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billypilgrim
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September 28th, 2014, 12:54 pm

Here's a good question - may be easy if you have enough time, but decent enough for an interview- Which is greater - e^(pi) or (pi)^e ? You don't know the value of ln(pi)
Last edited by billypilgrim on September 27th, 2014, 10:00 pm, edited 1 time in total.
 
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Cuchulainn
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September 28th, 2014, 4:29 pm

Doing a bit of real analysis on e^(x) - (x)^e for x >= 0 might throw up some results.
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CommodityQuant
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September 28th, 2014, 6:03 pm

Look at the logs of the two sides. Is pi > e * log(pi) ? In other words, is pi/log(pi) > e? Let's now differentiate x/log(x) and examine its value at pi. The derivative is (log(x) - 1)/log(x) ^ 2. The derivative is positive for x > e. Therefore pi/log(pi) > e/log(e) = e. Yes, pi > e * log(pi). This means that exp(pi) > pi ^ e.I'm not sure how to get Cuchulainn's more direct approach to work. When I tried that way, to analyse the derivative, I wanted to know the answer to the original problem, so I was going around in circles. (I've got a derivative, can't find the behaviour. I'm gonna try another approach. I've got a derivative, can't find the behaviour. I'm gonna try another approach. Will it go round in circles? Will it fly high like a bird up in the sky? etc.)CommodityQuant
 
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Cuchulainn
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September 28th, 2014, 6:45 pm

QuoteOriginally posted by: CuchulainnDoing a bit of real analysis on e^(x) - (x)^e for x >= 0 might throw up some results.When x = e we get equality. When x > e is it monotonically increasing?A graph is cheating but useful here (TYPE in exp(x) - x^271..)
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CommodityQuant
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September 28th, 2014, 6:55 pm

QuoteWhen x > e is it monotonically increasing? Yes.I shouldn't have said that this method is not straightforward. As a punishment to myself for that, I'll abstain from strawberry-banana smoothies for the next week.CommodityQuant
 
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Cuchulainn
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September 28th, 2014, 6:59 pm

QuoteOriginally posted by: CommodityQuantQuoteWhen x > e is it monotonically increasing? Yes.I shouldn't have said that this method is not straightforward. As a punishment to myself for that, I'll abstain from strawberry-banana smoothies for the next week.CommodityQuantDon't be hard on yourself.BTW why is it monotonically increasing (I have a feeling there's a long way and a short way ) =====The point x = e is a local minimum of exp(x) - x^e.
Last edited by Cuchulainn on September 28th, 2014, 10:00 pm, edited 1 time in total.
 
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billypilgrim
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September 28th, 2014, 10:22 pm

Well, this is how I thought of it - Raise both sides to the power of 1/(pi*e). So now you have, pi^(1/pi) and e^(1/e). That's x^(1/x). Find the maxima. Haven't actually solved the equation to get the answer. Will do it some time.
 
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billypilgrim
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September 28th, 2014, 10:42 pm

CQ's answer seems right too. Let me see if my solution matches. Btw, a friend was asked this at an interview (not a quant interview)Have to go now
 
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Cuchulainn
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September 29th, 2014, 7:22 am

I think you can prove pi/log pi > e/log e by using Tchebychev's theorem on the distribution of prime numbers:e^pi > pi^e<==>pi/log pi > e/log eDefine f(x) = x/log x Then f(x) is increasing for x > e. (df/dx = (log x - 1) /(log x)^2)Which is of course CQ's answer as well.QED // When x -> infinity, x/log x is the number of prime numbers that do not exceed x.
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CommodityQuant
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September 29th, 2014, 9:32 am

QuoteOriginally posted by: CuchulainnI think you can prove pi/log pi > e/log e by using Tchebychev's theorem on the distribution of prime numbers:QuoteCuchulainn, I strongly disagree with you here. Your proof (a more formal rewriting of my proof) makes no use whatsoever of the distribution of prime numbers and makes no use of the behaviour of x/log(x) as x tends to infinity.CommodityQuant
 
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billypilgrim
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September 29th, 2014, 9:34 am

Btw, I tried using what I had suggested before- x^1/x attains maxima at e. Hence e^1/e > pi^1/pi, and therefore, e^pi > pi^e
 
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Cuchulainn
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September 29th, 2014, 1:11 pm

QuoteOriginally posted by: CommodityQuantQuoteOriginally posted by: CuchulainnI think you can prove pi/log pi > e/log e by using Tchebychev's theorem on the distribution of prime numbers:QuoteCuchulainn, I strongly disagree with you here. Your proof (a more formal rewriting of my proof) makes no use whatsoever of the distribution of prime numbers and makes no use of the behaviour of x/log(x) as x tends to infinity.CommodityQuantI probably was not clear. The problem boils down to pi/log pi > e/log e which we see is true because there are more primes < pi than there are < e.
 
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Alan
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September 29th, 2014, 5:35 pm

x-1 > log x for all x > 1. Now let x = pi/e
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CommodityQuant
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September 29th, 2014, 6:31 pm

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: CommodityQuantQuoteOriginally posted by: CuchulainnI think you can prove pi/log pi > e/log e by using Tchebychev's theorem on the distribution of prime numbers:QuoteCuchulainn, I strongly disagree with you here. Your proof (a more formal rewriting of my proof) makes no use whatsoever of the distribution of prime numbers and makes no use of the behaviour of x/log(x) as x tends to infinity.CommodityQuantI probably was not clear. The problem boils down to pi/log pi > e/log e which we see is true because there are more primes < pi than there are < e.Cuchulainn,I think you were closer to being wrong than being unclear. The prime number theorem is about behaviour as n tends to infinity so you can't apply it in this case -- that's why your reasoning is simply wrong, unless you're using an idiosyncratic version of the prime number theorem.CommodityQuant
 
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savr
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September 30th, 2014, 8:18 am

Quote unless you're using an idiosyncratic version of the prime number theorem.Whereas the subprime number theorem had systemic underpinnings and implications.