CAMBRIDGE: Hyperbolic Functions

How on earth do you show:

\frac{x}{\sqrt{x^{2}+1}} < \sinh^{-1}x < x

I tried to approach the question by getting the inverse hyperbolic function into its logarithmic form:

\sinh^{-x}=\ln(x+\sqrt{x^{2}+1})

however, I can't seem to progress from here. Could anyone offer any help?

(edited 11 years ago)

No one?

i guess it is just for x>0

or not?

Reply 3

moritzplatz

i'll show one of the inequalities (even if it should be less or equal if you consider x in R as for x =0 it does not hold)

take exponential which preserve inequalities to get

x+ \sqrt{x^2 +1}\leq e^x

now it suffices to show

x+\sqrt{x^2+1} \leq 1+x+x^2/2

take squares

x^2+1 \leq 1+x^2 +x^4/4

and it's done

(edited 11 years ago)

Reply 4

SimonM

For the right hand side, I'd use that sinh is increasing and show that sinh x > x.

For the left hand side, I'd show that consider the function log(x+sqrt(1+x^2))-x/sqrt(1+x^2) and show that it's increasing.

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