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suppose I have a function f(x)=$\sqrt{x+1/x}$ - $\sqrt{x-1/x}$ where x $\ge{1}$ and want to find out its relative numeric condition for

a) $ x \rightarrow 1 $
b) $ x \rightarrow \infty $

Now the relative condition according to my materials is given by the following formula:

condRel(x) = (|f '(x)|*|x|) / |f(x)|

However f '(x) = $\dfrac{-(x^2+1)}{2*x^2*\sqrt{\dfrac{x^2-1}{x}}}$ - $\dfrac{\sqrt{x^2+1}}{2*\sqrt{1/x} * x^2}$ + $\dfrac{\sqrt{1/x}*x}{\sqrt{x^2+1}}$

and therefore f '(1) is not defined (f(x) is not differentiable for x = 1) (dividing by zero).

Does this simply mean that the problem is very bad conditioned for values close to 1? E.g. the condition seems to become bigger the close i get to 1

I am also unsure about values which go against infinity. Since the condition seems to stay around 1.5 for large values of x, does this mean that the problem is good (or at least extremeley bad) conditioned?

Thank you in advance for any help!

EDIT:

Can nobody help with how the problem is conditioned for $ x \rightarrow 1$ and $ x \rightarrow \infty $?

Since it is not differentiable for $ x \rightarrow 1$ I assume it is terribly / bad conditioned here and since it goes to 0 for $ x \rightarrow \infty $ can I assume it is good conditioned in this case?

Thx for any help!

Philipp
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1 Answers1

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I think you have made mistake in differentiation Actually $d/dx(\sqrt{(x+1/x)}+\sqrt{(x-1/x)}) = (1-1/x^2)/(2 \sqrt{(x+1/x)})+(1/x^2+1)/(2 \sqrt{(x-1/x)})$

So f(x) is not differentiable at 1 but its value comes down to 0 at infinity.

user58491
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  • Thank you very much. I think both differentiations are correct, since I calculated mine using mi ti voyage 200 and after rechecking I found that I get close to zero for values of x going against infinity as well. So, does this mean that my problem is well conditioned for x -> infinity?. And since its not differentiable for x -> 1 is the problem very bad conditioned for input values of x close to 1? E.g. can you assume that 1/0 = infinity here and the condition for x = 1 is infinity? – Philipp Sep 10 '13 at 08:46