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FP2 Maclaurin series help Watch

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    So in the Maclaurin expansion, it goes as follows

    f(x) = f(0) + f'(0)x + \frac {f''(0)}{2!} x^2 + .... + \frac {f^{r}(0)}{r!}x^r + ...

    The problem that I've encountered is that I don't know how to find f^{r}(x) so I can't complete the expansion or am I missing something? Perhaps I should be spotting a pattern?
    Name:  fp2 maclaurin.jpg
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    Like how do I get the general term as shown in the picture?
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    a) asks for series up to general x^r, so you need to spot a pattern b) asks you to write out the first few terms

    In a) its reasonable to spot r! multiplying the bracket to power -(r + 1). Sub in x = 0, you get f^r(0) = r!

    Corresponding term in the series is (r!/r!)x^r = x^r
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    (Original post by Physics Enemy)
    a) asks for the expansion up to general x^r, so you need to spot a pattern. b) just asks you to write out the first few terms.

    In a) its reasonable to spot r! multiplying the bracket to power -(r + 1). Sub in x = 0, you get f^r(0) = r!. Corresponding term in the McLaurin series is (r!/r!)x^r = x^r.
    hhmm ok I see where you're coming from. Also do you know what f^{(r)}(x) = r(r-1) .... is about?
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    (Original post by ManLike007)
    hhmm ok I see where you're coming from. Also do you know what f^{(r)}(x) = r(r-1) .... is about?
    That's r!, to justify spotting it they wrote a few factors. e.g) f^6(x) = 6 x 5 x ... 2(1 - x)^(-7) = 6!(1 - x)^(-7)
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    (Original post by Physics Enemy)
    That's r!, to justify spotting it they wrote out a few factors.
    So e.g) f^6(x) = 6 x 5 x ... 2(1 - x)^(-7) = 6!(1 - x)^(-7)
    Oh right I see now, never thought of it that like that.

    Also, why is x^r sometimes used and sometimes not? i.e in the photo I posted at the start of this thread, part a had an x^r in the expansion but not in part b? This also happens later in the exercise that I'm doing.
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    (Original post by ManLike007)
    So in the Maclaurin expansion, it goes as follows

    f(x) = f(0) + f'(0)x + \frac {f''(0)}{2!} x^2 + .... + \frac {f^{r}(0)}{r!}x^r + ...

    The problem that I've encountered is that I don't know how to find f^{r}(x) so I can't complete the expansion or am I missing something? Perhaps I should be spotting a pattern?
    Name:  fp2 maclaurin.jpg
Views: 18
Size:  53.4 KB
    Like how do I get the general term as shown in the picture?
    There's a mistake in the solution. The fourth line should read 6(1-x)^(-4)=3!(1-x)^(-4). We can see that f^2(x) = 2!(1-x)^(-3) and f'(x) = 1!(1-x)^(-2).

    If we carried on like that, we would get a pattern that f^n(x) = n!(1-x)^-(n+1). Using a Maclaurin expansion will then yield the required result.
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    (Original post by ManLike007)
    Oh right I see now, never thought of it that like that.

    Also, why is x^r sometimes used and sometimes not? i.e in the photo I posted at the start of this thread, part a had an x^r in the expansion but not in part b? This also happens later in the exercise that I'm doing.
    Probs because for some Qs they want/expect you to find the series up to general x^r, for others they don't (pattern may be too obscure / non existant).

    Desmos: there's no error in line 4 of their solution, 3.2 = 6 in terms of factors.
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    (Original post by Physics Enemy)
    Probably because for some Qs they want/expect you to find the series up to general x^r, for others they don't (pattern may be too obscure / non existant).

    Desmos - there's no error in line 4 of their solution, 3.2 = 6 in terms of factors.
    Perhaps yes I can see why you'd say that, I just found it odd why some series has x^r

    Anyway thanks for your help. I just noticed you edited one of your messages about the f^{(6)}(x) example. I get it now, simplifying the numbers into factorials to spot a pattern where possible.

    Massive thanks for your help!
 
 
 
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