Proving N-derivative test
Suppose $f:(a,b) rightarrow mathbb{R}$ is differentialbe on $(a,b)$ and $c in (a,b)$ has $f(c) = f'(c) = dots = f^{n-1}(c) = 0$ and $f^{(n)}(c) > 0$. If $n$ is even, then $f$ has a local min at $c$.
My attempt: Consider the interval $a < c < beta < B$, then by Taylor's Theorem, there exists $x_0 in (c, beta)$ such that $f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n$.
This is the part where I am stuck, I know that normally $f(c) not = 0$ where you can prove with $f(beta) - f(c) > 0$ but since $f(c) = 0$ in this case, I have no idea where to proceed.
real-analysis
asked Nov 19 '18 at 3:55
HD5450
64
add a comment |
Suppose $f:(a,b) rightarrow mathbb{R}$ is differentialbe on $(a,b)$ and $c in (a,b)$ has $f(c) = f'(c) = dots = f^{n-1}(c) = 0$ and $f^{(n)}(c) > 0$. If $n$ is even, then $f$ has a local min at $c$.
My attempt: Consider the interval $a < c < beta < B$, then by Taylor's Theorem, there exists $x_0 in (c, beta)$ such that $f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n$.
This is the part where I am stuck, I know that normally $f(c) not = 0$ where you can prove with $f(beta) - f(c) > 0$ but since $f(c) = 0$ in this case, I have no idea where to proceed.
real-analysis
asked Nov 19 '18 at 3:55
HD5450
64
How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17
add a comment |
Suppose $f:(a,b) rightarrow mathbb{R}$ is differentialbe on $(a,b)$ and $c in (a,b)$ has $f(c) = f'(c) = dots = f^{n-1}(c) = 0$ and $f^{(n)}(c) > 0$. If $n$ is even, then $f$ has a local min at $c$.
My attempt: Consider the interval $a < c < beta < B$, then by Taylor's Theorem, there exists $x_0 in (c, beta)$ such that $f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n$.
This is the part where I am stuck, I know that normally $f(c) not = 0$ where you can prove with $f(beta) - f(c) > 0$ but since $f(c) = 0$ in this case, I have no idea where to proceed.
real-analysis
asked Nov 19 '18 at 3:55
HD5450
64
Suppose $f:(a,b) rightarrow mathbb{R}$ is differentialbe on $(a,b)$ and $c in (a,b)$ has $f(c) = f'(c) = dots = f^{n-1}(c) = 0$ and $f^{(n)}(c) > 0$. If $n$ is even, then $f$ has a local min at $c$.
My attempt: Consider the interval $a < c < beta < B$, then by Taylor's Theorem, there exists $x_0 in (c, beta)$ such that $f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n$.
This is the part where I am stuck, I know that normally $f(c) not = 0$ where you can prove with $f(beta) - f(c) > 0$ but since $f(c) = 0$ in this case, I have no idea where to proceed.
real-analysis
real-analysis
asked Nov 19 '18 at 3:55
HD5450
64
asked Nov 19 '18 at 3:55
HD5450
64
asked Nov 19 '18 at 3:55
HD5450
64
asked Nov 19 '18 at 3:55
HD5450
64
asked Nov 19 '18 at 3:55
HD5450
64
64
How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17
add a comment |
How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17
How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17
add a comment |
1 Answer
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Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $delta > 0 $ such that $f^n(x) > 0 $ for all $ x in(c-delta , c+ delta)$. Now for all $ beta in (c-delta , c+ delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n $$ for some $ x_0 in (c-delta , c+ delta) $ this shows $ f(beta) geq 0 = f(c) $ .
answered Nov 19 '18 at 4:12
Red shoes
4,706621
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
add a comment |
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Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $delta > 0 $ such that $f^n(x) > 0 $ for all $ x in(c-delta , c+ delta)$. Now for all $ beta in (c-delta , c+ delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n $$ for some $ x_0 in (c-delta , c+ delta) $ this shows $ f(beta) geq 0 = f(c) $ .
answered Nov 19 '18 at 4:12
Red shoes
4,706621
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
add a comment |
Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $delta > 0 $ such that $f^n(x) > 0 $ for all $ x in(c-delta , c+ delta)$. Now for all $ beta in (c-delta , c+ delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n $$ for some $ x_0 in (c-delta , c+ delta) $ this shows $ f(beta) geq 0 = f(c) $ .
answered Nov 19 '18 at 4:12
Red shoes
4,706621
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
add a comment |
Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $delta > 0 $ such that $f^n(x) > 0 $ for all $ x in(c-delta , c+ delta)$. Now for all $ beta in (c-delta , c+ delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n $$ for some $ x_0 in (c-delta , c+ delta) $ this shows $ f(beta) geq 0 = f(c) $ .
answered Nov 19 '18 at 4:12
Red shoes
4,706621
Assuming the n-th derivative of $f$ is continuous, $f^n(c) >0$ implies there exist $delta > 0 $ such that $f^n(x) > 0 $ for all $ x in(c-delta , c+ delta)$. Now for all $ beta in (c-delta , c+ delta) $, by writing taylor expansion around $ x = c$ we get :
$$ f(beta) = frac{1}{n!}f^{(n)}(x_0)(beta - c)^n $$ for some $ x_0 in (c-delta , c+ delta) $ this shows $ f(beta) geq 0 = f(c) $ .
answered Nov 19 '18 at 4:12
Red shoes
4,706621
answered Nov 19 '18 at 4:12
Red shoes
4,706621
answered Nov 19 '18 at 4:12
Red shoes
4,706621
answered Nov 19 '18 at 4:12
Red shoes
4,706621
4,706621
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
add a comment |
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
so we are considering the interval within this delta neighbourhood? Also, how does that imply $f(c)$ is a local min.
– HD5450
Nov 19 '18 at 4:21
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
yes. because $f(c) = 0 $ but $f geq 0 $ on entire that interval .
– Red shoes
Nov 19 '18 at 4:25
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
Ah, I get it now. Thank you for the help.
– HD5450
Nov 19 '18 at 4:30
add a comment |
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How many times is $f$ differentiable? $n$ or $n+1$?
– Jimmy R.
Nov 19 '18 at 4:16
$f$ is $n$ times differentiable.
– HD5450
Nov 19 '18 at 4:17