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University of Warwick

Coventry

Lili!

oh ok..thanks a lot for your help. it really is appreciated! benwellsday, the only part of your workings for the second question that i dont understand is how you got the sum of 2^-n to be 2^-n/(1/2). surely you would use the geometric formula with a=r=1/2?

I used the sum to infinity which is like $S_{\infty} = \frac{a}{1-r}$ and using a = 2^{-N} and r = 1/2. a isn't 1/2 if you're starting at n=N.

benwellsday

Letting $q_1 = \frac{m_1}{n_1}$ and $q_2 = \frac{m_2}{n_2}$ I got

$\text{lcm} \{q_1 , q_2 \} = \frac { \text{lcm} \{ m_2 n_1 , m_1 n_2 \} }{n_1 n_2}$

I'm completely sure on my method but the formula works for all integers (let the n's be 1's) and it worked for two trivial cases which was good enough for me.

$\text{lcm} \{q_1 , q_2 \} = \frac { \text{lcm} \{ m_2 n_1 , m_1 n_2 \} }{n_1 n_2}$

I'm completely sure on my method but the formula works for all integers (let the n's be 1's) and it worked for two trivial cases which was good enough for me.

cheers!

benwellsday

Q3 part (iii) and (iv) need particular attention (at least for me) because I couldn't do part of them.

Q2.

c) Since both factors contain $\sqrt{2}$ as a coeffecient the polynomial cannot be factorised into two polynomials in $\mathbb{Q} [x]$ unless one is a constant.

hey, ours answers match for all of 1 and 2 so far, but i'm unsure about this it coudl just be me. i thought if its in Q, then all the coeffiecients were in Q, so how does the unless one is a constant bit work?

ok SO this will be my last question! i hope!

there have been a couple of these questions, both very similar..but i can't get them to work..

eg a(n) = 2 and a(n+1) = 1/2(a(n) + 2/a(n))

i) Calculate a(n+1)^2 - 2 to show that a(n)^2 > 2 for all n.

--i thought this sounded reasonably straightfoward but either i'm not realising something or have just made a stupid mistake! but yeah i got that it equals (a(n)^4 + 4 - 4(a(n)^2) / 4(a(n)^2) but then wasnt sure what to do with it..achh.

ii) Show a(n) is strictly decreasing

--calculating a(n+1) - a(n) = 1/2(a(n) + 2/a(n) - a(n-1) - 2/a(n-1)). how can i show this is less than zero?

there have been a couple of these questions, both very similar..but i can't get them to work..

eg a(n) = 2 and a(n+1) = 1/2(a(n) + 2/a(n))

i) Calculate a(n+1)^2 - 2 to show that a(n)^2 > 2 for all n.

--i thought this sounded reasonably straightfoward but either i'm not realising something or have just made a stupid mistake! but yeah i got that it equals (a(n)^4 + 4 - 4(a(n)^2) / 4(a(n)^2) but then wasnt sure what to do with it..achh.

ii) Show a(n) is strictly decreasing

--calculating a(n+1) - a(n) = 1/2(a(n) + 2/a(n) - a(n-1) - 2/a(n-1)). how can i show this is less than zero?

That's just a problem, it still means it's irreducible in Q[x] but for example $x^2 - 2 = \frac{1}{2} (2x^2 - 4)$ is a way to write it as a product of members of Q[x].

Lili!

ok SO this will be my last question! i hope!

there have been a couple of these questions, both very similar..but i can't get them to work..

eg a(n) = 2 and a(n+1) = 1/2(a(n) + 2/a(n))

i) Calculate a(n+1)^2 - 2 to show that a(n)^2 > 2 for all n.

--i thought this sounded reasonably straightfoward but either i'm not realising something or have just made a stupid mistake! but yeah i got that it equals (a(n)^4 + 4 - 4(a(n)^2) / 4(a(n)^2) but then wasnt sure what to do with it..achh.

ii) Show a(n) is strictly decreasing

--calculating a(n+1) - a(n) = 1/2(a(n) + 2/a(n) - a(n-1) - 2/a(n-1)). how can i show this is less than zero?

there have been a couple of these questions, both very similar..but i can't get them to work..

eg a(n) = 2 and a(n+1) = 1/2(a(n) + 2/a(n))

i) Calculate a(n+1)^2 - 2 to show that a(n)^2 > 2 for all n.

