STEP 3 paper help Watch

maths learner
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Hi, I'm new so apologies if this doesn't come out too well.

The question is as follows:
 A=\frac{1}{n}\displaystyle\sum_{  k=1}^n x_k
 B=\frac{1}{n}\displaystyle\sum_{  k=1}^n (x_k-A)^2

Show that B= \frac{1}{n}\displaystyle\sum_{k=  1}^n (x_k^2 -A^2) .

I've expanded out the binomial in B and simplified, but I'm stuck at the very end.

 B=\frac{1}{n}\displaystyle\sum_{  k=1}^n (x_k-A)^2 \Rightarrow \frac{1}{n}\displaystyle\sum_{k=  1}^n x_k^2 - \frac{2A}{n}\displaystyle\sum_{k  =1}^n x_k + A^2 . It's this last step, and i have a feeling i will be kicking myself when I see it or get a nudge in the right direction. Edit: The  x_k  are fixed numbers.
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davros
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(Original post by maths learner)
Hi, I'm new so apologies if this doesn't come out too well.

The question is as follows:
 A=\frac{1}{n}\displaystyle\sum_{  k=1}^n x_k
 B=\frac{1}{n}\displaystyle\sum_{  k=1}^n (x_k-A)^2

Show that B= \frac{1}{n}\displaystyle\sum_{k=  1}^n (x_k^2 -A^2) .

I've expanded out the binomial in B and simplified, but I'm stuck at the very end.

 B=\frac{1}{n}\displaystyle\sum_{  k=1}^n (x_k-A)^2 \Rightarrow \frac{1}{n}\displaystyle\sum_{k=  1}^n x_k^2 - \frac{2A}{n}\displaystyle\sum_{k  =1}^n x_k + A^2 . It's this last step, and i have a feeling i will be kicking myself when I see it or get a nudge in the right direction.
You just need to substitute back for the definition of A in the middle term on your RHS (i.e. write the sum of x_n in terms of A).
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maths learner
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(Original post by davros)
You just need to substitute back for the definition of A in the middle term on your RHS (i.e. write the sum of x_n in terms of A).
I feel dumb. I can't get it to cancel...
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davros
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(Original post by maths learner)
I feel dumb. I can't get it to cancel...
You have -2A^2 + A^2 so ...
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maths learner
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(Original post by davros)
You have -2A^2 + A^2 so ...
Where did the  -2A^2 come from? Sorry it's been a long day (I shouldn't try to do STEP when half asleep).
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DFranklin
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Note: it's more obvious if you write the middle term as \displaystyle 2A \frac{1}{n} \sum_{k=1}^n x_k
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DFranklin
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Note also that this is basically the proof that Var[X] = E[X^2] - E[X]^2 if you've covered that in stats.
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davros
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(Original post by maths learner)
Where did the  -2A^2 come from? Sorry it's been a long day (I shouldn't try to do STEP when half asleep).
It's (-2A/n)(nA).

(I was trying to hint at it subtly, but perhaps too subtly! Anyway, I'm off to watch the Dr Who thing now!)
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maths learner
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(Original post by DFranklin)
Note: it's more obvious if you write the middle term as \displaystyle 2A \frac{1}{n} \sum_{k=1}^n x_k
Ohhh yeah. Ugh I need to get use to writing my fractions in a form like that, I get tripped up like that quite often. Thanks! Also Yeah I have done the stats thing .
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maths learner
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(Original post by DFranklin)
Note: it's more obvious if you write the middle term as \displaystyle 2A \frac{1}{n} \sum_{k=1}^n x_k
Hi, I'm still struggling away at question 1 on the STEP 2010 paper, I was wondering if you would mind helping again...

we have the following sums:

 A=\frac{1}{n}\sum\limits_{k=1}^n  k_x
 B=\frac{1}{n}\sum\limits_{k=1}^n (x_k-A)^2
 C=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} x_k
 D=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} (x_k -C)^2

I need to express D in terms of B, A x_{n+1} and n. And then show that (n+1)D\geq nB for all values of  x_{n+1} but that D<B if and only if  A- \sqrt\frac{(n+1)B}{n} &lt; x_{n+1} &lt; A + \sqrt\frac{(n+1)B}{n} .

