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HSC 2015 Maths Marathon (archive) (1 Viewer)

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InteGrand

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Re: HSC 2015 2U Marathon

In the following answer, please pretend that the negative square root of two is actually positive on the second to last line of working. For some reason, if I remove the negative, LaTeX screws up. If anyone knows why, let me know.

Also, the question only asked about positive numbers, so we only needed to consider positive x (could just ignore the case).
 

Flop21

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Re: HSC 2015 2U Marathon

a) solve: log[x,4]=log[e,sqrt(3)]

I've put the base of the log first in the brackets, followed by the number of the log after - hopefully it makes sense.

And here's another....

b) solve: differentiate y=x^(2) log[e,2x]
 

InteGrand

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Re: HSC 2015 2U Marathon

a) solve: log[x,4]=log[e,sqrt(3)]

I've put the base of the log first in the brackets, followed by the number of the log after - hopefully it makes sense.

And here's another....

b) solve: differentiate y=x^(2) log[e,2x]












 

aanthnnyyy

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HSC 2015 2U Marathon

I don't know why I'm getting conf used about this.. Minds playing tricks on me today. Anyway to the question:
- find area of region bound by y=e^x, y axis and y=e

(SORRY SOLVED CARELESS MISTAKE) BUMP ^^
 
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Flop21

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Re: HSC 2015 2U Marathon

How would you do this...

Solve for x:

5^(x-1) = 3(x)

I know you go... (x-1)log5 = xlog3

Then... xlog5 -log5 = xlog3

Then what?

I got this in an exam today and know I screwed it up. I couldn't get the x values by themselves - actually I did, but they kept cancelling themselves out, which is confusing.

***

Another question also... when using simpson's rule... h/3 [(fo + fn) + 4(odds) + 2(evens)] What the hell do you do when the x values are like, 0, 1/3, 1/4, 1/5, 1. Fractions cannot be odd or even... so did I just screw that question up also because I memorised a crap version of the formula? I think I just labelled the values as 1, 2, 3, 4, 5 etc. then took the odds and evens of that.
 

VBN2470

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Re: HSC 2015 2U Marathon

^ To solve for x, you would do the following:

 
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braintic

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Re: HSC 2015 2U Marathon

The trapezoidal rule is more accurate than The Simpson's Rule when more subintervals are used.
Sorry for digging up a dead post, but I had to correct this statement.
In general, Simpson's rule converges much more quickly to the exact answer than does the trapezoidal rule.

The correct answer should be: when the function is linear
 

Flop21

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Re: HSC 2015 2U Marathon

Sorry for digging up a dead post, but I had to correct this statement.
In general, Simpson's rule converges much more quickly to the exact answer than does the trapezoidal rule.

The correct answer should be: when the function is linear
No, don't worry I was waiting for an approval or correction.


So the Simpson's rule is always more accurate even if you use say 100 sub intervals with the trapezoidal rule on a function, compared to 3 sub intervals with the Simpson's rule?
 

InteGrand

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Re: HSC 2015 2U Marathon

No, don't worry I was waiting for an approval or correction.


So the Simpson's rule is always more accurate even if you use say 100 sub intervals with the trapezoidal rule on a function, compared to 3 sub intervals with the Simpson's rule?
No, it's referring to when you use the same number of sub-intervals, using Simpson's Rule is generally more accurate than Trapezoidal Rule.
 

InteGrand

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Re: HSC 2015 2U Marathon

Sorry for digging up a dead post, but I had to correct this statement.
In general, Simpson's rule converges much more quickly to the exact answer than does the trapezoidal rule.

The correct answer should be: when the function is linear
When the function is linear, the trapezoidal rule and Simpson's rule give the same answer (which is the exact answer).
 

InteGrand

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Re: HSC 2015 2U Marathon

How would you do this...

Solve for x:

5^(x-1) = 3(x)

I know you go... (x-1)log5 = xlog3

Then... xlog5 -log5 = xlog3

Then what?

