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STEP Maths I, II, III 1994 Solutions

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Reply 20
Oh whoops :p:
I thought that since
P(r)(a)=0P^{(r)} (a) = 0 provided that 1 <=r <m
and since that part of the question only deals with 1 <=m <n,
it would be sufficient. Oh well, I'll be a bit more rigorous/specific next time :p:

Oh, I didn't see the degree of 'n' bit...
It says "show that" - would an explanation suffice? Something like (x²-1)^n begins with x2nx^{2n}, so that bracket is essentially a polynomial of degree "2n". Differentiating reduces the order by one, so differentiating it 'n' times reduces the degree to 2n-n = n?

(Also, I had a go at Q6, but I'm not sure about the last bit...stupid infinity. Hopefully someone will uninvent it for the next few months.)

Hm. I can barely do STEP II - am I even gonna get a single mark in STEP III? :frown:
Question 5 looks correct to me, I had done (i) which was really simple, but then I couldn't see what to do:frown:. Good work Rabite=)
For the 2nd part of STEP II, Q2,

I think what the examiners were intending was that you would use the linear transform Y=y/b,X=x/aY = y / b, X=x / a to transform
{(x,y):x2a2+y2b21,0y,0xc}\left\{(x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2} \le 1, 0\le y, 0\le x \le c\right\}
into {(X,Y):X2+Y21,0X,0Xc/a}\left\{(X,Y):X^2+Y^2 \le 1, 0\le X, 0\le X \le c/a \right\} and then reuse the first part. I think the integral was only introduced to give you a way (by substitution of X, Y) of getting the scaling factors right.

For the final part; calculate c=aba2+b2c = \frac{ab}{\sqrt{a^2+b^2}}. If we consider one quadrant, there's a "common" square to both ellipses (of area c^2), and two "extra bits", (one "extra bit" for each ellipse). The formula we've just calculated gives the area for the common square and an "extra bit". So using it twice will give us the total area + c^2 (because we're adding the common square twice, so we get an extra c^2). Finally, notice that we don't actually swap a,b in the formula, because we're also swapping the role of x,y (in terms of what c affects). So the area for a quadrent will be twice the value of the formula, minus c^2.

Now 1c2a2=a2c2a2=a2(a2b2)/(a2+b2)a2=a2a2+b2=c2b21-\frac{c^2}{a^2} = \frac{a^2 -c^2}{a^2} = \frac{a^2 - (a^2b^2)/(a^2+b^2)}{a^2} = \frac{a^2}{a^2+b^2}=\frac{c^2}{b^2} we get the area of a quadrant to be:

ab(cacb+sin1ca)c2ab \left(\frac{c}{a}\frac{c}{b} + \sin^{-1}\frac{c}{a}\right) - c^2
=sin1(ca)=sin1(ba2+b2) = \sin^{-1}(\frac{c}{a}) = \sin^{-1}\left(\frac{b}{\sqrt{a^2+b^2}}\right)

(Note that in fact, once you've got the idea that we can just apply some scaling, you don't even need to do this much work - instead of dividing the quadrant into a square and 2 extra bits, divide it in half along the line y=x. The area of each part is just a "stretched" sectors of a circle, and the very first argument for the first part of the question says the area of each sectors is just ab2sin1ca\frac{ab}{2}\sin^{-1}\frac{c}{a}. Then multiply by 8 for the 8 sectors ...)

[Incidentally, I didn't spot this when I did the question myself, because I split the original integral into a square and "extra bit" rather than a triangle and sector as Rabite did. Undoubtably Rabite's approach is what the examiners had in mind].
Reply 23
Ah hah.
Thanks for that! I simply knew it wasn't an integration question.
I started off down that transformation line, but I didn't know how to set it out, and for some reason, I wouldn't accept that the area would obey the rules. Even though it's probably obvious to...everyone else.
khaixiang
For question 2, the final result is correct, well done :biggrin: .. But you missed out the part where they asked you to show that P_n is a polynomial of degree n. Other than that, there's an incorrect assertion: "The first part of the question shows that (x^2-1) will be a factor of all derivatives",
You might well get away with this; I assumed rabite really just meant "all relevant derivatives"; the question is very clear about which derivatives vanish, so I assumed he knew perfectly well which do and do not vanish.

the very first result proven didnt show this, particularly it can be shown that the nth derivative and above of (x21)n(x^2-1)^n is not zero when x=±1x=\pm1. I have to prove all over again that f(r)(±1)=0 for n1r1 where f(x)=(x21)nf^{(r)}(\pm1)=0 \text{ for } n-1\geq r \geq 1 \text{ where } f(x)=(x^2-1)^n
Surely this does follow from the first part, since (x21)n=(x1)n(x+1)n(x^2-1)^n = (x-1)^n (x+1)^n?

