Completing the Square in Fourier Transform: Right or Wrong?

[ognik]

Hi - following an example in my book to calc. the Fourier Transform of the Gaussian. We need to complete the square to integrate the exponent, the power is $-a^2t^2+i\omega t$ ... - all good so far.

Trouble is when I complete the square I get \(\displaystyle -a^2(t-\frac{i\omega}{2a^2})^2 +\frac{\omega^2}{4a^4}\)

The book states \(\displaystyle -a^2(t-\frac{i\omega}{2a^2})^2 -\frac{\omega^2}{4a^2}\)

Could someone tell me which is right please?

 

[Prove It]

Hi - following an example in my book to calc. the Fourier Transform of the Gaussian. We need to complete the square to integrate the exponent, the power is $-a^2t^2+i\omega t$ ... - all good so far.

Trouble is when I complete the square I get \(\displaystyle -a^2(t-\frac{i\omega}{2a^2})^2 +\frac{\omega^2}{4a^4}\)

The book states \(\displaystyle -a^2(t-\frac{i\omega}{2a^2})^2 -\frac{\omega^2}{4a^2}\)

Could someone tell me which is right please?

If you expand your perfect square to check your answer, you will immediately see that the book's answer is right.

 

[MarkFL]

We begin with:

\(\displaystyle -a^2t^2+i\omega t\)

We can next factor out $-a^2$:

\(\displaystyle -a^2\left(t^2-\frac{i\omega}{a^2} t\right)\)

Next, take the coefficient of the linear term, divide by 2, square and the add inside the brackets and subtract on the outside:

\(\displaystyle -a^2\left(t^2-\frac{i\omega}{a^2} t+\left(\frac{i\omega}{2a^2}\right)^2\right)-\left(-a^2\left(\frac{i\omega}{2a^2}\right)^2\right)\)

Rewrite first term as square of binomial, and for the second term we have two negative signs and $i^2$, so it is negative:

\(\displaystyle -a^2\left(t-\frac{i\omega}{2a^2}\right)^2-a^2\left(\frac{\omega}{2a^2}\right)^2\)

Distribute into the second term:

\(\displaystyle -a^2\left(t-\frac{i\omega}{2a^2}\right)^2-\left(\frac{\omega}{2a}\right)^2\)

This is equivalent to what your book states. :)

 

[ognik]

What is bothering me is that I think the normal method to complete the square needs adjusting because of the $i^2$

If I expand the 1st term I get: $ \left( t-\frac{i\omega}{2a^2}\right)\left( t-\frac{i\omega}{2a^2}\right) = t^2-\frac{i\omega}{a^2} -\frac{w^2}{4a^4} $

The last term gets a minus sign because of the $i^2$ so we should add that last term back?

 

[MarkFL]

What is bothering me is that I think the normal method to complete the square needs adjusting because of the $i^2$

If I expand the 1st term I get: $ \left( t-\frac{i\omega}{2a^2}\right)\left( t-\frac{i\omega}{2a^2}\right) = t^2-\frac{i\omega}{a^2} -\frac{w^2}{4a^4} $

The last term gets a minus sign because of the $i^2$ so we should add that last term back?

Don't forget to give the middle term $t$ as a factor and then distribute the $-a^2$! :)

 

[Deveno]

What is bothering me is that I think the normal method to complete the square needs adjusting because of the $i^2$

If I expand the 1st term I get: $ \left( t-\frac{i\omega}{2a^2}\right)\left( t-\frac{i\omega}{2a^2}\right) = t^2-\frac{i\omega}{a^2} -\frac{w^2}{4a^4} $

The last term gets a minus sign because of the $i^2$ so we should add that last term back?

This may well be true, but you have a $-a^2$ factor overall, which will change the sign.

 

[ognik]

Ah, that's it, forgetting the minus from the $a^2$, thanks guys