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Worked example: coefficient in Taylor polynomial

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)
Finding the coefficient of the term containing (x+2)⁴ in the Taylor polynomial centered at x=-2 of x⁶-x³.

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  • starky ultimate style avatar for user LBKA
    Is there a video explaining how 0!=1!=1? Is that a definition?
    (8 votes)
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  • piceratops seed style avatar for user perry
    where is the x+2 coming from? Is it just part of the question/ is it arbitrary? Could the question be asking what the coefficient is if containing (x+19)^4 and have it be the same answer?
    (5 votes)
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  • blobby green style avatar for user Joshua Ronis
    Why would you ever make a series out of a function that is all ready a polynomial?
    (5 votes)
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  • blobby green style avatar for user Bahaa Soliman
    Hello, I recently started following your tutorials because my university lecturer does not explain appropriately. However, so far you are much better but in some questions in our notes, they ask for the 42nd term (really huge terms that cannot be done manually), and as soon as I went through the lecture notes, it appeared like black magic. Could you please do a tutorial about finding extremely far values for the Taylor/Maclaurin series? Also, how do you know if the centre is 0, or x itself is 0 if it does not specify it in the question?
    (2 votes)
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    • blobby green style avatar for user Keidar Gil
      It's good to note that some functions have repeating derivatives; for example, any derivative of e^x is simply e^x, or as another example, a derivative of sinx which is divisible by 4 is always going to be sinx (so, the 40th derivative of sinx would be sinx, and if the value were say 42 like in your case, the derivative could be represented as 40 +2, and since the 40th derivative is sinx, it is essentially just the second derivative of sinx, or -sinx.) I hope that helps; however, if the function is really messy, then you may be best off using something like Wolfram Alpha.
      (4 votes)
  • duskpin ultimate style avatar for user Sahel
    What is the point in finding an approximation to the function in this question? The actual function is a polynomial itself, and is a lot simpler than its approximation, which has infinite terms.
    (1 vote)
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  • spunky sam blue style avatar for user pittkor
    I started thinking that "what if we don`t know of original function f(x)"?
    (1 vote)
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  • leaf green style avatar for user abuasadsalah6
    the first practice in Maclaurin must x^3/6 not x^3/2
    (0 votes)
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Video transcript

- [Teacher] We're given an f of x and they say, what is the coefficient for the term containing x plus two to the fourth power in the Taylor polynomial, centered at x equals negative two of f? So like always, see if you can take a stab at this video on your own before we work through it together. Alright, now let's do this. So in general, our Taylor polynomial, p of x, it's going to have the form and remember, we're centering at x equals negative two so this means we're going to evaluate our function at where we're centering it. We are going to divide it by zero factorial which is just one. I'm just gonna write 'em all out just so you see the pattern and we could even say that's gonna be times x minus where we're centering it but if we're subtracting a negative two, it's gonna be x plus two and I could write to the zero power but once again, that's just going to be one so a lot of times, you won't see someone write this and this but I'm writing it just to show that there's a consistent pattern. So then you're gonna have plus the first derivative evaluated at negative two divided by one factorial which is still just one times x plus two to the first power plus the second derivative evaluated at negative two over two factorial times x plus two squared. I think you see where this is going and really all we care about is the one that has a fourth degree term and, well, actually, let me just write the third degree term too just so we get fluent doing this. So the third derivative evaluated at negative two over three factorial times x plus two to the third power and now, this is the part that we really care about, plus the fourth derivative. I could have just written a four there but I think you get what I'm saying. And then evaluate at x equals negative two divided by four factorial times x plus two to the fourth power. So what's the coefficient here? Well, the coefficient is this business. So we'd have to take the fourth derivative of our original function. We gotta take the fourth derivative of that original function, evaluate at negative two and divide it by four factorial so let's do that. So our function, so our first derivative, f prime of x is just going to be, just gonna use the power rule a lot, six x to the fifth minus three x squared. Second derivative is going to be equal to five times six is 30 x to the fourth. Two times three, minus six x to the first power. Third derivative. Third derivative of x is going to be equal to four times 30 is 120 x to the third power minus six and then the fourth derivative which is what we really care about is going to be three times 120 is 360 x to the second power and the derivative of a constant is just zero. So if we were to evaluate this at x equals negative two, so f, the fourth derivative, evaluated when x equals negative two is going to be 360 times negative two squared is four. I'm just gonna keep that as 360 times four. We can obviously evaluate that but we're gonna have to divide it by four factorial so the whole coefficient is going to be 360 times four which is the numerator here divided by four factorial, divided by four times three times two times one. Well, four divided by four, those are just gonna be one. 360 divided by three, maybe I'll think of it this way, 360 divided by six is going to be 60 and so that's all we have. We have 60 and then in the denominator we just have a one so this is going to simplify to 60. That's the coefficient for this term.