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Worked example: Taylor polynomial of derivative function

Rather than approximating a function, this time we are asked to approximate the derivative of a function.

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  • piceratops ultimate style avatar for user Travis Petersen
    Here's the step I don't understand: g(x) = g(2) + g'(2)(x - 2) + g''(2)(x - 2)^2 + ... When taking the derivative of g(x), why aren't the (x - 2) terms differentiated?
    (10 votes)
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  • leaf blue style avatar for user A D
    Anyone else think this question was ambiguous? I determined the second degree Taylor polynomial of g(x), then took the first derivative of that and evaluated at x=1. In other words, instead of not using the data g(2)=3 like Sal, I didn't use g'''(2)=2. Admittedly, what I did would probably give a less accurate approximation, but I think it still satisfies the wording of the question. If anyone wants to set me straight on this though, please do.
    (6 votes)
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    • piceratops ultimate style avatar for user Fai
      Hi!

      Thank you for your clarification. Yes, you could also go about it the way you did, and, as you correctly stated, your answer would be slightly less accurate. But you can always increase your accuracy by using the third derivative. So, just throw in an extra term g^(3)(x)*(x-2)^3/3!.

      This extra term will further increase the accuracy of your Taylor polynomial and you should end up with the same answer that they have gotten in the video above.
      (4 votes)
  • leafers seed style avatar for user Noah Schwartz
    It's only an approximation, so the answer will vary, but can't you also solve this by finding the Taylor series for g(x)? g(x) will approximately equal 3 + (x - 2) - 1/2(x-2)^2. Now, what does g'(1) mean? It means finding the slope of the tangent line at g(1). Therefore, if we take the derivative of our approximate function, we get 1 - (x-2) or 3 - x. Substituting 1 in for x, the approximation of the slope at g(1) becomes 2, or g'(1) approximately equals 2. It's not the exact answer Sal got, but since these both are approximations of the real answer, is this one equally valid?
    (6 votes)
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  • blobby green style avatar for user lazy111
    how come in none of the taylor/maclaurin vids has he gone into convergence and divergence and radius of convergence??
    (5 votes)
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  • leaf red style avatar for user Kody Kendall
    Where does the (x-2) term come from in the above example?
    (2 votes)
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    • leafers seed style avatar for user Travis Bartholome
      We have an (x-2) term because this particular Taylor polynomial is centered at x=2. Remember that in general, the formula for the nth order term of a Taylor polynomial is ( f^(n)[c] * (x-c)^n ) / n! where c is the center of our Taylor polynomial.

      Importantly, c is also the number at which the derivatives are evaluated to find the coefficients.

      Hope that helps.
      (5 votes)
  • blobby green style avatar for user razan
    i did the problem a slightly different way by calculating taylor series for g(x) to get the polynomial (-x^2+5x+1) then differentiated that one and solved for x=1 and got the same answer! is that okay?
    (3 votes)
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    • female robot grace style avatar for user tyersome
      Yes this is a valid, though somewhat more laborious, way to solve this problem. However, when I do the calculations this way, I get the same intermediate step as An Duy ...

      Note also that not simplifying the expression and just taking the derivatives of each term (using the power rule) is less painful!
      (2 votes)
  • leaf yellow style avatar for user Steven *
    So even though it says "centered at x=2", we actually make c=2 instead of x, and then treat the x as being whatever we want later? Is that correct?
    (2 votes)
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  • piceratops ultimate style avatar for user Nazariy Olegovich Vavryk
    What does second degree mean? does it mean that instead of starting with g(t) we start with the g'(t)? thanks
    (1 vote)
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    • aqualine ultimate style avatar for user Daniel
      It means that you'll get the Taylor polynomial up to the term where you use the second derivative and elevate (x-c) to the second power.
      For example if instead of the second degree polynomial he used the third degree it would add: (f'''(2)(x-2)^3)/3! to the Taylor Polynomial.
      (2 votes)
  • blobby green style avatar for user Ishina Ixox
    A function has an n th derivative evaluated at x = a that is proportional to n. What is the radius of convergence of the Taylor series of this function expanded about the point x = a?
    (1 vote)
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  • aqualine seedling style avatar for user Victoria
    How is it that Sal made g(x) = g'(2) ... and so forth?
    Shouldn't the equation look like g(s) = g'(2)... since every equation after it, the x is subsituted with 2?
    (1 vote)
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Video transcript

- So, let's say we've been given all this information about the function g and it's derivative evaluated at x equals two. We know g of two is equal to three. G prime of two is equal to one. The second derivative of g evaluated two is negative one. The third derivative of g evaluated at two is two. Given that, what we're being tasked with is we want to use the second degree Taylor polynomial centered at x equals two to approximate g prime of one. Not g of one, g prime of one and so I encourage you to pause this video and try to think about it on your own. I'm assuming you've had a go at it. Let's just remind ourselves what a second degree Taylor polynomial centered at x equals two would look like for a general function f of x. F of x would approximately be equal to, it would be f of two plus f prime of two times x minus two plus f prime prime of two times x minus two squared, all of that over two factorial. That would get us to a second degree place because it's x minus two squared. This is gonna give us a second degree polynomial. This is the general case. If we want to find the approximation for f centered at x equals but we're gonna do it for g prime. Let me write this down. Alright, so I'll do it in blue. So, we have g prime of x is what we're going to try and approximate and then we're going to evaluate it at x equals 1. G prime of x is going to be approximately equal to, well same thing, it's going to be the function that I'm going to try to approximate evaluated at two so g prime of two. Notice, so I'm approximate f of x, it's that function evaluated at two. If I'm approximating g prime of x, it's that function evaluated at two. Then, plus the first derivative of this thing which is the second derivative of g. G prime prime of two times x minus two and then plus the second derivative of the function that I'm trying to approximate but the second derivative of g prime is going to be the third derivative of g. It's going to be g prime prime prime of two times x minus two squared, all of that over two factorial. Now, they tell us what these things are. Let me use some new colors here. They tell us g of two is equal to three. This right here, oh, actually no, that's not what we're gonna wanna use. They tell us we're using g of two. They're telling us g prime of two is equal to one. G prime of two, so this right over here is equal to one. G prime prime of two is equal to negative one. This is negative one right over here and then finally the third derivative of g evaluated at two is two. Two over two factorial, two factorials is two times one or it's just two. So that and that cancel out. What are we left with for our approximation our second degree approximation of g prime of x centered at x equals two? We are left with g prime of x is approximately equal to one minus x minus two. One minus x minus two, I guess I could write that as plus two minus x so plus two minus x. The negative of x minus two is two minus x. Plus x minus two squared and obviously I could simplify this even more. This is three minus x plus x minus two squared and now I can evaluate it. If I wanna approximate g prime of one, I could say g prime of one is going to be approximately equal to, wherever I see the x is here I put in a one there, so it's going to be three minus one plus one minus two squared. Well, this is going to be two and then one minus two is negative one but then if I square it I get a positive one. So this whole thing is one. Two plus one is equal to three. Once again, this is an approximation for g prime of one. What did we do here? We found the Taylor series. The second degree Taylor series approximation for g prime of x centered around x equals two and then we evaluated that approximation at x equals one to approximate g prime of one. Anyway, hopefully, you found that fun.