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### Course: AP®︎/College Calculus BC>Unit 10

Lesson 11: Finding Taylor polynomial approximations of functions

# Taylor & Maclaurin polynomials intro (part 1)

A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Created by Sal Khan.

## Want to join the conversation?

• How could we apply this to a real world case?
• Have you ever wondered how a calculator is able to give you a value for sine at any value, while you're stuck memorizing just the important ones? You can use this concept to do things like approximate trig functions like sine or cosine for any value. Computers can find the sum for a polynomial series with, say, 1000 terms in a snap and give an accurate approximation of that function. I find Taylor series to be very clever.
• what dose a 3rd derivative represent? the first derivative is the slope of the tangent line. the second derivative is the degree that the tangent line of one point differs from the tangent line of a point next to it. so is there any basis for having a third derivative other then using it in a Maclauren series?
• The way I think of it is;
1st derivative: rate of change of the function. (which is equal to the slope in any given point on the graph)
2nd derivative: rate of change of the 1st derivative. (which, when focused on a specific point, is equal to the changing rate of the slope in that point which is what you said)
3rd derivative: rate of change of the 2nd derivative (I guess this would be the acceleration of the slope change in a point)
4th derivative: rate of change of the 3rd derivative (and here I just give up trying to relate the derivatives to the original function)
5th derivative: you guessed it - rate of change of the 4th derivative.

And so on... To me it's just one of those things you simply have to accept that you can't rely on your visualisation skills to understand it. Like a 5-dimensional space, try imagining what that looks like. Still, it's not difficult to calculate the distance between two points in it for example, by just accepting the space as "a space where a point requires 5 coordinates to specify it's location".
• Why at do we use n! in the denominator instead of just the n. I understood up to the second derivative why we introduce 1/2 but then on the third it I do not understand why we have 1/3! as this is 1/6 and not 1/3 which means that it will not cancel out. I hope you understand my questions.
Thank you very much for your help :)
• For the second derivative, it's not 1 / 2 it's really 1 / 2!
It's just that 2! happens to be equal to 2.

The factorial works well for repeated differentiation:
d/dx x^5 / 5! = x^4 / 4!
d/dx x^4 / 4! = x^3 / 3!
d/dx x^3 / 3! = x^2 / 2!
d/dx x^2 / 2! = x / 1!
d/dx x / 1! = 1 / 0!
d/dx 1 / 0! = 0
• at about , Sal adds an "x" after f '(0). Why?
• It's so p'(0) still equals f'(0) after he takes the derivative of his expression f(0)+f'(0)x. If the x wasn't there, then the f'(0) would end up equaling zero when he took the derivative, just as the f(0) did. Reviewing power rule may be helpful in clarifying this.
• At why is it (1/2) multiplied by (1/3)? According to Sal's logic from before, shouldn't it just have been (1/3)? I know there was an answer to this question later on, but I didn't understand at all from that... ><
• We want p´´´(c) to be equal to f´´´(c).
Let's look at that one term.
p(c) = terms + f´´´(c) * 1/(2*3) * x^3
Let's look at its first derivative.
p´(c) = terms + f´´´(c) * 1/(2*3) * 3 * x^2
p´(c) = terms + f´´´(c) * 1/2 * x^2
Let's take the second derivative.
p´´(c) = term + f´´´(c) * 1/2 * 2 * x^1
p´´(c) = term + f´´´(c) * x
Then let's take the third derivative.
p´´´(c) = f´´´(c) * 1 * x^0
p´´´(c) = f´´´(c)

You have to have both 1/2 and 1/3 so that the coefficients eventually cancel each other out.
• what is the difference between a polynomial and a function ? dont we write the function of x as a polynomial?
• If it is a polynomial, then it is a function. This is not true the other way around (If its a function, then it is a polynomial). However, the distinction Sal is trying to make in this video is that we can approximate a function using a talyor series which looks like a polynomial. The key here is that we can approximate functions such as sin(x) or cos(x), or ln(x). Clearly, these functions are not polynomials, but they can be written as a talyor series which when you "expand it" looks like a polynomial with x raised to some power, thus functions can be approximated to look like polynomial.

I hope this distinction makes a little more sense.
• i can't visualize why when we add new term of new polynomial then the new curve nearer the old curve !
• Well, by analogy, think of trying to identify a specific person without saying their name. Instead, you will just be listing off facts about that person. Each fact you list off will eliminate a certain number of persons, until you eventually have enough descriptions that only one human on earth meets all the qualifications.

Really, that is all we are doing here. We are adding more and more facts about the original function. And, we're doing it in such a way that it does not interfere with previously established facts. While with this method we might need infinitely many facts added together to get a perfect match, very often we are not interested in the entire domain of the function, but just a particular region. Given that, we only need enough facts for the estimate to be pretty good for the region we're interested in.
• I don't really understand the intuition of these series, I mean, the mechanisms seem perfect and comprehensible, but why can we be confident that P(x) which is expanded at the point zero, would fit into all x when n approaches infinity? we are just evaluating it at zero!