Question

Show that \(|e^{it} - 1 - it + \frac{t^2}{2}| \le t^2 \min(|t|,1)\) for all \(t\).

Answer 1

HI!!

Answer 2

i didn't try it, but is it a direct computation? i mean do you just take the absolute value and see what you get ?

Answer 3

Hey :)

Answer 4

no that doesnt seem to work hmm

Answer 5

I'm trying..
I've developed the cosine and sine function:
\(|e^{it}-1-it+t^2/2|\)
\( = |\cos(t)-(1-t^2/2) + i\sin(t) - it|\)
\( \le |\cos(t)-(1-t^2/2)| + |i\sin(t) - it|\)
\( = |\cos(t)-(1-t^2/2)| + |i||\sin(t) - t|\), and now,
\(= |R_c(t)| + |R_s(t)|\) where \(R_c(t)\) and \(R_s(t)\) are the rests in Taylor's series.

Answer 6

i think you have to square then

Answer 7

I'm bounding the rests and find \(|R_c(t)| \le t^3/6\) and \(|R_s(t)|\le \frac12t^2\).

Answer 8

i got this, not sure

t^4/4-t^2+t^2 cos(t)-2 sin(t)-2 cos(t)+3

Answer 9

Thanks. I don't see yet how to use the squared thing.

Answer 10

At the moment, I have \(|\dots| \le \frac{t^3}6 + \frac{1}{2}t^2 = t^2(\frac t6 + \frac12)\). That doesn't seem good enough.

Answer 11

t is a real number, right?

Answer 12

yes

Answer 13

I was noticing this similarity:

\[\frac{1}{2}(a+bi)^2 = \frac{a^2-b^2}{2} +iab\]
with
\[\frac{t^2}{2}-1-it\]

But didn't look very hard, might be something here though.

Answer 14

I was thinking using Pythagoras' thm for large t..

Answer 15

(I assume \(t>0\))
The point \((\frac{t^2}2, -t)\) has norm \(t\sqrt{\frac{t^2}4 + 1}\). I think it's "easy" to see that it's "very" smaller than \(t \times t = t^2\)...

Answer 16

That's ok for \(t\ge \sqrt{4/3}\). :/

Answer 17

Wait a sec, those look like the first 3 terms of the power series

Answer 18

Yes they are the first terms of the exponential, or, both the cosine and sine.. that's what I did above.. but I didn't get a bound sharp enough.

Answer 19

Since the LHS of the inequality amounts to the difference between the function \(e^{it}\) and its second order Taylor series about \(t=0\), you should be able to use Taylor's inequality, which in this case would suggest that
\[\left|e^{it}-\left(1+it-\frac{t^2}{2}\right)\right|\le\frac{M|t|^3}{3!}\]for some \(M\) that satisfies \(\left|\dfrac{\mathrm{d}^3}{\mathrm{d}t^3}e^{it}\right|\le M\). You can check that \(M=1\) in this case.

So now, you need to establish that
\[\frac{|t|^3}{6}\le t^2\min(|t|,1)\]which seems pretty easy.

Answer 20

For \(t\) in the unit disk, i.e. \(|t|<1\), the RHS reduces to \(|t|^3\), so the conclusion should be easy. The same goes for when \(t\) is outside of (or on) the unit disk, so that \(\min(|t|,1)=1\).

Answer 21

Thanks a lot. (Actually I didn't want to use the "complex" Taylor..)
But is \(\frac{|t|^3}6 \le t^2 \) for large \(t\)?

Answer 22

I suppose for that case, we're not really "near \(0\)", so Taylor's inequality need not apply. Forget what I said about \(|t|>1\)...

Answer 23

And you can ignore the fact that I'm using a complex series. The fact that the RHS has \(t^2\) without the absolute value totally slipped my mind.

Answer 24

\(|t|^3 + t^2 = t^2(|t|+1)\) .. and I first work with t>0, but you can see that the thing is completely symmetric for t<0. So, no problem with the absolute value.

Answer 25

Ok, maybe I can do this:
1) My geometric approach can deal with large \(t\): \(t\ge \sqrt{4/3}\), or say, \(t\ge 2\)...
2) @SithsAndGiggles 's answer is fine for \(|t|\) small. But thanks to the '6' in the denominator, as long as \(|t|\le 6\), I will have \(|t|/6 \le 1\). So!! Does it work when I stick everything together?

Answer 26

Okay, I got my way through it.

Answer 27

Thanks @SithsAndGiggles , @Kainui , @misty1212 :)