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The Proof is Trivial!

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Reply 780
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies

Spoiler


That's awesome man! Could you point me in the direction of some of the books/ online stuff that led you to knowing all this at 18!!!! :eek: (the middle two paragraphs confuse me in particular (apart from the sum of sines). If you have any good source for talking about links between integrals and summations that would be awesome swell!) :smile:
(edited 10 years ago)
Original post by Jkn
That's awesome man! Could you point me in the direction of some of the books/ online stuff that led you to knowing all this at 18!!!! :eek: (the middle two paragraphs confuse me in particular (apart from the sum of sines). If you have any good source for talking about links between integrals and summations that would be awesome swell!) :smile:


Spoiler that quote, it'll make the page too cluttered otherwise :biggrin: I'm afraid don't have any books or articles to link you to - just read stuff off Wikipedia and learn it etc. About the maths itself, integrals and sums can be linked in several different ways. Sometimes it is easier to integrate something by rewriting it as a series, where each term is easily integrated (note: swapping the sum and integral sign is not always allowed, but don't worry about that). Conversely, sometimes it is easier to evaluate a sum by writing it as an integral (if poss.). Below are two relatively easy examples you can try:

Spoiler

Reply 782
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
I'm afraid don't have any books or articles to link you to - just read stuff off Wikipedia and learn it etc.
I feared you'd say that :lol:. So do you dedicate time to learning this stuff or just flick through when you're bored?
About the maths itself, integrals and sums can be linked in several different ways. Sometimes it is easier to integrate something by rewriting it as a series, where each term is easily integrated (note: swapping the sum and integral sign is not always allowed, but don't worry about that). Conversely, sometimes it is easier to evaluate a sum by writing it as an integral (if poss.). Below are two relatively easy examples you can try:

Spoiler


Hmm thank you, I'll have a go at them at some point :smile:

Are you making use of maclaurins aswell?

Also, as a side note, could you put a teeny tiny link to my STEP thread in your one? :colondollar: I'm thinking I need to try and spark some interest early on or it will die out :lol:
Original post by Jkn
I feared you'd say that :lol:. So do you dedicate time to learning this stuff or just flick through when you're bored?


Ha, I'm very disorganised - the thought of a schedule is frightening. In other words, the latter.

Original post by Jkn
Are you making use of maclaurins aswell?


Of course, that's the trick!

Original post by Jkn
Also, as a side note, could you put a teeny tiny link to my STEP thread in your one?


Added to the OP.
Reply 784
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
Ha, I'm very disorganised - the thought of a schedule is frightening. In other words, the latter.

Added to the OP.

Well kudos for reaching a decent level in these complicated topics :smile: Are you going to have covered a lot of the course for next year, then? :tongue:

****ing lad!
Original post by Jkn
Well kudos for reaching a decent level in these complicated topics :smile: Are you going to have covered a lot of the course for next year, then? :tongue:

To be fair, the content of PIA seems relatively tame (I'm sure the problem sheets will be fiendishly difficult though); I've seen several people on TSR say they've covered all the syllabus already - it certainly isn't rare. What I plan on doing is going through this over the summer. Some people I've spoken to say it is the bible of calculus. :biggrin:
Reply 786
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Jkn
11
Problem 114​*

Find ∫tanx.dx\displaystyle\int\sqrt{tanx}.dx
Solution 114

t=tan⁡x:∫tan⁡x dx=∫2t2t4+1  dt=∫t2+1t4+1  dt+∫t2−1t4+1  dt\displaystyle t=\sqrt{\tan x}: \int \sqrt{\tan x}\,dx=\int \frac{2t^2}{t^4+1}\;dt=\int\frac{t^2+1}{t^4+1}\;dt+\int\frac{t^2-1}{t^4+1}\;dt

∙  ∫t2−1t4+1  dt=C.S. & P.F.122∫2t−2t2−2t+1−2t+2t2+2t+1  dt\bullet\;\displaystyle\int\frac{t^2-1}{t^4+1}\;dt\overset{\text{C.S. \& P.F.}}=\dfrac{1}{2 \sqrt{2}}\int\frac{2t-\sqrt{2}}{t^2-\sqrt{2}t+1}-\frac{2t+\sqrt{2}}{t^2+\sqrt{2}t+1}\;dt

    =122ln⁡(t2−2t+1t2+2t+1)+L\qquad\qquad\qquad\;\;= \displaystyle \dfrac{1}{2 \sqrt{2}}\ln\left(\frac{t^2-\sqrt{2}t+1}{t^2+\sqrt{2}t+1} \right)+\mathcal{L}

∙  ∫t2+1t4+1  dt=C.S. & P.F.∫1(2t+1)2+1+1(2t−1)2+1  dt\bullet\;\displaystyle\int\frac{t^2+1}{t^4+1}\;dt\overset{\text{C.S. \& P.F.}}=\int\frac{1}{(\sqrt{2}t+1)^2+1}+\frac{1}{( \sqrt{2}t-1)^2+1}\;dt