--i thought this sounded reasonably straightfoward but either i'm not realising something or have just made a stupid mistake! but yeah i got that it equals (a(n)^4 + 4 - 4(a(n)^2) / 4(a(n)^2) but then wasnt sure what to do with it..achh.

ii) Show a(n) is strictly decreasing

--calculating a(n+1) - a(n) = 1/2(a(n) + 2/a(n) - a(n-1) - 2/a(n-1)). how can i show this is less than zero?

Isn't it a(1) = 2.

For i) $a_{n+1}^2 - 2 = \frac {1}{4} (a_n ^2 + 4 + \frac{4}{a_n ^2}) = \frac{a_n ^4 + 4a_n ^2 + 4}{4a_n ^2}$

You can use that with induction to show a(n)^2 - 2 > 0 for all n.

For ii) why have you got a(n-1)'s, you should of left a(n) as it was, just as a(n) !

Yeah, that should work out alot easier.

benwellsday

Question time, analysis 2005, 4c)

Estimate the number of terms required to approximate pi within 10^-10 for each formula below

$\pi = 4 \sum ^{\infty}_{k=0} \frac{(-1)^k}{2k+1}$

and

$\pi = \sum ^{\infty}_{k=0} \frac{1}{16^k} (\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} -\frac{1}{8k+6})$

with the hint that the bracketed term is always between 0 and 4 (which need not be proved).

Estimate the number of terms required to approximate pi within 10^-10 for each formula below

$\pi = 4 \sum ^{\infty}_{k=0} \frac{(-1)^k}{2k+1}$

and

$\pi = \sum ^{\infty}_{k=0} \frac{1}{16^k} (\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} -\frac{1}{8k+6})$

with the hint that the bracketed term is always between 0 and 4 (which need not be proved).

Did anyone respond to the above??

Some 2005 questions that I'm not sure of:

Analysis

1(vi) If x and y are reals such that $x>y- \epsilon$ for all $\epsilon>0$ then x>y.

I put false since x=y is a possibility.

(x) If (a_n) is an increasing sequence which is bounded above then (a_n) tends to sup{a_n|n in N} as n tends to infinity.

False? I'm not sure because it doesn't say the a_n are real?

2(c) If a and b are real numbers with a<b write down an expression for c with a<c<b. Deduce that there are infintiely many rational numbers c.

Now I just wrote b-a>0 so $a< a+ \frac{b-a}{k} < b$ for $k>0, k \in N$.

Then I noticed it was 8 marks so i'm missing something major? So I wrote down something different using the integar function and 2^n, for n>N.

So what is the corret method and is anything wrong with the initial attempt?

2(c) above, quoted in Ben's post.

Foundations

2(c) Show that the $|P(X)| = 2^{|X|}$ (power set of X).

It's worth a hefty 6 marks marks and the mark scheme does it via induction-- you can do it in two lines using the 0 or 1 method used for each element in the set to derive all the combinations of sets, as done in the notes. Will that get full marks?

silent ninja

1(vi) If x and y are reals such that $x>y- \epsilon$ for all $\epsilon>0$ then x>y.

I put false since x=y is a possibility.

1(vi) If x and y are reals such that $x>y- \epsilon$ for all $\epsilon>0$ then x>y.

I put false since x=y is a possibility.

Yeah that works.

(x) If (a_n) is an increasing sequence which is bounded above then (a_n) tends to sup{a_n|n in N} as n tends to infinity.

False? I'm not sure because it doesn't say the a_n are real?

The a_n don't need to be real, for example consider a_1 = 3, a_2 = 3.1, a_3 = 3.14 etc... a_n tends to pi but a_n is always rational. I doubt they'd consider the comlpex numbers too.

2(c) If a and b are real numbers with a<b write down an expression for c with a<c<b. Deduce that there are infintiely many rational numbers c.

Now I just wrote b-a>0 so $a< a+ \frac{b-a}{k} < b$ for $k>0, k \in N$.

Then I noticed it was 8 marks so i'm missing something major? So I wrote down something different using the integar function and 2^n, for n>N.

So what is the corret method and is anything wrong with the initial attempt?