I think there're other more efficient methods to this question, as the question mentioned "however some other ways of doing this question do not use this [reduction] formula". Hopefully someone can spot it.
For what it's worth, I can't see any better method. The final result is fairly horrific, and by using the formula they give you, you're effectively being given most of it "for free".

I think you can get a reduction formula directly for In=11Pn(x)2dxI_n = \int_{-1}^1 P_n(x)^2 dx using the relation 2(n+1)(2n+1)Pn(x)=ddx{Pn+14n(n+1)Pn1}2(n+1)(2n+1) P_n(x) = \frac{d}{dx} \left\{ P_{n+1} - 4n(n+1)P_{n-1}\right\} (hopefully that's right, it was a bit fiddly deriving it), and then since the coefficient of x^n in P_n(x) is (2n!)/n! and the smaller coefficients of x^n vanish, we can relate I_n to the integral the question asks for. But I certainly don't believe it will be a shorter answer!

The P_n are actually the Legendre polynomials (although the Legendre polynomials are normally scaled to be Pn(x)=12nn!dndxn(x21)n\displaystyle P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^n. There's quite a bit of literature about them, and an important result is that 11Pn(x)Pn(x)=2/(2n+1)\int_{-1}^1 P_n(x)P_n(x) = 2/(2n+1) which is closely related to the result the question asks. And the usual method of proof uses that same formula for cos2n+1dx\displaystyle \int cos^{2n+1} dx. There's another method using generating functions, but I can't believe anyone would even think of that approach, let alone find and prove the generating function works under exam conditions.
Rabite
(Also, I had a go at Q6, but I'm not sure about the last bit...stupid infinity. Hopefully someone will uninvent it for the next few months.)I have to say, the end bit is not exactly convincing, though I don't know what they were expecting. You've certainly done 95% of the work, so I'd hope they would still give you nearly full marks for the question.

I think any approach that starts sinθθ\sin \theta \approx \theta is going to be difficult to make work here, because at the end of the day, the question is asking for an exact answer, not an approximation. Basically, once you've put in that \approx sign you're doomed, because you've got no way of showing it becomes an equal sign as n goes to infinity.

You are better off starting with limθ0sinθθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1, so limn2nθsinθ2n=1\lim_{n \to \infty}\frac{2^n}{\theta} \sin \frac{\theta}{2^n} = 1, so limn2nsinθ2n=θ\lim_{n \to \infty}{2^n}\sin \frac{\theta}{2^n} = \theta. And clearly limncosθ2n=1\lim_{n \to \infty} \cos \frac{\theta}{2^n} = 1, so dividing the 2nd result by the first gives the desired result.

Another approach you might want to take, is to use 'small o' notation: We write o(x)o(x) to indicate a term (usually an error in an approximation) where limx0o(x)x=0\lim_{x \to 0} \frac{o(x)}{x} = 0. Then you can write sinx=x+o(x) \sin x = x + o(x) (note the equals sign!), and then later on let x->0 to get rid of all the o(x) terms.

The nice thing about this approach is that you can do any approximations you like, and just hide them all in that o(x) term. Just make sure your approximation error really does satisfy the definition of o(x). (The standard approximations all do, as far as I know).
Reply 26
DFranklin
You might well get away with this; I assumed rabite really just meant "all relevant derivatives"; the question is very clear about which derivatives vanish, so I assumed he knew perfectly well which do and do not vanish.

Surely this does follow from the first part, since (x21)n=(x1)n(x+1)n(x^2-1)^n = (x-1)^n (x+1)^n?


Well, here's my problem, I don't really know how much I should write, or to what extent should I assume that the examiners' know about certain thing. I've read some STEP paper reports, and it seems that they're very particular about this kind of thing, conditions, etc.

I am actually aware that Rabite might just have meant "all relevant derivatives", and would more than happy to ignore about the conditions and the little details and assume the examiner knows I meant "all relevant derivatives", but as I said, I am quite confuse as to how much one should assume that the examiner knows.

And well, by proving all over again, I meant writing a few more lines showing how the result from the first part can imply that drdxr(x21)n\frac{d^{r}}{dx^{r}}(x^2-1)^n is zero, (i.e. by the way you've shown it), when x is -1,1 for the range of r mentioned in the first part. And here's my query, can I not prove this at all, since it would take up precious time and instead just write what Rabite had written (or not state anything at all), directly proceed to vanishing the "square bits" and simplfying my repeated integration by parts? Should I assume that the examiners consider this as trivial?

