Joshua Capel
Joshua is a recent PhD graduate from UNSW who enjoys learning and teaching mathematics. His research interests are mathematical physics with a focus on special functions.
Location School of Mathematics and Statistics, UNSW, Australia
Activity
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Joshua Capel made a comment
Hi Everyone, I am an Associate lecturer in the School of Mathematics and Statistics at UNSW, the University of New South Wales. I look forward to discussing the material with you.
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Ah I see. I think you're right, if I restrict 't' to be between 0 and 1 then I get a finite length curve (but I'm okay with that, my screen/graph paper is finite in size too). But the (partial) curve I get will _exacty_ match the curve y=x^2.
However, if you allow 't' to be any number (not just between zero and one) then you will get all of y=x^2
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You mean the previous weeks test?
I think that question is asking "Can you find control points that would construct y=x^2 as a de Casteljau Bezier curve?".
I believe the information to answer that is somewhere in articles. But if you want to be sure and actually do it, this is an interesting challenge. So let me know how you go.
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You might...
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Joshua Capel replied to Steven Philip
How can you defined the modulus (or absolute value) for a (complex?) number if you can't define that number.
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Joshua Capel replied to Steven Philip
Interesting point. I think we're just talking about different functions (my function g never told me what to do with negative numbers).
So I would say (f∘g)(x)=f(g(x)) is undefined for negative x since g(x) wasn't defined for negative numbers (here we are just discussion these as functions to real numbers). So how can you define f(y) when y was not...
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Thanks for spotting that, I've fixed the ratio in the answer to be the same as in the question.
Another interesting point about F/W ∝ r is that for small values of r the force due to gravity (W) is too strong for the buoyancy force (F) to overcome. So for a specific choice of materials there will be a minimum value of r needed to ensure we have a flyable...
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Joshua Capel replied to Ged Langosz
Interesting, I also thought about how much overall time I "lose" to exercise. But like you said, it's an excellent de-stresser, so that raises the quality of life for my non-exercise hours.
Something only briefly touched on in the article is that these are not hours per week, but MET-hours per week (https://en.wikipedia.org/wiki/Metabolic_equivalent). In...
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The question in the test asked what the slope is if you go up and an edge (where two of the triangular faces meet). The slope 1.27 here is the slope when you go directly up a face.
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Joshua Capel replied to Grahame Thorne
I think 'inverse relation' in this course is precisely relationships of the of the form y=a/x. For example Zipf's law says frequency (F) is inversely proportional to rank (R) ( F(R)=a/R ).
The precise statement of Benford's law is that the frequency of the digit given by the difference of logarithms (specifically: F(d)=(ln(d+1)-ln(d))/ln(10) ). And this...
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I think you mean x³ + y³ + 3xy = 1 gives you a straight line equivalent to y = -x+1 (or x+y-1=0).
But that's not 100% correct. If you play around you will find another solution at [x,y]=[-1,-1] (which is not on the line above).
To see why, you might be interested to note that x³ + y³ + 3xy - 1 factors as (x+y-1)(x²-xy+y²+x+y+1).
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Thanks for catching that. I'll fix that right now.
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Joshua Capel replied to Grahame Thorne
You might be thinking of Zipf's law (where frequency is proportional to rank). Trying re-reading the section **Relation with the y=1/x function** in the Benford's law step ( https://www.futurelearn.com/courses/maths-linear-quadratic/1/steps/43139).
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Joshua Capel replied to Rufaro D
Hi Rufaro,
Good work, the argument looks valid to me (although using the F^n for the matrix and F(n) for the Fibonacci number could potentially be a little confusing).
So you've proven that if it is true for F^n then it is true for F^(n+1). So since it is true for F^1 (i.e. true for just the original matrix F) then it is true for F^2, and since it is...
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Joshua Capel replied to Rufaro D
I think that's right. Do you think you could show that this should always happen?
I'll give you a hint, trying considering what happens for F^(n+1) = (F) (F^n). If it is true for the matrix F^n, then I think you might be able to show it is true for F^(n+1) as well.
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What would you like help in understanding?
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Joshua Capel replied to laura zh
Good effort, you're not far off (only a 5% error).
This would work but unfortunately one of the entries in your F32 is incorrect (it should be {{1346269,2178309},{2178309, 3524578}}).
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If you mean log to the base 10, then that would be the graph of log_10(x)=ln(x)/ln(10) (see: https://en.wikipedia.org/wiki/Logarithm#Change_of_base).
Unfortunately this would mean looking at the area under the graph y=1/(x ln(10)), which isn't as nice to look at as y=1/x.
You do need to need careful when dealing with log as some people mean ln(x) when...
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Hi Brian,
I think it helps to think of this as a shadow of a circle on the plane.
I found a nice picture on this web page: http://archiviomacmat.unimore.it/PAWeb/Sito/Inglese/247i.htm, that might help (take a look a the wooden model pictured as well).
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10x^2-3x+12=((20*x-3)^2+ 471)/40, which is always positive (and so has no zeroes).
