Mixing
Proving positivity of the exponential function
$begingroup$
Question. Without using the semigroup property ($mathrm{e}^{x}mathrm{e}^{y}=mathrm{e}^{x+y}$),
how can we show that $mathrm{e}^{x}>0$ for all $xinmathbb{R}$ only by using the series expansion?
Explanation.
From the series expansion of $mathrm{e}^{x}=sum_{k=0}^{infty}frac{x^{k}}{k!}$ for $xinmathbb{R}$, we see that $mathrm{e}^{x}>0$ for $xgeq0$.
Thus, if the series becomes negative, this can only happen for negative values of $x$.
So proving $mathrm{e}^{-x}$ for $x>0$ will complete the proof.
As the series converges uniformly on any compact interval $Isubsetmathrm{R}$, we can rearrange the terms of the series and write
$mathrm{e}^{-x}=lim_{ntoinfty}g_{n}(x)$ for $xgeq0$, where $g_{n}(x):=1+sum_{k=1}^{n}Big(frac{x^{2k}}{(2k)!}-frac{x^{2k-1}}{(2k-1)!}Big)$ for $xgeq0$ and $ninmathbb{N}$.
Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>frac{1}{mathrm{e}}$.
I need to prove the following.
Claim. There exists an increasing divergent sequence ${xi_{n}}subset(0,infty)$ such that $g_{n}$ is decreasing on $[0,xi_{n}]$ with $g_{n}(xi_{n})>0$ for $ninmathbb{N}$.
Strengthened Claim. $xi_{n}:=sum_{k=1}^{n}frac{1}{k}$ for $kinmathbb{N}$.
calculus real-analysis power-series special-functions exponential-function
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
$endgroup$
|
show 7 more comments
$begingroup$
Question. Without using the semigroup property ($mathrm{e}^{x}mathrm{e}^{y}=mathrm{e}^{x+y}$),
how can we show that $mathrm{e}^{x}>0$ for all $xinmathbb{R}$ only by using the series expansion?
Explanation.
From the series expansion of $mathrm{e}^{x}=sum_{k=0}^{infty}frac{x^{k}}{k!}$ for $xinmathbb{R}$, we see that $mathrm{e}^{x}>0$ for $xgeq0$.
Thus, if the series becomes negative, this can only happen for negative values of $x$.
So proving $mathrm{e}^{-x}$ for $x>0$ will complete the proof.
As the series converges uniformly on any compact interval $Isubsetmathrm{R}$, we can rearrange the terms of the series and write
$mathrm{e}^{-x}=lim_{ntoinfty}g_{n}(x)$ for $xgeq0$, where $g_{n}(x):=1+sum_{k=1}^{n}Big(frac{x^{2k}}{(2k)!}-frac{x^{2k-1}}{(2k-1)!}Big)$ for $xgeq0$ and $ninmathbb{N}$.
Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>frac{1}{mathrm{e}}$.
I need to prove the following.
Claim. There exists an increasing divergent sequence ${xi_{n}}subset(0,infty)$ such that $g_{n}$ is decreasing on $[0,xi_{n}]$ with $g_{n}(xi_{n})>0$ for $ninmathbb{N}$.
Strengthened Claim. $xi_{n}:=sum_{k=1}^{n}frac{1}{k}$ for $kinmathbb{N}$.
calculus real-analysis power-series special-functions exponential-function
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
$endgroup$
$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
1
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
1
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23
|
show 7 more comments
$begingroup$
Question. Without using the semigroup property ($mathrm{e}^{x}mathrm{e}^{y}=mathrm{e}^{x+y}$),
how can we show that $mathrm{e}^{x}>0$ for all $xinmathbb{R}$ only by using the series expansion?
Explanation.
From the series expansion of $mathrm{e}^{x}=sum_{k=0}^{infty}frac{x^{k}}{k!}$ for $xinmathbb{R}$, we see that $mathrm{e}^{x}>0$ for $xgeq0$.
Thus, if the series becomes negative, this can only happen for negative values of $x$.
So proving $mathrm{e}^{-x}$ for $x>0$ will complete the proof.
