# Inequality with e^x.

1. Sep 3, 2006

### MathematicalPhysicist

i need to prove that:
1+x+x^2/2!+...+x^n/n!<=e^x<=1+x+x^2/2!+...+x^n/n!+(e^x)x^(n+1)/(n+1)! for x>=0
without using the power sum of e^x.

the textbook hints that i should evaluate the integral $$\int_{0}^{x}e^udu$$ and then i should integrate over and over n times,
and obtain the upper and lower limits for this integral.
i did the first part for the lower limit and i got that:
1+x+x^2/2+...+x^n/n<=e^x
the lhs of the last inequality is ofcourse bigger or equal than 1+x+x^2/2!+...+x^n/n!
my problem is with the upper bound of this inequality.
how to obtain it without using the power sum?

2. Sep 3, 2006

### HallsofIvy

Staff Emeritus
Use repeated "integration by parts". In $$\int_0^x e^y dy$$ (I've changed u to y to be able to use u, v in the integration) let $u= e^y$, $dv= dy$. One each succeeding integration, let $u= e^y$, dv= whatever function is left times dy.

3. Sep 3, 2006

### MathematicalPhysicist

but how do i obtain the upper bound?
i already have lower bound, because i have e^x>=0
and thus if i keep integrating i get e^x-1>=0 and again e^x-x-1>=0 and thus up to n times integrating, but as i said this i did already my problem is with the upper bound.

4. Sep 3, 2006

### StatusX

Have you ever seen a proof of Taylor's theorem (about said series)? In the proof, you derive the first n terms of the Taylor series, plus an exact form for the remainder term (which is then usually approximated using the mean value theorem or something like that to show it goes to 0 as n goes to infinity). Here's one way to derive it for n=2:

$$f(x)=f(a) + \int_a^x f'(t) dt$$ (1)

$$f'(x)=f'(a) + \int_a^x f''(t) dt$$ (2)

plug (2) into (1):

$$f(x)=f(a) + \int_a^x \left(f'(a) + \int_a^x f''(t') dt' \right) dt$$

$$f(x)=f(a) + f'(a)(x-a) + \int_a^x \int_a^x f''(t') dt' dt$$

Last edited: Sep 3, 2006
5. Sep 3, 2006

### MathematicalPhysicist

this question shouldn't employ taylor's theorem, cause i still haven't got to it in this book.

6. Sep 3, 2006

### StatusX

Well the question is dangerously close to proving Taylor's theorem for f(x)=e^x. The proof I mentioned is simple enough that you could be expected to find it yourself without knowing Taylor's theorem, and you should be able to use it to recover those bounds.

7. Sep 3, 2006

### MathematicalPhysicist

but how do i use it here to get the upper bound?
i mean e^x=1+x+x^2/2!+...+x^n/n!
(but then im using taylor's theorem)
and thus i get that e^x<=e^x+e^x*x^(n+1)/(n+1)

so i think this question is actually using taylor's theorem in disguise.

8. Sep 3, 2006

### shmoe

From:

$$e^x=1+\int_{0}^{x}e^y dy$$

estimate the integral on the right using the fact that e^x is increasing, and you can get the bound for n=1. You can continue as StatusX has suggested.

I'd suggest induction on n though, you can avoid having to write multiple integrals this way. Just use the above relation and stick in the upper bound for n, you should be able to get the upper bound for n+1 this way.

9. Sep 3, 2006

### MathematicalPhysicist

but isn't this using taylor's theorem?
by the way the bound for $$\int_{0}^{1}e^udu<=e^x$$

10. Sep 3, 2006

### shmoe

Where? Nowhere in my post (or StatusX's for that matter)

Do you mean $$\int_{0}^{x}e^udu<=e^x$$? If so sure, but will this help you get e^x<=1+x*e^x? Can you think of another way to bound this integral using the fact that e^x is increasing?

11. Sep 3, 2006

### StatusX

"using Taylor's theorem" would mean saying "e^x = 1+x+x^2/2!+...+R(x), where R(x)=blah due to Taylor's theorem." All you're using is the fundamental theorem of calculus. And the idea I was suggesting for getting the upper bound is to replace:

$$\int_0^x dt_1 \int_0^x dt_2 ... \int_0^x dt_n \frac{d^n}{d{t_n}^n} \left(e^{t_n} \right) = \int_0^x dt_1 \int_0^x dt_2 ... \int_0^x dt_n e^{t_n}$$

with something at least as big. The idea that pops out from looking at the bound they want is to replace $e^{t_n}$ by its maximum on the integration range. As shmoe points out, you can clean up the notation here if you use induction instead.

Last edited: Sep 3, 2006
12. Sep 3, 2006

### MathematicalPhysicist

one thing that i do understand is that i get this $$/int_{0}^{x}e^udu+1=e^x$$
and i need to integrate over and over n times, thus i get that:
$$e^x=1+\int_{0}^{x}1+\int_{0}^{x}+...\int_{0}^{x}e^ududu'du''...du^n$$
but besides this i dnot know how to get a smaller bound than e^x to this.
btw, shmoe
for e^x<=1+xe^x= 1+x(1+x+...+x^n/n!)=1+x+x^2+...+x^(n+1)/n!
where i need to get 1+x+...+e^x*x^(n+1)/(n+1)!=1+x+...+(x^(n+1)/(n+1)!+...+x^(2n+1)/n!(n+1)!) and im not sure how exactly to estimate this.

13. Sep 3, 2006

### shmoe

Look, you've got:

$$e^x=1+\int_0^x e^u du$$

and you'd like to show $$e^x\leq 1+ x e^x$$. can you see no way to show:

$$\int_0^x e^u du\leq xe^x$$

Consider my hint e^x is increasing, draw the graph if necessary, but you should be able to get the last inequalty.

This isn't an equality.

14. Sep 3, 2006

### MathematicalPhysicist

but i dont understand why i need to get to this inequality
$$\int_0^x e^u du\leq xe^x$$
does this show the upper bound here?

edit:
i understand now you also integrate the rhs whilst you integrate n times the given integral.

ok thanks, btw, does my appraoch for the lower bound is correct?

15. Sep 3, 2006

### shmoe

What I just said. When n=0 you want to show

$$e^x\leq 1+xe^x$$

right??

You know:

$$e^x=1+\int_0^x e^u du$$

So bounding the integral above by xe^x will get the inequality you are after.

16. Sep 3, 2006

### shmoe

It looks like you claim to have proven:

1+x+x^2/2+...+x^n/n<=e^x

if so, this is false, just look at n=3 and x =1. Your bound would imply e>=1+1+1/2+1/3=2.8333...