If this is the question I think it is, you've read it wrong (like 90% of my analysis class did when we did it in class). Doesn't it ask for a rational c in the interval (a,b) where a and b are both rationals too.

In which case, take the average.

To show there are infinitely many, assume there aren't, hence there are a finite amount. "List them" and then show you can find another.

Foundations

2(c) Show that the $|P(X)| = 2^{|X|}$ (power set of X).

It's worth a hefty 6 marks marks and the mark scheme does it via induction-- you can do it in two lines using the 0 or 1 method used for each element in the set to derive all the combinations of sets, as done in the notes. Will that get full marks?

Perhaps Mond hadn't thought of the 0, 1 method back in 2005. Any method that works should be worth the marks, unless stated specifically otherwise (for example in a modular arithmetic question it said full marks are only available if you avoid a long calculation).

benwellsday

If anyone wants to check my Foundations 2008 answers it'd be appreciated

Q3 part (iii) and (iv) need particular attention (at least for me) because I couldn't do part of them.

Q1.

Q2.

Q3

Q4

Q3 part (iii) and (iv) need particular attention (at least for me) because I couldn't do part of them.

Q1.

Q2.

Q3

Q4

1(e) (x-2), (x+3)

(h) There exists an injection but no bijection (because no surjection), isn't it?

(i) I got the same but can you help me with this: I dont understand how that is a cycle because it keeps ending? I thought cycles repeated. When you hit 3 in alpha, there doesn't exist a 3 in beta so you're stuck?

(j) glad someone got the same answer!

2 I didn't do the division bit, i got bored

(d) I think you meant "No, polynomial is NOT irreducible."

I wrote every polynomail deg>1 is reducable in C[x] to a product of linear polynomials therefore no the polynomial is reducable.

I really need to learn my definitions though.

(e) didn't do, lazy again

3(ii)(b) I put fail on transitivity also.

(iii) How many equivalence classes are there? I put n. The equivalence relation given is basically modulo arithmetic, so you get from 0modn to (n-1)modn which makes n classes.

I'm not sure of the name of the set of equivalence classes...Z?

(iv) stuck on last bit too!

4 not done

benwellsday

Yeah that works.

The a_n don't need to be real, for example consider a_1 = 3, a_2 = 3.1, a_3 = 3.14 etc... a_n tends to pi but a_n is always rational. I doubt they'd consider the comlpex numbers too.

If this is the question I think it is, you've read it wrong (like 90% of my analysis class did when we did it in class). Doesn't it ask for a rational c in the interval (a,b) where a and b are both rationals too.

In which case, take the average.

To show there are infinitely many, assume there aren't, hence there are a finite amount. "List them" and then show you can find another.

The a_n don't need to be real, for example consider a_1 = 3, a_2 = 3.1, a_3 = 3.14 etc... a_n tends to pi but a_n is always rational. I doubt they'd consider the comlpex numbers too.

If this is the question I think it is, you've read it wrong (like 90% of my analysis class did when we did it in class). Doesn't it ask for a rational c in the interval (a,b) where a and b are both rationals too.

In which case, take the average.

To show there are infinitely many, assume there aren't, hence there are a finite amount. "List them" and then show you can find another.

Ah-ha, now that makes sense!

Thanks for input on all the others too.

Perhaps Mond hadn't thought of the 0, 1 method back in 2005. Any method that works should be worth the marks, unless stated specifically otherwise (for example in a modular arithmetic question it said full marks are only available if you avoid a long calculation).

hmm... So how did you do the one we covered above: $15^{22}mod11$

I wrote 15mod11= 4mod11 then proceeded by the squaring method. We got the same answer but is that the right method?

silent ninja

1(e) (x-2), (x+3)

(h) There exists an injection but no bijection (because no surjection), isn't it?

(i) I got the same but can you help me with this: I dont understand how that is a cycle because it keeps ending? I thought cycles repeated. When you hit 3 in alpha, there doesn't exist a 3 in beta so you're stuck?

(j) glad someone got the same answer!

(h) There exists an injection but no bijection (because no surjection), isn't it?

(i) I got the same but can you help me with this: I dont understand how that is a cycle because it keeps ending? I thought cycles repeated. When you hit 3 in alpha, there doesn't exist a 3 in beta so you're stuck?

