Mathematical headaches? Problem solved! My aim is to make maths easy for everyone. Any kind of question you have, ask away! Well, probably best to stick to maths.
I need to calculate $\int x^3 (x^3+1) (x^3 + 2)^{\frac 13} \dx$ and it’s giving me a headache! Can you help?
I’ve Blundered Using Parts, Rolled Out Fourier Expansions… Nothing!
Hi, IBUPROFEN, and thanks for your message! That’s a bit of a brute, but it can be done with a couple of substitutions.
What’s ugly?
One of the first questions I ask myself with something like this is: “What’s ugly?” Here, the $(x^3 + 2)^{\frac{1}{3}}$ isn’t very nice, so I pick the substitution $u = x^3 + 2$, which makes $\diff ux= 3x^2$ ; the middle bracket is $u-1$ and I’ve now got:
$\frac{1}{3}\int x (u-1) u^{\frac{1}{3}} \d u$.
… but I need to get rid of that $x$. I can write it as $(u-2)^{\frac{1}{3}}$, though:
Let’s let $v = u^2 – 2u$, so $\diff v u = 2u – 2$ — or $2(u-1)$1.
This becomes $\frac{1}{6} \int v^{\frac{1}{3}} \d v$, which is now pretty straightforward.
Aside
This works just fine – but looking at how it’s worked through, I suspect using the substitution $u = x^3 + 1$ would have led to a slightly more obvious second step. Both lead to the same answer, of course.
Tidying up
We get $\frac{1}{8} v^{\frac{4}{3}} + C$, or $\frac{1}{8} [u(u – 2)]^{\frac{4}{3}} + C$, or $\frac{1}{8} [x^3 (x+2)^3 ]^{\frac{4}{3}} + C$ or even – bringing the $x^3$ outside of the square bracket : $\frac{1}{8} x^4 (x+2)^{\frac{4}{3}} + C$.
Dave plugs Colin’s books. It takes us some time to recover from the shock.
Number of the podcast: $10^{6400} – 10^{6352} – 1$, a prime number consisting of 6,399 nines and an eight. Colin mentions the idea of Trinity Hall primes, although not by name.
Sam asked for ideas about what makes a good textbook – Dave has some.
Colin has a “$\pi$” day rant. Nobody writes a date as 3.14. We write it as 14/3, which is about 4.67, Feigenbaum’s constant.
Sam’s article is about the Menger sponge-cake she baked for Big MathsJam last year.
Sam also wrote songs for the MathsJam Jam (Ever Tried To Divide By Something You Shouldn’t’ve and Argand). Colin wrote The Signs They Need A-Changin’ about terrible British road signs for football stadiums (although @roice713 points out you can have such shapes in hyperbolic space.)
Colin is off to G4G next month but needs to crack a code first.
Dave liked Colin’s line: “Complex numbers are all fun and games until someone loses an $i$”.
Puzzle feedback: gold stars for @dragon_dodo (Dominika Vasilkova), @chrishazell72 (Chris Hazell) and @alephthought (Chris Hellings) who got the correct answer of 105,263,157,894,736,842.
Puzzle: There are eight winning lines in a 3×3 noughts-and-crosses grid, and 49 in a 3x3x3 grid. How many in a 3x3x3x3 grid? Can you generalise?
I enjoyed this one — no solution immediately jumped out at me, and I spend a great deal of time looking smugly at a way over-engineered circle theorems approach I can no longer remember.
Let’s label the apex of the triangle P, and the octagons points A to H, anticlockwise from the bottom left (so we’re looking for angle APD).
Now, APB is obviously $\piby 6$, and BPD is an isosceles triangle, so if we can find the angle PBD, we can easily finish the puzzle.
From there, it’s easy: HBDF is a square, and BDP is $\piby 3$, so PBD is $\piby 6$. (Another solution, by @remygarreau noted that AP and BD are parallel, which gives the same result.)
That makes the ‘base’ angles of the isosceles triangle each $\frac{5}{12}\pi$, and the angle required $\frac{7}{12}\pi$. Lovely!
I'm told that the three terms $a_1 = \log(2)$, $a_2 = \log(2\sin(x)-1)$ and $a_3 = \log(1-y)$ are in arithmetic progression and I need to find the range of possible values for $y$. I don't really know where to start!
– Logarithmic Arithmetic Progression Lacks A Clear Explanation
Hi, LAPLACE, and thanks for your message!
