mathsbrain
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I am fine for the first part, not sure how to do the HENCE part...it could have been easily done with letting f(x)=e^x-1-a^2 and showing its always positive...but the HENCE...
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help for proofs (1 Viewer)
- Thread starter mathsbrain
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I am fine for the first part, not sure how to do the HENCE part...it could have been easily done with letting f(x)=e^x-1-a^2 and showing its always positive...but the HENCE...
Where did you get this question?
You probably wouldn't be able to get this result from just^n}/{n!})
as
^n}{n!} \stackrel{?}{\ge}1+n^2)
is not actually true, even taking
. You'd have to at least start a bit further back.
I doubt this is the intended answer but one hence-y way could be to cook up another integral in a similar fashion that gives you the inequality you want, e.g.
^2}{1+x^2} \, \mathrm dx \ge 0.)
You probably wouldn't be able to get this result from just
is not actually true, even taking
I doubt this is the intended answer but one hence-y way could be to cook up another integral in a similar fashion that gives you the inequality you want, e.g.
I'm pretty sure that the required substitution is 
or )
in which case the result ^n}{e^n})
becomes 
. We then need to bring in the integral result or some modification (I tried trapezoidal and it doesn't get us there) to get the final result.
I agree with mathbrain, it is certainly a true result as = e^x - x^2 - 1)
has  > 0)
for all 
and  = 0)
so  \ge 0)
(and thus 
) for all .
I agree with mathbrain, it is certainly a true result as
mathsbrain
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damn...this is tough, can anyone help?I'm pretty sure that the required substitution is
I agree with mathbrain, it is certainly a true result as
Extending from my previous post and correcting an error...
We start with the proven result:
^n}{n!} \\ e^n n! &> (n + 1)^n \\ n! &> \frac{(n+ 1)^n}{e^n} \end{align*})
And use the substitution
:
^n}{e^n} \\ &> \frac{(e^a)^n}{e^n} \\ &> e^{a \times n - n} \\ n! &> e^{n(a - 1)} \\ \ln n! &> n(a - 1) \text{ . . . . . . (1)} \end{align*})
Now, returning to the starting inequality, we had:
\ln (n + 1) - n < \ln (n + 1)!)
Applying the substitution:
\ln (n + 1) - n < \ln (n+1)! \\ \ln n! &< e^a \ln e^a - n < \ln [(n+1) \times n!] \\ \ln n! &< ae^a - n < \ln e^a + \ln n! \\ \text{Yielding two results: } \ln n! &< ae^a - n \text{ . . . . . . (2A)} \\ \text{And} \ln n! &> ae^a - n - a \text{ . . . . . . (2B)} \end{align*})
Now, equations (1) and (2A) can be combined to conclude that
... which it true, but not terribly useful:
 \\ ae^a - n &> na - n \\ ae^a &> na \\ e^a &> n \\ e^a &> e^a - 1 \\ 0 &> -1 \end{align*})
The approach can be modified, however, increasing or decreasing one side of an inequality where the change cannot falsify the statement. This is akin to saying that if
and that 
then I can say with certainty that 
. In this case, I start by noting that 
allows me to alter {2B):
 = (n + 1 - 1)a \\ &> (e^a - 1)a \\ ae^a &> ae^a - a \\ a &> 0 \end{align*})
Other variants include modifying (1A)'s term)
like :
 \\ ae^a &> na - 2n + n = n(a - 1) \\ &> (e^a - 1)(a - 1) \\ ae^a &> ae^a - e^a - a + 1 \\ e^a &> 1 - a \end{align*})
(a - 1) \\ &> (e^a - 1 - 1)(a - 1) = (e^a - 2)(a - 1) \\ ae^a &> ae^a - 2a - e^a + 2 \\ 0 &> 2 - 2a - e^a \\ e^a &> 2 - 2a \end{align*})
These are getting closer, though the negative coefficient on the
is a problem. As an aside, we would be there if we can get 
as this could be integrated from 0 to to give 
, which is the required result.
(a - 1) \\ 2ae^a &> 2(n - 1)(a - 1) + n = 2an - 2a - 2n + 2 + n = n(2a - 1) - 2a + 2 \\ &> (e^a - 1)(2a - 1) - 2a + 2 \\ 2ae^a &> 2ae^a - 2a - e^a + 1 - 2a + 2 \\ 0 &> 3 - 4a - e^a \\ e^a &> 3 - 4a \end{align*})
(a - 1) \\ ae^a &> an - a^2 - n + a - n = n(a - 2) - a^2 - a \\ &> = (e^a - 1)(a - 2) - a^2 - a \\ ae^a &> ae^a - a - 2e^a + 2 - a^2 - a \\ 0 &> 2 - 2a - a^2 - 2e^a \\ e^a &> \frac{2 - a - a^2}{2} \end{align*})
I don't see how this approach is going to yield any result)
where )
has positive coefficients on any terms in . 
I've also been thinking about the geometric interpretations here:
We start with the proven result:
And use the substitution
Now, returning to the starting inequality, we had:
Applying the substitution
Now, equations (1) and (2A) can be combined to conclude that
The approach can be modified, however, increasing or decreasing one side of an inequality where the change cannot falsify the statement. This is akin to saying that if
Other variants include modifying (1A)'s term
These are getting closer, though the negative coefficient on the
I don't see how this approach is going to yield any result
I've also been thinking about the geometric interpretations here:
- Consider the function
plotted from
to
where
. Add divisions (for upper and lower rectangles) with width of 1 unit.
is the area of the resulting lower rectangles,
are the upper rectangles, and so the inequality
- Taking
-axis covering
and one side on the
-axis covering
- The integral exists entirely within the rectangle, so that the area of the rectangle (
) is divided into the part under the curve
) meaning the area above the curve (which is also the area under the curve against the
- This provides insight into some of the inequalities and the reason for the substitution, but does not yield a solution.
- I do wonder, though, if we need to work against the
- but while still using the original result to satisfy the "hence" requirement.