(j) glad someone got the same answer!

e) Might just be a copying up error.

h) Yeah I'm wrong, I just checked a finite case, i.e. {1} -> {1,2}

(d) I think you meant "No, polynomial is NOT irreducible."

I wrote every polynomail deg>1 is reducable in C[x] to a product of linear polynomials therefore no the polynomial is reducable.

I really need to learn my definitions though.

I just meant that there is no polynomial in C that is irreducible (i.e. every polynomial is reducible in C)

3(ii)(b) I put fail on transitivity also.

(iii) How many equivalence classes are there? I put n. The equivalence relation given is basically modn, so you get from 0modn to (n-1)modn which makes n classes.

I'm not sure of the name of the set of equivalence classes...Z?

(iv) stuck on last bit too!

4 not done

Ah that makes sense, and I'd guess $\mathbb{Z} / n$

And for $15^{22} mod 11$

benwellsday

Ah that makes sense, and I'd guess $\mathbb{Z} / n$

Ah yes, totally forgot about that Z/n notation thingy for modulo groups.

And for $15^{22} mod 11$

Ah that makes sense, and I'd guess $\mathbb{Z} / n$

Ah yes, totally forgot about that Z/n notation thingy for modulo groups.

And for $15^{22} mod 11$

Similar, but I just found that there is a little section on Fermat's Little Theorem in the notes. I didn't realise we covered that (serves me right for missing lectures).

Anyhow, using that it's slightly shorter since 11 is prime and doesn't divide 15 so

15^10= 1 mod11, 15^20=1mod11 then 15^2= 4^2mod11= 5mod11 so 15^22=1*5mod11

silent ninja

Similar, but I just found that there is a little section on Fermat's Little Theorem in the notes. I didn't realise we covered that (serves me right for missing lectures).

Anyhow, using that it's slightly shorter since 11 is prime and doesn't divide 15 so

15^10= 1 mod11, 15^20=1mod11 then 15^2= 4^2mod11= 5mod11 so 15^22=1*5mod11

Anyhow, using that it's slightly shorter since 11 is prime and doesn't divide 15 so

15^10= 1 mod11, 15^20=1mod11 then 15^2= 4^2mod11= 5mod11 so 15^22=1*5mod11

??

$15^{22}mod11=(15^2)^{11}mod11=225mod11=5?$

11*10=110

110*2=220... nice.

meh.

silent ninja

Did anyone respond to the above??

for the second one just work with some inequalities.

$\sum \frac{1}{16^k}BLAH\leq \sum \frac{1}{4^{2k-1}}$ and thats just a geometric sum. so find that and then you're done in a jiffy.

ther first one is different, since it is an alternating series you know that once it gets past a term which is less than 10^-10 then all the subsequent terms won't matter enough to get it out of that band. so just find the term which is less that 10^-10 and your done

amo17

hey guys,

just want to check sommat on the 2006 analysis q3, did you get all three series being convergent. i did, and i don't think somehow that they would set a question like that.

could someone please quickly check their answers to see what they got?

thanks

just want to check sommat on the 2006 analysis q3, did you get all three series being convergent. i did, and i don't think somehow that they would set a question like that.

could someone please quickly check their answers to see what they got?

thanks

For the first one I got -1/3 and for the 2nd I got 1. I never finished the 3rd one but by the looks of if I'd guess at divergence because for large n $\sqrt {n^3 + sin n} \approx n^{\frac{3}{2}}$

I guess my attempt at a proof would be

3 days, 1 hour, and 45 minutes of revision left....is anyone else not really worried?

I made the denominator larger (n > 2 eventually) just to divide out the 5n. It works without it but it doesn't look as good.

And how is it within the boundary, 3n^2 - 3n + 1 = 1 for n=1, and for n > 1, 3n^2 - 3n + 1 > 1. If it mattered I could change it to (n-2)^3 or (n-3)^3 or something. I guess there'd be problems when n = 1 then.

And how is it within the boundary, 3n^2 - 3n + 1 = 1 for n=1, and for n > 1, 3n^2 - 3n + 1 > 1. If it mattered I could change it to (n-2)^3 or (n-3)^3 or something. I guess there'd be problems when n = 1 then.

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