Could anybody help me to solve this, a question for brilliant guys and girls?

Question

Well, thanks for the people answer me the differentiation on f(x)=x!. I have been convinced with your answer that it is not a continuous function, thus can't be differentiated. so now please help me on the following question,

f(t) = (A)^t x tC(t/n)

where A<0, n>0
prove that when t approache to infinity, f(t) will approach to 0.

Note: tC(t/n) means the number of combination choosing t/n from t. t =kn where k is a integer (no decimal).

I will put your name into my paper if you can solve it. Thanks.

Sorry, my mistake.
the value of A is 0

Answer 1

Sorry I misunderstood your question at first. I don't know how you'd prove it, but I suspect that limit does not go to zero. Good luck with that!

Edit: I just thought of your solution. tC(t/n) will never approach zero because t/n will always be less than t (hence yielding a growing number of combinations of t/n things), and A^t will be all over the place as t->inf, not really approaching anything at all. Such a limit makes no sense anyway unless -1 infinity.

If I am not mistaken, tC(t/n) will not even get smaller as t->inf. The gap between the number of items t and the rate they are chosen (t/n) will grow as t->infinity, so the number of combinations will likewise grow and grow. Infinity is, after all, only a convenient idea to make generalizations for arbitrarily large numbers. We must consider infinity/n to be less than infinity because the difference between t and t/n is significant for the idea at hand.

Very interesting question. Good luck with the paper!

Answer 2

Firstly, the factorial per se is defined only on integers (hence discontinuous), but the gamma function, which evaluates to factorials for integer values, is continuous and differentiable. For any number n, its value is: definite integral from 0 to infinity of t^(n-1) times exp (-t) dt, where t is a dummy variable. (Isn't ASCII wonderful for mathematics? Not!) You should be able to differentiate under the integral sign to work out the gory details. As for the second issue, if I have read correctly what you are saying, I don't think that it does approach zero. Let n = 1, and play with it a bit and see what happens.

Answer 3

The greatest value of tC(t/n) can be tC(t/2)

where t/2 is the largest number smaller than or equal to t/2.

And A^t approaches 0 since A<

Answer 4

It is not possible as because (A)^t shall be a big number when t approach to infinity. Now tC(t/n) shall also be some number as it is the number of combination of t/n from t. Hence, multiplying both can never approach to 0.

Answer 5

The limit of tC(t/n) as t goes to infinity is infinity, since you can write tC(t/n) as t*(t-1)*...*(t-n+1)/n!. So if A < -1, then in absolute value, f(t) goes to infinity, and your assertion is incorrect.

Answer 6

R You In Summer School?

Answer 7

Start doing your own work. Go and research. It's best to find your own answers.

Answer 8

Congradulations! This question put me to sleep!

Good night.

Answer 9

ummm...2 complicated. im only going into 6th grade!!!

i just answered cuzz i wanted 2 points. :)

Answer 10

That is insane. No ******* thank you!