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The Proof is Trivial!

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Solution 344:

Spoiler

Solution 344

∑n≥014n+1−14n+2=∑n≥0∫01t4n(1−t) dt=∫01t−1t4−1 dt=ln⁡24+π8\displaystyle\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}
Reply 2242
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Zakee
Just by looking at the diverse ways people have approached Solution 344 validates the beauty of Mathematics.
Reply 2243
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Affine
Original post by Mladenov
Let C1C_{1} and C2C_{2} be externally tangent circles at MM and the radius of C2C_{2} is greater than the radius of C1C_{1}. Let AA be a point on C2C_{2} which does not lie on the line joining the centres of C1C_{1} and C2C_{2}. Let BB and CC be points on C1C_{1} such that ABAB and ACAC are tangent to C1C_{1}. The lines BMBM and CMCM intersect C2C_{2} again at EE and FF, respectively. Let DD be the point of intersection of the tangent to C2C_{2} at AA and the line EFEF. Prove that as AA varies, the locus of DD is a line.
Any sources/books on learning how to tackle questions like this, please? Thanks.
Original post by Lord of the Flies
Solution 344

∑n≥014n+1−14n+2=∑n≥0∫01t4n(1−t) dt=∫01t−1t4−1 dt=ln⁡24+π8\displaystyle\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}


Just out of interest when is switching the order of summations valid? Nice solution too. :smile:
Original post by bogstandardname
Just out of interest when is switching the order of summations valid? Nice solution too. :smile:


Thanks.

Spoiler

(edited 10 years ago)
Reply 2246
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henpen
Original post by Mladenov
Solution 158. Notice that the function is even.



Mladenov, which result obtains the second equality of the third line of solution 158? Is it some kind of elliptic integral?
Reply 2247
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henpen
Original post by Mladenov
Solution 158. Notice that the function is even.


In solving 158 (and similar problems), what quality of the integral made it obvious for you that Liebniz differentiation needed to be used, and how did you decide which constants to vary? I'm simply curious what to look out for.
Problem 345 **/***

Evaluate lim⁡x→0∫0tan⁡xtan⁡sin⁡t dt∫0sin⁡xln⁡(1+sin⁡t) dt\displaystyle \lim_{x \to 0} \frac{\displaystyle \int_{0}^{\tan x} \tan\sin t\ dt}{\displaystyle \int_{0}^{\sin x} \ln(1 + \sin t)\ dt}.
(edited 10 years ago)
Solution 345

L'H then tan⁡sin⁡tan⁡xln⁡(1+sin⁡sin⁡x)∼0tan⁡xln⁡(1+x)∼06x+2x36x−3x2→1\displaystyle\frac{\tan \sin \tan x}{\ln(1+\sin \sin x)}\sim_0 \frac{\tan x}{\ln (1+x)}\sim_0\frac{6x+2x^3}{6x-3x^2}\to 1
Original post by Lord of the Flies
Solution 345

L'H then tan⁡sin⁡tan⁡xln⁡(1+sin⁡sin⁡x)∼0tan⁡xln⁡(1+x)∼06x+2x36x−3x2→1\displaystyle\frac{\tan \sin \tan x}{\ln(1+\sin \sin x)}\sim_0 \frac{\tan x}{\ln (1+x)}\sim_0\frac{6x+2x^3}{6x-3x^2}\to 1


I don't like to question the master, but have you differentiated the integral correctly? Don't you have to include the derivative of the limit?
Original post by bananarama2
I don't like to question the master, but have you differentiated the integral correctly? Don't you have to include the derivative of the limit?


Yes, but this does not contribute to the limit, since it offers a factor sec⁡3x\sec^3 x which →1\to 1 as x→0x\to 0
Original post by Lord of the Flies
Yes, but this does not contribute to the limit, since it offers a factor sec⁡3x\sec^3 x which →1\to 1 as x→0x\to 0


Ahh I see. Thanks :smile:
How would you got about proving that the difference between the square numbers is a series of the odd numbers?
1 4 9 16 25 36...
+3 +5 +7 +9 +11...
Original post by Occams Chainsaw
How would you got about proving that the difference between the square numbers is a series of the odd numbers?
1 4 9 16 25 36...
+3 +5 +7 +9 +11...

∑n≥2n2−(n−1)2=∑n≥22n−1\displaystyle \sum_{n \geq 2} n^2 - (n-1)^2 = \sum_{n \geq 2} 2n - 1
(edited 10 years ago)
Original post by Felix Felicis
∑n≥1n2−(n−1)2=∑n≥12n−1\displaystyle \sum_{n \geq 1} n^2 - (n-1)^2 = \sum_{n \geq 1} 2n - 1


ffs. it's just induction :facepalm:
:colondollar:

I was trying Euler's :frown:
Original post by Occams Chainsaw
ffs. it's just induction :facepalm:
:colondollar:

I was trying Euler's :frown:


Huh? I don't quite see the induction used...
Original post by Occams Chainsaw
ffs. it's just induction :facepalm:
:colondollar:

I was trying Euler's :frown:

Simplicity is often the best. :wink: No induction though, just algebra, observe that n2−(n−1)2=2n−1n^2 - (n-1)^2 = 2n-1.

Lol - do you mean an Euler-esque method on how he solved the Basel problem?
(edited 10 years ago)
Original post by bananarama2
Huh? I don't quite see the induction used...

Felix didn't use induction. I saw his answer and realised what my friend was trying to get me to do...
We just started FP1 so he was trying to play with it!

(2n−1)=n2[br]n=1:1=1check![br][br](2k−1)=k2[br][br]k+1:2(k+1)−1=(k+1)2[br]2(k+1)−1=2(k+1)−1[br]k2+(2(k+1)−1)=(k+1)2[br]2(k+1)−1=(k+1)2(2n - 1) = n^2[br]n = 1: 1=1 check![br][br](2k - 1) = k^2 [br][br]k+1: 2(k + 1) - 1 = (k + 1)^2[br]2(k + 1) - 1 = 2(k + 1) - 1[br]k^2 + (2(k + 1) - 1) = (k + 1)^2 [br]2(k + 1) - 1 = (k + 1)^2
Not even sure if that makes sense right now. Brain = fried.
Original post by Felix Felicis
Simplicity is often the best. :wink: No induction though, just manipulation, observe that n2−(n−1)2=2n−1n^2 - (n-1)^2 = 2n-1.

Lol - do you mean an Euler-esque method on how he solved the Basel problem?

Gauss's method, not Euler. My lecturer got it wrong. I just had to google it. Arithmetic series! It's been a long, long day. Had lacrosse training early this morning and then rugby match before lunch! :redface:

EDIT: Felix, I am just not with it tonight. I'll come back tomorrow and try again! :biggrin:
(edited 10 years ago)
Original post by Lord of the Flies

Solution 345


Nice! You could also use L'hopital's twice. :tongue: (idk why the rhyme)

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