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Inverse hyperbolics in logarithmic form
By Anonymous on Thursday, February 1, 2001 - 12:39 pm:

Hi there,

I tried to get arcsinh(x) into logarithmic form. I also managed to get arccosh(x) into log form. But getting u=arctanh(x) into its log equilivant is turning into a mess, and I am a bit lost?

Another problem I am having is to get tanh(x) in terms of e, by using sinh(x) and cosh(x), in the form (e2x-1)/(e2x+1)?? I can only get as far as saying, Tanh(x) = Sinh(x)/Cosh(x) = (ex-e-x)/(ex+e-x)

Appreciate any light that can be thrown on this matter.

;)

By Pooya Farshim (P2572) on Thursday, February 1, 2001 - 01:02 pm:

For tanh(x) multiply top and bottom by ex and you'll get the result.

sinh(x)=(ex-e-x)/2
Let ex=t, have:
sinh(x)=(t-1/t)/2
so if sinh(x)=y (and so x=sinh-1(y)), say, we have:
2y=t-1/t => t2-2yt-1=0
The roots of the equation are:
t1,2=y±(y2+1)1/2
now ex=t => x=log (y ± (y2+1)1/2)
But (y2+1)1/2>y and and as log of -ve numbers are not real:
sinh-1(y)=log (y+(y2+1)1/2)


(All logs are to base e.)

By Hal 2001 (P3046) on Thursday, February 1, 2001 - 03:28 pm:

You can have it in this form:
arctanh(x)=½ln*((1+x)/(1-x)), |x|<1

[Just follow the same steps as for arcsinh above. - The Editor]

By Anonymous on Thursday, February 1, 2001 - 03:33 pm:

Thank you Pooya and Hal.

By Anonymous on Saturday, February 3, 2001 - 05:26 pm:

Hi,

if we know that arcsinh x = ln(x + (1+x2)0.5)
then how can we show that given x is large and positive,

arcsinh x = (roughly) ln2 + lnx + (1/4x2)

--------------
this is what i tried:
ln(x + (1+x2)0.5)
= ln( x + x(1 + x-2)0.5)
= ln(x(1 + (1 + x-2)0.5)
= lnx + ln(1 + (1 + x-2)0.5)
not sure where to go from here, appreciate it if you can show me the steps to get to the required answer. mine became very messy indeed.

Thank you.

By Kerwin Hui (Kwkh2) on Saturday, February 3, 2001 - 06:03 pm:

The next step is to write out the series expansion for (1+x-2)½=1+½x-2+...

Now substitute and you get

ln x + ln (2+½x-2+...) = ln 2 + ln x + ln (1 + x-2/4 + ...)

Now take the series expansion of ln (1+y) = y + higher order terms and you get the answer.

Kerwin

By Anonymous on Saturday, February 3, 2001 - 06:45 pm:

Kerwin, did you take a factor 2 out from that bit: ln(2 + 0.5x-2+....)?

Thanks.

By Kerwin Hui (Kwkh2) on Saturday, February 3, 2001 - 06:45 pm:

Yes.

By Anonymous on Saturday, February 3, 2001 - 07:39 pm:

OK, that's great!
Thanks for your help Kerwin.