Thread: Q4) h(x) = ax
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Old 04-29-2013, 02:37 AM
Ziad Hatahet Ziad Hatahet is offline
Join Date: Apr 2013
Location: San Francisco, CA
Posts: 23
Default Re: Q4) h(x) = ax

Originally Posted by vsuthichai View Post
How do you solve for the minimum a that produces the least squared error?
As far as I know, differentiate (ax_1-y_1)^2+(ax_2-y_2)^2 with respect to a, set it to 0, and solve for a.
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