hey!
ive put the question as a attachment as the equation is tricky to write out.
any help would be appreciated!
maths 5.JPG
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hey!
ive put the question as a attachment as the equation is tricky to write out.
any help would be appreciated!
maths 5.JPG
This may be a booger to evaluate because of the factorials. Maybe someone has a slick method. Using L'Hopital would lead into the derivatives of factorials. I believe that would entail the psi function
For instance, the derivative of (6n)! is $\displaystyle 6{\Psi}(6n+1)(6n)!$
I can tell you the limit is $\displaystyle \frac{1}{\sqrt{3{\pi}}}$.
But getting there may prove daunting.
Use Stirling Approximation
$\displaystyle \lim_{n\to \infty} \frac{n!}{\sqrt{2\pi n}\cdot \frac{n^n}{e^n}} = 1$
So we will replace,
$\displaystyle (6n)!$ by its asyomtomatic equal $\displaystyle \sqrt{12\pi n} \cdot \frac{(6n)^{6n}}{e^{6n}} = \sqrt{12 \pi n}\cdot \frac{6^{6n}\cdot n^{6n}}{e^{6n}}$
And,
$\displaystyle [(3n)!]^2$ is replaced by $\displaystyle 6\pi n \cdot \frac{3^{6n}n^{6n}}{e^{6n}}$
Thus,
$\displaystyle \frac{64^{-n} \cdot \sqrt{n} \sqrt{12\pi n} \cdot \frac{6^{6n}n^{6n}}{e^{6n}}}{6\pi n \cdot \frac{3^{6n}n^{6n}}{e^{6n}}} = \frac{64^{-n} n \sqrt{12\pi} 6^{6n} n^{6n}}{6\pi n 3^{6n} n^{6n}} = $$\displaystyle \frac{1}{\sqrt{3\pi}} \cdot \frac{64^{-n} 6^{6n}}{3^{6n}} = \frac{1}{\sqrt{3\pi}} \cdot 64^{-n} \cdot 2^{6n} \to \frac{1}{\sqrt{3\pi}}$
Very clever PH. That's just showin' off. :);)
How did you compute this limit? Did you guess it?Quote:
Originally Posted by galactus
No, I was lazy and just ran it through my TI-92. I started using L'Hopital and the psi function , but it turned quickly into a monster. I like your Sterling thing. I have heard of it, but never took the time to put it to application. Since your post, I looked it up in one of my math books. You learned me something, PH. Good egg.
I do not own any TI calculators, are they able to tell you what the limit is in terms of certain constants like $\displaystyle \pi$ and $\displaystyle e$?Quote:
Originally Posted by galactus
Some limits at least. I suspect that the algorithms that generate the solutions are limited by what the programmer thinks the user might input. And the TI series (at least) occasionally can't exactly solve some moderately complex problems that are simple to do by hand if you know the correct method. (They can, however do a nice job numerically on just about anything.)Quote:
Originally Posted by ThePerfectHacker
At the very least the TI series avoids any horrible functions such as the "polylog" function. (Mathematica used them to solve a differential equation for me once. I had a devil of a time looking up what they are and once I did I was left with no idea of what the solution meant.) The TIs can't handle anything nasty like that, but at least returns answers in terms of more common functions, or none at all.
-Dan
Mathematica (I do not own such a monster program) but I know that it can evaluate infinite integrals in terms of Zeta Functions, PolyGamma functions, ... and even give the actual values. :eek:Quote:
Originally Posted by topsquark
Yes, Maple does that also. Sometimes it'll give these wacky answers and I haven't a clue how to interpret them. I have to look up in it's database of terms to find out what they mean.
The TI-92 has been replaced by the Voyage 200. But I have had fun writing programs and such over the years.
These new calculators have 'pretty print'. It gives the answer algebraically, not just in a decimal form.
Although, the TI-92 won't give an indefinite integral if it flows outside elementary means, but it will give an answer if you enter in limits of integration.
For instance, suppose I want to find the indefinite integral of
$\displaystyle \int\frac{1}{\sqrt{1+x^{4}}}dx$, with no limits.
It'll just return $\displaystyle \int\frac{1}{\sqrt{1+x^{4}}}dx$
But if I enter limits, such as $\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x^{4}}}dx$, it returns the result 0.927037338651
