S.O.S. Mathematics CyberBoard

You know what is, not just and .

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Yes, φ(s)=f(x(s))=f((1-s) x_0 + s x_1). But how would that help?

First of all, am I on the right track by adding φ'(0)+φ''(0)≥0? If not, can you give me some hints on what to do next? It does not seem to lead to a contradiction.

Thanks.

Struggled for another day...still can't get the proof to work.

So I've shown that
0 = φ'(0) = x1 o Qx0 - x0 o Qx0 - b o (x1-x0)
and 0 ≤ φ''(0) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0

=> φ'(0)+φ''(0)≥0

I think from here I need to show that f(x1)-f(x0)≥0. This would be a contradiction.
But the problem is that the expression for φ'(0)+φ''(0) is a mess and doesn't seem to simplify to f(x1)-f(x0). What else can I do from here??

I know φ(s)=f(x(s))=f((1-s) x_0 + s x_1), but this may be negative or positive or zero.

Would someone be nice enough to help me out? Any input/comments is appreciated!

0 = φ'(0) = x1 o Qx0 - x0 o Qx0 - b o (x1-x0)
and 0 ≤ φ''(0) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0

=> φ'(0)+φ''(0)≥0

I think from here I need to show that f(x1)-f(x0)≥0. (which is a contradiction)


Will a Taylor expansion work? But the problem is I think this will introduce a remainder term...how can I get rid of the remainder?

Thanks.

Fill in the blank: is a (blank) function of s.

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Is it a convex function? How can this be proved?

Can you show the second derivative is always positive?

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By chain rule, φ''(s) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0, is this correct?

So in particular, φ''(0) = x1 o Qx1 - x1 o Qx0 - x0 o Qx1 + x0 o Qx0 ≥0 (since φ has a local min at s=0)

But now the expression φ''(s) does NOT depend on s, then can I conclude from here that φ''(s) ≥0 for ALL s??? Is this a valid argument?

If so, then this must be a major breakthrough for me.