Higher Level Integration Marathon & Questions (1 Viewer)

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: Extracurricular Integration Marathon

That's nice but it's not exactly an answer, since you haven't proven that the sum is the inverse hyperbolic sine of 1.
He used the Maclaurin series for the inverse hyperbolic sine (and the fact it holds at t = 1, which we can show for example by using Abel's theorem. We can show the series converges with the help of Stirling's approximation, for example).
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: Extracurricular Integration Marathon

One way (probably pretty inefficient so am not going to bother texing it) is to just differentiate with respect to the parameters.

I let x=a^2, y=b^2, and evaluated using the easy result that for a > 1.



(And chuck some absolute values around a and b if you want to allow them to be negative obviously.)
 
Last edited:

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: Extracurricular Integration Marathon

One way (probably pretty inefficient so am not going to bother texing it) is to just differentiate with respect to the parameters.

I let x=a^2, y=b^2, and evaluated using the easy result that for a > 1.



(And chuck some absolute values around a and b if you want to allow them to be negative obviously.)
You could combine that with the symmetry of the integral to cut through some of that I suppose....

I(a,b) = I(b,a), from reflection.
 

Mahan1

Member
Joined
Oct 16, 2016
Messages
87
Gender
Male
HSC
2014
Re: HSC 2017 MX2 Integration Marathon






 
Last edited:

pikachu975

Premium Member
Joined
May 31, 2015
Messages
2,739
Location
NSW
Gender
Male
HSC
2017
Re: HSC 2017 MX2 Integration Marathon

Add the two equations

B^3 + a^3 - 6B^2 - 6a^2 + 13B + 13a = 20
a^3 + B^3 - 6(B^2 + a^2) + 13(B+a) = 20
(a+B)(a^2 - aB + B^2) - 6(B^2 + a^2) + 13(B+a) = 20
(a+B)(a^2 - aB + B^2 - 6B^2 - 6a^2 + 13) = 20
(a+B)(-5a^2 - 5B^2 - aB + 13) = 20

Not sure what else to do but hopefully this helps someone but it probably won't
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: HSC 2017 MX2 Integration Marathon

Now that this is somewhere slightly more appropriate...

With the substitutions α = u+2, β = v+2, the following equations are obtained:

u³+u=-9, v³+v=9

f: x → x³+x is bijective, so there can only be one root.

Thus, u+v=0, since they have oppositely signed variables in the same equation.

Thus, α+β=4
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top