As the series converges uniformly on any compact interval $Isubsetmathrm{R}$, we can rearrange the terms of the series and write
$mathrm{e}^{-x}=lim_{ntoinfty}g_{n}(x)$ for $xgeq0$, where $g_{n}(x):=1+sum_{k=1}^{n}Big(frac{x^{2k}}{(2k)!}-frac{x^{2k-1}}{(2k-1)!}Big)$ for $xgeq0$ and $ninmathbb{N}$.
Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>frac{1}{mathrm{e}}$.
I need to prove the following.
Claim. There exists an increasing divergent sequence ${xi_{n}}subset(0,infty)$ such that $g_{n}$ is decreasing on $[0,xi_{n}]$ with $g_{n}(xi_{n})>0$ for $ninmathbb{N}$.
Strengthened Claim. $xi_{n}:=sum_{k=1}^{n}frac{1}{k}$ for $kinmathbb{N}$.
calculus real-analysis power-series special-functions exponential-function
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
$endgroup$
Question. Without using the semigroup property ($mathrm{e}^{x}mathrm{e}^{y}=mathrm{e}^{x+y}$),
how can we show that $mathrm{e}^{x}>0$ for all $xinmathbb{R}$ only by using the series expansion?
Explanation.
From the series expansion of $mathrm{e}^{x}=sum_{k=0}^{infty}frac{x^{k}}{k!}$ for $xinmathbb{R}$, we see that $mathrm{e}^{x}>0$ for $xgeq0$.
Thus, if the series becomes negative, this can only happen for negative values of $x$.
So proving $mathrm{e}^{-x}$ for $x>0$ will complete the proof.
As the series converges uniformly on any compact interval $Isubsetmathrm{R}$, we can rearrange the terms of the series and write
$mathrm{e}^{-x}=lim_{ntoinfty}g_{n}(x)$ for $xgeq0$, where $g_{n}(x):=1+sum_{k=1}^{n}Big(frac{x^{2k}}{(2k)!}-frac{x^{2k-1}}{(2k-1)!}Big)$ for $xgeq0$ and $ninmathbb{N}$.
Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>frac{1}{mathrm{e}}$.
I need to prove the following.
Claim. There exists an increasing divergent sequence ${xi_{n}}subset(0,infty)$ such that $g_{n}$ is decreasing on $[0,xi_{n}]$ with $g_{n}(xi_{n})>0$ for $ninmathbb{N}$.
Strengthened Claim. $xi_{n}:=sum_{k=1}^{n}frac{1}{k}$ for $kinmathbb{N}$.
calculus real-analysis power-series special-functions exponential-function
calculus real-analysis power-series special-functions exponential-function
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
edited Apr 12 '15 at 21:23
Yiorgos S. Smyrlis
63.3k1385163
63.3k1385163
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
asked Apr 5 '15 at 6:37
bkarpuzbkarpuz
500210
500210
$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
1
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
1
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23
|
show 7 more comments
$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
1
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
1
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23
$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
1
1
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
1
1
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23
|
show 7 more comments
4 Answers
4
active
oldest
votes
$begingroup$
A hyperbolic trigonometry approach. Set
$$
C(x)=sum_{k=0}^inftyfrac{x^{2k}}{(2k)!}quadtext{and}quad S(x)=sum_{k=1}^inftyfrac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $xinmathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
big(C^2(x)-S^2(x)big)'=2big(C(x)C'(x)-S(x)S'(x)big)=2big(C(x)S(x)-S(x)C(x)big)=0,
$$
and hence
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $xinmathbb R$,
$$
C(x)=sqrt{S^2(x)+1}>S(x).
$$
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
add a comment |
$begingroup$
Using termwise differentiation one finds that $exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $xmapsto e^x$, which is positive when $x=0$, is positive on its full domain ${mathbb R}$.
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
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In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
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I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
add a comment |
$begingroup$
Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$
Clearly $e^x = sum_k frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $mathbb{R}^+$. Consider the set $A = {x < 0 : e^x leq 0 }$ and assume that it is non empty.
