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Hi DAN7891,
Do you need an exact solution or just an approximation?
An approximation can be achieved quite easily with trial and improvement
To get an exact answer you need to solve a cubic equation in 3^x this can be done(as cubics go the equation is relatively easy to solve)
I can post further details of how to solve the equation to get an exact answer if that is what's required.
Hi all,
Found a nice generalisation of this problem stating that the sum of 2 or more consecutive integers cannot be a power of 2
Proof:
Let n, m and t be non-negative integers
Using that result that 1 + 2 + ... + n = 1/2n(n+1)
We have
(m+1) + (m+2) + ... + n = 1/2n(n+1) - 1/2m(m+1) = 1/2[n^2 + n - m^2 - m] = 1/2[(n+m)(n-m) + (n-m)] = 1/2(n-m)(m+n+1)
Now suppose
1/2(n-m)(m+n+1) = 2^t
Then (n-m)(m+n+1) =2^(t+1)
but one of (n-m) and (n+m+1) must be odd as their sum is odd and the only odd factor of 2^(t+1) is 1
This means either n = m = 0 (no sum at all) or n = m+ 1 (which means the sum only contains 1 number)
This contradiction proves the result for non-negative integers.
Including negative integers doesn't help matters since either the sum will be negative or we can cancel out all negative numbers because (-n) + n = 0
Hi yonski,
For the last 2 you need to use the cyclic symmetry of the scaler triple product (which you may need to prove):
a.(bxc) = b.(cxa) = c.(axb)
Hope this helps
Hi, I'm not too great on stats myself but here's how i think your problems can be solved:
a) Use the formula
b) This requires you to use the cdf of the normal distribution
In symbols
I won't solve the problem for you but here's a sequenc of steps to follow
The 25's look scary but it isn't actually that difficult to show that
Proof:
Let
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