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I need to prove that 1+x =< e^x for any x>=0.

  • Thread starter MikeyA
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Homework Statement



The question reads very simply, show that ex [tex]\geq[/tex]1+x [tex]\forall[/tex] x > 0


Homework Equations



None to speak of. I am not allowed to use calculus, and this is why I am having problems.

The Attempt at a Solution



I tried to break it up into cases:

When x=0, ex=e0=1 1+x=1+0=1

Hence ex=1+x.

Then I decided to take the natural log of both sides and try my hand at the other three cases (which I think should be 0<x<1, x=1, and x>1):


ln(1+x)[tex]\leq[/tex]lnex
<=> ln(1+x)[tex]\leq[/tex]x

ln2<x is easy, but I cannot even imagine how I would be able to solve the other cases.
 
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Answers and Replies

  • #2
eumyang
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Then I decided to take the natural log of both sides and try my hand at the other two cases (which I think should be 0<x<1, x=1, and x>1:


ln(1+x)[tex]\leq[/tex]lnex
<=> ln(1+x)[tex]\leq[/tex]x

ln2<e is easy, but I cannot even imagine how I would be able to solve the other cases.
Instead of doing this, try using the Mean Value Theorem.

EDIT: Oops, didn't notice the restriction of not using Calculus. I guess that using the Taylor series expansion is out, too?
 
  • #3
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Doing this with the MVT requires a bit of knowledge how calculus works, since this is in the Precalc forum I’d use the Intermediate value theorem instead and a proof by contradiction.

Suppose for contradiction sake that e^x -1 – x < 0. What does the IMV tell you about e^x -1 – x = 0?

If e^x -1 – x = 0 can you isolate e^x to one side, and take the ln of this equation giving you x = ln(1 +x). Is this possible for any x? Hint graph both of these equations on the same paper.

How does this imply a contradiction?
 
  • #4
Dick
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Doing this with the MVT requires a bit of knowledge how calculus works, since this is in the Precalc forum I’d use the Intermediate value theorem instead and a proof by contradiction.

Suppose for contradiction sake that e^x -1 – x < 0. What does the IMV tell you about e^x -1 – x = 0?

If e^x -1 – x = 0 can you isolate e^x to one side, and take the ln of this equation giving you x = ln(1 +x). Is this possible for any x? Hint graph both of these equations on the same paper.

How does this imply a contradiction?
I'm not sure that helps. The OP could have graphed e^x-1-x to begin with. How do you even define 'e' without calculus?
 
  • #5
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I'm not sure that helps. The OP could have graphed e^x-1-x to begin with. How do you even define 'e' without calculus?
I have absolutely no idea :confused:. I've been told I'm not allowed to use anything with haven't learned in the course to this point (although the Professor busted out the Triangle Inequality recently, which was definitely not covered in any lecture), and I would be willing to wager that the IMV is out as well. Is there some way I can state it simply and internally?

In our course so far, we've spent most of our time on the least upper bound property and basic properties of sequences. I'm really stumped :(.

Anyways, at this moment I've given up on an awesome solution, so I plan on handing something vaguely passable in, and here's where I am at for now:

e^x[tex]\geq[/tex]1+x for all x[tex]\geq[/tex]0

equivalent to e^x-1-x[tex]\geq[/tex]0 for all x[tex]\geq[/tex]0

Proposition i: There is no x[tex]\geq[/tex]0 s.t. e^x-1-x<0

Proposition ii: Proposition i is false:

Suppose Proposition ii is true.

Then there exists x greater than or equal to 0 such that e^x-1-x < 0

<=> e^x-x < 1
<=> ln(e^x-x) < ln(1)
<=> ln(e^x-x) < 0

ln(y)<0 does not exist. Thus proposition ii is false, and thus proposition i is true.




What do you think? I'm probably just going to hand this in, so if it is junk, I'd love to know about it, if only for future reference.
 
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  • #6
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I'm not sure that helps. The OP could have graphed e^x-1-x to begin with. How do you even define 'e' without calculus?
Graphing e^x -1 –x typically takes calculus.

He most likely has learned to graph an equation in the form of a^x, ln(x ), and how shifting works. I’m sure he’s also gone over the fact that two equations are equal exactly where their graphs cross.

you may want to comment somewhere on the differences between ln(e^x-x) and e^x-1-x domain.
 
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  • #7
rock.freak667
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I'm not sure that helps. The OP could have graphed e^x-1-x to begin with. How do you even define 'e' without calculus?
Graphing e^x -1 –x typically takes calculus.
It'd be easier to just draw y=ex and y=1+x on the same page and see (1+x) is below ex. That does not use calculus.
 
  • #8
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ln(y)<0 does not exist. Thus proposition ii is false, and thus proposition i is true.
There are certainly values for x such that ln(x)<0 since ln(x) diverges to negative infinity as x approaches 0 from the right. For instance ln(1/e) = -1 < 0.

Dick is right: calculus is necessary for all the ways I know how to define e. Anyway, another idea I had is to use the power series for [tex] e^x [/tex]. Really 1+x is just a truncation of this series. For any x[tex]\geq[/tex] 0,

[tex]e^x = \sum_{n=0}^\infty \frac{x}{n!} = 1 + x + \frac{x^2}{2} + \cdots \geq 1+x \, .[/tex]

Of course, as you said, it depends on whether or not you've seen this in class before!
 
  • #9
Redbelly98
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I'm not sure if this would help, but ex can be defined in terms of the limit (1+1/N)Nx, as N→∞. I would argue that is not calculus; even though it is a limit, it is not a derivative or integral.

It'd be easier to just draw y=ex and y=1+x on the same page and see (1+x) is below ex. That does not use calculus.
Nor is it a proof.
 

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