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Topic review (newest first)
- kylekatarn
- 2005-07-26 23:06:08
simply amazing!
thank you all
- MathsIsFun
- 2005-07-26 17:46:42
Magic, guys. Disproven by example.
- mathsyperson
- 2005-07-26 17:42:17
If you take n=1, then you can disprove both by counter-example.
Take a=1.2, b=0.00001, n=1
That gives (1.2+0.00001)^1.2*1</>(1.2^2)*1^0.00001, meaning 1.244...</>1.44.
So, in this case, a²*n^b is larger.
Almost all other cases result in (a+b)^na being larger.
e.g. (3+3)^3*1</>(3^2)*1^3, meaning 216</>9.
So, it can't be proven or disproven.
- ganesh
- 2005-07-26 17:33:58
The binomial expansion of the Left Hand Side
contains two terms, among many, which are
a^a*n and b^a*n which appear to make the LHS greater than the RHS,
but when we assign arbitrary values,
say a=10, b=1,000,000,000 and n=100
the LHS is (1,000,000,010)^1000, which would contain 9,001 digits;
the RHS becomes
100 x (100^1,000,000,000) which would contain more than 2 billion digits!
This happened because we assumed b>>n.
Otherwise, the LHS may be greater.
Say, when a=10, b=100, n=1000.
LHS would be 110^10,000 containing 20,414 digits and the RHS would be much smaller, viz. 100*(1000^100), containing approximately 300 digits!
- MathsIsFun
- 2005-07-26 17:32:47
n = a natural number?
Perhaps we could start off with a proof for n=1, then onto 2 or perhaps n+1
- kylekatarn
- 2005-07-26 15:10:40
can someone proove (or disproof) that:
(a+b)^(n*a)>(a^2)(n^b)
:|
conditions:
a>0
b>0
n>=1
any help would be great
thnkx in advnc.