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Cuchulainn
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exp(5) = [$]e^5[$]

September 21st, 2010, 9:30 am

And if all else fails, there is always the slide rule.
Last edited by Cuchulainn on September 20th, 2010, 10:00 pm, edited 1 time in total.
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Cuchulainn
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exp(5) = [$]e^5[$]

September 21st, 2010, 10:30 am

(double post)
Last edited by Cuchulainn on September 20th, 2010, 10:00 pm, edited 1 time in total.
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Cuchulainn
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exp(5) = [$]e^5[$]

September 21st, 2010, 10:31 am

Another way is to use Newton Raphsonexp(5) = x <==> log(x) = 5ThenTake Zerdna's x(0) = 120 x(n+1) = x(n) - x(n){log(x(n)) - 5}this converges in 3 iterations to 148.413. You need to have the log tables/slide rule at your side. Consult it three times.
Last edited by Cuchulainn on September 20th, 2010, 10:00 pm, edited 1 time in total.
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zerdna
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exp(5) = [$]e^5[$]

September 21st, 2010, 11:18 am

well, remembering that e^3 ~ 20 is a smaller trick than remembering bunch of logs of rational numbers for Newton Rapson. It's just some sugar coating on a very simple approach of expanding e^5 ~ 3^5*(1-0.1)^5 as Peniel pointed out. It's something though that is appreciated by the type of interviewer who asks this question. This is old school, you want numerical answer for an integral, a DE solution, etc while you are hiking in the forest with your buddies and drinking vodka. There is no pens and calculators -- you have advantage if you remember that 1/sqrt(2*pi)~ 0.4, e^3~ 20, what log(2) is, etc. It's some old physics guy who is asking this.
 
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Peniel
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exp(5) = [$]e^5[$]

September 21st, 2010, 11:24 am

outrun,e > 2.70 --> e^5 > 3^5 * (1 - 0.1)^5 ~ 145e < 2.73 --> e^5 < 3^5 * (1-0.09)^5 ~ 3^5*(1 - 0.45 + 0.081) ~ 153rounding half up -> 150rounding half down -> 140(...you still need to know that e ~ 2.71...)
 
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Cuchulainn
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exp(5) = [$]e^5[$]

September 21st, 2010, 11:28 am

QuoteOriginally posted by: zerdnawell, remembering that e^3 ~ 20 is a smaller trick than remembering bunch of logs of rational numbers for Newton Rapson. It's just some sugar coating on a very simple approach of expanding e^5 ~ 3^5*(1-0.1)^5 as Peniel pointed out. It's something though that is appreciated by the type of interviewer who asks this question. This is old school, you want numerical answer for an integral, a DE solution, etc while you are hiking in the forest with your buddies and drinking vodka. There is no pens and calculators -- you have advantage if you remember that 1/sqrt(2*pi)~ 0.4, e^3~ 20, what log(2) is, etc. It's some old physics guy who is asking this.You're kidding me when you remember 1/sqrt(2*pi)~ 0.4, e^3~ 20. Don't forget e = 2.71..... as well.There are millions of numbers like this. And we don't need the Ansatz e = 2.71 (kind of cheating, yes?)
Last edited by Cuchulainn on September 20th, 2010, 10:00 pm, edited 1 time in total.
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Peniel
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exp(5) = [$]e^5[$]

September 21st, 2010, 12:24 pm

Cuchulainn you're right, that's cheating.e > 1 + 1 + 1/2 + 1/6 + 1/24 = 2+17/27 > 2.708Let [$]R=\sum_{n=5}^{\infty} \frac{1}{n!}[$]We want R<10^-2[$]R = \frac{1}{120} \sum (1 + \frac{1}{6} + \frac{1}{6*7} +...)[$]
Last edited by Peniel on May 7th, 2013, 10:00 pm, edited 1 time in total.
 
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zerdna
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exp(5) = [$]e^5[$]

September 21st, 2010, 12:51 pm

QuoteYou're kidding me when you remember 1/sqrt(2*pi)~ 0.4, e^3~ 20. Don't forget e = 2.71..... as well.There are millions of numbers like this. And we don't need the Ansatz e = 2.71 (kind of cheating, yes?)I don't remember million numbers, about 10 numbers i guess -- Pi, e, sqrt(2), sqrt(3), log(2), log(10), this kind of things. If you traded derivatives or were a derivatives quant, you'd know that 1/sqrt(2*pi)~ 0.4. When you need to calculate Black Scholes approximately in your mind few times, you know. I am not that good with remembering logs, i remember log(2) ~ 0.7. To make sure i remember log(10) i use checksum log(2)+log(10) ~ 3. It's the same equality as the one that i used -- e^3~20, expresses the sum of two most important logs. As i said, there are indeed millions of numbers to remember in your approach with Newton Rapson, i advocate just remembering one or two.
 
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eugenek2010
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exp(5) = [$]e^5[$]

September 22nd, 2010, 4:51 am

dictionary.reference.com and any number of online sources define 'decimal place' as 'The position of a digit to the right of the decimal point'. The answer 14x.xx would have been sufficient if the task was to compute e^5 to two "significant digits". Like Cuchulainn said, the way to go is to start with e=2.7 1828 1828 and to do long multiplication. Cumbersome, but doable in 5 min with pen and paper.
 
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Cuchulainn
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exp(5) = [$]e^5[$]

September 22nd, 2010, 1:39 pm

QuoteOriginally posted by: Penieloutrun,e > 2.70 --> e^5 > 3^5 * (1 - 0.1)^5 ~ 145e < 2.73 --> e^5 < 3^5 * (1-0.09)^5 ~ 3^5*(1 - 0.45 + 0.081) ~ 153rounding half up -> 150rounding half down -> 140(...you still need to know that e ~ 2.71...)Maybe we can do it a bit differently. Let's say we have lost short term memory (like pin codes, e, pi, log(2) ) which can happen but we know the series expansion for e. Then we can reconstruct e to any accuracy using these approaches (you need up to e8)The calculation of e to many significant digitsBTW we need another budget for this because it will take > 5 minutes to do by long hand because of the many divisions (but they are by integers which is not too bad).
Last edited by Cuchulainn on September 21st, 2010, 10:00 pm, edited 1 time in total.
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
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AVt
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Joined: December 29th, 2001, 8:23 pm

exp(5) = [$]e^5[$]

September 22nd, 2010, 6:06 pm

So the actual task still is not formulated ... is exp(1) given or not?I have doubts that any company invests time to see several candidatesdoing lengthy, but senseless computations by hand.And if so: then at least HR would expect those to ask "WTF?" to avoidmindless robots.For me 'two decimal places' is more towards 'have a cross check for anerror of 1%', so perhaps even the question of 14* or 15* is irrelevant.PS: 'to the right of the decimal point' merely makes sense and I neverhave seen that, it is meant for the leading digits or what one shouldexpect for something like exp(41), even if ignoring rounding conventions?
 
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Cuchulainn
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exp(5) = [$]e^5[$]

September 22nd, 2010, 6:26 pm

QuoteSo the actual task still is not formulated ... is exp(1) given or not?If it is, then we are finished. If not we need a formula; here is a classic lim (1 +n)^1/n n-> 0So take n = 0.001 etc. QuoteIn that case, I showed that 2.70<e<2.73 and it implies that 145<e<153 (and no, exp(1) wasn't given).Good job they did not ask e(-5).
Last edited by Cuchulainn on September 21st, 2010, 10:00 pm, edited 1 time in total.
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
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