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Higher Level Integration Marathon & Questions (1 Viewer)
- Thread starter leehuan
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Paradoxica
-insert title here-
this feels like a meme
leehuan
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- 2015
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.
+a^2)\mathrm{d}x)
+2a}{1+2a\mathrm{cos}(x)+a^2}\mathrm{d}x = \frac{1}{a}\int_0^{\pi} (1-\frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2})\mathrm{d}x = \frac{\pi}{a}-\frac{1}{a}\int_0^{\pi} \frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2})
+a^2}\mathrm{d}x)
Let
+a^2}\mathrm{d}x = \int_0^{\infty} \frac{1-a^2}{1+2a\frac{1-t^2}{1+t^2}+a^2}\frac{2}{1+t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{1-a^2}{(1+2a+a^2)+(1-2a+a^2)t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{(1-a)(1+a)}{(1+a)^2+(1-a)^2t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{C}{C^2+t^2}\mathrm{d}t)
where
This is easily evaluated depending on the different cases:
If
or 
, we have 
, if 
then 
and 
, and if 
then 
and 
.
Case 1:
hence 
for some constant 
.
Evaluating the integral for
, we obtain that:
when .
Case or .
+a^2)\mathrm{d}x = \int_0^{\pi} \mathrm{ln}(1+2\frac{1}{a}\mathrm{cos}(x)+\frac{1}{a^2})\mathrm{d}x + \int_0^{\pi} 2\mathrm{ln}a\mathrm{d}x = 0 + 2\pi \mathrm{ln}(a))
(which follows from Case 1).
)
when or .
Case 3:
)\mathrm{d}x = \pi \mathrm{ln}2 + \int_0^{\pi} \mathrm{ln}(1-\mathrm{cos}(x))\mathrm{d}x)
Let)\mathrm{d}x)
By symmetry,)\mathrm{d}x+\int_0^{\frac{\pi}{2}} \mathrm{ln}(1+\mathrm{cos}(x))\mathrm{d}x=\int_0^{\frac{\pi}{2}} \mathrm{ln}(1-\mathrm{cos^2}(x))\mathrm{d}x=2\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(x))\mathrm{d}x = 2\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{cos}(x))\mathrm{d}x)
Let)\mathrm{d}x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(x))\mathrm{d}x)
\mathrm{sin}(x))\mathrm{d}x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2}\mathrm{sin}(2x))\mathrm{d}x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2})\mathrm{d}x+\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(2x))\mathrm{d}x + = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2})\mathrm{d}x+\frac{1}{2}\int_0^{\pi} \mathrm{ln}(\mathrm{sin}(u))\mathrm{d}u)
By symmetry,)\mathrm{d}u = 2\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(u))\mathrm{d}u)
Therefore + L)
, so )
.
Subbing the values back in, we obtain.
Case 4:
)\mathrm{d}x)
This can be done by the same method as case 3, and so.
Hence, when
, , and )
otherwise.
Let
where
This is easily evaluated depending on the different cases:
If
Case 1:
Evaluating the integral for
Case
Case 3:
Let
By symmetry,
Let
By symmetry,
Therefore
Subbing the values back in, we obtain
Case 4:
This can be done by the same method as case 3, and so
Hence, when
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Paradoxica
-insert title here-
Fourier Expansion
No idea what that is, I'm not all that far beyond HSC mathsFourier Expansion
Paradoxica
-insert title here-
Bonus Challenge: Prove the last two integrals are symmetric in (a,b) <=> (b,a) without evaluating it to the final result.
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Paradoxica
-insert title here-
For all positive real numbers, the iteration derived from Newton's Method applied to the equation q(x)=x²-2 results in the f(x) described above, and converges to the positive real root of the equation.
Shouldn't be hard assuming I didn't mess up typing the question.
Hence, fn(x) converges to √ ̅2 for x>0
The point x=0 can be discarded, as it is a set of zero measure.
Hence, the integral is equivalent to
My only problem with this question is that students don't know how to handle problematic points on the domain of an integral.