On 11/18/2010 07:32 PM, A. Jorge Garcia wrote:
> My Discrete Math students today came up with an observation: The
> average of any 3 consecutive terms of an arithmetic sequence equals
> the middle term. So, I thought we'd show off some CAS in SAGE (this
> class uses python primarily) and wrote the following in my SAGE
> notebook:
>
> var('c,d')
> term1 = c+(n-1)*d
> term2 = c+(n+1)*d
> sum = term1+term2
> avg = expand(sum/2)
> show(term1)
> show(term2)
> show(sum)
> show(avg)
>
> and got the following output
> c+(n-1)*d
> c+(n+1)*d
> 2*c+2*n*d
> c+n*d
>
> So, I thought we'd try something similar with geometric sequences:
> var('c,d')
> term1 =a*r**(n-1)
> term2 = a*r*(n+1)
> prod = term1*term2
> avg = simplify(sqrt(prod))
> show(term1)
> show(term2)
> show(sum)
> show(avg)
>
> however, I got the following output
> a*r**(n-1)
> a*r*(n+1)
> a**2*r**(2*n)
> sqrt(a**2*r**(2*n))
>
> I couldn't get the final result to reduce to a*r**n, is there a way to
> do this?
Evaluating
var('a n r')
assume(a > 0)
avg = sqrt(a**2 * r**(2 * n))
avg.simplify_radical()
I get
r^n*a
You can also use the simplify_full method.
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