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(edited 1 month ago)
Reply 1
Original post by username17263528
https://isaacphysics.org/questions/maclaurin_series_dipole12?stage=a_level
I’ve got part b, but I’m not sure how to solve part a as every time I remove a as a term, I get 0. For example, for the first derivative to try to get the Maclaurin series, I have 2q/(4pi Epsilon_0)(-2/(2z - a)^2 + 2/(2z + a)^2), which then equals zero. Would anyone please be able to help?

The key thing is that they want an expansion in |z|>>1 (say) rather than the usual |z|<1, so you have (simplified)
1/(z-1) - 1/(z+1)
so factor out a z on the (both) denominators and you should be able to use the usual binomial (maclaurin) results in 1/z. If you combined the terms before doing the expansion, it would be clear why its an expansion in even power terms (and its a bit simpler).
(edited 1 month ago)
Reply 2
Original post by username17263528
So should I be trying to do the Maclaurin expansion on q/(4 pi epsilon_0) (1/z) (1/(1 - 1/z) - 1/(1 + 1/z)), or have wrongly I interpreted what you are saying (highly possible)? Thanks

Its easier to just do a binomial and as youre after the first non-zero term combine the terms together first then just look at the constant term. So
1/(z-1) - 1/(z+1) = 2/(z^2 - 1) = 2/z^2 (1 - 1/z^2)^(-1) = ...
(edited 1 month ago)
Original post by mqb2766
Its easier to just do a binomial and as youre after the first non-zero term combine the terms together first then just look at the constant term. So
1/(z-1) - 1/(z+1) = 2/(z^2 - 1) = 2/z^2 (1 - 1/z^2)^(-1) = ...

Easier still to just treat as a GP, surely?
Reply 4
Original post by DFranklin
Easier still to just treat as a GP, surely?

Whichever way you look at it, its just .... as you just need the first term (constant).
Reply 5
Original post by username17263528
Sorry, I’m still a bit stuck. For my binomial expansion of 2/z^2(1 - 1/z^2)^-1, I’ve got 2/z^2 (1 + 1/z^2 + 1/z^4 + 1/z^6), and I’m not sure how to get a non-zero term out of this as none of the z bits are cancelling.
The first term in the brackets is 1 so the first nonzero term in the series is 2/z^2. Dfranklin was making the observation that 1/(1-(1/z^2)) is just a geometric series (infinite sum) with a=1 and r=(1/z^2) which is clear from your expansion, or you could expand the two individual fractions 1/(z-1) and 1/(z+1) in a similar fashion(s).

Note in the series there is no constant term (the value of the original function, when z is large, must be zero) and all the odd powers are zero, so the first nonzero term is 1/z^2
(edited 1 month ago)
Reply 6
Original post by username17263528
Thank you both, this all makes sense now and I can see where everything has been derived from. However, I still don’t have the correct answer as I have tried to multiply 1/z^2 by the constant given in the equation for V(t), q/(4 pi epsilon_0), to get the first non-zero term as q/(2 pi epsilon_0 z^2) and this is incorrect. I have also tried answers 1/z^2, 2/z^2 and q/(4 pi epsilon_0 z^2), which are incorrect as well.

Youd have to post your working, but obviously the 1/(z+1)-1/(z-1) was just to get the method right.
(edited 1 month ago)
Reply 7
Original post by username17263528
It’s not letting me attach a photo, but I’ve got
q/(4 pi epsilon_0) (1/(z - a/2) - 1/(z + a/2))
= q/(4 pi epsilon_0) ((z + a/2 - (z - a/2)/(z + a/2)(z-a/2))
= q/(4 pi epsilon_0) (a/(z^2 - a^2/4)), which is where I have got stuck, as surely removing a on the numerator just makes the second bracket, and therefore the whole thing, always equal to zero?

Its basically right, but I dont know what you mean by "removing a". You have
#*(a/$)
and you must be able to pull a outside? The numerator can be written as a*1, if you really need that hint. Then pull the z^2 out the denominator.

An alternative along Dfranklins hint is to pull the z^2 out the denomintor and rewrite it as
#*((a/z^2)/(1-a^2/(4z^2)))
and the first term in the geometric series can simply be read.
(edited 1 month ago)
As a general comment, I find it more intuitive in a question like this to write y = a/z and then rewrite in the form z^k * (expression only involving y) for a suitable choice of k.

In particular, expanding something of the form 1/(1+cy) or 1/(1-cy) should be automatic. You can expand more complicated expressions but it's harder and I feel there's a fair amount of confusion going on because of this.

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