EDIT: Also, I have read in the examiner's report for STEP papers, for "show that" question, examiners will usually demand more from the candidates in order to score the marks for that question. Since this is a "show that" question, shouldn't we take precaution in assuming that the examiners know what we meant?
Reply 27
Question 3, STEP I, 1994

(1x)n=(n0)(n1)x++(1)n(nn)xn\displaystyle (1-x)^{n}={n\choose 0} -{n\choose 1} x+\dots+(-1)^n{n\choose n}x^n
(1+x)n=(n0)+(n1)x++(nn)xn\displaystyle (1+x)^{n}={n\choose 0} +{n\choose 1} x+\dots+{n\choose n}x^n
(1x2)n=(n0)(n1)x2++(nn2)xn++(1)n(nn)x2n\displaystyle (1-x^2)^{n}={n\choose 0}-{n\choose 1} x^2+\dots+{n\choose \frac{n}{2}}x^n+\dots+(-1)^n{n\choose n}x^{2n}

First we consider n even.

Now collect the term xnx^{n} in the product of (1x)n(1+x)n(1-x)^{n}(1+x)^{n}
xn:2{(n0)(nn)xn(n1)(nn1)xn+}+(nn2)(nn2)xn\displaystyle x^{n}: 2\{{n\choose 0}{n\choose n}x^{n}-{n\choose 1}{n\choose n-1}x^{n}+\dots\}+{n\choose \frac{n}{2}}{n\choose \frac{n}{2}}x^n
But (nr)(nnr)=(nr)2=(nnr)2\displaystyle {n\choose r}{n\choose n-r}={n\choose r}^2={n\choose n-r}^2
Also, the term xnx^n in (1x2)n(1-x^2)^n is (nn2)xn{n\choose \frac{n}{2}}x^{n}
So then (nn2)xn=(n0)2xn(n1)2xn++(nn2)2xn++(nn)2xn\displaystyle {n\choose \frac{n}{2}}x^{n}={n\choose 0}^2x^{n}-{n\choose 1}^2x^{n}+\dots+{n\choose \frac{n}{2}}^2x^{n}+\dots+{n\choose n}^2x^{n}
 when n even (nn2)=(n0)2(n1)2++(nn2)2++(nn)2\displaystyle \therefore \text{ when n even } {n\choose \frac{n}{2}}={n\choose 0}^2-{n\choose 1}^2+\dots+{n\choose \frac{n}{2}}^2+\dots+{n\choose n}^2

For n odd, we want to simplify (n0)2(n1)2+(nn)2\displaystyle {n\choose 0}^2-{n\choose 1}^2+\dots-{n\choose n}^2
We can rewrite the sum as
{(n0)2(nn)2}+{(nn1)2(n1)2}++(1)n12{(nn12)2(nn+12)2}\displaystyle \{{n\choose 0}^2-{n\choose n}^2\}+\{{n\choose n-1}^2-{n\choose 1}^2\}+\dots+(-1)^{\frac{n-1}{2}}\{{n\choose \frac{n-1}{2}}^2-{n\choose \frac{n+1}{2}}^2\}
which is zero.
khaixiang
Well, here's my problem, I don't really know how much I should write, or to what extent should I assume that the examiners' know about certain thing. I've read some STEP paper reports, and it seems that they're very particular about this kind of thing, conditions, etc.

I am actually aware that Rabite might just have meant "all relevant derivatives", and would more than happy to ignore about the conditions and the little details and assume the examiner knows I meant "all relevant derivatives", but as I said, I am quite confuse as to how much one should assume that the examiner knows.Don't get me wrong, it's not 100% clear to me either.

I do think I have a fairly good idea how the degree papers at Cambridge get marked, and in that context, I don't think you'd lose more than a mark for a bit of carelessness like that. (Seeing as in my best two exam papers at Cambridge I dropped a total of 5 marks out of 280, it's not like I have an unrealistic belief about how you get full marks on a question). And yet the examiners' reports for STEP imply the marking for STEP is actually a lot stricter.

I don't know if it's partly the "public v.s private" face. Publicly, you don't want to tell candidates "oh, don't worry about a few odd mistakes"; you want them to try to produce the best answers possible. Privately, you know mistakes are going to happen. Are you really going to dock several marks for a slip of the pen/tongue? I don't know the answer, but I would hope not.