Let $(x_n)_{nin mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L in A$ by continuity of $f$
$$ f(L) = f(lim_{n to infty} x_n) = lim_{n to infty} f(x_n) leq 0.$$
Therefore $a := sup A in A$ and $a < 0$ and $e^a leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = {c' leq a ; vert forall x in (c',a], ; ;f(x) = 0 }.$$ If $inf C = k > - infty$, then $exists delta > 0$ s.t. the interval $[k- delta, k + delta]$ around $k$ is such that $f > 0$ on $[k- delta, k)$ or $f < 0$ on $[k- delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-delta,k$). In both cases by the mean value theorem, $exists alpha in (k-delta/2,k)$ s.t.
$$ f'(alpha) = frac{f(k) - f(k- delta/2)}{delta/2} = - frac{f(k-delta/2)}{delta/2} $$
This is a contradiction since $f'(alpha) = f(alpha)$ and $f(k- delta/2)$ have the same sign and are both not equal to zero.
If $inf C' = - infty$ then $e^x = 0 forall x leq a.$ We must proceed differently:
Consider the function $F : mathbb{R}^+ rightarrow mathbb{R}: x mapsto F(x) = int_{a+x}^0 e^t ;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get
$$ F(x) = int_{x+a}^0 e^t dt = int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 ; forall t > a$ the exponential function is increasing on $(a, + infty)$ so
$$a + x > 0 Rightarrow e^{a + x} > e^{0} = 1 iff F(x) = 1 - e^{a+x} < 0.$$
But $ a + x > 0 Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = { x < 0 : e^x leq 0 } = varnothing$ and the exponential function is positive everywhere.
answered Jan 10 at 19:48
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
The series expansion is
$$
e^x=sum_{n=0}^inftyfrac{x^n}{n!}=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots
$$
For $xge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
$endgroup$
1
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
|
show 1 more comment
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
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active
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active
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$begingroup$
A hyperbolic trigonometry approach. Set
$$
C(x)=sum_{k=0}^inftyfrac{x^{2k}}{(2k)!}quadtext{and}quad S(x)=sum_{k=1}^inftyfrac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $xinmathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
big(C^2(x)-S^2(x)big)'=2big(C(x)C'(x)-S(x)S'(x)big)=2big(C(x)S(x)-S(x)C(x)big)=0,
$$
and hence
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $xinmathbb R$,
$$
C(x)=sqrt{S^2(x)+1}>S(x).
$$
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
add a comment |
$begingroup$
A hyperbolic trigonometry approach. Set
$$
C(x)=sum_{k=0}^inftyfrac{x^{2k}}{(2k)!}quadtext{and}quad S(x)=sum_{k=1}^inftyfrac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $xinmathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
big(C^2(x)-S^2(x)big)'=2big(C(x)C'(x)-S(x)S'(x)big)=2big(C(x)S(x)-S(x)C(x)big)=0,
$$
and hence
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $xinmathbb R$,
$$
C(x)=sqrt{S^2(x)+1}>S(x).
$$
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
add a comment |
$begingroup$
A hyperbolic trigonometry approach. Set
$$
C(x)=sum_{k=0}^inftyfrac{x^{2k}}{(2k)!}quadtext{and}quad S(x)=sum_{k=1}^inftyfrac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $xinmathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
big(C^2(x)-S^2(x)big)'=2big(C(x)C'(x)-S(x)S'(x)big)=2big(C(x)S(x)-S(x)C(x)big)=0,
$$
and hence
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $xinmathbb R$,
$$
C(x)=sqrt{S^2(x)+1}>S(x).
$$
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
$endgroup$
A hyperbolic trigonometry approach. Set
$$
C(x)=sum_{k=0}^inftyfrac{x^{2k}}{(2k)!}quadtext{and}quad S(x)=sum_{k=1}^inftyfrac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $xinmathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
big(C^2(x)-S^2(x)big)'=2big(C(x)C'(x)-S(x)S'(x)big)=2big(C(x)S(x)-S(x)C(x)big)=0,
$$
and hence
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $xinmathbb R$,
$$
C(x)=sqrt{S^2(x)+1}>S(x).
$$
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
answered Apr 5 '15 at 23:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.3k1385163
63.3k1385163
add a comment |
add a comment |
$begingroup$
Using termwise differentiation one finds that $exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $xmapsto e^x$, which is positive when $x=0$, is positive on its full domain ${mathbb R}$.
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
add a comment |
$begingroup$
Using termwise differentiation one finds that $exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $xmapsto e^x$, which is positive when $x=0$, is positive on its full domain ${mathbb R}$.
