Mixing
Rudin - Principles of math analysis, ex. 15 chapt. 5
$begingroup$
$newcommand{RR}{mathbb{R}}$
Suppose $a in RR, D = (a,+infty), f : D rightarrow RR$ differentiable two times on its domain, $M_0, M_1, M_2 in RR : |f(x)| leq M_0, |f'(x)| leq M_1, |f''(x)| leq M_2, forall x in D$. Prove that $M_1^2 leq 4 M_0 M_2$.
For Taylor theorem, if I choose $x_0 in D$ there exists $xi in D $ s.t. $ forall x in D$,
$$
-M_1 leq f'(x) = f'(x_0) + f''(xi)(x-x_0) leq M_1 + M_2(x-x_0)
$$
But if I choose $x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2 $ ($x, x_0 in D$ since I can shift $x, x_0$ by an arbitrary constant without changing their difference) I obtain the thesis.
Does this proof make sense? Is there any flaw?
real-analysis calculus proof-verification
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
$endgroup$
add a comment |
$begingroup$
$newcommand{RR}{mathbb{R}}$
Suppose $a in RR, D = (a,+infty), f : D rightarrow RR$ differentiable two times on its domain, $M_0, M_1, M_2 in RR : |f(x)| leq M_0, |f'(x)| leq M_1, |f''(x)| leq M_2, forall x in D$. Prove that $M_1^2 leq 4 M_0 M_2$.
For Taylor theorem, if I choose $x_0 in D$ there exists $xi in D $ s.t. $ forall x in D$,
$$
-M_1 leq f'(x) = f'(x_0) + f''(xi)(x-x_0) leq M_1 + M_2(x-x_0)
$$
But if I choose $x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2 $ ($x, x_0 in D$ since I can shift $x, x_0$ by an arbitrary constant without changing their difference) I obtain the thesis.
Does this proof make sense? Is there any flaw?
real-analysis calculus proof-verification
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
$endgroup$
add a comment |
$begingroup$
$newcommand{RR}{mathbb{R}}$
Suppose $a in RR, D = (a,+infty), f : D rightarrow RR$ differentiable two times on its domain, $M_0, M_1, M_2 in RR : |f(x)| leq M_0, |f'(x)| leq M_1, |f''(x)| leq M_2, forall x in D$. Prove that $M_1^2 leq 4 M_0 M_2$.
For Taylor theorem, if I choose $x_0 in D$ there exists $xi in D $ s.t. $ forall x in D$,
$$
-M_1 leq f'(x) = f'(x_0) + f''(xi)(x-x_0) leq M_1 + M_2(x-x_0)
$$
But if I choose $x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2 $ ($x, x_0 in D$ since I can shift $x, x_0$ by an arbitrary constant without changing their difference) I obtain the thesis.
Does this proof make sense? Is there any flaw?
real-analysis calculus proof-verification
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
$endgroup$
$newcommand{RR}{mathbb{R}}$
Suppose $a in RR, D = (a,+infty), f : D rightarrow RR$ differentiable two times on its domain, $M_0, M_1, M_2 in RR : |f(x)| leq M_0, |f'(x)| leq M_1, |f''(x)| leq M_2, forall x in D$. Prove that $M_1^2 leq 4 M_0 M_2$.
For Taylor theorem, if I choose $x_0 in D$ there exists $xi in D $ s.t. $ forall x in D$,
$$
-M_1 leq f'(x) = f'(x_0) + f''(xi)(x-x_0) leq M_1 + M_2(x-x_0)
$$
But if I choose $x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2 $ ($x, x_0 in D$ since I can shift $x, x_0$ by an arbitrary constant without changing their difference) I obtain the thesis.
Does this proof make sense? Is there any flaw?
real-analysis calculus proof-verification
real-analysis calculus proof-verification
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
edited Feb 1 at 9:06
YuiTo Cheng
2,3694937
2,3694937
asked Mar 10 '13 at 18:31
user32847
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1 Answer
1
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oldest
votes
$begingroup$
By Taylor's formula you have $$
f(x+2h)=f(x)+2hf'(x)+2h^{2}f''(w)$$ with $x<w<x+2h$.
