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Prove positivity of $f$ when $f'' > f$ [duplicate]
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If $,f''(x) ge f(x)$, for all $xin[0,infty),$ and $,f(0)=f'(0)=1$, then is $,f(x)>0$?
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I'm given that $f:mathbbR to mathbbR$ is twice differentiable and $f(0) = f'(0) =1$. Assuming that $f''(x) > f(x)$ everywhere show that $f(x) > 0$ for all $x$.
I know that $f$ and $f’$ are continuous ($f’’$ exists). Since $f(0) = f’(0)= 1$ there is some $delta > 0$ where $f(x), f’(x) > 0$ for $-delta leq x leq delta$. I tried using the second-order Taylor approximation to extend the interval but I cannot see how to show $f(x) > 0$ for all $x < -delta$.
calculus
asked Jul 20 at 20:52
WoodWorker
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Jul 21 at 0:47
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This question already has an answer here:
If $,f''(x) ge f(x)$, for all $xin[0,infty),$ and $,f(0)=f'(0)=1$, then is $,f(x)>0$?
1 answer
I'm given that $f:mathbbR to mathbbR$ is twice differentiable and $f(0) = f'(0) =1$. Assuming that $f''(x) > f(x)$ everywhere show that $f(x) > 0$ for all $x$.
I know that $f$ and $f’$ are continuous ($f’’$ exists). Since $f(0) = f’(0)= 1$ there is some $delta > 0$ where $f(x), f’(x) > 0$ for $-delta leq x leq delta$. I tried using the second-order Taylor approximation to extend the interval but I cannot see how to show $f(x) > 0$ for all $x < -delta$.
calculus
asked Jul 20 at 20:52
WoodWorker
42128
marked as duplicate by amWhy calculus
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Jul 21 at 0:47
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up vote
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up vote
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down vote
favorite
This question already has an answer here:
If $,f''(x) ge f(x)$, for all $xin[0,infty),$ and $,f(0)=f'(0)=1$, then is $,f(x)>0$?
1 answer
I'm given that $f:mathbbR to mathbbR$ is twice differentiable and $f(0) = f'(0) =1$. Assuming that $f''(x) > f(x)$ everywhere show that $f(x) > 0$ for all $x$.
I know that $f$ and $f’$ are continuous ($f’’$ exists). Since $f(0) = f’(0)= 1$ there is some $delta > 0$ where $f(x), f’(x) > 0$ for $-delta leq x leq delta$. I tried using the second-order Taylor approximation to extend the interval but I cannot see how to show $f(x) > 0$ for all $x < -delta$.
calculus
asked Jul 20 at 20:52
WoodWorker
42128
This question already has an answer here:
If $,f''(x) ge f(x)$, for all $xin[0,infty),$ and $,f(0)=f'(0)=1$, then is $,f(x)>0$?
1 answer
I'm given that $f:mathbbR to mathbbR$ is twice differentiable and $f(0) = f'(0) =1$. Assuming that $f''(x) > f(x)$ everywhere show that $f(x) > 0$ for all $x$.
I know that $f$ and $f’$ are continuous ($f’’$ exists). Since $f(0) = f’(0)= 1$ there is some $delta > 0$ where $f(x), f’(x) > 0$ for $-delta leq x leq delta$. I tried using the second-order Taylor approximation to extend the interval but I cannot see how to show $f(x) > 0$ for all $x < -delta$.
This question already has an answer here:
If $,f''(x) ge f(x)$, for all $xin[0,infty),$ and $,f(0)=f'(0)=1$, then is $,f(x)>0$?
1 answer
calculus
asked Jul 20 at 20:52
WoodWorker
42128
asked Jul 20 at 20:52
WoodWorker
42128
asked Jul 20 at 20:52
WoodWorker
42128
asked Jul 20 at 20:52
WoodWorker
42128
42128
marked as duplicate by amWhy calculus
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1 Answer
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A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^-x$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) geqslant 1$, which implies $f(x) geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^-x = [f'(x) - f(x)] e^x cdot e^-2x $.
We have, since $f'' >f$,
$$fracddxleft([f'(x) - f(x)] e^xright) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
answered Jul 20 at 21:07
RRL
43.7k42260
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^-x$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) geqslant 1$, which implies $f(x) geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^-x = [f'(x) - f(x)] e^x cdot e^-2x $.
We have, since $f'' >f$,
$$fracddxleft([f'(x) - f(x)] e^xright) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
answered Jul 20 at 21:07
RRL
43.7k42260
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
add a comment |Â
up vote
3
down vote
accepted
A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^-x$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) geqslant 1$, which implies $f(x) geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^-x = [f'(x) - f(x)] e^x cdot e^-2x $.
We have, since $f'' >f$,
$$fracddxleft([f'(x) - f(x)] e^xright) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
answered Jul 20 at 21:07
RRL
43.7k42260
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^-x$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) geqslant 1$, which implies $f(x) geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^-x = [f'(x) - f(x)] e^x cdot e^-2x $.
We have, since $f'' >f$,
$$fracddxleft([f'(x) - f(x)] e^xright) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
answered Jul 20 at 21:07
RRL
43.7k42260
A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^-x$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) geqslant 1$, which implies $f(x) geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^-x = [f'(x) - f(x)] e^x cdot e^-2x $.
We have, since $f'' >f$,
$$fracddxleft([f'(x) - f(x)] e^xright) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
answered Jul 20 at 21:07
RRL
43.7k42260
edited Jul 20 at 21:12
answered Jul 20 at 21:07
RRL
43.7k42260
answered Jul 20 at 21:07
RRL
43.7k42260
answered Jul 20 at 21:07
RRL
43.7k42260
43.7k42260
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
add a comment |Â
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
this is nice and I appreciate the intuition given in the "clue," +1
– qbert
Jul 20 at 21:12
add a comment |Â