Deriving the equation

**noob mathematician** · September 24th 2010, 11:46 PM

Hi all!

I have the following general equation which is: $s(s+t)=((1-t)s(x)^\alpha + ts(x+1)^\alpha)^\frac{1}{\alpha}$ . where s is survival function

It is told to me that when I substitute $\alpha =0$ , I will be able to obtain

$log(s(x+t))=(1-t)log (s(x))+(t) log (s(x+1))$ .

But how do I do that since when $\alpha=0$ , $1/\alpha$ goes to infinity??

**mr fantastic** · September 25th 2010, 02:01 AM

Originally Posted by noob mathematician

Hi all!

I have the following general equation which is: $s(s+t)=((1-t)s(x)^\alpha + ts(x+1)^\alpha)^\frac{1}{\alpha}$ . where s is survival function

It is told to me that when I substitute $\alpha =0$ , I will be able to obtain

$log(s(x+t))=(1-t)log (s(x))+(t) log (s(x+1))$ .

But how do I do that since when $\alpha=0$ , $1/\alpha$ goes to infinity??

Since direct substitution leads to the indeterminant form $1^{\infty}$ you would be expected to take the limit $\alpha \to 0$ .

**noob mathematician** · September 25th 2010, 02:34 AM

Originally Posted by mr fantastic

Since direct substitution leads to the indeterminant form $1^{\infty}$ you would be expected to take the limit $\alpha \to 0$ .

Thanks for the hint.
But I still finding it hard to take the limit to 0. How do I go about doing it? Substitute $\alpha$ with a very small number? 0.000001 etc
Or do I differentiate the equation?

**mr fantastic** · September 25th 2010, 03:01 AM

Originally Posted by noob mathematician

Thanks for the hint.
But I still finding it hard to take the limit to 0. How do I go about doing it? Substitute $\alpha$ with a very small number? 0.000001 etc Mr F says: NO. That's not how you take a limit.
Or do I differentiate the equation?

The technique can be found here: Indeterminate form - Wikipedia, the free encyclopedia

**noob mathematician** · September 25th 2010, 04:09 AM

Oh great! L'Hôpital's rule.. Thanks so much

**matheagle** · September 25th 2010, 08:10 PM

take the logarithm first

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