If three terms are in arithmetic progression, it means that the differences between consecutive terms is always the same. That is to say:
There are further restrictions on what can go into $\log$ as an argument: $1-y$ and $2\sin(x)-1$ must both be positive. Put another way, $y < 1$ and $\sin(x) > \frac{1}{2}$.
Rearranging (*) and getting rid of logs gives us $(2\sin(x)-1)^2 = 2(1-y)$. However, we know that $(2\sin(x)-1)$ is greater than zero and – by definition – no greater than 1. That means $0 < (2\sin(x)-1)^2 \le 1$ – and the middle of that inequality could just as well be $2(1-y)$, because they're equal.
$0 < 2(1-y) \le 1$
That rearranges to $-1 < -y \le -\frac{1}{2}$, which means that $\frac{1}{2} \le y \lt 1$.
Double-checking, that satisfies the original inequality we had for $y$, so it's our final answer.
When RITANGLE advises you to use technology to answer a question, you know it’s going to get messy.
So, with some trepidation, here goes:
(As usual, everything below the line may contain spoilers.)
It’s easy enough to do this in Geogebra – but somehow a little bit unsatisfactory to move the points around until they work. It is possible to do analytically – but only if you can solve quartics!
I can’t solve quartics. However, Wolfram|Alpha can – and I think that’s an acceptable use of technology in this case.
Here’s how I solved it.
Using the sine rule on the big triangle BAD gives $\frac{\sin(\alpha)}{2}=\frac{\sin\br{\frac{5}{6}\pi}}{1 + CD}$, which leads to $\sin(\alpha)=\frac{1}{1+CD}$.
In triangle ACD, angle ACD is $2\alpha$, and angle ADC is $\piby 6 – \alpha$. Using sine rule on this gives $\frac{\sin(2\alpha)}{2} = \frac{\sin\br{\piby 6 – \alpha}}{1}$.
We can tidy this up: $\sin(\alpha)\cos(\alpha) = \frac{1}{2}\cos(\alpha) – \frac{\sqrt{3}}{2}\sin(\alpha)$.
However, we know what $\sin(\alpha)$ is, so we can throw it in. Doubling everything: $2\frac{\cos(\alpha)}{1+CD} = \cos(\alpha) – \frac{\sqrt{3}}{1+CD}$.
Rearranging to make $\cos(\alpha)$ the subject leaves us with: $\cos(\alpha)\br{1 – \frac{2}{1+CD}}= \frac{\sqrt{3}}{1+CD}$.
Multiply both sides by $(1+CD)$ to get $\cos(\alpha)= \frac{\sqrt{3}}{CD-1}$.
Now we’re getting somewhere! We have expressions in terms of $CD$ for $\cos(x)$ and $\sin(x)$, which means we can turn to the identity $\sin^2(x) + \cos^2(x)\equiv 1$.
This is going to be messy, but no matter: $\frac{1}{(1+CD)^2} + \frac{3}{(CD-1)^2}=1$
Multiplying up, and letting $x=CD$ (as we should have done earlier, but I can’t be bothered going back to change it now), we get $(x-1)^2 + 3(1+x)^2 = (1+x)^2(1-x)^2$.
That leads to $4x^2 + 4x + 4 = x^4 – 2x^2 + 1$, or $0=x^4 – 6x^2 – 4x – 3 = 0$.
According to the Wolf, this has real solutions around -2.19 (which makes no sense in this context) and 2.80, which is the answer we’re after.
And, for some reason, we have to multiply it by 100, to which I say “shan’t”.
I was wondering: given a quadratic function with real coefficients, what complex arguments lead to real answers?
– Researching Equations And Lines
Hi, REAL, and thanks for your message!
This turns out to be simpler than I expected: if you have a quadratic $f(z) = az^2 + bz + c$ with real coefficients $a$, $b$ and $c$, and $z = x + y\i$ (with $x$ and $y$ real), you can simply find the complex part as $f_i(z) = 2axy + by$.
For $f(z)$ to be real, $f_i(z)$ must be zero, which gives $y(2ax + b)=0$.
If $y=0$, we have a real number to start with (which I think we already knew would give real answers); alternatively, if $x = \frac{-b}{2a}$, we get a real answer. This is neat: it’s the line of complex numbers perpendicular to the real axis, passing through the vertex of the (real) graph.
What about complex coefficients?
That leads to a natural extension: what if we don’t assert that the coefficients are real?
This gets a bit messy, but let’s roll with it: $f_i(z)=a_i (x^2 – y^2) + 2a_r xy + b_i x + b_r y + c_i$. Where is this zero? It looks for all the world like a hyperbola!
However, I’ll leave the properties of it as something for you to research once you’ve finished with equations and lines!