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
add a comment |
$begingroup$
Using termwise differentiation one finds that $exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $xmapsto e^x$, which is positive when $x=0$, is positive on its full domain ${mathbb R}$.
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
$endgroup$
Using termwise differentiation one finds that $exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $xmapsto e^x$, which is positive when $x=0$, is positive on its full domain ${mathbb R}$.
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
answered Apr 5 '15 at 11:15
Christian BlatterChristian Blatter
173k7113326
173k7113326
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
add a comment |
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
That's perfectly correct. However, I cannot use this for my purpose as my functions do not satisfy a homogeneous linear equation. But as I have mentioned before, the proof is correct.
$endgroup$
– bkarpuz
Apr 5 '15 at 11:18
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
In order to prove the Uniqueness part of Picard-Lindelof you NEED the semigroup properties of the exponential!
$endgroup$
– Yiorgos S. Smyrlis
Apr 7 '15 at 7:31
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
$begingroup$
I think that we can use the Uniqueness result by Peano, which requires the right-hand side function to be nonincreasing in $y$. As $mathrm{e}^{x}geq1$ is obvious for $xgeq0$, we need to show that $y:=mathrm{e}^{-x}>0$ for $xgeq0$. Note that $y^{prime}=-y$ (decreasing in $y$) and $y(0)=1$.
$endgroup$
– bkarpuz
Apr 7 '15 at 18:28
add a comment |
$begingroup$
Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$
Clearly $e^x = sum_k frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $mathbb{R}^+$. Consider the set $A = {x < 0 : e^x leq 0 }$ and assume that it is non empty.
Let $(x_n)_{nin mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L in A$ by continuity of $f$
$$ f(L) = f(lim_{n to infty} x_n) = lim_{n to infty} f(x_n) leq 0.$$
Therefore $a := sup A in A$ and $a < 0$ and $e^a leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = {c' leq a ; vert forall x in (c',a], ; ;f(x) = 0 }.$$ If $inf C = k > - infty$, then $exists delta > 0$ s.t. the interval $[k- delta, k + delta]$ around $k$ is such that $f > 0$ on $[k- delta, k)$ or $f < 0$ on $[k- delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-delta,k$). In both cases by the mean value theorem, $exists alpha in (k-delta/2,k)$ s.t.
$$ f'(alpha) = frac{f(k) - f(k- delta/2)}{delta/2} = - frac{f(k-delta/2)}{delta/2} $$
This is a contradiction since $f'(alpha) = f(alpha)$ and $f(k- delta/2)$ have the same sign and are both not equal to zero.
If $inf C' = - infty$ then $e^x = 0 forall x leq a.$ We must proceed differently:
Consider the function $F : mathbb{R}^+ rightarrow mathbb{R}: x mapsto F(x) = int_{a+x}^0 e^t ;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get
$$ F(x) = int_{x+a}^0 e^t dt = int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 ; forall t > a$ the exponential function is increasing on $(a, + infty)$ so
$$a + x > 0 Rightarrow e^{a + x} > e^{0} = 1 iff F(x) = 1 - e^{a+x} < 0.$$
But $ a + x > 0 Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = { x < 0 : e^x leq 0 } = varnothing$ and the exponential function is positive everywhere.
answered Jan 10 at 19:48
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$
Clearly $e^x = sum_k frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $mathbb{R}^+$. Consider the set $A = {x < 0 : e^x leq 0 }$ and assume that it is non empty.
Let $(x_n)_{nin mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L in A$ by continuity of $f$
$$ f(L) = f(lim_{n to infty} x_n) = lim_{n to infty} f(x_n) leq 0.$$
Therefore $a := sup A in A$ and $a < 0$ and $e^a leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = {c' leq a ; vert forall x in (c',a], ; ;f(x) = 0 }.$$ If $inf C = k > - infty$, then $exists delta > 0$ s.t. the interval $[k- delta, k + delta]$ around $k$ is such that $f > 0$ on $[k- delta, k)$ or $f < 0$ on $[k- delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-delta,k$). In both cases by the mean value theorem, $exists alpha in (k-delta/2,k)$ s.t.
$$ f'(alpha) = frac{f(k) - f(k- delta/2)}{delta/2} = - frac{f(k-delta/2)}{delta/2} $$
This is a contradiction since $f'(alpha) = f(alpha)$ and $f(k- delta/2)$ have the same sign and are both not equal to zero.
