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March 29, 2024, 12:36:07 am

Author Topic: 3U Maths Question Thread  (Read 1230257 times)  Share 

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annabeljxde

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Re: 3U Maths Question Thread
« Reply #3990 on: March 11, 2019, 07:27:19 pm »
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Jamon took the above two I believe? So I'll take this one :P

When you're dealing with volumes, in your first line I can already see you've made a mistake loads of people make. The key thing here is that you need to take top curve squared minus bottom curve squared, not top curve minus bottom curve all squared.



What you've done is great, you've just started with the wrong integral!

Hope this helps :)

Ohhh, that makes so much more sense! Thank you~~

Oh and I don't think Jamon got around to answering my second question (the one with the cos^3 theta). If you have the time, it would be greatly appreciated if you could help me approach that question as well. But nevertheless thank you for that last one! I was stuck on that one for so long and it was really gnawing at me!!
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fun_jirachi

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Re: 3U Maths Question Thread
« Reply #3991 on: March 11, 2019, 08:32:08 pm »
+1
You should get the result if you substitute in the result Jamon derived in his reply; but there is another way to do it :)

In the same way that you derive the sine result that Jamon derived, you can also derive a similar result for cosine (I'm going to skip the working here for the sake of time saving when typing it up, but it involves much the same thing (lots of double angle and expansion results)) (ask if you want it though!) ie.


Hopefully this makes sense, and ask again if you need clarification!

Hope this helps! :)

EDIT: When subbing in the result Jamon derived, it's slightly off!
Over there it says

Which is correct, but the line two lines under it should say


And when you sub that result in for the sin theta cubed in your answer, it should turn out okay,
« Last Edit: March 11, 2019, 08:36:58 pm by fun_jirachi »
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annabeljxde

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Re: 3U Maths Question Thread
« Reply #3992 on: March 11, 2019, 09:25:39 pm »
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Hey guys! I have a question about induction:

If the answer for an induction question involving nQ (where Q is an integer) includes k in the Q expression, is the answer valid?? Also, I sometimes end up with different answers than the answers stated in the marking criteria for past exam questions, but I still end up proving my answer. Does this make my answer incorrect or are there different methods/ways to proving a statement by induction?
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annabeljxde

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Re: 3U Maths Question Thread
« Reply #3993 on: March 11, 2019, 09:26:42 pm »
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You should get the result if you substitute in the result Jamon derived in his reply; but there is another way to do it :)

In the same way that you derive the sine result that Jamon derived, you can also derive a similar result for cosine (I'm going to skip the working here for the sake of time saving when typing it up, but it involves much the same thing (lots of double angle and expansion results)) (ask if you want it though!) ie.


Hopefully this makes sense, and ask again if you need clarification!

Hope this helps! :)

EDIT: When subbing in the result Jamon derived, it's slightly off!
Over there it says

Which is correct, but the line two lines under it should say


And when you sub that result in for the sin theta cubed in your answer, it should turn out okay,

Yes! I tried out this method and it worked! Thank you again~~  :) :) :) :)
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fun_jirachi

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Re: 3U Maths Question Thread
« Reply #3994 on: March 11, 2019, 09:39:43 pm »
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Hey guys! I have a question about induction:

If the answer for an induction question involving nQ (where Q is an integer) includes k in the Q expression, is the answer valid?? Also, I sometimes end up with different answers than the answers stated in the marking criteria for past exam questions, but I still end up proving my answer. Does this make my answer incorrect or are there different methods/ways to proving a statement by induction?

Hey there,

Do you have an example for this? Finding it a little difficult to explain without an example :)
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Jefferson

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Re: 3U Maths Question Thread
« Reply #3995 on: March 15, 2019, 05:57:12 pm »
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For the bisection (halving the interval) approximation method, if the root is a turning point (double root) on the x axis, this method would fall apart right?
Since there is a root despite having no change in sign on either side.

So for questions that says, e.g.

"show that there are no roots, or if there are roots, between x = 3 and x = 4, getting two same or opposing signs respectively on either side isn't enough and we'll need to differentiate to confirm?" (unless we know what the graph looks like, of course).

Also, what does Maths In Focus means when they say
"If we halve the interval several times, the approximation to the root will USUALLY, but NOT always, become MORE accurate.
« Last Edit: March 15, 2019, 06:02:54 pm by Jefferson »

RuiAce

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Re: 3U Maths Question Thread
« Reply #3996 on: March 15, 2019, 06:03:40 pm »
+2
For the bisection (halving the interval) approximation method, if the root is a turning point (double root) on the x axis, this method would fall apart right?
Since there is a root despite having no change in sign on either side.

So for questions that says, e.g.

"show that there are no roots, or if there are roots, between x = 3 and x = 4, getting two same or opposing signs respectively on either side isn't enough and we'll need to differentiate to confirm?" (unless we know what the graph looks like, of course).
Yep.

Double roots can be viewed as the "enemy" of the bisection method. So can any root that involves some kind of "turning point" behaviour.

In general, to show that no roots lie between a certain interval, more work is necessary. We need to compute the corresponding function values of course, but we also need something stronger (like say monotonic increasing on said interval).

