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# Taylor & Maclaurin polynomials intro (part 1)

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.A (LO)
,
LIM‑8.A.1 (EK)
,
LIM‑8.A.2 (EK)
,
LIM‑8.B (LO)
,
LIM‑8.B.1 (EK)
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?

• •   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.
• Why is it that we use the factorial? I thought we would divide on the number of the grade so it would cancel later with the derivative :/
Thank you! •   "Why is it that we use the factorial? I thought we would divide on the numberS of the gradeS so it would cancel later with the derivativeS :/ "
We aren't taking one derivative, we're taking the Nth derivative of the highest-degree monomial, each time, the exponent would "come down" to the coefficient and cancel with whatever's down your denominator.

For example, to get the third derivative of p(x), we need to take the derivative 3 times, each time the so-called "grade" comes down and becomes the coefficient of the derivative.
p(x) = x^3
p'''(x) = 3*2*1 = 6, not 3.
Say we know f'''(x) = 1, how do we model p(x) so that p'''(x) = f'''(x) = 1?
Exactly, we let p(x) = (1/6) * x^3, or generally, p(x) = (1/3!)*x^3
• this series have to be evaluated only in p(0) or can be calculated in any point of the graphic? •   This is the Maclaurin Series (a Taylor Series evaluated at zero). A Taylor Series can be evaluated at any point.
• Why start taylor series by talking about maclauren series, it seems to add another condition ( centered around) which seem to complicate things when trying to understand the concept behind taylor series? •   The video is entitled "Maclauren and Taylor Series INTUITION." I would argue that the purpose of the video isn't to give a rigorous introduction to Taylor series but rather to introduce a related but simpler concept. Thus the "Intuition." Starting with Maclauren gives people a conceptual foundation from which they can generalize.

As to the contention that this is like teaching 1+0=1 to a first grader, the fact is that these videos are aimed at people with some degree of math background. Once you have the basic pattern it is easy for people familiar with graphs to perform a simple transformation on the Maclauren Series to get the Taylor Series.

What is better to teach first: X^2 is a parabola or (x-a)^2+b is a parabola shifted a units horizontally and b units vertically? I would say the former.
• 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 :) • at about , Sal adds an "x" after f '(0). Why? • 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.
• 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. 