I want to know how to calculate this finite sum
sum of i=1 to 52 of (i^2/3)*1/60
what method can I use for this?
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...
$\displaystyle S= \sum_{i=1}^{52} i^{\frac{2}{3}$ (1)
.... then, if You transform the (1) in
$\displaystyle S=1+\sum_{i=2}^{n} i^{\frac{2}{3}} = 1 + \sum_{k=1}^{n-1}(1+k)^{\frac{2}{3}}$ (2)
..., apart the 'direct computation' of course, You can consider to use of the 'Euler-McLaurin formula'...
$\displaystyle \sum_{k=1}^{n-1} f(k)= \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} + \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} -$
$\displaystyle -\frac{1}{720}\ \{f^{(3)}(n)- f^{(3)}(0)\} + \frac{1}{30240}\ \{f^{(5)}(n)-f^{(5)}(0)\} -$
$\displaystyle - \frac{1}{1209600}\ \{f^{(7)}(n)-f^{(7)}(0)\} +...$
$\displaystyle ...+ \frac{B_{2k}}{(2k)!}\ \{f^{(2k-1)}(n)-f^{(2k-1)} (0)\} +...$ (1)
... where the $\displaystyle B_{2k}$ are the 'Bernoulli's numbers'. In Your case is $\displaystyle f(t)= (1+t)^{\frac{2}{3}}$ and $\displaystyle n=52$. The computation of the $\displaystyle f^{(2k-1)}(*)$ 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
$\displaystyle \chi$ $\displaystyle \sigma$
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.
If you don't use $\displaystyle \LaTeX$ , be careful with brackets
sum of i=1 to 52 of (i^2/3)*1/60 means $\displaystyle \sum_{i=1}^{52}\frac{i^2}{3}\cdot \frac{1}{60}$
sum of i=1 to 52 of ( i^(2/3) )*1/60 means $\displaystyle \sum_{i=1}^{52}i^{\frac{2}{3}}\cdot \frac{1}{60}$
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...
$\displaystyle S=\sum_{i=1}^{52} i^{\frac{2}{3}}$ (1)
... You have two possibilities...
a) if a computer is available for You the 'brute force sum' gives as result...
$\displaystyle S_{52,BF}= 441.492660573...$
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 $\displaystyle S_{n, EMc}$ obtained with the Euler-McLaurin formula 'breaked up' to the third term...
$\displaystyle S_{n,EMc}= 1 + \int_{0}^{n} f(t)\ dt -\frac{1}{2}\ \{f(0)+f(n)\} +$
$\displaystyle + \frac{1}{2}\ \{f^{'}(n)-f^{'}(0)\} - \frac{1}{720}\ \{f^{'''}(n)-f^{'''}(0)\} $ (2)
... where is $\displaystyle f(t)=(1+t)^{\frac{2}{3}}$. If You insert in (2) n=52 You obtain...
$\displaystyle S_{52,EMc}= 441.333255853...$
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...
$\displaystyle S_{100,BF}= 1303.28975964...\ ,\ S_{100,EMc}= 1303.10747032...$
$\displaystyle S_{1000,BF}= 60049.8503587...\ ,\ S_{1000,EMc}= 60049.6170634...$
$\displaystyle S_{10000,BF}= 2.78518522699... 10^{6}\ ,\ S_{10000,EMc}= 2.78518496989... 10^{6}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$