The hardest binomial expansion question i could find help

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psycholeo
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Keep in mind this is the hardest question I could find for A-Level maths. But I got so impossibly stuck. I'd be shocked if this isn't A* level but I have no idea for what to do from b) onwards. I know student room is used for hints but I was wondering if someone could work out the answer and then work through how they got it. Thanks!!

18. In special relativity the energy of an object with mass m and speed v is given by:

E = (mc^2)/(1-(v^2)/c^2)^0.5
where c = 3 x 10^8 ms^-1 is the speed of light.

a) Find the first three non-zero terms of the binomial expansion, in increasing powers of v, stating the range of which it it valid.
b) Let E2 be the expansion containing two terms and E be the expansion containing three terms. By what percentage if E3 bigger than E2 if v is:
i) 10% of the speed of light
ii) 90% of the speed of light.
c) Prove E > E3 > E2
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RDKGames
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(Original post by psycholeo)
Keep in mind this is the hardest question I could find for A-Level maths. But I got so impossibly stuck. I'd be shocked if this isn't A* level but I have no idea for what to do from b) onwards. I know student room is used for hints but I was wondering if someone could work out the answer and then work through how they got it. Thanks!!

18. In special relativity the energy of an object with mass m and speed v is given by:

E = (mc^2)/(1-(v^2)/c^2)^0.5
where c = 3 x 10^8 ms^-1 is the speed of light.

a) Find the first three non-zero terms of the binomial expansion, in increasing powers of v, stating the range of which it it valid.
b) Let E2 be the expansion containing two terms and E be the expansion containing three terms. By what percentage if E3 bigger than E2 if v is:
i) 10% of the speed of light
ii) 90% of the speed of light.
c) Prove E > E3 > E2
OK so I'm sure you got the following for part (a)

E = mc^2 + \dfrac{m}{2}v^2 + \dfrac{3m}{8c^2} v^4 + O(v^6)

So, then we have

E_2 = mc^2 + \dfrac{m}{2}v^2

E_3 = mc^2 + \dfrac{m}{2}v^2 + \dfrac{3m}{8c^2} v^4

Now if v= 0.1c, then

E_2 = \dfrac{mc^2}{10} + \dfrac{mc^2}{2000} = \dfrac{201}{2000} mc^2

E_3 = \dfrac{mc^2}{10} + \dfrac{mc^2}{2000} + \dfrac{3mc^2}{800000} = \dfrac{80403}{800000} mc^2

So determine the % increase.
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psycholeo
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(Original post by RDKGames)
OK so I'm sure you got the following for part (a)

E = mc^2 v + \dfrac{m}{2}v^3 + \dfrac{3m}{8c^2} v^5 + O(v^7)

So, then we have

E_2 = mc^2 v + \dfrac{m}{2}v^3

E_3 = mc^2 v + \dfrac{m}{2}v^3 + \dfrac{3m}{8c^2} v^5

Now if v= 0.1c, then

E_2 = \dfrac{mc^3}{10} + \dfrac{mc^3}{2000} = \dfrac{201}{2000} mc^3

E_3 = \dfrac{mc^3}{10} + \dfrac{mc^3}{2000} + \dfrac{3mc^3}{800000} = \dfrac{80403}{800000} mc^3

So determine the % increase.
Not sure if that's correct, think E = mc^2 + 0.5mv^2 + (3mv^4)/8c^2. And I've tried to work out the percentage increase, E3/E2. I ended up with 1 + (6v^4)/(8c^2v^2 + 8c^4), but wasn't sure how to simplify further
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RDKGames
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(Original post by psycholeo)
Not sure if that's correct, think E = mc^2 + 0.5mv^2 + (3mv^4)/8c^2.
Yeah you're right, I just copy and pasted your formula into Wolfram Alpha and it decided to slap on an extra v at the end for some reason.
Post amended, barely anything changes.

And I've tried to work out the percentage increase, E3/E2. I ended up with 1 + (6v^4)/(8c^2v^2 + 8c^4), but wasn't sure how to simplify further
Well I'm not sure how you've arrived at that nonsense. The v's shouldn't even be there -- look at what I did.

EDIT: OK, I suspect you did the division before the substitution. In which case,

\dfrac{E_3}{E_2} = \dfrac{mc^2 + \dfrac{m}{2}v^2 + \dfrac{3m}{8c^2} v^4}{mc^2 + \dfrac{m}{2}v^2} = 1 + \dfrac{\dfrac{3}{8c^2} v^4}{c^2 + \dfrac{1}{2}v^2} = 1+\dfrac{3v^4}{8c^4 + 4c^2v^2}
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psycholeo
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How did you get the last part? I got an 8 instead of a 4 on the bottom
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RDKGames
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(Original post by psycholeo)
How did you get the last part? I got an 8 instead of a 4 on the bottom
You didn't multiply correctly through.

Take the big fraction after the 2nd equality sign, and multiply the numerator and denominator through by 8c^2
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psycholeo
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Oh right my bad I got the b) I) correct but what did you get for b) ii)?
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RDKGames
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(Original post by psycholeo)
Oh right my bad I got the b) I) correct but what did you get for b) ii)?
Approx. 17.5%
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psycholeo
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(Original post by RDKGames)
Approx. 17.5%
That's what I got but the textbook says 17.7
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RDKGames
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(Original post by psycholeo)
That's what I got but the textbook says 17.7
Seem close enough to me.
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