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Find condition on $t$ such that $ |xi(t) |leq 1$

0

$begingroup$

Let $a, b in mathbb C$, $x,theta in mathbb R$, $x>0$

I'm looking for a (strict, if possible) condition on $t > 0$ such that the number $|xi(t)| = |1 - frac2atx^2sin^2(fractheta2)+bt|$ will be no greater than $1$ for all $theta$.

What I tried:

Writing $a = a_R + ia_I, b = b_R + ib_I$ we thus have

$|xi(t)|^2 = |1 - frac2(a_R + ia_I)tx^2sin^2(fractheta2)+(b_R + ib_I)t|^2 = |(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)+i(-frac2a_Itx^2sin^2(fractheta2)+b_It)|^2=(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)^2+(-frac2a_Itx^2sin^2(fractheta2)+b_It)^2 = dots$

I got to a very ugly term with $t$ and $t^2$ and there's no obvious way to proceed.

calculus inequality complex-numbers

share|cite|improve this question

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

$endgroup$

add a comment | 

0

$begingroup$

Let $a, b in mathbb C$, $x,theta in mathbb R$, $x>0$

I'm looking for a (strict, if possible) condition on $t > 0$ such that the number $|xi(t)| = |1 - frac2atx^2sin^2(fractheta2)+bt|$ will be no greater than $1$ for all $theta$.

What I tried:

Writing $a = a_R + ia_I, b = b_R + ib_I$ we thus have

$|xi(t)|^2 = |1 - frac2(a_R + ia_I)tx^2sin^2(fractheta2)+(b_R + ib_I)t|^2 = |(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)+i(-frac2a_Itx^2sin^2(fractheta2)+b_It)|^2=(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)^2+(-frac2a_Itx^2sin^2(fractheta2)+b_It)^2 = dots$

I got to a very ugly term with $t$ and $t^2$ and there's no obvious way to proceed.

calculus inequality complex-numbers

share|cite|improve this question

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

$endgroup$

add a comment | 

$begingroup$

Let $a, b in mathbb C$, $x,theta in mathbb R$, $x>0$

I'm looking for a (strict, if possible) condition on $t > 0$ such that the number $|xi(t)| = |1 - frac2atx^2sin^2(fractheta2)+bt|$ will be no greater than $1$ for all $theta$.

What I tried:

Writing $a = a_R + ia_I, b = b_R + ib_I$ we thus have

$|xi(t)|^2 = |1 - frac2(a_R + ia_I)tx^2sin^2(fractheta2)+(b_R + ib_I)t|^2 = |(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)+i(-frac2a_Itx^2sin^2(fractheta2)+b_It)|^2=(1-frac2a_Rtx^2sin^2(fractheta2)+b_Rt)^2+(-frac2a_Itx^2sin^2(fractheta2)+b_It)^2 = dots$

I got to a very ugly term with $t$ and $t^2$ and there's no obvious way to proceed.

calculus inequality complex-numbers

share|cite|improve this question

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

$endgroup$

share|cite|improve this question

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

share|cite|improve this question

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

asked Apr 8 at 20:03

Rick JokerRick Joker

283214

asked Apr 8 at 20:03

Rick JokerRick Joker

283214