If $inf C' = - infty$ then $e^x = 0 forall x leq a.$ We must proceed differently:
Consider the function $F : mathbb{R}^+ rightarrow mathbb{R}: x mapsto F(x) = int_{a+x}^0 e^t ;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get
$$ F(x) = int_{x+a}^0 e^t dt = int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 ; forall t > a$ the exponential function is increasing on $(a, + infty)$ so
$$a + x > 0 Rightarrow e^{a + x} > e^{0} = 1 iff F(x) = 1 - e^{a+x} < 0.$$
But $ a + x > 0 Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = { x < 0 : e^x leq 0 } = varnothing$ and the exponential function is positive everywhere.
answered Jan 10 at 19:48
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$
Clearly $e^x = sum_k frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $mathbb{R}^+$. Consider the set $A = {x < 0 : e^x leq 0 }$ and assume that it is non empty.
Let $(x_n)_{nin mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L in A$ by continuity of $f$
$$ f(L) = f(lim_{n to infty} x_n) = lim_{n to infty} f(x_n) leq 0.$$
Therefore $a := sup A in A$ and $a < 0$ and $e^a leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = {c' leq a ; vert forall x in (c',a], ; ;f(x) = 0 }.$$ If $inf C = k > - infty$, then $exists delta > 0$ s.t. the interval $[k- delta, k + delta]$ around $k$ is such that $f > 0$ on $[k- delta, k)$ or $f < 0$ on $[k- delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-delta,k$). In both cases by the mean value theorem, $exists alpha in (k-delta/2,k)$ s.t.
$$ f'(alpha) = frac{f(k) - f(k- delta/2)}{delta/2} = - frac{f(k-delta/2)}{delta/2} $$
This is a contradiction since $f'(alpha) = f(alpha)$ and $f(k- delta/2)$ have the same sign and are both not equal to zero.
If $inf C' = - infty$ then $e^x = 0 forall x leq a.$ We must proceed differently:
Consider the function $F : mathbb{R}^+ rightarrow mathbb{R}: x mapsto F(x) = int_{a+x}^0 e^t ;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get
$$ F(x) = int_{x+a}^0 e^t dt = int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 ; forall t > a$ the exponential function is increasing on $(a, + infty)$ so
$$a + x > 0 Rightarrow e^{a + x} > e^{0} = 1 iff F(x) = 1 - e^{a+x} < 0.$$
But $ a + x > 0 Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = { x < 0 : e^x leq 0 } = varnothing$ and the exponential function is positive everywhere.
answered Jan 10 at 19:48
DigitalisDigitalis
528216
$endgroup$
Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$
Clearly $e^x = sum_k frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $mathbb{R}^+$. Consider the set $A = {x < 0 : e^x leq 0 }$ and assume that it is non empty.
Let $(x_n)_{nin mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L in A$ by continuity of $f$
$$ f(L) = f(lim_{n to infty} x_n) = lim_{n to infty} f(x_n) leq 0.$$
Therefore $a := sup A in A$ and $a < 0$ and $e^a leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = {c' leq a ; vert forall x in (c',a], ; ;f(x) = 0 }.$$ If $inf C = k > - infty$, then $exists delta > 0$ s.t. the interval $[k- delta, k + delta]$ around $k$ is such that $f > 0$ on $[k- delta, k)$ or $f < 0$ on $[k- delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-delta,k$). In both cases by the mean value theorem, $exists alpha in (k-delta/2,k)$ s.t.
$$ f'(alpha) = frac{f(k) - f(k- delta/2)}{delta/2} = - frac{f(k-delta/2)}{delta/2} $$
This is a contradiction since $f'(alpha) = f(alpha)$ and $f(k- delta/2)$ have the same sign and are both not equal to zero.
