Proof of a simple inequality

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Homework Statement


Prove [tex] 1 - x \leq e^{-x}[/tex] for [tex]0 \leq x \leq 1[/tex] by calculus.

The Attempt at a Solution


Sketching shows that the statement seems to be true.

Let's assume that
[tex]f(x) = x + e^{-x} -1[/tex], and
[tex]f(x) \geq 0[/tex].

If the function is continuous and increasing, then the function gets all the
values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.
Thus, the initial assumptions are true.

Please, point out any mistakes.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Prove [tex] 1 - x \leq e^{-x}[/tex] for [tex]0 \leq x \leq 1[/tex] by calculus.

The Attempt at a Solution


Sketching shows that the statement seems to be true.

Let's assume that
[tex]f(x) = x + e^{-x} -1[/tex], and
[tex]f(x) \geq 0[/tex].
Here's your basic mistake- you are assuming what you want to prove!
[tex]f(x)\ge 0[/tex] is the same as [tex]x+ e^{-x}- 1\ge 0[/itex] which is the same as [tex]x- 1\ge -e^{-x}[/tex] and, multiplying both sides by -1, the same as [tex]1- x\le e^{-x}[/tex]

If the function is continuous and increasing, then the function gets all the
values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.
Thus, the initial assumptions are true.

Please, point out any mistakes.
You need to learn that your whole concept here is WRONG! You cannot assume the conclusion and then prove that the assumptions are true.
 
  • #3
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You need to learn that your whole concept here is WRONG! You cannot assume the conclusion and then prove that the assumptions are true.

I changed the proof.

The Attempt at a Solution


We can write that
[tex]f(x) = x + e^{-x} -1[/tex].

We want to prove that
[tex]f(x) \geq 0[/tex].

If the function is continuous and increasing, then the function gets all the
values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.

Thus, the statement is true if we can prove the continuity and the fact the
function increases.

The function is continuous.

Proof:
At any point [tex] c \in \Re, f(c)[/tex] is defined.
The limit of [tex]f(x)[/tex] exists as [tex]x[/tex] approaches [tex]c[/tex]
either from the left or from the right, and it equals [tex]f(c)[/tex].

The function is clearly increasing, since [tex]f(0) < f(1)[/tex], and any point [tex]k[/tex] in the
interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex].

Thus, the statement must be true.

Please, point out any mistakes.
 
  • #4
epenguin
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The function is clearly increasing, since [tex]f(0) < f(1)[/tex], and any point [tex]k[/tex] in the
interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex].

Thus, the statement must be true.
You have not proved that as far as I can see.

You were asked to prove it by calculus.

You have not used any, but doing so would make it easy.
 
  • #5
HallsofIvy
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I changed the proof.

The Attempt at a Solution


We can write that
[tex]f(x) = x + e^{-x} -1[/tex].

We want to prove that
[tex]f(x) \geq 0[/tex].

If the function is continuous and increasing, then the function gets all the
values between the interval [0, [tex]e^{-1}[/tex]], by Boltzman's Min-Max theorem.

Thus, the statement is true if we can prove the continuity and the fact the
function increases.

The function is continuous.

Proof:
At any point [tex] c \in \Re, f(c)[/tex] is defined.
The limit of [tex]f(x)[/tex] exists as [tex]x[/tex] approaches [tex]c[/tex]
either from the left or from the right, and it equals [tex]f(c)[/tex].
Have you proved that limit? It might be that you are allowed to assert that ex is continuous and so f(x) is continuous. If not, you wil have to give a more detailed proof than simply stating that the limit is the value of the function. What, exactly, are you allowed to assume about ex for this problem?

The function is clearly increasing, since [tex]f(0) < f(1)[/tex], and any point [tex]k[/tex] in the
interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex].
Again, you cannot simply assert that "any point [tex]k[/tex] in the interval [0, 1] is less or equal to f(1) that is [tex]f(k) \leq f(1)[/tex]" Nor would proving that statement tell you anything useful; f might start out at f(0)= 0 drop below 0 before going back up to [itex]e^{-1}[/itex]. Simply showing that "f(0)< f(1) and f(x)< f(1) for all x in [0, 1]" does NOT show that the function is increasing.

What is f'(x)? What is true about f'(x) for x between 0 and 1? What does that tell you?

Thus, the statement must be true.

Please, point out any mistakes.
 
  • #6
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What is f'(x)? What is true about f'(x) for x between 0 and 1? What does that tell you?

I get
[tex]f'(x) = 1 - e^{-x}[/tex]

I see that
[tex]f'(0) = 0[/tex] and

[tex]f'(1) = 1 - e^{-1}[/tex],
which is greater than 0.

I note that f'(x) is the interval (0, 1] greater than zero.
This suggests me that the function is increasing.
(Is it this enough to prove that the function increases?)

Have you proved that limit? It might be that you are allowed to assert that ex is continuous and so f(x) is continuous. If not, you wil have to give a more detailed proof than simply stating that the limit is the value of the function.

It seems to be hard to prove for the general event.
I will try nevertheless.

We have a problem
[tex] lim_{x -> c} f(x) [/tex],
where [tex]c[/tex] is a real number and belongs to the interval [0, 1].

Proof:

Let [tex]\epsilon > 0[/tex].
We need to find [tex]\delta > 0[/tex] such that
[tex] 0 < |x - c| < \delta => |(x + e^{-x} -1) - C| < \epsilon[/tex],
where [tex]f(c) = C[/tex].

The choice is of [tex]\delta[/tex] is hard because we have a general proof.
Perhaps, we should prove first that the minimum point has a limit.
Then, we could similarly say that the maximum point has a limit.
Thus, it may be possible to say that each point in the interval has a limit.

Please, pinpoint my mistakes.
 
  • #7
epenguin
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I note that f'(x) is the interval (0, 1] greater than zero.
This suggests me that the function is increasing.
(Is it this enough to prove that the function increases?)

In a word, yes.
At least as long as the function exists and is continuous in the interval of interest.

Something starts as 0 and increases ever after, it must be > 0 ever after surely QED? - I cannot see your problems unless it is that you have to express that simple idea in some difficult language, in which case I have butted in, but I hope the difficult language does not stop you seeing the simple ideas. :smile:
 
  • #8
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I cannot see your problems unless it is that you have to express that simple idea in some difficult language - -.

Please, let me know how I should do the proof correctly.
Do I need to use epsilon and delta in the proof of the limit?
 
  • #9
epenguin
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Please, let me know how I should do the proof correctly.
Do I need to use epsilon and delta in the proof of the limit?

I don't see you need epsilon, delta, or limit!
You are only trying to prove something is bigger than 0. If something increases it gets bigger.

Say your proof in your own words. If those are not the words you are supposed to say let your Prof. or someone here correct them.
 
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