leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

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leehuan

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Seems a bit rigorous though to test for values... even though yes this is an MX2 question...
 

InteGrand

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Seems a bit rigorous though to test for values... even though yes this is an MX2 question...
They probably wouldn't mind in MX2 if you just assumed it was monotonically increasing, unless that Q. was worth a lot of marks.
 

leehuan

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f(x)sin(x) = c for some constant c is not doable algebraically right? (for f(x) is some polynomial)
 

seanieg89

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Another MX2 problem


Progress:





By symmetry about the y-axis, it suffices to show the circle just passes through S.

Since AB is the diameter of our circle, it suffices to show ASB is a right angle, which you can do just by multiplying gradients of AS and BS.

This does it v.quickly.
 
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leehuan

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By symmetry about the y-axis, it suffices to show the circle just passes through S.

Since AB is the diameter of our circle, it suffices to show ASB is a right angle, which you can do just by multiplying gradients of AS and BS.
Of course! I was hoping that someone would find a circle geometry trick for me. Thanks!
 

seanieg89

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Edit: Note that we have skirted over the issue of existence and uniqueness of a solution, as we are not given enough information about S or boundary conditions at infinity for this PDE to be well posed. (Note that adding to our solution will give us another solution for instance. This integral formula is the natural approach though if S is nice enough for our problem to be well posed, as the resulting solution will also vanish at infinity, and it will be the unique such solution.)
 
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leehuan

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I can't quit BoS I need seanieg and InteGrand and co.









Is this correct? If so, is there a neater way to express this?
 
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