# please i need help here

### please i need help here

who can help me with this
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Guest

### Re: please i need help here

There appear to be some mistakes in your attachment. Assuming your question is:
minimize $$J = (0.18-b)^2+(0.33-b(1-a))^2+(0.45-b(1-a+a^2))^2$$
$$a=-0.819145705784351526557732...$$
$$b= 0.180862291009977588342398...$$

By differentiating $$J$$ with respect to $$a$$ you get
$$2(0.33-b(1-a))b+2(0.45-b(1-a+a^2))b(1-2a)$$
which should be $$0$$ at the minimum.
Rearranging gives:
$$2b(0.78-0.9a-b(2-4a+3a^2-2a^3)) = 0$$
So either $$b=0$$ or
$$b = (0.78-0.9a)/(2-4a+3a^2-2a^3)$$
(It is easy to check at the end that $$b=0$$ doesn't give the global minimum.)

Similarly by considering $$dJ/db=0$$ we get
$$b=(0.96-0.78a+0.45a^2)/(3-4a+4a^2-2a^3+a^4)$$.
Which means

$$(0.78-0.9a)/(2-4a+3a^2-2a^3) = (0.96-0.78a+0.45a^2)/(3-4a+4a^2-2a^3+a^4)$$
Which rearranges and simplifies to
$$11a^4-30a^3+6a^2+14a-14 = 0$$

Quartic equations can be solved exactly
https://en.wikipedia.org/wiki/Quartic_f ... _for_roots
however the exact solution is very complicated. The numerical answer you get is
$$a=-0.819145705784351526557732...$$ which means $$b= 0.180862291009977588342398...$$
(or $$a=2.3654...$$ which you can easily check does not give the global minimum).

Hope this helped,

R. Baber.
Guest

### Re: please i need help here

Sorry, but I'm unable to understand that what you have written in the attachment.

leesajohnson

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