Rewrite this as $$f'(x)=frac{f(x+2h)-f(x)}{2h}-hf''(w)$$
Take absolute values we then have
$$M_{1}<2M_{0}/2h+hM_{2}$$
But the function $g(h)=M_{0}/h+hM_{2}$ obtains maximum at $M_{2}=M_{0}/h^{2}$. This force $h^{2}=frac{M_{0}}{M_{2}}$. And the maximum is obtained at $2sqrt{M_{0}M_{2}}$.
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
$endgroup$
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
By Taylor's formula you have $$
f(x+2h)=f(x)+2hf'(x)+2h^{2}f''(w)$$ with $x<w<x+2h$.
Rewrite this as $$f'(x)=frac{f(x+2h)-f(x)}{2h}-hf''(w)$$
Take absolute values we then have
$$M_{1}<2M_{0}/2h+hM_{2}$$
But the function $g(h)=M_{0}/h+hM_{2}$ obtains maximum at $M_{2}=M_{0}/h^{2}$. This force $h^{2}=frac{M_{0}}{M_{2}}$. And the maximum is obtained at $2sqrt{M_{0}M_{2}}$.
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
$endgroup$
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
add a comment |
$begingroup$
By Taylor's formula you have $$
f(x+2h)=f(x)+2hf'(x)+2h^{2}f''(w)$$ with $x<w<x+2h$.
Rewrite this as $$f'(x)=frac{f(x+2h)-f(x)}{2h}-hf''(w)$$
Take absolute values we then have
$$M_{1}<2M_{0}/2h+hM_{2}$$
But the function $g(h)=M_{0}/h+hM_{2}$ obtains maximum at $M_{2}=M_{0}/h^{2}$. This force $h^{2}=frac{M_{0}}{M_{2}}$. And the maximum is obtained at $2sqrt{M_{0}M_{2}}$.
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
$endgroup$
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
add a comment |
$begingroup$
By Taylor's formula you have $$
f(x+2h)=f(x)+2hf'(x)+2h^{2}f''(w)$$ with $x<w<x+2h$.
Rewrite this as $$f'(x)=frac{f(x+2h)-f(x)}{2h}-hf''(w)$$
Take absolute values we then have
$$M_{1}<2M_{0}/2h+hM_{2}$$
But the function $g(h)=M_{0}/h+hM_{2}$ obtains maximum at $M_{2}=M_{0}/h^{2}$. This force $h^{2}=frac{M_{0}}{M_{2}}$. And the maximum is obtained at $2sqrt{M_{0}M_{2}}$.
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
$endgroup$
By Taylor's formula you have $$
f(x+2h)=f(x)+2hf'(x)+2h^{2}f''(w)$$ with $x<w<x+2h$.
Rewrite this as $$f'(x)=frac{f(x+2h)-f(x)}{2h}-hf''(w)$$
Take absolute values we then have
$$M_{1}<2M_{0}/2h+hM_{2}$$
But the function $g(h)=M_{0}/h+hM_{2}$ obtains maximum at $M_{2}=M_{0}/h^{2}$. This force $h^{2}=frac{M_{0}}{M_{2}}$. And the maximum is obtained at $2sqrt{M_{0}M_{2}}$.
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
answered Mar 11 '13 at 11:21
Bombyx moriBombyx mori
13.1k63076
13.1k63076
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
add a comment |
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
Yes, this is true, and I have already seen this proof. My question was about the way I tried to solve the problem choosing $ x - x_0 = 4 M_0 - dfrac{2M_1}{M_2} - M_1 ^2$. Does it makes sense? Or why it shouldn't?
$endgroup$
– user32847
Mar 11 '13 at 22:58
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
$begingroup$
The maximum may not be obtained, but the inequality would hold nevertheless.
$endgroup$
– Bombyx mori
Mar 12 '13 at 1:53
add a comment |
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