    =12[arctan⁡(2t+1)+arctan⁡(2t−1)]+S\qquad\qquad\qquad\;\;= \displaystyle \frac{1}{\sqrt{2}}\left[\arctan (\sqrt{2}t+1)+\arctan (\sqrt{2}t-1)\right]+\mathcal{S}

I=12(arctan⁡(2tan⁡x+1)+arctan⁡(2tan⁡x−1)\displaystyle I=\frac{1}{\sqrt{2}}\Bigg( \arctan (\sqrt{2\tan x}+1)+ \arctan (\sqrt{2\tan x}-1)

      +ln⁡tan⁡x−2tan⁡x+1tan⁡x+2tan⁡x+1)+D\displaystyle \;\;\;+\ln \sqrt{\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2 \tan x}+1}}\Bigg)+\mathcal{D}

Seriously Jkn? :biggrin:
Reply 788
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
To be fair, the content of PIA seems relatively tame (I'm sure the problem sheets will be fiendishly difficult though); I've seen several people on TSR say they've covered all the syllabus already - it certainly isn't rare. What I plan on doing is going through this over the summer. Some people I've spoken to say it is the bible of calculus. :biggrin:

The problem sheets look tame! Look at numbers and sets! :bhangra:

It's the content that looks pretty fiendish I'd say! (i.e. new things like vector space, etc...)

That looks like a brick bro! :lol: Good luck! I think I may go for some of the "Intro to analysis"-type textbooks, bit of set-theory crap, Markov chains, etc..

Try my problem :colone:
Reply 789
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies

Spoiler


Seriously Jkn? :biggrin:

For ****s sake! I'm impressed you typed the latex that fast. As for the maths, you're a nutter! :lol: You better not have seen it before............. -.-
Original post by Jkn
For ****s sake! I'm impressed you typed the latex that fast. As for the maths, you're a nutter! :lol: You better not have seen it before............. -.-


I think I type LaTeX faster than I type plain English :biggrin: As for the integral: it is obvious what to do at each step, simply a matter of mechanically splitting things up. More of an exercise than a problem, in my opinion.
Reply 791
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Jkn
11
Original post by Lord of the Flies
I think I type LaTeX faster than I type plain English :biggrin: As for the integral: it is obvious what to do at each step, simply a matter of mechanically splitting things up. More of an exercise than a problem, in my opinion.

What about plain french? :pierre:

Hmm, I've never thought about the difference between problems and exercises before, hmm...

True, haha :tongue: A lot of people would feel pretty lost though, on account of its awkwardness!

Lets talk STEP then. Have you ever found a question you couldn't do before, are you aiming for 360 and how often do you try questions? :smile: (sorry for the interview, lmao! :lol:)
Problem 115**

Evaluate
Unparseable latex formula:

I=\displaystyle\int\Big{(}\int_{-x}^{x}sec(t) \ dt \Big{)}sin(x)\ dx

Solution 115

⋯=∫2(∫0xsec⁡t dt) sin⁡x dx=−2cos⁡x∫0xsec⁡t dt+∫2 dx=2x−2cos⁡xln⁡(tan⁡x+sec⁡x)+C\displaystyle\begin{aligned} \cdots &= \int 2\left(\int_0^x \sec t \, dt\right) \,\sin x\,dx\\&=-2\cos x\int_0^x \sec t\,dt+\int 2\, dx\\&=2x-2\cos x\ln (\tan x+\sec x)+\mathcal{C}\end{aligned}
(edited 10 years ago)
Reply 794
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
Solution 115

⋯=∫2(∫0xsec⁡t dt) sin⁡x dx=−2cos⁡x∫0xsec⁡t dt+∫2 dx=2x−2cos⁡xln⁡(tan⁡x+sec⁡x)\displaystyle\begin{aligned} \cdots &= \int 2\left(\int_0^x \sec t \, dt\right) \,\sin x\,dx\\&=-2\cos x\int_0^x \sec t\,dt+\int 2\, dx\\&=2x-2\cos x\ln (\tan x+\sec x)\end{aligned}

...+c :pierre:
Original post by Jkn
...+c :pierre:


I claim that c = 0. Come at me bros.
Reply 796
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Jkn
11
Original post by Lord of the Flies
I claim that c = 0. Come at me bros.

Like ****. Do you even lift brah? :pierre:
Original post by Jkn
Like ****. Do you even lift brah?


Bitch please, I recite π\pi backwards in the shower.
Reply 798
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
Bitch please, I recite π\pi backwards in the shower.

Disproven by contradiction brah :pierre: I can do the first hundred or so as it happens brah:penguin:
Reply 799
Avatar for james22
james22
16
Original post by Jkn
Problem 116

Prove that 229801∑n=0∞(4n)!(1103+26390n)(n!)43964n=1π\displaystyle\frac{2\sqrt{2}}{9801}\sum_ {n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}=\frac{1}{ \pi}.

Also, find using wolfram alpha (or whatever), the approximate difference between π\pi and the approximation given by the first 10 terms of the sequence (to one decimal place).


Does this require spotting some complicated foruier series or is there an elementary solution?

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