Before anything else, let’s get our terms straight. The 27 cubes can be grouped into four classes: vertices, of which there are eight; edges, of which there are 12; centres, of which there are six; and one in the very middle.
I’ll also label the six faces front, back, left, right, top and ground (rather than bottom, so that they all have different initial letters.)
“It can’t possibly be done!”
My first systematic thought was “what happens if I put this colour in one of the corners? Where can its friends go?”
The answer to that is “not in many places.” The corner lies on three faces, leaving only eight possible places for the remaining two cubes to go – three centres, four edges and the middle.
Here, I blundered: I reckoned that the only possible way to pick two of those eight without a clash on any face was one on an edge and one at a centre – and given that there are only six centres available, it couldn’t possibly be done. There would be two corners left over.
Hang on a second
What about the middle? The middle could be the same colour as two opposite corners, and this would solve that counting problem: two of the eight corners could be the same colour, and the remaining six would all be different colours.
We’d also use up six of the edges, leaving six more unspecified. That works out as two per face (because each edge lives on two sides), and I’d hope to be able to colour those appropriately.
Making it work
I did do this using the widget, but I think it might be clearer and easier to follow1 using numbers on a grid. Throughout, the left-hand grid represents the ground layer and the right-hand grid the top layer. The middle one, unsurprisingly, is in between. The left- and right-most columns of each grid are the left and right faces, and the top and bottom rows the back and front faces, respectively.
The first step is to put the middle in place, with its two friends arranged in opposite corners:
Now, that is interesting. The first three colours line up in columns – a different column in each grid, and each a rotation of the one before. Let’s move on to the next corner, and stick a 4 in the ground-back-right. I’ll pick the top-front edge and left centre to go with it, like so:
Doing it colour-wise, I was crossing my fingers at this point that it would all work, but with numbers it’s pretty obvious where the remaining cubes go – the 6s where the columns with 4 and 5 and rows with 3 and 9 meet, and the 8s where the 7/9 columns meet 2/5.
And there we are! The top and ground grids have each digit once, as do the left and right ones, and the front and back.
But there’s more
As an added bonus, so do each of the middle slices – which means you could cut off any face and move it to the other side of the cube, as often as you liked, and it would still work. You could also swap any pair of parallel slices, as often as you liked.
In the representation I’ve gone for, each horizontal slice is the same as the one before it, shifted forward one and right one (wrapping around in an obvious way).
There are plenty of questions I don’t have answers to, though, the main one being: is this the only solution (up to symmetry)? I’ll leave that as an exercise for the interested reader.
I’m told that a rectangular box has a surface area of 64cm2, and I have to find the maximum possible volume. How would I do that?
– Can Uncle Bring Obviousness Into Differentiation?
Hi, CUBOID, and thanks for your message – I certainly hope I can!
We have two relevant equations here: if the sides of the box are $x$, $y$, and $z$, then the volume is $V = xyz$ and the surface area is $2xy + 2yz + 2zx = 64$ – better still, $xy + yz + zx = 32$.
I’m going to try to do this without resorting to vector calculus (which would probably be simplest.)
Let’s start by assuming – for the moment – that $z$ is a fixed, known value. We can then say: $y(x+z) = 32 – xz$, so $y = \frac{32-xz}{x+z}$.
Now we can write the volume as $V = \frac{xz(32-xz)}{x+z}$.
You thought we were going to differentiate? Well, so did I, but I threw myself a symmetry dummy: we could just as easily have made $x$ the subject and got $V=\frac{yz(32-yz)}{y+z}$, which is identical: this means that for any fixed value of $z$, $x=y$ at the maximum.
That means, for any given value of $z$, we have $V=x^2z$ and a surface area of $x^2 + 2xz=32$.
But, we can rearrange that last relationship: $z = \frac{32-x^2}{2x}$. Putting that into the $V$ equation gives $V = \frac{x\br{32-x^2}}{2}$. Differentiating gives $\diff Vx = \frac{32 – 3x^2}{2}$, which has a maximum at $x= \sqrt{\frac{32}{3}}$.
That gives us $V = \frac{1}{2}\sqrt{\frac{32}{3}} \frac{64}{3}$, or $\frac{128}{9}\sqrt{6}$.1
Hope that helps!
– Uncle Colin
It turns out, to the surprise of nobody who’s played with this before, that the maximum volume comes from a cube.
If you flick to the back of an old A-level formula sheets, you might spot a list of random digits like this one from an MEI book:
A random number table (MEI formula book)
Why on earth would you want random numbers?