Don't get me wrong, your comments are entirely reasonable, because when you're practising, you do want to get into the practice of avoiding silly mistakes. But it's easy to look at a question where you made 3 little mistakes and think "well, that's a complete disaster"; my experience suggests that it's probably not that bad.

And well, by proving all over again, I meant writing a few more lines showing how the result from the first part can imply that drdxr(x21)n\frac{d^{r}}{dx^{r}}(x^2-1)^n is zero, (i.e. by the way you've shown it), when x is -1,1 for the range of r mentioned in the first part. And here's my query, can I not prove this at all, since it would take up precious time and instead just write what Rabite had written (or not state anything at all), directly proceed to vanishing the "square bits" and simplfying my repeated integration by parts? Should I assume that the examiners consider this as trivial?
Again, from experience, there's a bit of psychology going on here. If you come across as "I'm really good and I'm trying to get 6 full questions out in 3 hours", the examiners will tend to assume you missed out putting down steps because you were trying to save writing time, not because you didn't realise they were needed in the first place. If you're clearly struggling, they are more likely to assume the worst.

At the end of the day it's all speculation. You have to balance the risk of getting dinged a few marks against the benefit of saving some time.
Reply 29
STEP I - Question 6

All parts now complete - thanks to DFranklin on part iii.

i)
f(x)=ix1ix+1f(x)=\frac{ix-1}{ix+1}

f(x)=(ix1)2(ix+1)(ix1)f(x)=\frac{(ix-1)^2}{(ix+1)(ix-1)}

f(x)=x2ix+1x21f(x)=\frac{-x-2ix+1}{-x^2-1}

f(x)=(x1)+2ixx2+1f(x)=\frac{(x-1)+2ix}{x^2+1}

Hence: Ref(x)=x1x2+1Re f(x)=\frac{x-1}{x^2+1}

and:Imf(x)=2xx2+1Im f(x)=\frac{2x}{x^2+1}

ii)
f(x)=ix+1ix1f(x)^*=\frac{ix+1}{ix-1}

f(x)f(x)=(ix1)(ix+1)(ix+1)(ix1)f(x)f(x)^*=\frac{(ix-1)(ix+1)}{(ix+1)(ix-1)}

f(x)f(x)=1f(x)f(x)^*=1

iii)
f(f(x))=i(ix1ix+1)1i(ix1ix+1)+1f(f(x))=\frac{i({\frac{ix-1}{ix+1}})-1}{i({\frac{ix-1}{ix+1}})+1}

f(f(x))=i(ix1)(ix+1)i(ix1)+(ix+1)f(f(x))=\frac{i(ix-1)-(ix+1)}{i(ix-1)+(ix+1)}

f(f(x))=x+ix+i+1xix+i1f(f(x))=\frac{x+ix+i+1}{x-ix+i-1}

f(f(x))=x(1+i)+(i+1)x(1i)+(i1)f(f(x))=\frac{x(1+i)+(i+1)}{x(1-i)+(i-1)}

f(f(x))=x(1+i)+(i+1)x(1i)+(i1)=(x+1)(1+i)(x1)(1i)=x+1x11+i1i=i(x+1x1)f(f(x))=\frac{x(1+i)+(i+1)}{x(1-i)+(i-1)} = \frac{(x+1)(1+i)}{(x-1)(1-i)} = \frac{x+1}{x-1}\frac{1+i}{1-i} = i\left(\frac{x+1}{x-1}\right)

So we can now say:

Ref(f(x))=0Re f(f(x))=0

Imf(f(x))=x+1x1Im f(f(x))=\frac{x+1}{x-1}

iv)

f(f(f(x)))=(ix1)(ix+1)(1+i)+(i+1)(ix1)(ix+1)(1i)+(i1)f(f(f(x)))=\frac{{\frac{(ix-1)}{(ix+1)}}(1+i)+(i+1)}{{\frac{(ix-1)}{(ix+1)}}(1-i)+(i-1)}

f(f(f(x)))=(ix1)(1+i)+(i+1)(ix+1)(ix1)(1i)+(i1)(ix+1)f(f(f(x)))=\frac{(ix-1)(1+i)+(i+1)(ix+1)}{(ix-1)(1-i)+(i-1)(ix+1)}

f(f(f(x)))=2ix2x2i2f(f(f(x)))=\frac{2ix-2x}{2i-2}

f(f(f(x)))=x(i1)i1f(f(f(x)))=\frac{x(i-1)}{i-1}

f(f(f(x)))=xf(f(f(x)))=x
zrancis
f(f(x))=x(1+i)+(i+1)x(1i)+(i1)f(f(x))=\frac{x(1+i)+(i+1)}{x(1-i)+(i-1)}

