# Thread: find m and b for min. expression.

1. ## find m and b for min. expression.

Find all real values of $m$ and $b$ for which $\sum_{j=1}^{50}(j^2+mj+b)^2$ is Minimum.

2. Originally Posted by jacks
Find all real values of $m$ and $b$ for which $\sum_{j=1}^{50}(j^2+mj+b)^2$ is Minimum.
First you will need to remember how to sum the the first n integers and their squares and the cubes.

$\displaystyle \sum_{j=1}^{n}j=\frac{n(n+1)}{2}$

$\displaystyle \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle \sum_{j=1}^{n}j^3=\left[\frac{n(n+1)}{2}\right]$

Now the function you want to minimize is

$\displaystyle f(m,b)=\sum_{j=1}^{50}(j^2+mj+b)^2$

$\displaystyle \frac{df}{dm}=\sum_{j=1}^{50}2(j^2+mj+b)j$

$\displaystyle \frac{df}{db}=\sum_{j=1}^{50}2(j^2+mj+b)$

$\displaystyle \frac{d^2f}{dm^2}=\sum_{j=1}^{50}2j^2$

$\displaystyle \frac{d^2f}{db^2}=\sum_{j=1}^{50}2$

$\displaystyle \frac{d^2f}{dmdb}=\frac{d^2f}{dbdm} = \sum_{j=1}^{50}2j$

Can you finish using the 2nd derivative test from here?

3. Originally Posted by TheEmptySet
First you will need to remember how to sum the the first n integers and their squares and the cubes.

$\displaystyle \sum_{j=1}^{n}j=\frac{n(n+1)}{2}$

$\displaystyle \sum_{j=1}^{n}j^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle \sum_{j=1}^{n}j^3=\left[\frac{n(n+1)}{2}\right]$

Now the function you want to minimize is

$\displaystyle f(m,b)=\sum_{j=1}^{50}(j^2+mj+b)^2$

$\displaystyle \frac{df}{dm}=\sum_{j=1}^{50}2(j^2+mj+b)j$

$\displaystyle \frac{df}{db}=\sum_{j=1}^{50}2(j^2+mj+b)$

$\displaystyle \frac{d^2f}{dm^2}=\sum_{j=1}^{50}2j^2$

$\displaystyle \frac{d^2f}{db^2}=\sum_{j=1}^{50}2$

$\displaystyle \frac{d^2f}{dmdb}=\frac{d^2f}{dbdm} = \sum_{j=1}^{50}2j$

Can you finish using the 2nd derivative test from here?
I should probably mention that all the derivatives involved are partial derivatives.