functions to be the same at [math]x = a[/math]. That is, I want their first derivatives to be equal at [math]x = a[/math]:
[math]p^{\prime}(a) = f^{\prime}(a)[/math]
I can take the derivative of [math]p(x)[/math] easily enough, and then plug in [math]a[/math] for [math]x[/math]. Again, most of the terms become zero and we are left with
[math]\begin{align} p^{\prime}(x) &= c_1 + 2 c_2 \cdot (x-a) + 3 c_3 \cdot (x-a)^2 + 4 c_4 \cdot (x-a)^3 + \dots \\ p^{\prime}(a) &= c_1 + 2 c_2 \cdot (a-a) + 3 c_3 \cdot (a-a)^2 + 4 c_4 \cdot (a-a)^3 + \dots \\ &= c_1 + 2 c_2 \cdot (0) + 3 c_3 \cdot (0)^2 + 4 c_4 \cdot (0)^3 + \dots \\ &= c_1 \\ \end{align}[/math]
In other words, I can assure that [math]p(x)[/math] and [math]f(x)[/math] have the same slope at [math]x = a[/math] by setting [math]c_1[/math] to the value of [math]f^{\prime}(a)[/math]:
[math]c_1 = f^{\prime}(a)[/math]
again remembering that [math]f^{\prime}(a)[/math] is a number, not a function.
I want to keep going. I will now insist that the concavity of [math]p(x)[/math] match the concavity of [math]f(x)[/math] at [math]x = a[/math], which amounts to their second derivatives being equal:
[math]p^{\prime \prime}(a) = f^{\prime \prime}(a)[/math]
I repeat my process, taking the second derivative of [math]p(x)[/math] and plugging in [math]a[/math] for [math]x[/math]:
[math]\begin{align} p^{\prime \prime}(x) &= 2 c_2 + 2 \cdot 3 c_3 \cdot (x-a) + 3 \cdot 4 c_4 \cdot (x-a)^2 + \dots \\ p^{\prime \prime}(a) &= 2 c_2 + 2 \cdot 3 c_3 \cdot (a-a) + 3 \cdot 4 c_4 \cdot (a-a)^2 + \dots \\ &= 2 c_2 + 2 \cdot 3 c_3 \cdot (0) + 3 \cdot 4 c_4 \cdot (0)^2 + \dots \\ &= 2 c_2 \\ \end{align}[/math]
This time I ended up with [math]2c_2[/math] as the value of [math]p^{\prime \prime}(a)[/math] instead of just [math]c_2[/math]. Why is that?
If I look back, I can see it happened when I took the first derivative of the [math]c_2 \cdot (x – a)^2[/math] term. When I differentiated that term, I ended up placing a coefficient of [math]2[/math] in front of that term as dictated by the Power Rule.
So I will have to set [math]2c_2[/math] equal to [math]f^{\prime \prime}(a)[/math]:
[math]2c_2 = f^{\prime \prime}(a)[/math]
[math]c_2 = \dfrac{f^{\prime \prime}(a)}{2}[/math]
I continue my process, matching the third derivatives, fourth derivatives, fifth, sixth, and so on. Each time, I take another derivative of [math]p(x)[/math] and then plug [math]a[/math] in for [math]x[/math], which eliminates all but one of the terms, and set it equal to the next derivative of [math]f[/math]. As with the second derivative term, all of the terms accumulate additional factors in their coefficients as they are differentiated again and again. For instance, if I look at the term that originally was [math]c_3 \cdot (x – a)^3[/math], it becomes [math](2 \cdot 3 \cdot c_3)[/math], or [math](3! \cdot c_3)[/math] after three differentiations (where the exclamation point indicates a factorial). After four differentiations, the term that originally was [math]c_4 \cdot (x – a)^4[/math] becomes [math](2 \cdot 3 \cdot 4 \cdot c_4)[/math], or [math](4! \cdot c_4)[/math], and so on.
[math]3! \cdot c_3 = f^{\prime \prime \prime}(a) \qquad c_3 = \dfrac{f^{\prime \prime \prime}(a)}{3!}[/math]
[math]4! \cdot c_4 = f^{(4)}(a) \qquad c_4 = \dfrac{f^{(4)}(a)}{4!}[/math]
(note that [math]f^{(4)}(a)[/math] means, fourth derivative of [math]f(x)[/math], evaluated at [math]x = a[/math])
[math]5! \cdot c_5 = f^{(5)}(a) \qquad c_5 = \dfrac{f^{(5)}(a)}{5!}[/math]
Once I have computed as many coefficients as I want, I can put them all together into the original expression for the power series:
[math]p(x) = f(a) + f^{\prime}(a) \cdot (x-a) + \dfrac{f^{\prime \prime}(a)}{2!} \cdot (x-a)^2 + \dfrac{f^{\prime \prime \prime}(a)}{3!} \cdot (x-a)^3 + \dots[/math]
where the [math]f(a)[/math], [math]f^{\prime}(a)[/math], etc. are the values that I computed for all the coefficients [math]c_0[/math], [math]c_1[/math], etc.
Let’s recap. What I have done is to specify all the coefficients for a power series [math]p(x)[/math] in such a way that this power series perfectly mimics the original function and all its derivatives at [math]x = a[/math]. If you know summation notation, then you would probably agree that I could write the whole series as
[math]\sum\limits_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!} \cdot (x-a)^n[/math]
Where the [math]\dfrac{f^{(n)}(a)}{n!}[/math] are the coefficients of each of the terms of the infinite polynomial.
Hopefully this makes the formulation of Taylor series less of a mystery! If you would like more help in Calculus, I would be happy to provide you with 1:1 tutoring. Please see my website at https://andrewjeungtutoring.com .