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Proof of e^x

Does anyone know how to prove that for all x>0,

e^x > 1+x

using just alevel maths and further maths knowledge. Ive seen proofs which consider f(x) = e^x-x-1 and show that it is an increasing function etc... But they then go on to use the mean-value theorem which i am not familiar with.

[Btw i am writing this from memory so could anyone clarify if it is either of these or could you have both really]:
e^x> 1+x for x>0
or e^x> or = to 1+x, for x> or equal to 0 (with equality at x=0).

Also, is this true for all x (so now considering negatives; on a quick google search people specify x>0 but e^x gets small very quick for as x gets more negative but is always greater than 0 so i can see it clearly applying for x<-1.What about -1<x<0 then.

I can see in the increasing fucntion proof that f(x) is only increasing for x>0 so i get that restriction but is there another proof which doesnt require that, because i thought it would be true for all x.
(edited 9 years ago)

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Reply 1
Consider the power series for e^x.
Reply 2
Also, this is true for all x. Another way to prove it is to consider the function y=e^x-x-1, note that this is positive as x goes to +/-infinity so it's minimum value is either positive (at infinity) or at a turning point. It is easy to calculate teh runing points of this function and show that they are not negative.
Reply 3
Original post by newblood
Does anyone know how to prove that for all x>0,

e^x > 1+x

using just alevel maths and further maths knowledge. Ive seen proofs which consider f(x) = e^x-x-1 and show that it is an increasing function etc... But they then go on to use the mean-value theorem which i am not familiar with.

[Btw i am writing this from memory so could anyone clarify if it is either of these or could you have both really]:
e^x> 1+x for x>0
or e^x> or = to 1+x, for x> or equal to 0 (with equality at x=0).

Also, is this true for all x (so now considering negatives; on a quick google search people specify x>0 but e^x gets small very quick for as x gets more negative but is always greater than 0 so i can see it clearly applying for x<-1.What about -1<x<0 then.

I can see in the increasing fucntion proof that f(x) is only increasing for x>0 so i get that restriction but is there another proof which doesnt require that, because i thought it would be true for all x.


Try plotting y=e^x and y=x+1 on a graph and, the y=e^x line should be above the other one. You might want to look at the points where the lines meet and points near to them.
By looking at the gradients of the line (dy/dx), if the y=e^x line, has a gradient steeper than the other line (e^x > 1) and it starts off above the other line, the 2 lines will never meet and one will always be above the other one.
Mean Value Theorem is degree level mathematics in England. It's actually quite an intuitive theorem. This diagram from Wiki basically shows it:
Attachment not found

It's just saying there'll always be a point on the curve where its derivative is the same as that of a straight line joining the end-points. Its applications aren't always quite as intuitive.

As james22 said, use the power series of e^x and the fact that x>0. That is probably the "cleanest" proof available to you. Consideration of the graphs and using gradients with some case analysis will also work.
Reply 5
Original post by james22
Consider the power series for e^x.


Original post by Erebusaur
Mean Value Theorem is degree level mathematics in England. It's actually quite an intuitive theorem. This diagram from Wiki basically shows it:
Attachment not found

It's just saying there'll always be a point on the curve where its derivative is the same as that of a straight line joining the end-points. Its applications aren't always quite as intuitive.

As james22 said, use the power series of e^x and the fact that x>0. That is probably the "cleanest" proof available to you. Consideration of the graphs and using gradients with some case analysis will also work.


With the power series though, (and i now remember that it was for all x as i initially thought of the power series method and became stuck) i would need to also prove that:

x^2/2!+x^4/4!+......>x^3/3!+x^5/5!+....

which i dont know how to
Reply 6
Original post by Erebusaur
Mean Value Theorem is degree level mathematics in England. It's actually quite an intuitive theorem. This diagram from Wiki basically shows it:
Attachment not found

It's just saying there'll always be a point on the curve where its derivative is the same as that of a straight line joining the end-points. Its applications aren't always quite as intuitive.

