- Define a function:
Prove that for all and for all . Prove the following inequalities are valid for all ,
- Using integration and part (a) prove
- Using integration and part (a) prove
- State and prove a generalization of the inequalities above.
- First, we take the derivative of ,
Since is strictly increasing everywhere (since for all ) we have for and for . Thus, has a minimum at and so for all . Therefore, when ; hence,
Furthermore, when we have
Therefore, if then and so we have
- Using the inequalities in part (a) we integrate over the interval to ,
For the other inequality we proceed similarly,
- We use the same strategy as before, starting with the inequalities we established in part (b),
Similarly, for the other inequality,
- Claim: The following general inequalities hold for all :
Proof. The proof is by induction. We have already established the inequalities for the case for all three inequalities. Now, to prove the first inequality assume
Then we have
This establishes the first inequality. For the second inequality we have proved the case already. Assume
Since is odd we may write for some nonnegative integer . We want to show that the inequality must hold for the next odd integer, . We have,
We want to show that the inequality holds for so we integrate both sides again,
Hence, if the inequality holds for odd , then it also holds for the next odd number, . Hence, it holds for all positive, odd integers.
The exact same induction argument works for all of the even integers (starting with the case we proved in part (c))
It’s not possible that e^x is strictly larger than 1 + x + x^2 because that implies e^0=1>1. I think Apostol made a mistake. Also, regarding the indefinite integral inequality, the RH integral which includes the constant C, if C is less then 1 how the inequality is still valid after adding more than C to the LH side? I was thinking of taking derivatives which would yield the original inequality, then since E(0) equals the polynomial @0 then things make sense but this seems like an inductive proof which seems out of context of the problem. Hopefully I’m not missing something here…
Sorry, Apostol specified x>0. So lower limit of indefinite integral = 0?
Hi. So, yes Apostol specified x > 0 so there is no error from him. I made a huge mistake in the approach to the problem. Taking indefinite integrals and looking at the value of the constant was just completely wrong. I’ve fixed it now… We take definite integrals from 0 to $x$ and evaluate everything and it works out. I’m sure I’ve introduced new errors in fixing the whole thing. (Basically all of parts (b) – (d) were wrong so I rewrote them completely.)
(Also, I went ahead and incorporated your edits into your first comment and deleted the correction. Apparently, people cannot edit their own comments. I’ll try to work on the technology for that at some point, but definitely no promises on when that might get changed.)