I am stuck here! Not too sure how to extract the real and imaginary parts, I don't want to rationalise but I have a sinking feeling that I might have to...Factorize the top and bottom:

x(1+i)+(i+1)x(1i)+(i1)=(x+1)(1+i)(x1)(1i)=x+1x11+i1i=i(x+1x1)\frac{x(1+i)+(i+1)}{x(1-i)+(i-1)} = \frac{(x+1)(1+i)}{(x-1)(1-i)} = \frac{x+1}{x-1}\frac{1+i}{1-i} = i\left(\frac{x+1}{x-1}\right)

Incidentally, maybe it's just me, but I found it easier to rearrange f(z)=iz1iz+1f(z)=\frac{iz-1}{iz+1} as z+izi\frac{z+i}{z-i} (multiplying and top/bottom by -i).
Reply 31
DFranklin
Factorize the top and bottom:

x(1+i)+(i+1)x(1i)+(i1)=(x+1)(1+i)(x1)(1i)=x+1x11+i1i=i(x+1x1)\frac{x(1+i)+(i+1)}{x(1-i)+(i-1)} = \frac{(x+1)(1+i)}{(x-1)(1-i)} = \frac{x+1}{x-1}\frac{1+i}{1-i} = i\left(\frac{x+1}{x-1}\right)

Incidentally, maybe it's just me, but I found it easier to rearrange f(z)=iz1iz+1f(z)=\frac{iz-1}{iz+1} as z+izi\frac{z+i}{z-i} (multiplying and top/bottom by -i).


Thanks for that. Not too sure how I missed that simple rearrangement of f(z)!

So we can now say:

Ref(f(x))=0Re f(f(x))=0

Imf(f(x))=x+1x1Im f(f(x))=\frac{x+1}{x-1}
Reply 32
Is there anything more to this question? Everything about this question is so A-level-ish (not saying it's easy though). I spent some time looking for something that I might have missed, especially at the significance of f(x).f(x)*=1 on computing f(f(x)) and f(f(f(x))), but couldn't find any.
Reply 33
I wouldn't say so. Although for STEP I, imaginary numbers aren't on the syllabus anymore as they are not met at single maths A-level.
khaixiang
Is there anything more to this question? Everything about this question is so A-level-ish (not saying it's easy though). I spent some time looking for something that I might have missed, especially at the significance of f(x).f(x)*=1 on computing f(f(x)) and f(f(f(x))), but couldn't find any.
Don't think so, it's just an algebra grind. Q8 on the STEP III paper is like this as well (only worse, I'd say).

I'd forgotten this, but looking on Wiki, it does actually turn out that you can compose two transforms (az+b)/(cz+d), (ez+f)/(gz+h) by just multiplying the (complex) matrices

Unparseable latex formula:

\left(\begin{array}{cc}{a&b\\c&d}\end{array}\right)\left(\begin{array}{cc}{e&f\\g&h}\end{array} \right)



which would make this question come out a lot quicker. But I'm not convinced quoting this without proof would impress the examiners.

Seeing as we've been talking about marking, the "word of mouth" at Cambridge was that these kinds of questions (involving nothing but a lot of manipulation) tend to be marked a lot more harshly than the ones that require "ideas". If that is accurate, it would be worth spending a little time checking your work on this kind of question.
Reply 35
Question 6, STEP III, 1994

z2=(z1a)i+az3=(z2b)i+bz3=(a+bz1)+(ab)iz4=(z3c)i+cz4=(a+bcz1)i+(b+ca)z5=(z4d)i+dz5=(z1ab+c+d)+(a+b+cd)i\\z_{2}=(z_{1}-a)i+a\\z_{3}=(z_{2}-b)i+b\\z_{3}=(a+b-z_{1})+(a-b)i\\z_{4}=(z_{3}-c)i+c\\z_{4}=(a+b-c-z_{1})i+(b+c-a)\\z_{5}=(z_{4}-d)i+d\\z_{5}=(z_{1}-a-b+c+d)+(-a+b+c-d)i

z1=z5 if and only if a+bcd=(a+b+cd)i(ac)+(ac)i=(bd)i(bd)(ac)=(bd){1+i1+i}ac=(bd)i\\z_{1}=z_{5} \text{ if and only if } a+b-c-d=(-a+b+c-d)i\\(a-c)+(a-c)i=(b-d)i-(b-d)\\(a-c)=(b-d)\{\frac{-1+i}{1+i}\}\\ \therefore a-c=(b-d)i