As james22 said, use the power series of e^x and the fact that x>0. That is probably the "cleanest" proof available to you. Consideration of the graphs and using gradients with some case analysis will also work.


i get what your saying about the mean value theorem but that link for the diagram doesnt work, could you put another maybe?
Original post by newblood
With the power series though, (and i now remember that it was for all x as i initially thought of the power series method and became stuck) i would need to also prove that:

x^2/2!+x^4/4!+......>x^3/3!+x^5/5!+....

which i dont know how to

I see what you're saying and I was thinking about this for a while too. But you don't need to show that because you have an extra condition.

e^x = 1 + x + x^2/2! + x^3/3! +...

So, e^x - x - 1 = x^2/2! + x^3/3! +... Now, if x is negative you have that all of the odd powers are negative and even powers are positive, which poses a problem. This seems like too much work for A-level, so you think about what else you know or are told. I'm trying not to make it too obvious.

Original post by newblood
i get what your saying about the mean value theorem but that link for the diagram doesnt work, could you put another maybe?




So, for some value between a and b, call it c, the gradient at c of f is equal to the gradient of that line.

More formally, if a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then c(a,b)\exists c \in (a, b) (there exists a value c between a and b) such thatf(c)=f(b)f(a)ba. f'(c) = \frac{f(b) - f(a)}{b-a} \, .
Reply 8
Original post by newblood
Does anyone know how to prove that for all x>0,
e^x > 1+x


For all x>0 the proof is simple

ex=n=0xnn!=1+x+x22!+x33!+.....>1+x=n=01xnn!\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+..... >1+x=\sum_{n=0}^{1} \frac{x^n}{n!}
(edited 9 years ago)
I would use gradients. The power series works for numbers greater than zero, but gradients seem to work for when x is less than zero as well.

The mean value theorem isn't really needed IMO :smile:

Posted from TSR Mobile
I made a post to reply, but it's pending approval from a moderator (I guess it's the picture URL I included).

You're right, majmuh24, but then the question only asks for the positive case (which is what I delicately tried to point towards in my pending post). I'm not entirely sure how I'd phrase a solution involving the graphs and gradients, though.
Reply 11
Original post by Erebusaur
I made a post to reply, but it's pending approval from a moderator (I guess it's the picture URL I included).

You're right, majmuh24, but then the question only asks for the positive case (which is what I delicately tried to point towards in my pending post). I'm not entirely sure how I'd phrase a solution involving the graphs and gradients, though.


yes thats true but if we do consider negative x as well then, the power series case wouldnt be possible due to what i mentioned above (well you would need to prove the extra bit)
Original post by newblood
yes thats true but if we do consider negative x as well then, the power series case wouldnt be possible due to what i mentioned above (well you would need to prove the extra bit)


Well, I'm not sure about how you would go about phrasing it, but we know we have equality of the functions and the gradients of both of them at x=0. From here on, the gradient of e^x slowly approaches 0 while the gradient of 1+x stays constant at 1, so logically it should be 'above', I'm not sure about how you would prove it but I hope you understand what I'm saying :smile:

Posted from TSR Mobile
Reply 13
Original post by newblood
yes thats true but if we do consider negative x as well then, the power series case wouldnt be possible due to what i mentioned above (well you would need to prove the extra bit)


For x > 0 the power series will do nicely to show that e^x > 1 + x.

As you have noticed, when x = 0 we get equality.

If x <= -1 then the RHS <= 0 and we know that e^x can never be <= 0 so we also have e^x > 1 + x.

This leaves -1 < x < 0 as the only tricky area to consider. There's probably a slick way of sorting out this case too, but it's a bit late in the evening for me to see it, so I'll leave you to ponder it :biggrin:

(If this is still going in the morning, I'll take a proper look :smile: )
Original post by newblood
With the power series though, (and i now remember that it was for all x as i initially thought of the power series method and became stuck) i would need to also prove that:

x^2/2!+x^4/4!+......>x^3/3!+x^5/5!+....