There are all sorts of reasons one might want a set of random numbers: perhaps you have a large set of data and want to pick out a random subsample of it – you need a source of random numbers. Perhaps you have a simulation you want to run, with random parameters – you need a source of random numbers. Perhaps you want to play a game of Snakes and Ladders but your three-year-old has eaten the dice…
Hang on a second. Be right back.
Nowadays, of course, you could just go to random.org and ask for a bespoke selection of artisanal random numbers, such as a selection of $n$ from a list with or without replacement, lottery numbers, and so on. Generated by fluctuations in the atmosphere, I’m told. Haughty sniff. Fine, if you want to get random number from… physics, go ahead, be my guest.
A battery of tests
I don’t know how MEI (or anyone else) generated their random digits – but I do know that they are submitted to a rigorous series of tests to make sure they’re properly random.
Which is a bit of a head-scratcher: after all, the set of digits 1111…. is just as likely as any other set of the same length. However, there are tests for what one would expect a large number of random digits to look like. For example:
Frequency: Every digit would appear roughly the same number of times.
Serial: Every pair of digits appears roughly the same number of times.
Poker: The patterns in sets of five digits appear in roughly the right frequency.
Gap: The gaps between zeros (or any other digit) are distributed correctly.
There are many others, of increasing sophistication, with names like Kolmogorov-Smirnov, Wald-Wolfowitz, and The Diehard Tests – but at a minimum, a useful set of random numbers will pass the four tests above.
Selecting a sample
Suppose we want a sample of ten people from a population of 450 conveniently-labelled individuals (from 0 to 449). Doing it the old-fashioned way, we could use the random number table to generate the appropriate labels.
Given a big table of random numbers, you should pick your starting position at random. Because it is currently 4:04, I shall pick the fourth box in the fourth column as my starting point.
I then go down, row by row, taking the first three digits of each row, rejecting any that are at least 450, until I have ten:
165
991 (reject)
183
450 (reject)
424
943 (reject)
082
377
328
783 (reject)
543 (reject)
334
899 (reject)
455 (reject)
687 (reject and start the next colum)
244
172
(… the next eight require rejecting …)
407
So, the selection for our sample is 165, 183, 424, 082, 377, 328, 334, 244, 172 and 407. Don’t try to tell me you’d rather have a website do that for you!
Kish grids
An interesting problem comes up in telephone surveys: if you always interview the person who answers the phone, you incur a bias towards phone-answerers.
A method developed to avoid this is the Kish grid, which works as follows.
For your $k$th call in the survey:
Ask how many eligible people live in the household (call it $n$).
Assign each person a number in order of age, starting with one.
For each house, calculate $k \pmod{n}$. (For the purposes of the survey, a remainder of 0 counts as $n$).
Interview that numbered person in the list.
(Telephone surveyors, however, tended to have a grid available that would tell them which person to ask for – apparently not everyone likes doing division on the fly mid-call1 !)
This is quite neat – in the long term, every member of an $n$-person household is equally likely to be picked, although there is a slight bias in favour of young people.
So there you go. Hopefully that counts as an interesting post about random numbers tables!
Am I unusual? I would be more freaked out by the phone call than the division.
I was asked to complete the square on $f(x) = 2x^2 + 13x + 20$. I started by halving everything, which makes it cleaner, but the solution manual disagrees. What gives?
– Have Always Loathed Functions
Hi, HALF, and thanks for your message!
Dividing a quadratic by the $x^2$ coefficient — here, 2 — to make the numbers easier is often a good strategy, but you have to be careful. In particular, here you have an equation: $f(x) = 2x^2 + 13x + 20$ – and you can’t just halve one side of it unless you halve the other.
So, you can tackle it that way, as long as you account for the two.
You can do $\frac{1}{2}f(x) = x^2 + \frac{13}{2}x + 10$, then:
When you worked through, presumably you got an answer of $\br{x+\frac{13}{4}}^2 – \frac{9}{16}$. You can see that this isn’t the same as $f(x)$ by substituting a value into both – for example, $0$.
In the original function, $f(0)=20$. In the halved answer, $\br{\frac{13}{4}}^2 – \frac{9}{16} = 10$, which is not the same.
Since the idea of completing the square here is to write the same function in a different way, it’s no good to have different answers for the same value! The two things aren’t equivalent, so something has gone wrong.
When you can halve it
You can halve the expression without any penalty if you know it’s equal to zero – for example, if you’re trying to find the roots. In that case, halving the other side of the equation doesn’t change it, and everything is hunky-dory.