(a-c)=(b-d)i represents the geometric condition that the line joining A and C is perpendicular to the line joining B and D in an argand diagram.


z2=(z1a)eiθ+az3=(z1a)e2iθ+(ab)eiθ+bz4=(z1a)e3iθ+(ab)e2iθ+(bc)eiθ+c\\z_{2}=(z_{1}-a)e^{i\theta}+a\\z_{3}=(z_{1}-a)e^{2i\theta}+(a-b)e^{i\theta}+b\\z_{4}=(z_{1}-a)e^{3i\theta}+(a-b)e^{2i\theta}+(b-c)e^{i\theta}+c
Now since a,b and c are distinct points on the unit circle in the complex plane, let
a=eiαb=eiβc=eiγ\\a=e^{i\alpha}\\b=e^{i\beta}\\c=e^{i\gamma}
Rearranging and factorising:

z4=z1e3iθ+(1eiθ)(eiγ+ei(β+θ)+ei(α+2θ))\\z_{4}=z_{1}e^{3i\theta}+(1-e^{i\theta})(e^{i\gamma}+e^{i(\beta+\theta)}+e^{i(\alpha+2\theta)})
For Z1 to always coincide with Z4,

e3iθ=1,θ=2kπ3e^{3i\theta}=1, \therefore \theta=\frac{2k\pi}{3} Where k is natural numbers which are non-multiple of 3
Also,
(1eiθ)(eiγ+ei(β+θ)+ei(α+2θ))=0\\(1-e^{i\theta})(e^{i\gamma}+e^{i(\beta+\theta)}+e^{i(\alpha+2\theta)})=0
1eiθ0 since θ2mπ,m=0,1,2,1-e^{i\theta}\neq0 \text{ since } \theta\neq2m\pi, m=0,1,2,\dots

eiγ+ei(β+θ)+[br]ei(α+2θ)=0\\e^{i\gamma}+e^{i(\beta+\theta)}+[br]e^{i(\alpha+2\theta)}=0

So the arguments of a,b and c must satisfy the following for a given angle of rotation.

eiθ=eiβ±e2iβ4ei(β+γ)2eiα\displaystyle \\e^{i\theta}=\frac{-e^{i\beta}\pm\sqrt{e^{2i\beta}-4e^{i(\beta+\gamma)}}}{2e^{i\alpha}}

Well, is that all? Not sure whether I got it right. Will appreciate if someone can check for me.
I think you can go further for the last part. We have:

eiγ(1+ei(θ+βγ)+ei(2θ+αγ))=0e^{i\gamma}(1+e^{i(\theta+\beta-\gamma)}+e^{i(2\theta+\alpha -\gamma)}) = 0. Since eiγ0e^{i\gamma} \neq 0 we must have 1+ei(θ+βγ)+ei(2θ+αγ))=01+e^{i(\theta+\beta-\gamma)}+e^{i(2\theta+\alpha -\gamma)}) = 0 (*)

Write u, v to be the principal arguments of ei(θ+βγ),ei(2θ+αγ)e^{i(\theta+\beta-\gamma)}, e^{i(2\theta+\alpha-\gamma)} respectively. Considering the imaginary part of (*), we see Im[eiu]=Im[eiv]Im[e^{iu}] = -Im[e^{iv}]. Considering the real part, we see eiu,eive^{iu}, e^{iv} must both have real part <= 0. So we must in fact have u = -v, which in tern implies 2cos(u)= -1. So we finally deduce that u=θ,v=θu =\theta, v=-\theta or vice versa.

Now u=θ    βγ=0u=\theta \implies \beta-\gamma = 0 contradicting the uniqueness of B,C. So u=θ(2θ(mod2π))u=-\theta (\equiv 2\theta \pmod{2\pi}) and so θ+βγ=θ(mod2π)    β=γ+θ(mod2π)\theta+\beta-\gamma = -\theta \pmod{2\pi} \implies \beta = \gamma + \theta \pmod{2\pi}.

Similarly the value for u gives 2θ+αγ=θ(mod2π)    α=γ+2θ(mod2π)2\theta+\alpha-\gamma = \theta \pmod{2\pi} \implies \alpha = \gamma + 2\theta \pmod{2\pi}.

