Login

Welcome, Guest. Please login or register.

March 29, 2024, 10:29:09 am

Author Topic: 3U Maths Question Thread  (Read 1230432 times)  Share 

0 Members and 5 Guests are viewing this topic.

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #2685 on: August 23, 2017, 10:30:51 pm »
+4
Would love some help with q5d) thank you!!


EDIT: Am also unsure of how to properly get 6b) if we weren't given part a) for guidance. In particular, the RHS coefficient, I got the right answer but am not sure what it means (I was using part a) as a model, but my understanding isn't there yet)

(Image removed from quote.)

Hey! The RHS of the thing we need to prove is \(\binom{2n}{n+1}\), the coefficient of \(x^{n+1}\) on the RHS of that given consideration. So that bit is done already. On the left, consider:



Let's pair off coefficients to get powers of \(x^{n+1}\). First pair, \(\binom{n}{1}\binom{n}{n}\). Next, \(\binom{n}{2}\binom{n}{n-1}\).

Try and finish it off from there - There's only one more step before equating them! Remember the vital property of symmetry for binomial coefficients!


bsdfjnlkasn

  • Forum Obsessive
  • ***
  • Posts: 417
  • Respect: +28
Re: 3U Maths Question Thread
« Reply #2686 on: August 24, 2017, 07:20:35 am »
0

Hey there!

For 7a) I'm not so sure how they're getting the bottom values for r in the coefficients... also, don't understand the restriction on r especially as I thought it could only be equal to 4.


jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #2687 on: August 24, 2017, 11:22:28 am »
+5
Hey there!

For 7a) I'm not so sure how they're getting the bottom values for r in the coefficients... also, don't understand the restriction on r especially as I thought it could only be equal to 4.

(Image removed from quote.)

So this is the same trick as in the previous question, but this we are considering some unknown power of \(x\), \(x^r\). On the LHS, the coefficient is simply:



Now consider the right hand side:



So how do we get a power of \(x^r\)? Take each term from the right bracket and pair it off: \(1\) pairs with \(\binom{n}{r}\). \(4\) pairs with \(\binom{n}{r-1}\), and so on. You can see the result form from there.

As for the restriction, clearly \(r\le n+4\), just by the nature of the binomial expansion, so that is a start. However, for this result to be valid, \(r\) needs to be high enough that the bottom term of each of those binomial coefficients isn't negative - That is, we can't have \(\binom{n}{-2}\), for example. So, \(r\ge4\).


bsdfjnlkasn

  • Forum Obsessive
  • ***
  • Posts: 417
  • Respect: +28
3U Maths Question Thread
« Reply #2688 on: August 24, 2017, 07:22:52 pm »
0
Hey there! Can I get some help with 11a) thank you :)

And for part b) what am I meant to be finding the coefficient for and how do I get that result?




« Last Edit: August 24, 2017, 08:41:31 pm by bsdfjnlkasn »

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #2689 on: August 24, 2017, 07:32:05 pm »
+5
Hey there! Can I get some help with 11a) thank you :)

And for part b) what am I meant to be finding the coefficient for and how do I get that result?

(Image removed from quote.)


bsdfjnlkasn

  • Forum Obsessive
  • ***
  • Posts: 417
  • Respect: +28
Re: 3U Maths Question Thread
« Reply #2690 on: August 24, 2017, 08:41:41 pm »
0
Hi Rui!

Thanks for that, with Q11, I was mostly having difficulty with the LHS as i'm not sure how to use the LHS of the expression in part a) or if i'm meant to at all....

Also, I would really appreciate some help with Q17 here, am happy to post my progress. Not sure what to do about the RHS in my last line/if there's a more efficient way of proving this.



« Last Edit: August 24, 2017, 08:47:08 pm by bsdfjnlkasn »

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #2691 on: August 24, 2017, 09:27:11 pm »
+5




RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #2692 on: August 24, 2017, 09:37:40 pm »
+5
Hi Rui!

