[Dog4444]

(Original post by gff)

(Original post by TheMagicMan)

How to generate this things?

[gff]

(Original post by Dog4444)
How to generate this things?

Spoiler:

Show

Goldbach proved that no polynomial with integral coefficients can produce a prime for all integers.
There are some other polynomials here: Prime-Generating Polynomials.

The one that I posted is due to Legendre, and the MagicMan's one is due to Euler.
However, a result due to Euler showed that generates primes only for .

[TheMagicMan]

(Original post by Dog4444)
How to generate this things?

(Original post by gff)

Spoiler:

Show

Goldbach proved that no polynomial with integral coefficients can produce a prime for all integers.
There are some other polynomials here: Prime-Generating Polynomials.

The one that I posted is due to Legendre, and the MagicMan's one is due to Euler.
However, a result due to Euler showed that generates primes only for .

just to add to this, there exists a formula (in 26 unknowns) whereby all real outputs of the formula, given integer inputs, are prime, and all primes are represented by some choice of the 26 unknowns (i.e. the real outputs of the formula form a bijection with the primes)

[ben-smith]

Anyone want to give me a hint on how to do this integral?

Having one of those days...

[fruktas]

(Original post by TheUltimateProof)
29/120

I'm somewhat still worried that I won't do it with zero revision so I think I'll go through one of the edexcel revision guides 2 weeks before the exam and hit up the past papers for peace of mind

I'm assuming that's what you need to get an A overall? I need 33/120 myself, but I really want to get the A* and I can't decide what to do

[Blutooth]

Hmm, me thinks an x^3 +y^3 factorisation might help. But i'm not sure haing not done the question.

Further hint possibly.

Spoiler:

Show

(s^3+ c^3)^2=((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2= (1+(2sc))*(1-sc)^2
then espress in sin(2thetas)?

[TheMagicMan]

(Original post by ben-smith)
Anyone want to give me a hint on how to do this integral?

Having one of those days...

A coupl of potential lines of attack that I would consider:

half tan substitution

conversion to multiple angles (i.e cos(nx) etc.)

note the denominator is a square...quotient rule inversion perhaps?

[ben-smith]

(Original post by Blutooth)
Hmm, me thinks an x^3 +y^3 factorisation might help. But i'm not sure haing not done the question.

Further hint possibly.

Spoiler:

Show

(s^3+ c^3)^2=((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2= (1-(2sc))*(1-sc)^2
then espress in sin(2thetas)?

I've tried that. don't end up with anything nice that I can see.

(Original post by TheMagicMan)
A coupl of potential lines of attack that I would consider:

half tan substitution

conversion to multiple angles (i.e cos(nx) etc.)

note the denominator is a square...quotient rule inversion perhaps?

I tried half tan, the large powers make it look disgusting. The fact that the denominator is a square is due to this being an integral in polar coordinates but I hadn't thought of quotient rule inversion so cheers for that.

[Dog4444]

(Original post by Blutooth)
Further hint possibly.

Spoiler:

Show

(s^3+ c^3)^2=((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2= (1-(2sc))*(1-sc)^2
then espress in sin(2thetas)?

How did you get it? And it doesn't work on my calculator.

[ben-smith]

(Original post by Dog4444)
How did you get it? And it doesn't work on my calculator.

s^2+c^2=1...

[Dog4444]

(Original post by ben-smith)
s^2+c^2=1...

I'm not sure what do you mean.

http://www.wolframalpha.com/input/?i...%5E2%29%29%5E2

http://www.wolframalpha.com/input/?i...1-xy%29%29%5E2

[ben-smith]

(Original post by Dog4444)
I'm not sure what do you mean.

http://www.wolframalpha.com/input/?i...%5E2%29%29%5E2

http://www.wolframalpha.com/input/?i...1-xy%29%29%5E2

I thought, from your post, that you didn't understand how he made the step:
((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2

If you do, ignore me

[Dog4444]

(Original post by ben-smith)
I thought, from your post, that you didn't understand how he made the step:
((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2

If you do, ignore me

You're right I didn't understand it. And I'm not sure it's right; I can't get it on my calc and looking at wolframs contour plots, it doesn't look like they're the same.

[TheMagicMan]

(Original post by ben-smith)
I thought, from your post, that you didn't understand how he made the step:
((s+c)(s^2 -sc+c^2))^2= ((s+c)(1-sc))^2

If you do, ignore me

You can turn it into

If you then take numerator and distribute it (separate the -1 from the cos(4x), make the substitution and multiply out the multiple angles, then make the half angle substitution, it works

I've written about ten pages of solid algebra to get to the answer so it's a bit of a slog and I'm not writing out the whole thing in latex

Much harder than any step integral imo

(the answer is evaluated between 0 and pi/4 I think which is 3)

Wolfram agrees numerically thankfully

[Blutooth]

(Original post by Dog4444)
You're right I didn't understand it. And I'm not sure it's right; I can't get it on my calc and looking at wolframs contour plots, it doesn't look like they're the same.

Sorry there is an aberrant- sign. The line 1-2sc should be 1+2sc.

[ben-smith]

(Original post by TheMagicMan)
You can turn it into

If you then take numerator and distribute it (separate the -1 from the cos(4x), make the substitution and multiply out the multiple angles, then make the half angle substitution, it works

I've written about ten pages of solid algebra to get to the answer so it's a bit of a slog and I'm not writing out the whole thing in latex

Much harder than any step integral imo

(the answer is evaluated between 0 and pi/2 I think)

the awkward moment when you realise tan isn't defined at pi/2...
edit: either you edited or I can't read

[Dog4444]

(Original post by Blutooth)
Sorry there is an aberrant- sign. The line 1-2sc should be 1+2sc.

EDIT: Forgot it is sines and cosines. Sorry.

(Original post by TheMagicMan)
I've written about ten pages of solid algebra to get to the answer so it's a bit of a slog and I'm not writing out the whole thing in latex

You're damn quick.

[TheMagicMan]

(Original post by ben-smith)
the awkward moment when you realise tan isn't defined at pi/2...
edit: either you edited or I can't read

I edited now...forgot to change the limits in my last sub....

[Blutooth]

(Original post by Dog4444)
The step before.

x^3+y^3:
http://www.livephysics.com/ptools/on...2By%5E3%29%5E2

http://www.livephysics.com/ptools/on...%5E2%29%29%5E2

http://www.livephysics.com/ptools/on...-x*y%29%29%5E2



You're damn quick.

I'm just using the fact that (x^3+y^3)=(x+y)(x^2-xy+y^2)
However, I'm not sure that's the right approach anyway.

[TheMagicMan]

(Original post by Dog4444)



You're damn quick.

STEP has really sharpened up my speed on integrals...and it was big writing

anyway I have managed to get WA to throw up a step by step solution by putting in one of my early integrands...if you go down and press show steps here in the integral bit...it's scary but overkill as always with WA...I love how they use as a substitution (I only had 5 in mine)