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# Taylor polynomial remainder (part 1)

The more terms we have in a Taylor polynomial approximation of a function, the closer we get to the function. But HOW close? Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Created by Sal Khan.

## Want to join the conversation?

• Why wouldn't the n+1'th derivative of the function be 0 as well? •   Remember also that the function f(x) isn't necessarily a polynomial. p(x) is its polynomial representation, but f(x) can be any function - sin(x), e^x, sqrt(x), etc. Those are examples of functions that will never go to 0 no matter how many times you take the derivative.
• when he derives p(a), and says it is equal to f'(a), how is that possible when the term that has f'(x) is multiplied by (x-a), so at (a) this term should also be zero? • I know this was posted four years ago, but I may as well write this in case it helps anyone. The reason p'(a) = f'(a) (and p''(a) = f''(a), etc) is because of the following:

We are given:
p(x)=f(a)+f'(a)(x-a)+f''(a)((x-a)^2)/2!+...

To find p'(x), we have to take the derivative of each term in p(x). Since f(a) is a constant (since a is just a number that the function is centered around), the derivative of that would be 0. When taking the derivative of the second term, ie f'(a)(x-a), using the product rule (u'v +uv') gives you f''(a)(x-a)+f'(a)(1), which, when evaluated at a, is f''(a)(a-a)+f'(a)(1) which simplifies to f'(a) since f''(a)(a-a)=0. All of the terms after that are 0 since the product rule doesn't lead to the (x-a)^n part of each term 'cancelling out' as it did here. You can verify this last bit by trying the product rule on them yourself. This whole concept applies to every derivative of the polynomial up through the degree of the polynomial, as far as I know. Hope this helps someone!
• What do you mean by bound toward the end of the video? • Being "bound" means that you know that a value is definitely between two limits. For instance, you might be interested in knowing that your approximation is good to 1% or to 0.01% or to one part in a million.

With calculators, it's often easy enough to add another term and get your error to be bound to any level you want. However, when math meets the physical world, you could get a scenario where you have 3000 cables supporting a load on a bridge. The math of exactly what's happening on each cable is too hard so you turn to approximations. If you can show your approximation is valid (or bound) to within 2% and you know that all materials have a 25% safety tolerance, you don't need to spend any more time or money on figuring out the details of your cable loads more precisely.
• Is this related to Big O Notation?

And, by the way, what is "Big O Notation"? • Big O notation is computer science term, It is used to describe the processing complexity of the algorithm.

For example, If you have n random numbers and you want to find minimum number out of it then Big O for it will be n. Which means that we will have to check every number once to find minimum number out of it.
• Well, there's the obvious problem that the degree of functions such as the sine or e^x is infinite. Does this mess up the principle of bounding the error function? Why or why not? • At , When we want to find the N+1 dervivative of the error (E(X)), why is the N+1 derivative of theTaylor Polynomial (P(x)) equals to zero, and not the N+1 derivative of the function, Why is the last on equal to the N+1 derivative of the error? • Remember that P(x) is an nth polynomial if you try to figure out the 3rd derivative of x^2 you will get zero, In fact if you have a polynomial function with highest degree n and you get the (n+1)th derivative you get zero that is because every time you take the derivative you apply the power rule where you decrease the power by one until it becomes 0 in which case you have a constant function ex: f(x) = 5 x^0 = 5 and the derivative of a constant function will equal zero, point is that take the derivative of a polynomial enough times you will have a constant function take the derivative one more time you will get 0.
one more thing is that we are assuming we don't know what kind of function is f(x) it could be a polynomial in which case f'^(n+1)(x) will equal zero or it could be an exponential function like 2e^x in which case f'^(n+1)(x) will equal 2e^a at x=a so we actually don't know which one could it be so we leave it as it is and we write it f'^(n+1)(x).
I hope I was helpful to you.
Good luck and never give up :)
• since this approximation completely matched f(x), why is the (n+1)th derivative of P(x) zero since it goes on and on to infinity? Which means, why is P(x) confined to the nth term? • Or another way of saying it is that you assume you've created an approximation of order n. When you take the (n+1)th derivative you get 0 and I assume the video explains that well enough.

Why, then, would you not do an infinite number of terms? Practical considerations. Calculators used to use Taylor series expansions to calculate sin, cos, and e^x. (I think most use lookup tables and interpolation now.) However, your calculator can't do an infinite number of terms. Instead, they know they're going to show the answer to, say, 8 decimal places. How many terms do you need for your sin(x) approximation in order to know that your Taylor's series expansion is good to at least 8 decimal places? That's an example of bounding your error AND of why you'd stop at some point.
• At , is there any proof that the derivative of E is also equal to the difference of derivatives of f and P? I mean, we only defined E(x) as the difference of a function's value and the value of its taylor polynomial at some x. But how can we be sure about the derivatives too? • Mr. Sal says that Taylor series is useful because the derivative at a is equal to the derivative of the function at a... But I don't understand why we need to use Taylor series in the first place. Why not just use the function?

In other words.. what's the point of approximating a function when you actually have the"real" function?  