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- Thread starter Beamsbox
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So

\sech \left(\sech^{-1} x\right) = x

- #3

Beamsbox

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\sech \left(\sech^{-1} x\right) = x

Sorry, I'm not quite sure what you've written here...

Perhaps it's the definition of inverse, that I'm not 'seeing'...

Say, for

For, x+y/z, the inverse would be (z(y-x))...? Exactly which calculations do you 'invert', all of them?

I'm going to look more into this, @:

http://en.wikipedia.org/wiki/Inverse_function

Last edited:

- #4

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So the approach to find the inverse of y=f(x) is to express x=g(y). Can you do that ?

- #5

Beamsbox

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I thik that once I have a definition for sech

- #6

Beamsbox

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Define sech in terms of e

No u-substitution necessary, but the quadratic equation comes into play.

Thanks for your help, bigubau.

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