hi,
there is a way in Prime to do this sum?
\[ \sum_{k=1}^{\infty}{\frac {(-1)^{k+1}}{k^{2}} } \]
the value is \( \frac {π^{2}}{12} \)
HP Prime gives symbolic form, not the value of the sum...
Thanks
Salvo
Same problem in xcas and Maxima. Wolframalpha gives the right answer.
(02-06-2015 04:47 PM)retoa Wrote: [ -> ]Same problem in xcas and Maxima. Wolframalpha gives the right answer.
yes, in fact!
As I like much more Prime (and HP 50g), I wonder why they don't...
(02-06-2015 02:26 PM)salvomic Wrote: [ -> ]hi,
there is a way in Prime to do this sum?
\[ \sum_{k=1}^{\infty}{\frac {(-1)^{k+1}}{k^{2}} } \]
the value is \( \frac {π^{2}}{12} \)
HP Prime gives symbolic form, not the value of the sum...
Thanks
Salvo
You can do
\[ \sum_{k=1}^{\infty}{\frac {-1}{(2*k)^{2}} } + \sum_{k=1}^{\infty}{\frac {1}{(2*k-1)^{2}} } \]
By the way I get the correct answer on the HP50G but my Prime seems unable to calculate Psi(1/2,1) in a numeric value.
I get :
1/4*Psi(1/2,1)-Pi²/24
Same on 50G then ->NUM returns 0.8224...
On the Prime ~ don't 'solve' Psi(0.5,1) . Strange ...
I also tried to decompose it in
\( \sum_{k=1}^{\infty}(\frac{1}{(2k-1)^2}-\frac{1}{(2k)^2}) \)
to avoid the (-1)^(k+1), but I did not get the wanted result. Still the Psi(1/2,1)
(02-06-2015 05:08 PM)Gilles Wrote: [ -> ]You can do
\[ \sum_{k=1}^{\infty}{\frac {-1}{(2*k)^{2}} } + \sum_{k=1}^{\infty}{\frac {1}{(2*k-1)^{2}} } \]
By the way I get the correct answer on the HP50G but my Prime seems unable to calculate Psi(1/2,1) in a numeric value.
thanks a lot, Gilles,
yes I see that Prime don't approx Psi1/2,1); my HP50 does it.
Hope in a next firmware to have the symbolic result (π^2/12), more interesting than Psi()
Indeed, for the approx value of Psi(x,1), Xcas calls the GSL, that is not available on the Prime.
(02-06-2015 06:55 PM)parisse Wrote: [ -> ]Indeed, for the approx value of Psi(x,1), Xcas calls the GSL, that is not available on the Prime.
I understand.
There is no other way to approximate Psi on Prime?
thank you
the problem is now solved with the firmware 7820!
Answer: -π/12