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# FP2 Maclaurin series help watch

1. So in the Maclaurin expansion, it goes as follows

The problem that I've encountered is that I don't know how to find so I can't complete the expansion or am I missing something? Perhaps I should be spotting a pattern?

Like how do I get the general term as shown in the picture?
2. 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
3. (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 is about?
4. (Original post by ManLike007)
hhmm ok I see where you're coming from. Also do you know what 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)
5. (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 sometimes used and sometimes not? i.e in the photo I posted at the start of this thread, part a had an in the expansion but not in part b? This also happens later in the exercise that I'm doing.
6. (Original post by ManLike007)
So in the Maclaurin expansion, it goes as follows

The problem that I've encountered is that I don't know how to find so I can't complete the expansion or am I missing something? Perhaps I should be spotting a pattern?

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.
7. (Original post by ManLike007)
Oh right I see now, never thought of it that like that.

Also, why is sometimes used and sometimes not? i.e in the photo I posted at the start of this thread, part a had an 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.
8. (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

Anyway thanks for your help. I just noticed you edited one of your messages about the example. I get it now, simplifying the numbers into factorials to spot a pattern where possible.

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