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How can I prove $frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$ using Rolle's theorem?
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For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.
analysis rolles-theorem
edited yesterday
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onzepattes
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up vote
0
down vote
favorite
For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.
analysis rolles-theorem
edited yesterday
Nosrati
25.8k62252
asked yesterday
onzepattes
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.
analysis rolles-theorem
edited yesterday
Nosrati
25.8k62252
asked yesterday
onzepattes
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For all $t>0$, there exists $cin]0,t[$ such that
$$frac{t-arctan(t)}{t^3}=frac{1}{3(1+c^2)}$$
Thanks in advance.
analysis rolles-theorem
analysis rolles-theorem
edited yesterday
Nosrati
25.8k62252
asked yesterday
onzepattes
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Nosrati
25.8k62252
asked yesterday
onzepattes
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Nosrati
25.8k62252
edited yesterday
Nosrati
25.8k62252
edited yesterday
Nosrati
25.8k62252
25.8k62252
asked yesterday
onzepattes
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
onzepattes
11
asked yesterday
onzepattes
11
11
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
onzepattes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday
add a comment |
What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday
What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday
What have you already tried? Try editing your question to show this in order to get the most appropriate answers
– MRobinson
yesterday
add a comment |
2 Answers
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Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.
answered yesterday
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Hint:
If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.
answered yesterday
ajotatxe
52.1k23688
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.
answered yesterday
Nosrati
25.8k62252
add a comment |
up vote
1
down vote
Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.
answered yesterday
Nosrati
25.8k62252
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.
answered yesterday
Nosrati
25.8k62252
Hint: I think you mean Cauchy mean value theorem
$$dfrac{f(b)-f(a)}{g(b)-g(a)}=dfrac{f'(c)}{g'(c)}$$
let $f(t)=t-arctan t$ and $g(t)=t^3$.
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
25.8k62252
add a comment |
add a comment |
up vote
0
down vote
Hint:
If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.
answered yesterday
ajotatxe
52.1k23688
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
add a comment |
up vote
0
down vote
Hint:
If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.
answered yesterday
ajotatxe
52.1k23688
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.
answered yesterday
ajotatxe
52.1k23688
Hint:
If $f$ is differentiable in $[0,t]$ then define $g(x)=x[f(x)-f(t)]$ for $xin[0,t]$. Then you can apply Rolle's theorem to $g$.
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
answered yesterday
ajotatxe
52.1k23688
52.1k23688
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
add a comment |
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
This would imply that $f(x)=$frac{x-arctan(x)}{x^3}. Bu this is precisely the function we want to get rid off in the first place!!
– onzepattes
yesterday
add a comment |
onzepattes is a new contributor. Be nice, and check out our Code of Conduct.
onzepattes is a new contributor. Be nice, and check out our Code of Conduct.
onzepattes is a new contributor. Be nice, and check out our Code of Conduct.
onzepattes is a new contributor. Be nice, and check out our Code of Conduct.
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– MRobinson
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