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
asked Jan 3 at 20:27
Paul Keseru
494
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
asked Jan 3 at 20:27
Paul Keseru
494
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
asked Jan 3 at 20:27
Paul Keseru
494
$f(z) = arcsin(frac{2z}{(1+z^2)}) + 2arctan(z)$
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
asked Jan 3 at 20:27
Paul Keseru
494
asked Jan 3 at 20:27
Paul Keseru
494
edited Jan 3 at 20:49
asked Jan 3 at 20:27
Paul Keseru
494
asked Jan 3 at 20:27
Paul Keseru
494
asked Jan 3 at 20:27
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
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
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?
answered Jan 3 at 21:07
Sauhard Sharma
74516
add a comment |
Your Answer
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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
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
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
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
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
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
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
$$f(2019)=f(1)=$$
$$arcsin(1)+2arctan(1)=pi$$
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
edited Jan 3 at 21:09
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
answered Jan 3 at 20:35
hamam_Abdallah
38k21634
38k21634
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
add a comment |
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
and how does the derivative helps me in this case ?
– Paul Keseru
Jan 3 at 20:44
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
@PaulKeseru If it is zero, the function is constant.
– hamam_Abdallah
Jan 3 at 20:45
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?
answered Jan 3 at 21:07
Sauhard Sharma
74516
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?
answered Jan 3 at 21:07
Sauhard Sharma
74516
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?
answered Jan 3 at 21:07
Sauhard Sharma
74516
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 Jan 3 at 21:07
Sauhard Sharma
74516
answered Jan 3 at 21:07
Sauhard Sharma
74516
answered Jan 3 at 21:07
Sauhard Sharma
74516
answered Jan 3 at 21:07
Sauhard Sharma
74516
74516
add a comment |
add a comment |
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