simple proof $f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
So the thing that I need to do is to show that $f(2019) = pi$.
So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.
Any tips?
analysis functions
add a comment |
$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
So the thing that I need to do is to show that $f(2019) = pi$.
So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.
Any tips?
analysis functions
add a comment |
$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
So the thing that I need to do is to show that $f(2019) = pi$.
So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.
Any tips?
analysis functions
$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
So the thing that I need to do is to show that $f(2019) = pi$.
So the thing that I have tried is to calculate $f(0), f(1)$, and to see some kind of connection (or recurrence relation), so I can easily calculate $f(2019)$, but I couldn't find any.
Any tips?
analysis functions
analysis functions
edited 2 days ago
asked 2 days ago
Paul Keseru
494
494
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
hint
For $z>1$,
$$f'(z)=$$
$$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+infty)$ and
$$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$
or
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
add a comment |
Start by writing
$$z=tan(theta)$$
So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.
Now the second term will become
$$=2arctan(tan(theta))=2theta$$
This is allowed as $theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
$$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$
Since $pi/4lttheta lt pi/2$,
$pi/2lt2theta lt pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$
Therefore
$$arcsinleft(sin(2theta)right)=pi - 2theta$$
Adding the first and second terms together
$$=pi - 2theta + 2theta = pi$$
I am attaching a link for more explanation
$arcsin(sin x)$ explanation?
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',
autoActivateHeartbeat: false,
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%2f3060967%2fsimple-proof-fz-arcsin-frac2z1z2-2-arctanz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
hint
For $z>1$,
$$f'(z)=$$
$$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+infty)$ and
$$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$
or
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
add a comment |
hint
For $z>1$,
$$f'(z)=$$
$$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+infty)$ and
$$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$
or
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
add a comment |
hint
For $z>1$,
$$f'(z)=$$
$$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+infty)$ and
$$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$
or
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
hint
For $z>1$,
$$f'(z)=$$
$$frac{1}{sqrt{1-(frac{2z}{1+z^2})^2}}frac{2(1+z^2)-4z^2}{(1+z^2)^2}+frac{2}{1+z^2}=0$$
thus $f$ is constant at $[1,+infty)$ and
$$f(2019)=lim_{zto+infty}f(z)=0+2frac{pi}{2}=pi$$
or
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
edited 2 days ago
answered 2 days ago
hamam_Abdallah
37.9k21634
37.9k21634
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
add a comment |
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
and how does the derivative helps me in this case ?
– Paul Keseru
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
2 days ago
add a comment |
Start by writing
$$z=tan(theta)$$
So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.
Now the second term will become
$$=2arctan(tan(theta))=2theta$$
This is allowed as $theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
$$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$
Since $pi/4lttheta lt pi/2$,
$pi/2lt2theta lt pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$
Therefore
$$arcsinleft(sin(2theta)right)=pi - 2theta$$
Adding the first and second terms together
$$=pi - 2theta + 2theta = pi$$
I am attaching a link for more explanation
$arcsin(sin x)$ explanation?
add a comment |
Start by writing
$$z=tan(theta)$$
So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.
Now the second term will become
$$=2arctan(tan(theta))=2theta$$
This is allowed as $theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
$$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$
Since $pi/4lttheta lt pi/2$,
$pi/2lt2theta lt pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$
Therefore
$$arcsinleft(sin(2theta)right)=pi - 2theta$$
Adding the first and second terms together
$$=pi - 2theta + 2theta = pi$$
I am attaching a link for more explanation
$arcsin(sin x)$ explanation?
add a comment |
Start by writing
$$z=tan(theta)$$
So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.
Now the second term will become
$$=2arctan(tan(theta))=2theta$$
This is allowed as $theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
$$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$
Since $pi/4lttheta lt pi/2$,
$pi/2lt2theta lt pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$
Therefore
$$arcsinleft(sin(2theta)right)=pi - 2theta$$
Adding the first and second terms together
$$=pi - 2theta + 2theta = pi$$
I am attaching a link for more explanation
$arcsin(sin x)$ explanation?
Start by writing
$$z=tan(theta)$$
So, for tan$(theta)=2019$, $theta$ will lie somewhere in between $pi /4$ and $pi/2$.
Now the second term will become
$$=2arctan(tan(theta))=2theta$$
This is allowed as $theta$ falls within the range of arctan$(x)$
Now taking the first term
$$=arcsinleft(frac{2tan(theta)}{1+tan^2(theta)}right)$$
$$=arcsinleft(2sin(theta)cos(theta)right)=arcsinleft(sin(2theta)right)$$
Since $pi/4lttheta lt pi/2$,
$pi/2lt2theta lt pi$
This falls outside the range of arcsin$(x)$
Now by symmetry around $pi/2$ you can see that for it to come back within the range, for such $x$ we use $pi - x$
Therefore
$$arcsinleft(sin(2theta)right)=pi - 2theta$$
Adding the first and second terms together
$$=pi - 2theta + 2theta = pi$$
I am attaching a link for more explanation
$arcsin(sin x)$ explanation?
answered 2 days ago
Sauhard Sharma
73216
73216
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%2f3060967%2fsimple-proof-fz-arcsin-frac2z1z2-2-arctanz%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