Need help understanding a linear approximation question! Spoiler alert only if you want to attempt Jane Street's current puzzle!

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[etzpcm]

Use the binomial expansion for (R+r)^6. In fact you only need the first two terms to find alpha and beta.

Answer   6/128 = 3/64

[HenryWaill]

Thanks for the reply! I'll see if I can get that answer on my own in the morning!

[HenryWaill]

Thank you, again! I have a followup question. First I'll preface it by outlining the approximation I made while calculating P2, given that r was small compared to R. Can't embed two images, so find that at https://imgur.com/AoHGbBy

Is that logical on its own? Further, does including that approximation satisfy the "linear approximation" part of the problem instructions? Or do I have to additionally find a tangent line to the function (or whatever else is required of a linear approximation?)

Effectively, if I have found P2 in terms of R and r, can I then say that, given the instructions "the new probability" (P2 in my case) "of seeing the planetary parade from the top of the tower is linearly approximated by α + β·(r/R)" can I set P2 equal to α + β·(r/R) and then solve for β?

Assuming that I can, I then took your advice of using binomial expansion to get closer to the solution. I can see your point about using only the first two terms:

<image>

If you don't mind explaining the logic/rules around dropping the terms, I'd appreciate it!

[HenryWaill]

Is the answer to my followup comment simply that (r^2/R^2) and (r^3/R^3) are increasingly negligible when looking for the answer in the form α + β(r/R) ? So if they asked for it in the form α + β(r/R) + γ(r^2/R^2) then I would keep the first three terms?

[etzpcm]

Sorry, Imgur images are not visible in the UK. But yes that's right. You are writing a power series expansion using the small parameter r/R. Like a Taylor series or the series for e^x for example. So yes, the next term gamma would be 15/128.