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I want to take the limit of log(-(5^(1/2)*((-347922205179541/562949953421312)^n - (910872158600853/562949953421312)^n as n tends to infinity, but the answer in matlab always return the the whole thing,can anyone tell me how to get the exact answer?
The command i put in was
limit(log(-(5^(1/2)*((-347922205179541/562949953421312)^n - (910872158600853/562949953421312)^n))/5)/n,Inf)

And the answer i got from matlab was
ans =

limit(log(-(5^(1/2)*((-347922205179541/562949953421312)^n - (910872158600853/562949953421312)^n))/5)/n, n = Inf)
kanezila
Newcomer

Posts: 10
Joined: Sun Nov 20, 2011 8:38 pm

Your first formula doesn't have matching parentheses, so I don't know what it's supposed to mean. Generally, it looks like you're basically taking the limit of log(a^n - b^n) with a few extra frills going on. This will be the same as the limit of log(a^n) as long as a > b, since the other factor will add at most a hyperbolic correction factor. Of course this can be simplified to n log a.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/pub/pari ... -2-6-2.exe

CRGreathouse
Global Moderator

Posts: 12566
Joined: Sat Nov 25, 2006 9:52 am
Location: UTC -5

I understand what you mean, but can you tell me how to get an exact answer, like ans=2134 or something?
kanezila
Newcomer

Posts: 10
Joined: Sun Nov 20, 2011 8:38 pm

I don't remember MatLab well enough to do that, sorry. It's been years since I last used it, and I wasn't even all that good then.
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/pub/pari ... -2-6-2.exe

CRGreathouse
Global Moderator

Posts: 12566
Joined: Sat Nov 25, 2006 9:52 am
Location: UTC -5

Perhaps, you have compiled the matlab software incorrectly.
http://www.topckit.com