00:01
For this exercise, we have q of x is equal to lambda sub 1, c squared sub 1, plus lambda sub 2, c squared sub 2, plus, plus lambda p, c squared sub p, plus lambda p plus lambda p plus 1, c squared p plus 1, c squared p plus 1, plus dot dot dot and then we have plus lambda r c squared sub r dot dot dot plus lambda sub r c squared r plus sub r plus sub r plus lambda sub n c squared sub n and so we have q of x is equal to lambda sub 1, c squared up 1, plus lambda sub 2, c squared, sub 2, plus da, dot, dot, dot, plus lambda sub p, c squared sub p, plus lambda sub p plus lambda sub r c squared sub r for this exercise we have furthermore we now have that q c sub 1 w vector sub 1 plus c sub 2 w vector sub 2 plus dot dot dot plus c sub n w vector sub in this is equivalent to q c sub 1 minus vector v sub 1 divided by the root lambda 1.
02:56
And this is just going to continue to be to c sub in vector v sub in.
03:06
And then here this is going to equal q of c sub 1 over root.
03:15
And this is the absolute value of lambda 1, times vector v sub 1 plus plus and this is c sub in vector v sub in and this is equal to lambda sub 1 c sub 1 over the root of lambda sub 1 squared plus dot dot dot and this will go all the way to 0 times c squared sub r plus 1 plus dot, dot, plus zero times c squared, sub in...