Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to simplify the following equation, by completing the square: $$\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$$

As $\displaystyle\frac{\left(x-y\right)^{2}}{2(t-s)}+C$.

How can I do this? I can't seem to be able to deal with the fractions effectively when completing this square...

share|improve this question
    
There is no need to add evaluations like your parenthetical remark to question titles... –  Mariano Suárez-Alvarez Feb 14 '11 at 3:30
    
I'm sorry... I'm new to Stack Exchange. –  squ1d Feb 14 '11 at 3:31
    
@squid: If I am not wrong you are trying to do something with the normal distribution, you do not need to actually complete the square and work out the integrals and stuff. All you need to recognize is that you can add the expectation and variance of these independent normal random variables. –  user17762 Feb 14 '11 at 3:40
    
Yes Sivaram - that is correct. And I would have done that, but the professor has explicitly said to calculate it by direct integration - to my demise. I basically end up with two lines of a letter-salad and can't get anywhere. By the way, I don't have to turn this in since it was due last week, but it is annoying the hell out of me. –  squ1d Feb 14 '11 at 3:42
add comment

1 Answer 1

up vote 4 down vote accepted

The only way to have $$\frac{\left(x-z\right)^{2}}{2(u-s)}+\frac{(z-y)^{2}}{2(t-u)}$$ equal to $$\frac{\left(x-y\right)^{2}}{2(t-s)}+C$$ is to let $$C=\frac{(-s y+s z+t x-t z-u x+u y)^2}{2 (s-t) (s-u) (t-u)}.$$ I'm not sure this is completing a square, though.

For extra fun, the numerator in $C$ is exactly $$\left|\begin{array}{ccc}1&1&1\\s&t&u\\x&y&z\end{array}\right|^2,$$ while the denominator is $$-2\left|\begin{array}{ccc}1&1&1\\s&t&u\\s^2&t^2&u^2\end{array}\right|.$$

share|improve this answer
    
Thanks. This is what I had - but I have to integrate the exponential function to the power of this, and I really didn't want to... :/ –  squ1d Feb 14 '11 at 3:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.