I'm on the expressing D in terms of the other sums part I've got the following so far:

 D=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} x_k^2 -C^2 (By the same logic as how B was expressed above).
 D=\frac{1}{n+1}(n(B+A^2)+x_{n+1}  ^2) - \frac{(nA+x_{n+1})^2}{(n+1)^2} . Which was found by rearranging the B term and rewriting C in terms of A. Now i'm stuck...
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DFranklin
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I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
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maths learner
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(Original post by DFranklin)
I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
I'll have a go now. Thanks for the reply. I don't see from first glance how we just get nB though. Opps missed you wrote and other terms XD!
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brianeverit
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(Original post by maths learner)
Hi, I'm still struggling away at question 1 on the STEP 2010 paper, I was wondering if you would mind helping again...

we have the following sums:

 A=\frac{1}{n}\sum\limits_{k=1}^n  k_x
 B=\frac{1}{n}\sum\limits_{k=1}^n (x_k-A)^2
 C=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} x_k
 D=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} (x_k -C)^2

I need to express D in terms of B, A x_{n+1} and n. And then show that (n+1)D\geq nB for all values of  x_{n+1} but that D<B if and only if  A- \sqrt\frac{(n+1)B}{n} &lt; x_{n+1} &lt; A + \sqrt\frac{(n+1)B}{n} .

I'm on the expressing D in terms of the other sums part I've got the following so far:

 D=\frac{1}{n+1}\sum\limits_{k=1}  ^{n+1} x_k^2 -C^2 (By the same logic as how B was expressed above).
 D=\frac{1}{n+1}(n(B+A^2)+x_{n+1}  ^2) - \frac{(nA+x_{n+1})^2}{(n+1)^2} . Which was found by rearranging the B term and rewriting C in terms of A. Now i'm stuck...
I almost agree with you but I have
D=\frac{1}{n+1}B+\frac{n}{(n+1)^  2}A^2 for the first two terms, otherwise I agree with you.
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maths learner
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(Original post by DFranklin)
I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
Okay so I've simplified down, and gotten to:

 (n+1)D=nB+\frac{n}{n+1}(A-x_{n+1})^2. Which is what is gotten in the STEP solution guide. But then I don't know how you actually prove that  (n+1)D\geq nB for all values of  x_{n+1}.

I think I'm getting there. For the first part of the inequality can I simply say that  nD\geq B for all values since it's obvious by looking at it... I'm not really sure how you prove it. For the second part I have this:

 D= \frac{nB}{n+1}+\frac{n}{(n+1)^2}  (A-x_{n+1})^2. Now this is less than B iff  \frac{nB}{n+1} + \frac{n}{(n+1)^2}(\frac{-1}{n+1}B)&lt;B. But then I'm not too sure. I know if I expanded out rather than re wrote the term containing A in it I can start to get some form of quadratic which I presume I'll need, so that could be an approach?
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maths learner
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(Original post by brianeverit)
I almost agree with you but I have
D=\frac{1}{n+1}B+\frac{n}{(n+1)^  2}A^2 for the first two terms, otherwise I agree with you.
hmm. Maybe I made a slip up or something, but anyway I end up with what they get in the solution, I just don't know how you prove the inequalities :/.
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brianeverit
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(Original post by maths learner)
hmm. Maybe I made a slip up or something, but anyway I end up with what they get in the solution, I just don't know how you prove the inequalities :/.
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maths learner
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(Original post by brianeverit)
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Ah brilliant thank you. I'll work through it and see if I can follow the logic. I hate proving bi-conditional statements. Find them tricky.
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