hmm seems I'm getting an incorrect answer. Can I split i^2/3 as (i^2)^1/3 and then use the i^2 formula? I don't know if that is the same. I figure my wrong answer is because I used the i^2 formula, plugged in n as 52 and cubed rooted that number, then divided by 60.
If the sum to be computed is...
(1)
.... then, if You transform the (1) in
(2)
..., apart the 'direct computation' of course, You can consider to use of the 'Euler-McLaurin formula'...
(1)
... where the are the 'Bernoulli's numbers'. In Your case is and . The computation of the is left to You. It is difficult to extablish for what n the Euler-McLaurin formula [which of course is 'approximative'...] is 'competitive' with the direct computation...
Kind regards
I still don't get it...do I have to use the euler formula for this? Wlfram alpha gave me the answer but my manual calculation is incorrect.
help? This is what I did
1/60 [(n)(n+1)(2n+1)÷6]^1/3 and plug n in as 52.
Yes it is the second one. I can't use the i^2 formula on it. How do I solve this?? I've been trying to figure out if fits into some sort of series like power series but all of those are infinite that involve convergence/divergence. What should I do?!
Why are You so discouraged?... Your task is the computation of...
(1)
... You have two possibilities...
a) if a computer is available for You the 'brute force sum' gives as result...
b) if only a [scientific...] calculator and limited time is available for You, the Euler-McLaurin formula, that I described few post ago, may be useful...
Now we write the value of obtained with the Euler-McLaurin formula 'breaked up' to the third term...
(2)
... where is . If You insert in (2) n=52 You obtain...
If more precision is requested, You can use more terms of the Euler-McLaurin formula. May be interesting to compare the 'brute force sum' and the 'Euler-McLaurin third degree sum' for greater values of n...
Kind regards