Thanks for that, with Q11, I was mostly having difficulty with the LHS as i'm not sure how to use the LHS of the expression in part a) or if i'm meant to at all....

Also, I would really appreciate some help with Q17 here, am happy to post my progress. Not sure what to do about the RHS in my last line/if there's a more efficient way of proving this.

(Image removed from quote.)

(Image removed from quote.)



I'll let you try from here. Remember to use the symmetry identity \( \binom{N}{R} = \binom{N}{N-R} \) somehow
« Last Edit: August 24, 2017, 09:48:25 pm by RuiAce »

samuels1999

  • Adventurer
  • *
  • Posts: 21
  • "Expectation is the root of all heartache." -W.S.
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #2693 on: August 25, 2017, 10:26:16 pm »
0
I have a question. That has been troubling me.

Thanks,
Samuel

"If you can't explain it simply, you do not understand it well enough" -Albert Einstein
Year 12 2017
Subjects: Adv. English, Mathematics, Mathematics Ext 1, Modern History, Physics, Design and Technology
....yeah....its a pretty odd bunch

jamonwindeyer

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 10150
  • The lurker from the north.
  • Respect: +3108
Re: 3U Maths Question Thread
« Reply #2694 on: August 25, 2017, 11:20:54 pm »
+6
I have a question. That has been troubling me.

Thanks,
Samuel

Hey! So there could be a trick to this, but when I see this, my immediate response is to just solve that first equation to find \(x\). So let's do that. You use the auxillary angle transformation:



I've gotten that with the normal auxillary angle process, let me know if you needed help with that bit! From there:



So now, just go to \(\cos{2x}\):



Might be a little error up there, did this quick, but this is roughly the process ;D

samuels1999

  • Adventurer
  • *
  • Posts: 21
  • "Expectation is the root of all heartache." -W.S.
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #2695 on: August 26, 2017, 06:42:17 pm »
0
Thanks Jamon!  :D
I think the pi/6 should have been pi/3. But other than that it really helped.
"If you can't explain it simply, you do not understand it well enough" -Albert Einstein
Year 12 2017
Subjects: Adv. English, Mathematics, Mathematics Ext 1, Modern History, Physics, Design and Technology
....yeah....its a pretty odd bunch

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #2696 on: August 26, 2017, 06:47:09 pm »
+3
Thanks Jamon!  :D
I think the pi/6 should have been pi/3. But other than that it really helped.

samuels1999

  • Adventurer
  • *
  • Posts: 21
  • "Expectation is the root of all heartache." -W.S.
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #2697 on: August 26, 2017, 10:17:18 pm »
0
Oh yes...I understand. I'm wrong.

Thanks.
"If you can't explain it simply, you do not understand it well enough" -Albert Einstein
Year 12 2017
Subjects: Adv. English, Mathematics, Mathematics Ext 1, Modern History, Physics, Design and Technology
....yeah....its a pretty odd bunch

epherbertson

  • Trailblazer
  • *
  • Posts: 25
  • Respect: 0
Re: 3U Maths Question Thread
« Reply #2698 on: August 31, 2017, 08:12:07 am »
0
Hey

Not sure if i have already asked this question but I am struggling with this question...Probably something super easy that my brain hasn't recognised but any help would be appreciated.

Thank You :)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: 3U Maths Question Thread
« Reply #2699 on: August 31, 2017, 09:36:07 am »
+5
Hey

Not sure if i have already asked this question but I am struggling with this question...Probably something super easy that my brain hasn't recognised but any help would be appreciated.

Thank You :)



_______________________________________________________


We know this, because if a line has gradient 1, it must be parallel to y=x. Essentially, the tangent here is 'steeper' than the line y=x. Note that the line y=x joins the points (0,0) and (1,1) so we can sketch this as well
« Last Edit: September 17, 2017, 04:32:39 pm by RuiAce »