Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$
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Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$ , Prove for every $n,minmathbbN^+$, there exists $xi=xi_n,min(0,1)$ such that
$$ncdotfracf'(xi)f(xi)=mcdotfracf'(1-xi)f(1-xi)$$
calculus
asked Aug 22 at 4:38
Jaqen Chou
3378
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up vote
2
down vote
favorite
Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$ , Prove for every $n,minmathbbN^+$, there exists $xi=xi_n,min(0,1)$ such that
$$ncdotfracf'(xi)f(xi)=mcdotfracf'(1-xi)f(1-xi)$$
calculus
asked Aug 22 at 4:38
Jaqen Chou
3378
Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$ , Prove for every $n,minmathbbN^+$, there exists $xi=xi_n,min(0,1)$ such that
$$ncdotfracf'(xi)f(xi)=mcdotfracf'(1-xi)f(1-xi)$$
calculus
asked Aug 22 at 4:38
Jaqen Chou
3378
Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)ne 0,forall xin(0,1)$ , Prove for every $n,minmathbbN^+$, there exists $xi=xi_n,min(0,1)$ such that
$$ncdotfracf'(xi)f(xi)=mcdotfracf'(1-xi)f(1-xi)$$
calculus
asked Aug 22 at 4:38
Jaqen Chou
3378
asked Aug 22 at 4:38
Jaqen Chou
3378
asked Aug 22 at 4:38
Jaqen Chou
3378
asked Aug 22 at 4:38
Jaqen Chou
3378
3378
Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08
add a comment |Â
Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08
Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
MVT method:
Consider an auxiliary function
$$
F(x) =f(x)^n f(1-x)^m, quad x in [0,1],
$$
then Rolle's theorem achieves the goal.
answered Aug 22 at 5:22
xbh
2,627114
add a comment |Â
up vote
1
down vote
Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.
Let $h(x) = ln(f(x)) + fracmn ln(f(1 - x))$, defined over $(0, 1)$. Then,
$$h'(x) = fracf'(x)f(x) - fracmn fracf(1 - x)f'(1 - x)$$
so $h'(x) = 0$ if and only if $x$ is a suitable choice of $xi$. Since $lim_x to 0^+ f(x) = 0$ and $lim_x to 1^- f(x) = f(1) > 0$, we have that
$$lim_x to 0^+ h(x) = lim_x to 1^- h(x) = -infty.$$
It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $xi$ you're looking for.
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
MVT method:
Consider an auxiliary function
$$
F(x) =f(x)^n f(1-x)^m, quad x in [0,1],
$$
then Rolle's theorem achieves the goal.
answered Aug 22 at 5:22
xbh
2,627114
add a comment |Â
up vote
5
down vote
accepted
MVT method:
Consider an auxiliary function
$$
F(x) =f(x)^n f(1-x)^m, quad x in [0,1],
$$
then Rolle's theorem achieves the goal.
answered Aug 22 at 5:22
xbh
2,627114
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
MVT method:
Consider an auxiliary function
$$
F(x) =f(x)^n f(1-x)^m, quad x in [0,1],
$$
then Rolle's theorem achieves the goal.
answered Aug 22 at 5:22
xbh
2,627114
MVT method:
Consider an auxiliary function
$$
F(x) =f(x)^n f(1-x)^m, quad x in [0,1],
$$
then Rolle's theorem achieves the goal.
answered Aug 22 at 5:22
xbh
2,627114
answered Aug 22 at 5:22
xbh
2,627114
answered Aug 22 at 5:22
xbh
2,627114
answered Aug 22 at 5:22
xbh
2,627114
2,627114
add a comment |Â
add a comment |Â
up vote
1
down vote
Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.
Let $h(x) = ln(f(x)) + fracmn ln(f(1 - x))$, defined over $(0, 1)$. Then,
$$h'(x) = fracf'(x)f(x) - fracmn fracf(1 - x)f'(1 - x)$$
so $h'(x) = 0$ if and only if $x$ is a suitable choice of $xi$. Since $lim_x to 0^+ f(x) = 0$ and $lim_x to 1^- f(x) = f(1) > 0$, we have that
$$lim_x to 0^+ h(x) = lim_x to 1^- h(x) = -infty.$$
It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $xi$ you're looking for.
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
add a comment |Â
up vote
1
down vote
Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.
Let $h(x) = ln(f(x)) + fracmn ln(f(1 - x))$, defined over $(0, 1)$. Then,
$$h'(x) = fracf'(x)f(x) - fracmn fracf(1 - x)f'(1 - x)$$
so $h'(x) = 0$ if and only if $x$ is a suitable choice of $xi$. Since $lim_x to 0^+ f(x) = 0$ and $lim_x to 1^- f(x) = f(1) > 0$, we have that
$$lim_x to 0^+ h(x) = lim_x to 1^- h(x) = -infty.$$
It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $xi$ you're looking for.
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.
Let $h(x) = ln(f(x)) + fracmn ln(f(1 - x))$, defined over $(0, 1)$. Then,
$$h'(x) = fracf'(x)f(x) - fracmn fracf(1 - x)f'(1 - x)$$
so $h'(x) = 0$ if and only if $x$ is a suitable choice of $xi$. Since $lim_x to 0^+ f(x) = 0$ and $lim_x to 1^- f(x) = f(1) > 0$, we have that
$$lim_x to 0^+ h(x) = lim_x to 1^- h(x) = -infty.$$
It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $xi$ you're looking for.
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.
Let $h(x) = ln(f(x)) + fracmn ln(f(1 - x))$, defined over $(0, 1)$. Then,
$$h'(x) = fracf'(x)f(x) - fracmn fracf(1 - x)f'(1 - x)$$
so $h'(x) = 0$ if and only if $x$ is a suitable choice of $xi$. Since $lim_x to 0^+ f(x) = 0$ and $lim_x to 1^- f(x) = f(1) > 0$, we have that
$$lim_x to 0^+ h(x) = lim_x to 1^- h(x) = -infty.$$
It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $xi$ you're looking for.
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
answered Aug 22 at 5:16
Theo Bendit
12.3k1844
12.3k1844
add a comment |Â
add a comment |Â
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Actually I think this form is about the application of Mean Value Theorem.
– Jaqen Chou
Aug 22 at 4:40
Yeah, you are right, but what have you tried?
– xbh
Aug 22 at 5:05
Apply Cauchy's Mean Value Theorem.
– Anik Bhowmick
Aug 22 at 6:08