Really Hard harder 3unit :O (1 Viewer)

Affinity

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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
 
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conics2008

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

Trebla

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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 Bn = Bn(1) = Bn(0), which has a rather neat property that Bn = 0 for n=3,5,7,9,...etc. I actually learnt this in first year uni SSP and it was quite interesting lol.
 
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Affinity

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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

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conics2008 said:
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
 

conics2008

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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
 

Trebla

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(ii)
Since dBn(x)/dx = nBn-1(x)
01dBn(x) = n∫01Bn-1(x) dx
but ∫01Bn(x)dx = 0
Replacing n with n - 1:
01Bn-1(x)dx = 0
Also: ∫01dBn(x) = Bn(1) - Bn(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.x1 - 1
= 1
.: LHS=RHS hence statement is true for n=1
Assume true for n=k
Bk(x + 1) - Bk(x) = kxk-1
Need to prove it is true for n=k+1
Bk+1(x + 1) - Bk+1(x) = (k+1)xk
Bk+1(x + 1) - Bk+1(x)
= (k+1)∫Bk(x+1)dx - (k+1)∫Bk(x)dx
= (k+1)∫{Bk(x+1)dx - Bk(x)}dx
= (k+1)∫kxk-1dx by assumption
= (k+1)xk + c
i.e. Bk+1(x + 1) - Bk+1(x) = (k+1)xk + c
sub x = 0: Bk+1(1) - Bk+1(0) = c
=> c = 0 from (ii) where n = k+1
.:Bk+1(x + 1) - Bk+1(x) = (k+1)xk
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
Bn(m + 1) - Bn(m) = nmn-1
Sum the m's (i.e. x's) from 0 to some integer k.
n∑km=0mn-1 = ∑m=0k{Bn(m + 1) - Bn(m)}
The RHS is a collapsing sum:
RHS = [Bn(1) - Bn(0)] + [Bn(2) - Bn(1)] - [Bn(3) - Bn(2)] + ...... + [Bn(k) - Bn(k - 1)] + [Bn(k + 1) - Bn(k)]
= Bn(k + 1) - Bn(0)
Hence:
n∑km=0mn-1 = Bn(k + 1) - Bn(0)

(v)
Let k = 135 and n = 5
5∑135m=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∑135m=0m4 = B5(136) - B5(0)]
=>∑135m=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 ;)
 
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