So in fact, A, B, C are the vertices of an equilateral triangle.
STEP III Q5

(I hope I don't have one of those with mixed up naming, because this question seem unusually easy to me for being STEP III...)

f(x)=arcsin(x)f(x)=11x2f(x)=x(1x2)32=x3(1x2)2f(x)=arcsin(x)\newline f'(x)=\frac{1}{\sqrt{1-x^2}}\newline f''(x)=x(1-x^2)^{-\frac{3}{2}}=\frac{x}{^3\sqrt(1-x^2)^2}

Prove (1x2)f(x)xf(x)=0(1-x^2)f''(x) - xf'(x)=0 Substituting gives:
(1x2)x(1x2)32x(1x2)12=x(1x2)12x(1x2)12=0\frac{(1-x^2)x}{(1-x^2)^{\frac{3}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}=\frac{x}{(1-x^2)^{\frac{1}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}=0

Prove that (1x2)f(n+2)(x)(2n+1)xf(n+1)(x)n2fn(x)=0n>0(1-x^2)f^{(n+2)}(x) - (2n+1)xf^{(n+1)}(x) - n^2f^n(x)=0 \forall n>0
The case n=0 is already proven, now assume true for n=k
(1x2)f(k+2)(x)(2k+1)xf(k+1)(x)k2fk(x)=0k>0(1-x^2)f^{(k+2)}(x) - (2k+1)xf^{(k+1)}(x) - k^2f^k(x)=0 \forall k>0
Differentiate w.r.t. x
(f(k+3)(x)x2f(k+3)(x)2xf(k+2)(x))(2kf(k+1)(x)+2kxf(k+2)(x)+f(k+1)(x)+xf(k+2))k2f(k+1)(x)=(1x2)f(k+3)(x)(2kx+3x)f(k+2)(x)(2k+k2+1)f(k+1)(x)(f^{(k+3)}(x)-x^2f^{(k+3)}(x)-2xf^{(k+2)}(x))-(2kf^{(k+1)}(x)+2kxf^{(k+2)}(x)+f^{(k+1)}(x)+xf^{(k+2)})-k^2f^{(k+1)}(x) = \newline (1-x^2)f^{(k+3)}(x)-(2kx+3x)f^{(k+2)}(x)-(2k+k^2+1)f^{(k+1)}(x)
Simplifying this gives:
(1x2)f(k+3)(x)(2(k+1)+1)xf(k+2)(x)((k+1)2)f(k+1)(x)(1-x^2)f^{(k+3)}(x)-(2(k+1)+1)xf^{(k+2)}(x)-((k+1)^2)f^{(k+1)}(x)
This is the same as what we get when substituting k+1 into the assumption, so by mathematical induction P(k)    P(k+1)\bf{P}(k) \implies \bf{P}(k+1)
Q.E.D.

Now, to express this as a MacLaurin series we can first see it is an odd function (if not convinced just substitute x=0 into what just was proven...) so no even derivatives.
f(0)=0
f'(0)=1

So we have(10)2f2k+3(0)(2k+1)2f2k+1(0)=0(1-0)^2f^{2k+3}(0)-(2k+1)^2f^{2k+1}(0)=0. Meaning: f2(k+1)+1(0)=(2k+1)2f2k+1f^{2(k+1)+1}(0)=(2k+1)^2f^{2k+1} Where k= 0, 1, 2...
So the coefficients of the terms are
a1=1a3=13!a5=325!=340a7=52×327!=5×37×6×4×2a_1=1 \newline a_3=\frac{1}{3!} \newline a_5=\frac{3^2}{5!}=\frac{3}{40} \newline a_7=\frac{5^2\times3^2}{7!}=\frac{5\times 3}{7\times6\times4\times2}
etc. so the MacLaurin expansion is arcsin(x)=x+16x3+340x5+5196x7+...arcsin(x)=x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{196}x^7+...


To find the power series expansion forg(x)=Ln(1+x1x)g(x)=Ln(\sqrt{\frac{1+x}{1-x}}) we can see it as 12[Ln(1+x)Ln(1+(x))]\frac{1}{2}[Ln(1+x)-Ln(1+(-x))] for which we can use the McLaurin series for Ln(1+x).
Ln(1+x)=xx22+x33...+(1)n+1xnnLn(1+x)= x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{x^n}{n}
So: 12[(xx22+x33+...+(1)n+1xnn)(xx22+x33...+(1)n+1(x)nn)]\frac{1}{2}[(x - \frac{x^2}{2}+\frac{x^3}{3}+...+(-1)^{n+1}\frac{x^n}{n})- (-x - \frac{x^2}{2}+\frac{x^3}{3}-...+(-1)^{n+1}\frac{(-x)^n}{n})]
Even terms cancel and leaves 12×2(x+x33+x55+...+x2n+12n+1)\frac{1}{2}\times 2 (x+\frac{x^3}{3}+\frac{x^5}{5}+...+\frac{x^{2n+1}}{2n+1})
This means the the power series for g(x) is n=0x2n+12n+1\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1} and thus the coefficient is 12n+1\frac{1}{2n+1}
Comparing coefficients of the series for f(x) and g(x) we can see that g(x)>f(x). For example the coefficients of x7x^7 which are17>5×37×6×4×2\frac{1}{7}>\frac{5\times 3}{7\times6\times4\times2} for g(x) and f(x) respectively.


Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).
nota bene
Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).

a2n+1a2n1=(2n1)22n(2n+1)\displaystyle\frac{a_{2n+1}}{a_{2n-1}} = \frac{(2n-1)^2}{2n(2n+1)}

Let bnb_n be the corresponding coefficients for g, so that g(x)=bnxng(x) = \sum b_n x^n. Then b2n+1b2n1=2n12n+1\displaystyle\frac{b_{2n+1}}{b_{2n-1}} = \frac{2n-1}{2n+1}

So if cn=an/bnc_n = a_n / b_n, then c2n+1c2n1=(2n1)2(2n+1)(2n1)2n(2n+1)=2n12n<1\displaystyle\frac{c_{2n+1}}{c_{2n-1}} = \frac{(2n-1)^2(2n+1)}{(2n-1)2n(2n+1)} = \frac{2n-1}{2n} < 1

So since a3<b3a_3 < b_3, and the an,bna_n,b_n are all non-negative, we have an<bna_n<b_n for all (odd) n3n \ge 3.

(N.B. Am using my wife's laptop as my machine is chkdsk'ing a 250GB drive right now, so I don't have the question in front of me)
Reply 39
STEP I - Question 4

i)

cos(2α)=2cos2α1cos(2\alpha)=2cos^2\alpha-1

cosα=12(cos(2α)+1)cos\alpha=\sqrt{\frac{1}{2}(cos(2\alpha)+1)}

Similarily:

sinα=12(cos(2α)1)sin\alpha=\sqrt{\frac{1}{2}(cos(2\alpha)-1)}

Hence:

tan(α2)=12(cosα1)12(cosα+1)tan(\frac{\alpha}{2})=\frac{\sqrt{\frac{1}{2}(cos \alpha-1)}}{\sqrt{\frac{1}{2}(cos\alpha+1)}}

tan(α2)=12(1cosα)12(1cosα)tan(\frac{\alpha}{2})=\frac{\sqrt{\frac{1}{2}(1-cos \alpha)}}{\sqrt{\frac{1}{2}(-1-cos\alpha)}}
Rationalising:

tan(α2)=(1cosα)2(1cosα)(1cosα)tan(\frac{\alpha}{2})=\frac{\sqrt{(1-cos\alpha)^2}}{\sqrt{(-1-cos\alpha)(1-cos\alpha)}}

tan(α2)=(1cosα)(sin2α)tan(\frac{\alpha}{2})=\frac{(1-cos\alpha)}{\sqrt{(sin^2\alpha)}}

tan(α2)=1cosαsinαtan(\frac{\alpha}{2})=\frac{1-cos\alpha}{sin\alpha}

ii)

01sinα12xcosα+x2dx\int^1_0\frac{sin\alpha}{1-2xcos\alpha+x^2}dx

01sinα(xcosα)2+1cos2αdx\int^1_0\frac{sin\alpha}{(x-cos\alpha)^2+1-cos^2\alpha}dx

01sinα(xcosα)2+sin2ααdx\int^1_0\frac{sin\alpha}{(x-cos\alpha)^2+sin^2\alpha\alpha}dx

Using the result given:

=[sinα.1sinαtan1xcosαsinα]01=[sin\alpha.\frac{1}{sin\alpha}tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0

=[tan1xcosαsinα]01=[tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0

=[tan11cosαsinα][tan1cosαsinα]=[tan^{-1}\frac{1-cos\alpha}{sin\alpha}]-[tan^{-1}\frac{-cos\alpha}{sin\alpha}]
From part i)

tan(α2)=cosα+1sinαtan(\frac{\alpha}{2})=\frac{cos\alpha+1}{sin\alpha}

α2=tan1cosα+1sinα\frac{\alpha}{2}=tan^{-1}\frac{cos\alpha+1}{sin\alpha}

By also noting that tan1(cotα)=απ2tan^{-1}(-cot\alpha)=\alpha-\frac{\pi}{2}

=[tan1xcosαsinα]01=[α2][απ2]=[tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0=[\frac{\alpha}{2}]-[\alpha-\frac{\pi}{2}]

=12(πα)=\frac{1}{2}(\pi-\alpha)

QED

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