- October 26th, 2010, 8:05 am
- Forum: Careers Forum
- Topic: Is the crisis really that bad?
- Replies:
**29** - Views:
**26511**

<t>I found a job after looking for a few months. I don't know where you're based, but I found it really tough to get an interview (almost all the headhunters were useless, a few on-line applications worked).Anyway, as some people said, you have a very diverse experience, and I believe this works aga...

- September 22nd, 2010, 6:35 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

That was with a well-known American bank in London.No, I didn't get the job, so I'm still looking...

- September 22nd, 2010, 6:26 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

<t>Just to clarify: when I did this test, I think I misunderstood the question: I thought they expected 148.41** .But I don't think it's sensible (nor very clever) to ask for a such accuracy, I think the idea was just to show that you can come up quickly with a rough approximation of e^5.I believe t...

- September 21st, 2010, 12:24 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

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[$]R = \frac{1}{120} \sum (1 + \frac{1}{6} + \frac{1}{6*7} +...)[$]

- September 21st, 2010, 11:24 am
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

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...)

- September 21st, 2010, 9:19 am
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

<t>Thanks you for your answers guys.English isn't my mother tongue, so maybe AVt is right -> 2 decimal places means 14*.*** (and not 148.41***).In this case, I like Zerdna's answer (you need to know that e^3~20, though), or I guess you can just take e~2.7 and compute directly e^5 ~ (2.7)^5.Or someth...

- September 19th, 2010, 5:02 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

Thanks AVt for your answers and you're right: this is probably "first-year undergrad stuff".Still, the answer is 148.41 and I don't know how to get there in 5mins

- September 19th, 2010, 4:17 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

OK, a little misunderstanding here, I thought you meant the exponential base.Anyway, the question is: how to compute exp(5) in 5mins with a pen and a piece of paper?

- September 19th, 2010, 1:56 pm
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

<t>Sorry for being unclear. Forget the two digits, I meant two decimal places.This was actually from a quant test (with an American bank), so pen and paper. Expected time to do it: 5-7mins.The base is exp(1).I thought about using Taylor series directly (something like exp(5) = exp(1)^5 = (1+1/2+1/6+...

- September 19th, 2010, 10:43 am
- Forum: Brainteaser Forum
- Topic: exp(5) = [$]e^5[$]
- Replies:
**546** - Views:
**127635**

How to compute [$]e^5[$] with two decimal places?

- May 26th, 2009, 10:00 am
- Forum: Brainteaser Forum
- Topic: Binomial Sum
- Replies:
**4** - Views:
**46061**

OK:p=5

- May 22nd, 2009, 2:03 pm
- Forum: Brainteaser Forum
- Topic: Binomial Sum
- Replies:
**4** - Views:
**46061**

What does "a numerator of sum(....)" mean in this case?

- July 22nd, 2008, 4:59 pm
- Forum: Brainteaser Forum
- Topic: Expectation at Infinity
- Replies:
**15** - Views:
**53293**

another try:with:u_1 = 2u_{k+1} = a*u_k + 1

- July 22nd, 2008, 4:18 pm
- Forum: Brainteaser Forum
- Topic: Expectation at Infinity
- Replies:
**15** - Views:
**53293**

I have a try: if a=1 then E[X_i]=(1/2)*(1/3)*...*(1/i)=1/i! --> 0 when i->+infinity <edit: mistake corrected>

- January 6th, 2008, 4:49 pm
- Forum: Brainteaser Forum
- Topic: Fibonacci(i+j-1)
- Replies:
**7** - Views:
**67190**

It definitely helps.Seems I'm more comfortable with big summations than with basic arithmetic.Anyway, it's time to tackle the case k>1.

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