Really Hard harder 3unit :O (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/Bernoullipolynomials.jpg

Assuming B0(x) = 1
I integrate it from 0 to 1, obtaining x + c. But the integration of that = 0, so therefore c = -1. BUT the question states -1/2.
I tested for the other values but HOW the truck do they get the constants ? :O

I hope affinity can explain to me
Thanks in advance

 

[Affinity]

you aren't reading carefully enough..

You were supposed to make integral of B_1 0, what you did was setting the value of B_1 at 1 to 0

 

[namburger]

K i got it

 

[conics2008]

where the fuck do you shit out your questions from... seriously they aren't even hsc style question nor past trials question...

 

[Trebla]

This question is from Sydney Grammar Trial 1999

For (i):
B'₄(x) = 4B₃(x)
=> B₄(x)= x⁴ - 2x³ + x² + c
But ∫₀¹B₄(x) dx = 0, so
[x⁵/5 - x⁴/2 + x³/3 + cx]₀¹ = 0
1/5 - 1/2 + 1/3 + c = 0
=> c = - 1/30
Thus: B₄(x)= x⁴ - 2x³ + x² - 1/30
= x²(x - 1)² - 1/30

As an aside, the constants are called Bernoulli numbers B_n = B_n(1) = B_n(0), which has a rather neat property that B_n = 0 for n=3,5,7,9,...etc. I actually learnt this in first year uni SSP and it was quite interesting lol.

 

[Affinity]

They are classic questions.. very classic...

Usually what they do for q8 is to take something classic like this and specialise it to some case...

 

[namburger]

where the fuck do you shit out your questions from... seriously they aren't even hsc style question nor past trials question...

HSC questions are just a more simpler version of what i post up.

Anyway, im not gona be posting anything up for harder 3u as they are so random and you can't prepare for them haha

 

[hopethisworks]

classy

 

[conics2008]

hey namburger, seriousy those question dont even like the question we do at school nor the trials nor hsc..

I'm sticking to hsc papers question. those are just insane xD

 

[u-borat]

solid question this; got it all out but it took a while.

 

[Undermyskin]

Gosh, help me with the last 3 parts!

 

[Trebla]

(ii)
Since dB_n(x)/dx = nB_n-1(x)
∫₀¹dB_n(x) = n∫₀¹B_n-1(x) dx
but ∫₀¹B_n(x)dx = 0
Replacing n with n - 1:
∫₀¹B_n-1(x)dx = 0
Also: ∫₀¹dB_n(x) = B_n(1) - B_n(0)
Hence the result follows.

(iii)
For n=1
LHS = B₁(x + 1) - B₁(x)
= x + 1 - 1/2 - (x - 1/2)
= 1
RHS = 1.x^{1 - 1}
= 1
.: LHS=RHS hence statement is true for n=1
Assume true for n=k
B_k(x + 1) - B_k(x) = kx^k-1
Need to prove it is true for n=k+1
B_k+1(x + 1) - B_k+1(x) = (k+1)x^k
B_k+1(x + 1) - B_k+1(x)
= (k+1)∫B_k(x+1)dx - (k+1)∫B_k(x)dx
= (k+1)∫{B_k(x+1)dx - B_k(x)}dx
= (k+1)∫kx^k-1dx by assumption
= (k+1)x^k + c
i.e. B_k+1(x + 1) - B_k+1(x) = (k+1)x^k + c
sub x = 0: B_k+1(1) - B_k+1(0) = c
=> c = 0 from (ii) where n = k+1
.:B_k+1(x + 1) - B_k+1(x) = (k+1)x^k
The statement is true for n = k+1, if true for n = k, but since it is true for n = 1, it is true for all n ≥ 1.

(iv)
Using the fact that
B_n(m + 1) - B_n(m) = nm^n-1
Sum the m's (i.e. x's) from 0 to some integer k.
n∑^k_m=0m^n-1 = ∑_m=0^k{B_n(m + 1) - B_n(m)}
The RHS is a collapsing sum:
RHS = [B_n(1) - B_n(0)] + [B_n(2) - B_n(1)] - [B_n(3) - B_n(2)] + ...... + [B_n(k) - B_n(k - 1)] + [B_n(k + 1) - B_n(k)]
= B_n(k + 1) - B_n(0)
Hence:
n∑^k_m=0m^n-1 = B_n(k + 1) - B_n(0)

(v)
Let k = 135 and n = 5
5∑¹³⁵_m=0m⁴ = B₅(136) - B₅(0)
We need the Bernoulli polynomial B₅(x):
B'₅(x) = 5B₄(x)
=> B₅(x) = ∫[5x⁴ - 10x³ + 5x² - 1/6]dx
B₅(x) = x⁵ - (5/2)x⁴ + (5/3)x³ - x/6 + c
But ∫₀¹B₅(x)dx = 0
=> ∫₀¹[x⁵ - (5/2)x⁴ + (5/3)x³ - x/6 + c]dx = 0
=> 1/6 - 1/2 + 5/12 - 1/12 + c = 0
=> c = 0
Hence B₅(x) = x⁵ - (5/2)x⁴ + (5/3)x³ - x/6
5∑¹³⁵_m=0m⁴ = B₅(136) - B₅(0)]
=>∑¹³⁵_m=0m⁴ = (1/5)[B₅(136) - B₅(0)
But:
B₅(0) = 0
and
(1/5)B₅(136) = 9 134 962 308
Hence we have the result.

Have fun deciphering that! LOL