which i dont know how to


You only need to prove that if you are doing it for all x, that's why I showed the other method.
I can't think of an easy way to use the power series to show that exx+1,xe^x \geq x + 1, \forall x. Sorry!
For x0x \geq 0, use the power series as above. For x1 x \leq -1, it's trivial, since exe^x is negative. So you need only consider x(1,0)x \in (-1,0).
But let
[br]f:[1,0]R[br]f: [-1,0] \to \mathbb{R}
:xexx1[br]\,: x \mapsto e^x - x - 1[br]

Then f(x)=ex1f'(x) = e^x - 1 for all x[1,0]x \in [-1,0], and since xexx \mapsto e^x is monotonic increasing, ff' is likewise, so f(x)f(0)=11=0f'(x) \leq f'(0) = 1 - 1 = 0. Thus, ff is monotonic decreasing on this interval, so f(x)f(0)=11=0f(x) \geq f(0) = 1 - 1 = 0, giving the result.
(edited 9 years ago)
Original post by davros
For x > 0 the power series will do nicely to show that e^x > 1 + x.

As you have noticed, when x = 0 we get equality.

If x <= -1 then the RHS <= 0 and we know that e^x can never be <= 0 so we also have e^x > 1 + x.

This leaves -1 < x < 0 as the only tricky area to consider. There's probably a slick way of sorting out this case too, but it's a bit late in the evening for me to see it, so I'll leave you to ponder it :biggrin:

(If this is still going in the morning, I'll take a proper look :smile: )


Okay, now you can use the power series. You pair up terms like this:
x2n(2n)!+x2n+1(2n+1)!\frac{x^{2n}}{(2n)!} + \frac{x^{2n+1}}{(2n+1)!}. Since -1 < x < 0:

Unparseable latex formula:

1 > -x \rightarrow x^{2n} > -x^{2n+1} \rightarrow \frac{x^{2n}}{(2n+1)!} > -\frac{x^{2n+1}}{(2n+1)!}}



So, we have:
x2n(2n)!>x2n(2n+1)!>x2n+1(2n+1)!x2n(2n)!>x2n+1(2n+1)!\frac{x^{2n}}{(2n)!} > \frac{x^{2n}}{(2n+1)!} > -\frac{x^{2n+1}}{(2n+1)!} \rightarrow \frac{x^{2n}}{(2n)!} > -\frac{x^{2n+1}}{(2n+1)!}

Finally:
x2n(2n)!+x2n+1(2n+1)!>0\frac{x^{2n}}{(2n)!} + \frac{x^{2n+1}}{(2n+1)!} > 0.

Then you sum this from n=1 to infinity and you prove what was required for -1 < x < 0. I think that's right.
This has problems I believe:

Set ex=x+1.e^x = x + 1.

Then, 1+x+x22!+x33!+...=1+x1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = 1 + x

So, x22!+x33!+...=0\frac{x^2}{2!} + \frac{x^3}{3!} + ... = 0

This only has solution x = 0 (this is where the problem potentially is). Hence, either exx+1,xe^x \geq x + 1, \forall x or exx+1,xe^x \leq x + 1, \forall x because otherwise there would be two points of intersection.

Since ex>x+1e^x > x + 1 for x=2x = -2, exx+1xe^x \geq x + 1 \forall x.

Alternatively, the fact that exe^x is convex and meets x + 1 (which is not convex) at only one point works too.
Reply 19
Original post by BlueSam3
For x0x \geq 0, use the power series as above. For x1 x \leq -1, it's trivial, since exe^x is never negative. So you need only consider x(1,0)x \in (-1,0).
But let
[br]f:[1,0]R[br]f: [-1,0] \to \mathbb{R}
:xexx1[br]\,: x \mapsto e^x - x - 1[br]

Then f(x)=ex1f'(x) = e^x - 1 for all x[1,0]x \in [-1,0], and since xexx \mapsto e^x is monotonic increasing, ff' is likewise, so f(x)f(0)=11=0f'(x) \leq f'(0) = 1 - 1 = 0. Thus, ff is monotonic decreasing on this interval, so f(x)f(0)=11=0f(x) \geq f(0) = 1 - 1 = 0, giving the result.


I think you meant to say this :smile:

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