The Proof is Trivial!

Original post by Smaug123

Spoiler

Reply 1966

Original post by joostan

I think you mean rational coefficients :tongue:

Took a crack at the limits one but copied out the question wrong. :facepalm:

Back to square one.

I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong? :wink:

Reply 1967

Original post by Jkn

<this was spoilered>Oo is that that 'Variational Principles' stuff? :tongue:

</spoilered>

Yep, it is - the stuff done in Part IB is pretty basic, but I just can't get the Euler-Lagrange equations to stick in my head, and they take me about two minutes to derive (during which time I could have solved a question if I knew the E-L equations). I'm resorting to Anki to learn them by heart :P

Reply 1968

Original post by Jkn

I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong? :wink:

Maybe joostan thought you meant "monic polynomial" or something :smile:

Reply 1969

joostan

13

Original post by Jkn

I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong? :wink:

Good point . . . :redface:

Original post by Smaug123

Maybe joostan thought you meant "monic polynomial" or something

That's what I was thinking, but I wasn't really thinking

(edited 10 years ago)

Reply 1970

Hasufel

16

Original post by bogstandardname

Is the supposed to be a natural log?

Yup! (we could, if we wished here, use the Gamma function)

Reply 1971

Original post by Smaug123

Yep, it is - the stuff done in Part IB is pretty basic, but I just can't get the Euler-Lagrange equations to stick in my head, and they take me about two minutes to derive (during which time I could have solved a question if I knew the E-L equations). I'm resorting to Anki to learn them by heart :P

Hahaha #CambridgeMathsLAD :lol:

No idea what any of that stuff is bro!

Anki? :tongue:

Original post by Hasufel

Yup! (we could, if we wished here, use the Gamma function)

Spoiler

Reply 1972

Original post by Jkn

<spoilered>Oo is that that 'Variational Principles' stuff? :tongue:

</spoilered>

The highlight of the course is when we prove that in the problem of finding the function that minimises the distance between two points, a straight line extremises that distance, and it is a local minimum. I don't think we ever proved that it was a global minimum.

Reply 1973

Original post by Jkn

Hahaha #CambridgeMathsLAD :lol:

No idea what any of that stuff is bro!

Anki? :tongue:

Euler-Lagrange is the differential equation which you use to extremise a functional (in the same way as you use the differential equation

\dfrac{df}{dx} = 0

to extremise a function).
Seriously, IB Variational Principles contains about five theorems, and the rest of it is examples :P
Anki is a spaced-repetition program (like Memrise, if you ever used that) - it's designed for optimal rote-learning of small facts. I've used it to rote-learn a proof of the Sylow theorems, for example (by splitting the three theorems up into 51 short linked facts).
I hasten to add that I'm not one of these people that finds Cambridge maths basic in general - I struggle just as much as anyone else :smile:

from the look of your preparation, the first year will essentially be entirely stuff you already know (possibly excepting groups)…

Reply 1974

Hasufel

16

Original post by Jkn

Hahaha #CambridgeMathsLAD :lol:

No idea what any of that stuff is bro!

Anki? :tongue:

Spoiler

Yes, indeed.

Reply 1975

joostan

13

Original post by Hasufel

PROBLEM 277

Derive (pardon the pun) a series representation, excluding the constant of integration, for:

\displaystyle \int x^{n}(In(x))^{m}dx

(Hint: one way is to incorporate derivatives into your answer)

Not at all sure on this one, I had to look up a few properties and what not, which I don't think I can prove but:
As that's apparently

\ln(x)

Solution 277:

Spoiler

Where m is a positive integer:

Spoiler

(edited 10 years ago)

Reply 1976

FireGarden

14

Solution 285

Firstly

\dfrac{d}{dx}(x^n) = nx^{n-1}

Repeating,

\dfrac{d^a}{dx^a}(x^n) = \dfrac{n!}{(n-a)!}x^{n-a}

Generalise for any real a with gamma function;

\dfrac{d^a}{dx^a}(x^n) = \dfrac{\Gamma(n+1)}{\Gamma(n-a+1)}x^{n-a}

Set

a = \frac{1}{2}

and we get

H(x^n) = \dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})}x^{n- \frac{1}{2}}

Repeating

H

twice gives;

Unparseable latex formula:

H^2(x^n) = H(\dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})}x^{n- \frac{1}{2}}) [br][br]\ \\[br][br]= \dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})} H(x^{n-\frac{1}{2}})[br][br]\ \\[br][br]= \dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})}\dfrac{\Gamma(n+ \frac{1}{2})}{\Gamma(n)}x^{n-1}[br][br]

=nx^{n-1}

And so we find

H^2(x^n) = D(x^n)

(edited 10 years ago)

Reply 1977

Problem 287 */***

Prove that

\displaystyle 2(ab+bc+ca)^4=a^4(b-c)^4+b^4(c-a)^4+c^4(a-b)^4

when

a+b+c=0

.

Another problem that came from an interesting place and has a rather insane generalisation. As usual, the tall tale will follow a correct proof! :biggrin:

Original post by Smaug123

Euler-Lagrange is the differential equation which you use to extremise a functional (in the same way as you use the differential equation

\dfrac{df}{dx} = 0

to extremise a function).
Seriously, IB Variational Principles contains about five theorems, and the rest of it is examples :P
Anki is a spaced-repetition program (like Memrise, if you ever used that) - it's designed for optimal rote-learning of small facts. I've used it to rote-learn a proof of the Sylow theorems, for example (by splitting the three theorems up into 51 short linked facts).
I hasten to add that I'm not one of these people that finds Cambridge maths basic in general - I struggle just as much as anyone else :smile:

from the look of your preparation, the first year will essentially be entirely stuff you already know (possibly excepting groups)…

Sounds a bit brutal, I may take a peak later in the summer.

Oh right, yeah I've used similar things before. They tended to help with boring slogs (spanish vocal, stats definitions, physics definitions, etc..), I used several iPhone apps. :tongue:

Hmm, I doubt that! I still have {groups}, matrices, a lot of vector stuff, applications of s. relativity to solving problems, all of probability, etc.. from 1A that I haven't really scratched the surface on! :redface:

My fear is that this would mean I have to spend my whole first year focusing on the areas of maths I disliked so much that I didn't touch on it over the summer! :frown:

..like I doubt I will have much fun with integration over the next year or two! :lol:

Original post by joostan

I = -\left(\dfrac{-1}{n+1} \right)^{m+1}\Gamma (m+1,t)

Nice stuff man.

Ah, that's what I got! :smile:

I = \displaystyle\sum_{k=0}^m \left( \dfrac{(-1)^k m!\ln(x)^{m-k}x^{n+1}}{(m-k)!(n+1)^{k+1}} \right)

Looks similar to the series I derived using IBP => recurrence relation (functional equation) => solve. Probably the same but my notebooks/ridiculous piles of paper are not beside me atm! :lol:

Reply 1978

joostan

13

Original post by Jkn

Nice stuff man.

Ah, that's what I got! :smile:

Looks similar to the series I derived using IBP => recurrence relation (functional equation) => solve. Probably the same but my notebooks/ridiculous piles of paper are not beside me atm! :lol:

Cheers

Yeah I did it by IBP at first (hence deriving them is a positive integer series) but then I realised it was for all values, so along came the gamma :cool:

(edited 10 years ago)

Reply 1979