1)Given a function defined on [a,b], we say it is null iff INT(a,b) f^2(x) dx=0.
Note, f(x)=0 is null on any interval however the converse is not true.
Show that if g(x) is any function defined on [a,b] and f(x) is a null function then,
INT(a,b)g(x)f(x) dx=0

2)Show that the diameter is the longest chord. (Not necessarily hard I want to see how many reasons you can come up with).

3)Let a,b be positive integers. Given necessary and sufficient conditions so that,
(ab)/(a+b)
Is a positive integer.

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And he (Elisha) went up from thence unto Bethel: and as he was going up by the way, there came forth little children out of the city, and mocked him, and said unto him, "Go up, thou bald head"; "go up, thou bald head". And he turned back, and looked on them, and cursed them in the name of the Lord. And there came forth two she-bears out of the wood, and tore up forty and two children of them.
Second Kings 2: 23-24

2)
angle subtended by it is greatest at the centre
half of diameter is greaterst(=radius, for rest chords, it is sqrt(r^2-a^2), where a is not zero)

2)
a=k
b=k(k-a)
where k-a+1 divides k

Keep Smiling
Malay

(1)

It seems false the way you have stated the problem. For consider (ix + n). It can easily be shown that

Int(a,b) (ix + n)^2 dx = 0

for

[a,b] = [-sqrt(3n^2), sqrt(3n^2)]

But with (g(x) = x) then

Int(a,b) x(ix + n) dx = 2in^3 * sqrt(3)


(2)

Center the circle on the origin and use a simple max value formula:

(d/dx) chord length = 0 or undefined

(d/dx) 2 * sqrt(r^2 - x^2) = 0 or undefined

-2x * (1 / sqrt(r^2 - x^2)) = 0 or undefined

x = -r, 0, r

Plug in the values

2 * sqrt(r^2 - (-r)^2) = 0

2 * sqrt(r^2 - 0) = 2*r

2 * sqrt(r^2 - r^2) = 0

So, the max value is obviously the line (x = 0) which passes through origin, a.k.a the circle's center.

And finally:

Chord that passes through center of circle = Diameter

Done-and-done.

I believe he's referring to real valued functions.

-Dan

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"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert

Oh. Well, if so, then the only valid f(x) would be (f(x) = 0). Because the square of any real function is going to be always positive, thus the only way to have zero area under the curve would be for the function to be a constant zero. And (0 * g(x) = 0).

I assumed his question was more complicated than that.

That's because you assumed the functions involved are continuous! It's pretty trivial in that case

If however they are assumed only integrable (as they should - ps. I love Hacker's problems, coz they are almost never "well posed", and that's exactly what makes a problem interesting ) than we can just apply Cauchy-Schwartz:



Q(uite)E(asily)D(oneth).


For 2, this came to my mind. Consider a tangent to the circle. Rotating this by the touching point, for an angle s ε (0,π) we obtain a chord of the circle, of length say l(s)>0. This function is defined on (0,π), is continuous, and can be extended to a function continuous on [0,π] by the (obvious) l(0)=l(π)=0. So it must attain a maximum in (0,π). We now see it is increasing for s<π/2 and decreasing for s>π/2, so the maximum is attained for s=π/2 - this chord is exactly the diameter.

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The only sure thing in this world
is the principle of uncertainty...

My PDE's are never well posed, however my problems are.
Exactly! My solution as well.

__________________
And he (Elisha) went up from thence unto Bethel: and as he was going up by the way, there came forth little children out of the city, and mocked him, and said unto him, "Go up, thou bald head"; "go up, thou bald head". And he turned back, and looked on them, and cursed them in the name of the Lord. And there came forth two she-bears out of the wood, and tore up forty and two children of them.
Second Kings 2: 23-24