Make sure you don't just randomly pick numbers (it's not a fun challenge if you can't guarantee a solution).
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This challenge is a little mean. This can't be factored! (over real numbers)
The best I can do is rewrite it in the form : 14x^2+15x+24=((28 x+15)^2+1119)/56. Notice that it always positive (so it has no zeroes).
Make sure you don't just randomly pick numbers (it's not a fun challenge if you can't guarantee a solution)
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Joshua Capel replied to Josie Wright
It doesn't become 3(r-2)-1. If it helps, try to think of 3((r+2)-2)-1 like 3("new r" - 2)-1, where "new r"=r+2.
What we're trying to show here is that, if we know [r,s] is on the line y=3x-1, then we also know that [r+2,s] is on the line y=3(x-2)-1. Rewriting r=(r+2)-2 is just a clever way to quickly see how that "new r" variable would appear in this equation.
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You are correct. I looked it up and it is 1.2 kPa per 100m (not 1.2kPa per 1000m). I'll try to fix these up. Thanks.
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The reflection in the line y=x is given by the transformation T([x,y])=[y,x].
Perhaps I should give one little warning about using rotations in equations (since this is something I often need to stop and think about myself). The rotation Ged provided
x' = x cos θ + y sin θ
y' = -x sin θ + y cos θ
is a clockwise rotation by the angle θ, and it tells... -
Joshua Capel replied to [Learner left FutureLearn]
I agree, although it's not wrong, it feels like an unnecessary distraction (so I'll change it). Thanks.
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I can't know for sure, but I think choosing x=3 was just a case of trial and error (we tried it and saw helpful about the result).
There are plenty of other guesses you could take. For example try x=1, this gives 1-4y^2=3 or y^2=-1/2 (a negative number). Since a square is always positive this has no solution (i.e. no point on this curve is has coordinate...
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I notice some other participants mention an online plotter called Desmos (https://www.desmos.com/calculator). It seems nice, so you might want to give that a try too.
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Hi John,
I think these numbers should be taken with a large grain of salt. I found another testing website that allowed me to pick server locations and I was getting 679.8 Mbps for a Melbourne (Australia) Server, but only 29.1 Mbps for a London (Great Britain) server.
Perhaps unsurprisingly, the other speeds I found seemed to match the shape of the...
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Nice explanation John. That's also the way I try to think about it.
If you want to think about this in terms of the transformations we saw before, I have sometimes found it is useful to denote the new value with a different symbol. For example, using a star: [x*,y*]=T([x,y]).
Let's say we translate to the right by 5 and down by 2, the new coordinates...
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Nice! Using the fact that (y+f(y))/2 = 3 is very creative.
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Nice investigation. I just wanted you to know that the order of the transformations matters (so you just might need to double check you are getting what you wanted).
Let's try and work out the formula for your transformations. If we first use the reflection then we take the point [x,y] to the point [-x,y]. And now the translation will take this point (ie...
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Joshua Capel replied to Suzanne Robson
Do you mean 'Length of candle burnt', or 'Length of candle remaining'?
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Joshua Capel replied to Norma Cobián
Australia has a tax free threshold. So, at least here, it's not quite a direct proportionality (and different tax brackets would also complicate things).
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Joshua Capel replied to Sophie Plunkett
For a fixed distance? Try thinking about what happens when your double velocity (does time double?).
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Joshua Capel replied to Graham McDougall
That's not quite right. Doubling the flow doesn't doesn't double the time it takes. Try thinking about 'Volume' = 'Flow rate' × 'Time' (V = F × T).
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That sounds correct to me. I like the way you broke it down into two parts. But the final answer (a reflection in the line x=1) might be easier to understand than having to consider two transformations.
Maybe something interesting to think about. You do need to be careful which order you do these transformations. It has to be your reflection first followed...
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Joshua Capel replied to Rufaro D
I think 'Volume of water' would be directly proportional to 'Time to empty' if we can somehow guarantee that water is flowing out at a constant rate (maybe by using a constant speed pump?).
How about filling the bath instead? My taps usually have a constant rate of flow, so now 'Volume of water in the bath' and 'Time spent filling the bath' should be...
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Joshua Capel replied to Josie Wright
If you mean a direct proportionality then I have to ask: Do you feel twice as tired when you sleep twice as long?
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Joshua Capel replied to Mike Cragg
For a direct proportionality you should have the quantities scaling at the same rate. But if you double the diameter of a circle its area will increase increases by a factor of four. So this new pipe would carry four times as much water (if the water is moving at the same speed in both cases).
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Joshua Capel replied to Rufaro D
If we double the flow rate then the bath will empty in half the time. So 'Flow rate' and 'Time to empty' aren't an example of a directly proportionality. Can you see another quantity in your example which would be directly proportional to 'Time to empty'? (try thinking of a bath with a fixed flow rate).
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Joshua Capel made a comment
Hi Everyone, I also want to say welcome to the course. I'm sure this course will lead to a lot of interesting discussions.