Jefferson

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Re: 3U Maths Question Thread
« Reply #3997 on: March 15, 2019, 06:19:50 pm »
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Yep.

Double roots can be viewed as the "enemy" of the bisection method. So can any root that involves some kind of "turning point" behaviour.

In general, to show that no roots lie between a certain interval, more work is necessary. We need to compute the corresponding function values of course, but we also need something stronger (like say monotonic increasing on said interval).

Oh alright, thank you for clarifying!

spnmox

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Re: 3U Maths Question Thread
« Reply #3998 on: March 16, 2019, 06:13:00 pm »
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I'm having some trouble simplifying this proof for maths induction:

Prove that the sum from r=1 to n of 3^(-r) = ((3^n)-1)/2(3^n) for all positive integers n.

For prove that n=k+1, I'm not sure how to simplify ((3^k)-1)/2(3^k) + 3^(-k-1) to obtain the RHS.

Jefferson

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Re: 3U Maths Question Thread
« Reply #3999 on: March 16, 2019, 08:50:25 pm »
+1
I'm having some trouble simplifying this proof for maths induction:

Prove that the sum from r=1 to n of 3^(-r) = ((3^n)-1)/2(3^n) for all positive integers n.

For prove that n=k+1, I'm not sure how to simplify ((3^k)-1)/2(3^k) + 3^(-k-1) to obtain the RHS.

Did took a longer route at some stage, but wasn't bothered to go back and change it.
Just power rules mostly in my approach.

Edit: I just saw where you were up to, so I'll do the solutions again, in black (attachment 2).

« Last Edit: March 16, 2019, 09:00:47 pm by Jefferson »

spnmox

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Re: 3U Maths Question Thread
« Reply #4000 on: March 17, 2019, 09:31:21 pm »
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Did took a longer route at some stage, but wasn't bothered to go back and change it.
Just power rules mostly in my approach.

Edit: I just saw where you were up to, so I'll do the solutions again, in black (attachment 2).

Thank you! Just for future reference, what is the best way to simplify difficult induction proofs or is it a case of getting better with practice?

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Re: 3U Maths Question Thread
« Reply #4001 on: March 19, 2019, 10:15:29 pm »
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How do I find the range of 1/(sqrt(-x^2+7x)

david.wang28

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Re: 3U Maths Question Thread
« Reply #4002 on: March 24, 2019, 05:25:09 pm »
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Hello,
I am stuck on a challenging Newton's method of approximation question. Can anyone please help me out? Thanks :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #4003 on: March 24, 2019, 05:34:11 pm »
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Hello,
I am stuck on a challenging Newton's method of approximation question. Can anyone please help me out? Thanks :)
It's actually just applying Newton's method the usual way, but an interesting thing happens.
\[ \text{Taking }f(x) = 8x^3 - 16x^2 -25\\ \text{we have }f^\prime(x) = 24x^2 - 32x \]
\[ \text{If }x_1 = 2.5\text{ is our first approximation, then}\\ x_2 = 2.5 - \frac{f(2.5)}{f^\prime(2.5)}\]
\[ \text{But plugging }2.5 - \frac{8(2.5)^3 - 16(2.5)^2 - 25}{24(2.5)^2 - 32(2.5)} \text{ into our calculator}\\ \text{we see that }x_2 = 2.5\]
\[ \textbf{This is not a coincidence.}\\ \text{Note that }f(2.5) = 8(2.5)^3 - 16(2.5)^2 - 25 = 0.\\ \text{Therefore }x=2.5\text{ is actually }\textbf{the root of the original equation we're interested in!}\]
\[ \text{So if }x_1 = 2.5\text{, it follows that }x_2 = 2.5.\\ \text{If }x_2 = 2.5\text{, it follows that }x_3 = 2.5.\\ \text{Hence in general, }x_n = 2.5\text{ for every integer }n.\]

david.wang28

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Re: 3U Maths Question Thread
« Reply #4004 on: March 24, 2019, 05:54:57 pm »
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It's actually just applying Newton's method the usual way, but an interesting thing happens.
\[ \text{Taking }f(x) = 8x^3 - 16x^2 -25\\ \text{we have }f^\prime(x) = 24x^2 - 32x \]
\[ \text{If }x_1 = 2.5\text{ is our first approximation, then}\\ x_2 = 2.5 - \frac{f(2.5)}{f^\prime(2.5)}\]
\[ \text{But plugging }2.5 - \frac{8(2.5)^3 - 16(2.5)^2 - 25}{24(2.5)^2 - 32(2.5)} \text{ into our calculator}\\ \text{we see that }x_2 = 2.5\]
\[ \textbf{This is not a coincidence.}\\ \text{Note that }f(2.5) = 8(2.5)^3 - 16(2.5)^2 - 25 = 0.\\ \text{Therefore }x=2.5\text{ is actually }\textbf{the root of the original equation we're interested in!}\]
\[ \text{So if }x_1 = 2.5\text{, it follows that }x_2 = 2.5.\\ \text{If }x_2 = 2.5\text{, it follows that }x_3 = 2.5.\\ \text{Hence in general, }x_n = 2.5\text{ for every integer }n.\]
Ahhh, way easier than I thought it will be. Thank you! :)
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