Problem 235

suspect42 · Post by **suspect42** » Tue Mar 10, 2009 10:03 am

I found a closed-form for the sum, and tried to solve the equation that results for r. Using a few different solvers (Maxima and a hand-coded Newton-method), I get the same result, but it seems not to be the correct answer.

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Is this close to the correct answer?

TripleM · Post by **TripleM** » Tue Mar 10, 2009 10:47 am

suspect42 · Post by **suspect42** » Tue Mar 10, 2009 12:16 pm

Thanks for replying!

Got the answer. I'd made a mistake copying the expression from notebook to program.

Very embarrassing

Smaug · Post by **Smaug** » Sat Dec 11, 2010 10:22 pm

I have a result to 12 decimal places, but it's not accepted. It's definitely right, as well, says Mathematica when I substitute the result back in. It's a little greater than 1.00. Does this sound roughly right?

TripleM · Post by **TripleM** » Sat Dec 11, 2010 10:52 pm

Yes, but if Project Euler says wrong answer then it is definitely the wrong answer

If you want to PM me the value you come up with I can tell you how many decimal places it is correct to (but it won't be 12).

maomaoloverose · Post by **maomaoloverose** » Fri Mar 04, 2011 1:59 am

I have a result. it's a little greater than 1.001, and Abs(s(5000) - (-600000000000)) < 0.000000046, but it's not accepted.

i made a stupid mistake!

TripleM · Post by **TripleM** » Fri Mar 04, 2011 2:03 am

That's because it's wrong

The correct answer doesn't begin with 1.001.. if that's what you were implying by 'a little more'.

mdean · Post by **mdean** » Sun Oct 23, 2011 5:16 am

Before I get started on this, is it fair to ask if this number to that many decimal places will fit in a C++ long double? Somehow I get the horrible feeling even if it fits, the limit of its precision is going to be seriously pushed...

thundre · Post by **thundre** » Sun Oct 23, 2011 6:46 am

mdean wrote:Before I get started on this, is it fair to ask if this number to that many decimal places will fit in a C++ long double? Somehow I get the horrible feeling even if it fits, the limit of its precision is going to be seriously pushed...

The result fits in a Java double, which is smaller than a C++ long double.

mdean · Post by **mdean** » Sun Oct 23, 2011 9:25 am

Ah. Good to know that shouldn't be a problem then. Thanks.

mdean · Post by **mdean** » Wed Oct 26, 2011 7:19 am

How many roots are there? I get one around -1 that isn't being accepted.

TripleM · Post by **TripleM** » Wed Oct 26, 2011 8:16 am

Your answer is incorrect

(You may want to re-read the problem in case you missed something.)

mdean · Post by **mdean** » Wed Oct 26, 2011 8:55 am

Nope. Didn't miss anything, just reversed a sign.

I saw 2 roots before the change. Right now, I just see one obvious one.

enderw88 · Post by **enderw88** » Mon Mar 05, 2012 8:37 pm

Driving me crazy. I have the same answer to 14 significant figures from two separate sources (Mathematica and some custom Lisp) At 12 significant figures I can't get |600,000,000 + s(r)|< 1.03

At 14 significant figures I can drive the difference to below 1x10^-4. I must be making the same mistake twice...The answer is of the form 1.abc def ghi jkl correct?

jaap · Post by **jaap** » Mon Mar 05, 2012 10:43 pm

enderw88 wrote:600,000,000

Count the zeroes in that number.

enderw88 · Post by **enderw88** » Tue Mar 06, 2012 1:37 am

jaap wrote:
enderw88 wrote:600,000,000
Count the zeroes in that number.

Thank you. That's what I get for doing a problem on a trans-atlantic flight...

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