If $inf C' = - infty$ then $e^x = 0 forall x leq a.$ We must proceed differently:
Consider the function $F : mathbb{R}^+ rightarrow mathbb{R}: x mapsto F(x) = int_{a+x}^0 e^t ;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get
$$ F(x) = int_{x+a}^0 e^t dt = int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 ; forall t > a$ the exponential function is increasing on $(a, + infty)$ so
$$a + x > 0 Rightarrow e^{a + x} > e^{0} = 1 iff F(x) = 1 - e^{a+x} < 0.$$
But $ a + x > 0 Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = { x < 0 : e^x leq 0 } = varnothing$ and the exponential function is positive everywhere.
answered Jan 10 at 19:48
DigitalisDigitalis
528216
edited Jan 11 at 14:10
answered Jan 10 at 19:48
DigitalisDigitalis
528216
answered Jan 10 at 19:48
DigitalisDigitalis
528216
answered Jan 10 at 19:48
DigitalisDigitalis
528216
528216
add a comment |
add a comment |
$begingroup$
The series expansion is
$$
e^x=sum_{n=0}^inftyfrac{x^n}{n!}=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots
$$
For $xge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
$endgroup$
1
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
|
show 1 more comment
$begingroup$
The series expansion is
$$
e^x=sum_{n=0}^inftyfrac{x^n}{n!}=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots
$$
For $xge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
$endgroup$
1
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
|
show 1 more comment
$begingroup$
The series expansion is
$$
e^x=sum_{n=0}^inftyfrac{x^n}{n!}=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots
$$
For $xge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
$endgroup$
The series expansion is
$$
e^x=sum_{n=0}^inftyfrac{x^n}{n!}=1+x+frac{x^2}{2!}+frac{x^3}{3!}+frac{x^4}{4!}+cdots
$$
For $xge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
edited Apr 5 '15 at 13:49
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
answered Apr 5 '15 at 8:14
Alice RyhlAlice Ryhl
5,99011235
5,99011235
1
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
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I cannot say that the proof is rigorious.
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– bkarpuz
Apr 5 '15 at 12:37
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
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– bkarpuz
Apr 5 '15 at 13:59
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
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– Alice Ryhl
Apr 5 '15 at 14:02
|
show 1 more comment
1
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Put $x:=-20$ in your second displayed formula!
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– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
1
1
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
Put $x:=-20$ in your second displayed formula!
$endgroup$
– Christian Blatter
Apr 5 '15 at 9:41
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
@ChristianBlatter Right, okay, if $x<-1$ then you should group them with $1$ seperate, good point.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 9:45
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
$begingroup$
I cannot say that the proof is rigorious.
$endgroup$
– bkarpuz
Apr 5 '15 at 12:37
1
1
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
$begingroup$
Note that you have used the semi group property $mathrm{e}^{x}mathrm{e}^{-x}=mathrm{e}^{0}=1$, which is an infraction of the rule.
$endgroup$
– bkarpuz
Apr 5 '15 at 13:59
1
1
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
$begingroup$
@bkarpuz It becomes difficult to rearrange the terms for $x<-1$. It would be easier to prove that $e^xe^{-x}=1$ using the series expansion.
$endgroup$
– Alice Ryhl
Apr 5 '15 at 14:02
|
show 1 more comment
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$begingroup$
Do you have a special reason for not using that property?
$endgroup$
– PhoemueX
Apr 5 '15 at 6:39
$begingroup$
@PhoemueX Exactly, the functions I am working are of the series form and they do not have such nice properties. When I plot their graphics, they seem to be positive everywhere but I could not handle it. This pushed me back to the exponential function.
$endgroup$
– bkarpuz
Apr 5 '15 at 6:42
$begingroup$
Then say what functions you are working on! What is the point of asking us to do something else that you don't actually want?
$endgroup$
– user21820
Apr 5 '15 at 6:44
1
$begingroup$
With $p_n(x)=sum_{k=0}^n frac{x^k}{k!}$, it is true that $p_n$ is always positive when $n$ is even while $p_n$ has a unique (negative) root when $n$ is odd (this can be proved by induction). If you can show that that negative root tends to $-infty$, I think you're done. The identities $p_n(x) = frac{x^n}{n!}+p_{n-1}(x)$ and $p'_n(x) = p_{n-1}(x)$ are useful here.
$endgroup$
– Greg Martin
Apr 5 '15 at 19:58
1
$begingroup$
Fair point. But this isn't too hard: for a fixed negative $x$, the sequence ${p_n(x)colon n$ odd$}$ is increasing for $n>|x|$. So if one of the terms is strictly positive, so is the limit.
$endgroup$
– Greg Martin
Apr 12 '15 at 19:23