Every x is making f(x) positive
- Thread starter shahar
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You could use the start of the expansion of [MATH]e^{x^2}[/MATH]
[MATH]e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...[/MATH][MATH]\therefore e^{x^2}=1+x^2+\frac{x^4}{2!}+...+\frac{x^{2n}}{n!}+.[/MATH]..
(if you'll pardon the informal notation)
Since all the powers are even, it is clear that [MATH]e^{x^2}[/MATH] is indeed positive - in fact every term is positive when x≠0, so:
[MATH]e^{x^2} > 1+x^2[/MATH], when x≠0
See how you go from there.
[MATH]e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...[/MATH][MATH]\therefore e^{x^2}=1+x^2+\frac{x^4}{2!}+...+\frac{x^{2n}}{n!}+.[/MATH]..
(if you'll pardon the informal notation)
Since all the powers are even, it is clear that [MATH]e^{x^2}[/MATH] is indeed positive - in fact every term is positive when x≠0, so:
[MATH]e^{x^2} > 1+x^2[/MATH], when x≠0
See how you go from there.
Dr.Peterson
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Have you tried finding the minimum and maximum?
The reason the denominator is positive is that it is a power of a positive constant. You omitted the most important word.
Dr.Peterson
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Well, x can be any real number; that's not what you're asking. Are you saying that y can be 0? That would contradict what you want to show; and it isn't true.
Please show HOW you came to that conclusion. It certainly isn't true that (e^0 - 0) = (1 - 1) !
Dr.Peterson
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Nonsense. All you've found is that one value of y is 1!
Please try actually finding the minimum; you do know some calculus, right?
Or try a different method if you see one.
Maybe tell how you find the minimum and maximum of the function. Like, show the steps that you did.
After that, it is pretty easy to see.
For the sake of completeness, I'll post the rest of my solution.
(I don't think this method will be considered in any case).
[MATH]e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...[/MATH]
[MATH]\therefore e^{x^2}=1+x^2+\frac{x^4}{2!}+...+\frac{x^{2n}}{n!}+.[/MATH]..
Since all the powers are even, it is clear that [MATH]e^{x^2}[/MATH] is indeed positive - in fact every term is positive when x≠0, so:
[MATH]e^{x^2} > 1+x^2[/MATH], when x≠0
So the top line: [MATH]e^{x^2}-2x > 1+x^2 -2x =(1-x)^2\ge 0[/MATH], when x≠0
[MATH]\therefore[/MATH] top line > 0 when x≠0
and top line =1 when x=0, so top line is positive for all x, and therefore f(x)>0 for all x.
(I don't think this method will be considered in any case).
[MATH]e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...[/MATH]
[MATH]\therefore e^{x^2}=1+x^2+\frac{x^4}{2!}+...+\frac{x^{2n}}{n!}+.[/MATH]..
Since all the powers are even, it is clear that [MATH]e^{x^2}[/MATH] is indeed positive - in fact every term is positive when x≠0, so:
[MATH]e^{x^2} > 1+x^2[/MATH], when x≠0
So the top line: [MATH]e^{x^2}-2x > 1+x^2 -2x =(1-x)^2\ge 0[/MATH], when x≠0
[MATH]\therefore[/MATH] top line > 0 when x≠0
and top line =1 when x=0, so top line is positive for all x, and therefore f(x)>0 for all x.