I got this in an exam today and know I screwed it up. I couldn't get the x values by themselves - actually I did, but they kept cancelling themselves out, which is confusing.

***

Another question also... when using simpson's rule... h/3 [(fo + fn) + 4(odds) + 2(evens)] What the hell do you do when the x values are like, 0, 1/3, 1/4, 1/5, 1. Fractions cannot be odd or even... so did I just screw that question up also because I memorised a crap version of the formula? I think I just labelled the values as 1, 2, 3, 4, 5 etc. then took the odds and evens of that.
For the Simpson's rule, we are meant to label the values as 0,1,2, etc, rather than starting at 1, and then we can apply the rule you quoted.

Also, the x-values we use for Simpson's rule should be evenly spaced. The one you gave isn't evenly spaced (or in the right order, since 1/3 > 1/4, etc.). Were the x-values in the Q actually 0, 0.25. 0.5, 0.75, 1 (i.e. dividing the interval [0,1] into two sub-intervals)?
 

Flop21

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Re: HSC 2015 2U Marathon

For the Simpson's rule, we are meant to label the values as 0,1,2, etc, rather than starting at 1, and then we can apply the rule you quoted.

Also, the x-values we use for Simpson's rule should be evenly spaced. The one you gave isn't evenly spaced (or in the right order, since 1/3 > 1/4, etc.). Were the x-values in the Q actually 0, 0.25. 0.5, 0.75, 1 (i.e. dividing the interval [0,1] into two sub-intervals)?
No they weren't the exact intervals, sorry cannot remember them. But I was given them, so I'm assuming they were evenly spaced, from 0 - 1. I think it had 5.

Damn, I cannot remember if I labeled it 0,1,2 or 1,2,3.


Can I ask one more question?

You know how we find h for that formula, (b-a)/n, what is n? It is the number of subintervals right? How do you work that out? Say you had 5 function values (to sub into x).

Thanks for your help!
 

Drsoccerball

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Re: HSC 2015 2U Marathon

No they weren't the exact intervals, sorry cannot remember them. But I was given them, so I'm assuming they were evenly spaced, from 0 - 1. I think it had 5.

Damn, I cannot remember if I labeled it 0,1,2 or 1,2,3.


Can I ask one more question?

You know how we find h for that formula, (b-a)/n, what is n? It is the number of subintervals right? How do you work that out? Say you had 5 function values (to sub into x).

Thanks for your help!
n is number of subintervals yes
5 function values = 4 subintervals
2 applications of trapezoidal rule = 4 subintervals
1 application of simpsons rule = 3 subintervals
correct me if im wrong havnt done this in ages
 

braintic

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Re: HSC 2015 2U Marathon

n is number of subintervals yes
5 function values = 4 subintervals
2 applications of trapezoidal rule = 4 subintervals
1 application of simpsons rule = 3 subintervals
correct me if im wrong havnt done this in ages
Just clarifying that:

n is the number of sub-intervals if you are using the h/6 version of the rule.
When using the h/3 version , n is the number of applications of the rule.
[Or it can vary depending on whether you use h=(b-a)/n or h=(b-a)/2n ]

And your numbers are not correct:
2 apps of trap rule = 2 sub-intervals
1 app of simpson's rule = 2 sub-intervals (3 function values)

For Trap Rule:
n apps = n sub-intervals = (n+1) function values

My recommendation for Simpson's rule:
n apps = 2n subintervals = (2n+1) function values, h=(b-a)/2n, Integral = h/3 (....)
 
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braintic

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Re: HSC 2015 2U Marathon

When the function is linear, the trapezoidal rule and Simpson's rule give the same answer (which is the exact answer).
Yep .... silly statement.

