Question

prove that the inverse tangent of x ...
tan^-1(x) = 1/2 [ln ( (1+x)/(1-x) )...
we can assume the 0 13 years ago

Answer 1

is that = sign or less than sign in the equation

Answer 2

tan^-1(x) equals 1/2 [ln ( (1+x)/(1-x) )

Answer 3

we cud use the mean value theorem

Answer 4

This relation is not true.
arctan(1/sqrt(3)) of course = pi/6
but x = 1/sqrt(3) in the ln expression is not equal to this.

Answer 5

cud you pls recheck your question

Answer 6

omg...i'm so sorry...that wasn't tan^-1x...
it should have been tanh^_1x...
tanh^-1(x) = 1/2 [ln ( (1+x)/(1-x) )
hyperbolic inverse tan of x...sorry everyone!

Answer 7

it s ok

Answer 8

So start with x = tanh(y) using the definition of tanh and then manipulate it to solve for y

Answer 9

so first wats \[\tanh x=\frac{\sinh x}{\cosh x}=\frac{ e^x-e^{-x} }{ e^x+e^{-x} }\]

Answer 10

as @JamesJ suggested

Answer 11

if i have ln(something/something), how do you get rid of the ln ?

Answer 12

well that is the same as\[\ln \frac{a}{b}=\ln a-\ln b\]
getting rid of ln completly NOT SURE

Answer 13

take the exp of both sides

Answer 14

to get you started ...

if x = tanh(y), then
x = (e^y - e^-y)/(e^y + e^-y)

(e^y + e^-y)x = e^y - e^-y

e^y(x-1) = - e^-y . (x+1)

thus

e^2y = ....

Answer 15

no...its inverse tanhx

Answer 16

Yes, I'm starting with tanh(y) and solving for y, because
if x = tanh(y), then
y = invtanh(x)

Answer 17

tanh^1(x) = 1/1-x^2

Answer 18

it looks like you are deriving why?

Answer 19

ok...i'm trying to follow...thanks for your help btw...i'm having a hard time with this one...

Answer 20

you know what Jonask, you're correct...i didn't mean to take the derivative...i'm sorry

Answer 21

if
x = tanh(y)
then
y = invtanh(x)

Now if x = tanh(y), then
x = (e^y - e^-y)/(e^y + e^-y)
and by the algebra abovee we have
e^2y = (1+x)/(1-x)
thus

2y = ....

and hence

y = ....

Answer 22

Hi James...is this leading me to prove that...
tanh^-1(x) = 1/2 [ln ( (1+x)/(1-x) ) ???
I guess I don't understand why I am solving for "y"

Answer 23

y is just a dummy variable. On the one hand we have
y = invtanh(x)

and on the other we are developing an expression for y in terms of x. If you follow what I have written for you above you will end up showing exactly that

\[ y = \frac{1}{2} \ln \frac{1+x}{1-x} \]

Answer 24

I will try

Answer 25

@JamesJ since the relatioship between x and y is\[x^2-y^2=1\]
\[\tanh y=\tan (x^2-1)\]

Answer 26

is this correct to say

Answer 27

no, that's not correct

Answer 28

\[\tanh \sqrt{x^2-1}\]

Answer 29

...at least as we have set up x and y to date.

There are a few ways to prove this identity, but the way I have laid out here is the most direct.

Answer 30

I need walked through this one step by step...I can't follow the logic explained above unfortunately...
if I start on the left side, tanh^-1x = (e^x - e^-x)/(e^x + e^-x)...
from here i need simple direction...thanks

Answer 31

Ok. Let's start again.

Suppose y = tanh^-1 (x)

Then x = tanh(y), which means by the definition of hyperbolic tangent, tanh, that
\[ x = \frac{e^y - e^{-y}}{e^y + e^{-y}} \]

What you want to do now is manipulate that expression to solve for y.

Does this much make sense?

Answer 32

i would like to get away from the logic of soling for "y" if possible...i just wanna make the left side look like the right going from (e^x - e^-x)/(e^x + e^-x)...
...do i need to rid of the denominator somehow or multiply by a fancy form of "1" or what?

Answer 33

I can't decide on my next move

Answer 34

You can also do this.

tanh^-1(x) = (1/2).ln( (1+x)/(1-x) ), call this second expression just s for simplicity.

Then because tanh(tanh^-1 x) = x, by definition of an inverse function, it must also be the case that tanh(s) = x.

So proving that tanh(s) = x is another strategy for solving the problem.

Answer 35

[btw, I don't know why you didn't like the other strategy. The basic idea is very simple, isn't it? We have that y = tanh^-1(x) and we also show that y = (1/2)ln((1+x)/(1-x)) hence it must be that tanh^-1(x) = (1/2)ln((1+x)/(1-x))

But whatever floats your boat.]

Answer 36

haha...i'm sorry but I just don't understand "Mathese"...
I appreciate the help, I really do, but I just can't follow your logic unfortunately James...sorry but thanks

Answer 37

if someone could just tell me the very next step without all the precursors and reasoning, I will probably remember what to do

Answer 38

But that's the thing. This requires some thinking. It's not a grade 10 level problem.

Answer 39

James, I am well adept at "thinking"...I have a 4.0 gpa over 105 credit hours in the STEM tract...I don't appreciate the grade 10 remark...did it ever occur to you that your way of explaining perhaps doesn't work for everyone?
I'm not going to get in a tiff over this...thanks but you can go help someone else now