''If the second derivative of f exists and f''(x)≥0 for all x ε I , then f is convex on the interval I "
though this is not the definition of convex function it is good enough to prove that a function is convex. No?
1 . What is a " convex " function ?
2 . Prove that " y = x 2 " is a convex function .
3 . Out of 15 persons , 6 are girls while 9 are boys . We want the boys to form a line with increasing heights , and also the girls , but the heights of the girls and the boys are not to be compared . Find the number of such arrangements .
I did all of them ! YAY !
''If the second derivative of f exists and f''(x)≥0 for all x ε I , then f is convex on the interval I "
though this is not the definition of convex function it is good enough to prove that a function is convex. No?
The word ''sufficient'' has got some mathematical meaning..
however i am sorry if i hurt your emotions for ISI.
@nasiko.. yes it is sufficient .. but that won't fetch a ISI feel. :)
why ?
f(x) = x2 f"(x) = 2 is not this sufficient to conclude f is convex?
f(x) = x4, f'(x) = 3x3 , f'(x) is increasing , is it not suff. to conclude thatf is convex?
''If the second derivative of f exists and f''(x)≥0 for all x ε I , then f is convex on the interval I "______Titu andreescu
'' if f is differentiable and f' is increasing then f is convex '' ___ Michael Spivak
In the given case, proving f"(x)≥0 will do.
But then in the prev question you would have to prove that this is equivalent to proving convexity
Take -
f ( x ) = x 4
It is most certainly a convex function . But is its double derivative always greater than zero ?
from wiki-pedia
"
The distinction between convex, strictly convex, and strongly convex can be subtle at first glimpse. If f is twice continuously differentiable and the domain is the real line, then we can characterize it as follows:
f\,\! convex if and only if f''(x) \ge 0 for all x\,\!
f\,\! strictly convex if f''(x) > 0 \,\! for all x\,\! (note: this is sufficient, but not necessary)
f\,\! strongly convex if and only if f''(x) \ge m > 0 for all x\,\!
"
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Interviews are this much easy?? Or these are just the bait towards the trench? [3]
Kreyszig , that is not the definition of a convex function .
Nasiko , your proof is correct . I did this one by showing that the area under the graph of the given parabola and any of its chord is greater than zero .
we have the inequality 2(x2+y2)≥(x+y)2
or , ((x+y)/2)2≤(x2+y2)/2
so?? [5]
Yeah , it is alright kreyszig . By the way , how would you attempt qs. no. 2 ?