Converting $tanh^{-1}{x}$ to an expression involving the natural logarithm
up vote
2
down vote
favorite
I know how to convert $sinh^{-1}{x}$ and $cosh^{-1}{x}$ to $ln{|x+sqrt{x^2 pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$tanh^{-1}{frac{x}{2}}$$
Can someone please show me how to convert it to a $ln$ form? thanks!
calculus hyperbolic-geometry
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
add a comment |
up vote
2
down vote
favorite
I know how to convert $sinh^{-1}{x}$ and $cosh^{-1}{x}$ to $ln{|x+sqrt{x^2 pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$tanh^{-1}{frac{x}{2}}$$
Can someone please show me how to convert it to a $ln$ form? thanks!
calculus hyperbolic-geometry
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know how to convert $sinh^{-1}{x}$ and $cosh^{-1}{x}$ to $ln{|x+sqrt{x^2 pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$tanh^{-1}{frac{x}{2}}$$
Can someone please show me how to convert it to a $ln$ form? thanks!
calculus hyperbolic-geometry
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
I know how to convert $sinh^{-1}{x}$ and $cosh^{-1}{x}$ to $ln{|x+sqrt{x^2 pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$tanh^{-1}{frac{x}{2}}$$
Can someone please show me how to convert it to a $ln$ form? thanks!
calculus hyperbolic-geometry
calculus hyperbolic-geometry
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
edited Jan 15 '12 at 14:19
user38268
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
asked Jan 15 '12 at 12:36
yotamoo
1,13371837
1,13371837
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$
Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
add a comment |
up vote
0
down vote
$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :
$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$
$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
add a comment |
up vote
0
down vote
Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=tanh^{-1}x$
$tanh y=x$
$frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=frac{1+x}{1-x}$
$y=frac{1}{2}ln(frac{1+x}{1-x})$
edited Nov 22 at 14:28
Community♦
1
answered Apr 30 '17 at 5:49
Daniel
1
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f99254%2fconverting-tanh-1x-to-an-expression-involving-the-natural-logarithm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$
Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
add a comment |
up vote
1
down vote
accepted
Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$
Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$
Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$
Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$
Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
answered Jan 15 '12 at 12:46
Ragib Zaman
28.4k34889
28.4k34889
add a comment |
add a comment |
up vote
0
down vote
$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :
$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$
$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
add a comment |
up vote
0
down vote
$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :
$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$
$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
add a comment |
up vote
0
down vote
up vote
0
down vote
$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :
$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$
$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :
$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$
$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
answered Jan 15 '12 at 13:25
Peđa Terzić
7,88022570
7,88022570
add a comment |
add a comment |
up vote
0
down vote
Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=tanh^{-1}x$
$tanh y=x$
$frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=frac{1+x}{1-x}$
$y=frac{1}{2}ln(frac{1+x}{1-x})$
edited Nov 22 at 14:28
Community♦
1
answered Apr 30 '17 at 5:49
Daniel
1
add a comment |
up vote
0
down vote
Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=tanh^{-1}x$
$tanh y=x$
$frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=frac{1+x}{1-x}$
$y=frac{1}{2}ln(frac{1+x}{1-x})$
edited Nov 22 at 14:28
Community♦
1
answered Apr 30 '17 at 5:49
Daniel
1
add a comment |
up vote
0
down vote
up vote
0
down vote
Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=tanh^{-1}x$
$tanh y=x$
$frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=frac{1+x}{1-x}$
$y=frac{1}{2}ln(frac{1+x}{1-x})$
edited Nov 22 at 14:28
Community♦
1
answered Apr 30 '17 at 5:49
Daniel
1
Why is people not answering?
Look, it's quite intuitive so I'll just show you the step by step:
$y=tanh^{-1}x$
$tanh y=x$
$frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=x$
$e^{y}-e^{-y}=xe^{y}+xe^{-y}$
$e^{2y}-1=xe^{2y}+x$
$e^{2y}(1-x)=1+x$
$e^{2y}=frac{1+x}{1-x}$
$y=frac{1}{2}ln(frac{1+x}{1-x})$
edited Nov 22 at 14:28
Community♦
1
answered Apr 30 '17 at 5:49
Daniel
1
edited Nov 22 at 14:28
Community♦
1
edited Nov 22 at 14:28
Community♦
1
edited Nov 22 at 14:28
Community♦
1
1
answered Apr 30 '17 at 5:49
Daniel
1
answered Apr 30 '17 at 5:49
Daniel
1
answered Apr 30 '17 at 5:49
Daniel
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f99254%2fconverting-tanh-1x-to-an-expression-involving-the-natural-logarithm%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