And I'll take this opportunity to correct a misunderstanding many people have:
Many people learn that Simpson's rule gives the exact value for the area of a parabolic region. This is only true if the area is external to the parabola (unless calculated indirectly by subtraction)
 

InteGrand

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Re: HSC 2015 2U Marathon

No they weren't the exact intervals, sorry cannot remember them. But I was given them, so I'm assuming they were evenly spaced, from 0 - 1. I think it had 5.

Damn, I cannot remember if I labeled it 0,1,2 or 1,2,3.


Can I ask one more question?

You know how we find h for that formula, (b-a)/n, what is n? It is the number of subintervals right? How do you work that out? Say you had 5 function values (to sub into x).

Thanks for your help!
In that formula, n is the number of sub-intervals, you're right. Sub-intervals refers to how many points we are using. Basically, is our interval is [a, b], we can split it up evenly using the following x-values:

[x0, x1, x2, ..., xn – 1, xn], where

x0 = a (the starting point of our interval where we're integrating)

xn = b (our ending point).

Each 'sub-interval' refers to the following intervals: [x0, x1], [x1, x2], ..., [xn – 1, xn]. As you can see, there are n sub-intervals, because we have a total of n + 1 points being used.

Since we have equal spacing, the width of each sub-interval is h = (total width of intervals)/(number of intervals) = (ba)/n.

Now for Simpson's rule, we actually apply each approximation by spanning over two sub-intervals. E.g. if we had our original interval of integration as [0, 1], we'd need to split it up into at least two sub-intervals, e.g. via the partition {0,½,1} (here, we're using 3 x-values, and the two sub-intervals are [0, ½], [½, 1]).

Then what Simpson's rule would do is it essentially finds the unique quadratic (or line if the function we're integrating is linear) that goes through the points (0, f(0)), (½, f(½)), and (1, f(1)) and computes the integral of this quadratic (or line), and gives us this value as an approximation to the integral of our original function. So you can see that for Simpson's rule to work, we'll need at least 3 x-values. If we want more x-values, we can add two more (so 5 x-values), but we can't add just one more, since Simpson's rule requires a multiple of 2 to be the number of sub-intervals. E.g. if we tried using the x-values {0, ⅓, ⅔, 1} (i.e. 4 x values, which is 3 sub-intervals), our Simpson's rule would first be applied over the interval [0, ⅔] (using the y-values at 0, ⅓, and ⅔), but for the next interval, we couldn't use Simpson's rule, since we don't have 3 remaining x values.

So in general, for Simpson's rule, n (the no. of sub-intervals) must be even, because we use two sub-intervals for each 'use' of Simpson's rule.

(Sorry if that's hard to understand without a picture. Here's a picture: http://www.mathwords.com/s/simpsons_rule.htm ; Simpson's rule would be used on the interval [x0, x2] (as 3 x-values are used), and then the next one would use the interval [x2, x4], so even number of sub-intervals required (a sub-interval is just between adjacent x values)

So if we had 5 x values to use {x0 = a, x1, x2, x3, x4 = b}, we'd have n = 4, and Simpson's rule would be used on [x0, x2] and [x2, x4], and we'd sum up the results. (By using Simpson's rule on these intervals, I mean the rule for using it when there's just 3 values.) If you do this, you see that the 'odd' x's (i.e. the x's with odd-numbered subscripts) get counted four times, and the ones with even subscripts get counted twice (and the end points of the entire interval get counted once).
 
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InteGrand

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Re: HSC 2015 2U Marathon

Yep .... silly statement.

And I'll take this opportunity to correct a misunderstanding many people have:
Many people learn that Simpson's rule gives the exact value for the area of a parabolic region. This is only true if the area is external to the parabola (unless calculated indirectly by subtraction)
What do you mean by external?
 

InteGrand

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Re: HSC 2015 2U Marathon

Are you sure you haven't already guessed what I mean?
I thought what most people learn is that Simpson's Rule gives the following integral exactly: , where is a quadratic.

This is indeed correct, so I'm guessing I'm not sure what you're referring to by what many people have a misunderstanding with.
 
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