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Reply 120

EnglishMuon

Original post by Zacken

There's only one uni and I'm not sure they even do maths :erm:

Original post by physicsmaths

Go to a local unis lectures or somehong. The lecturer must like maths a little bit 👀

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http://www.uom.ac.mu/fos/index.php/mathematics
The name of Jayrani CHEENEEBASH is almost as good as King's Herbert Huppert :tongue:

Reply 121

theDanIdentity

Original post by EnglishMuon

A great book in my opinion is Mathematical Analysis by K.G. Binmore- Its an old one but contains some really interesting questions and answers to all of them. Covers a great amount of stuff too. http://www.amazon.co.uk/Mathematical-Analysis-A-Straightforward-Approach/dp/0521288827
(Its unbelievable how much some of these books cost, I just go for second hand nearly always now being the cheapo I am :tongue:

)

Original post by physicsmaths

A first course in Mathematical analysis. Pdf online aswell.

Original post by Zacken

Oh gosh, thank you so much!! :biggrin:

what texts did you guys use to get such an understanding of STEP maths? or any of the above mathematical derivations you've blurted all out on TSR? aside from the 'mathematical analysis by burkill' one i need all texts you've used; cheers

i'm interested in learning this stuff and doing an exam on it soon aha

Reply 122

Zacken

Original post by theDanIdentity

Those textbooks won't help for STEP, at all. You just need to get really comfortable with the A-Level stuff.

Reply 123

EnglishMuon

Original post by theDanIdentity

Yeah probably these specific ones wont help, unless they happen to ask a question on one of these 'obscurer topics' youve seen before, but for STEP, older textbooks seem to be the best, and cover some very STEP-like questions. For example, "A concise Introduction to Pure Mathematics" is very good, and the Pure Mathematics (and Applied) by Bostock and Chandler cover all the content very well.
https://www.amazon.co.uk/Concise-Introduction-Mathematics-Third-Chapman/dp/1439835985
https://www.amazon.co.uk/Pure-Mathematics-1-v/dp/0859500926/ref=sr_1_2?s=books&ie=UTF8&qid=1461759217&sr=1-2&keywords=bostock+and+chandler+mathematics
https://www.amazon.co.uk/Pure-Mathematics-2-v/dp/0859500977/ref=pd_bxgy_14_img_2?ie=UTF8&refRID=1Q2YZJF8S1D3BDSYMVRS

Reply 124

EnglishMuon

Original post by Zacken

I wasn't helping, it was more of a joint effort thing. :tongue:

(I really liked this, I don't often get to do maths with others 'cause solo-teaching on an island and shiz) - I've got something by the way (I haven't gotten around to doing it myself yet)):

A Ford circle is a circle tangent to the

x

-axis at a point

\left(\frac{p}{q}, 0\right)

for coprime integers

p, q

with radius

\frac{1}{2q^2}

. Show that, whilst some Ford circles touch, none of them intersect.

This is an interesting question, have you got a solution? Im in the middle of trying to form a contradiction, but nothing conclusive yet.

Reply 125

Zacken

Original post by EnglishMuon

This is an interesting question, have you got a solution? Im in the middle of trying to form a contradiction, but nothing conclusive yet.

I don't have one, haven't tried the problem out yet - but it's a fairly well known thing; even has its own Wiki page, so you should be able to look it up online if need be.

Reply 126

username1162972

Original post by EnglishMuon

This is an interesting question, have you got a solution? Im in the middle of trying to form a contradiction, but nothing conclusive yet.

This was the configuration in this years Q1 BMO2 :smile:

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Reply 127

EnglishMuon

Original post by physicsmaths

This was the configuration in this years Q1 BMO2 :smile:

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lol nice! Isnt that rule 34 of the internet? if there is a question online, there exists a BMO question related to it :wink:

Reply 128

EnglishMuon

Day 25-28 Summary
One point that I seem to have overlooked (or at least not appreciated enough) a few times now is the constant of integration when integrating series.
Consider the expansion of

(1-kx)^{-1/2}= 1+ \displaystyle\sum_{n=0}^ \infty \dfrac{(2n)!}{4^{n}(n!)^2} k^{n}x^{n}

(which may be derived via binomial theorem).
A common way of then finding new expressions for infinite series can come from integrating either side.
So we end up with

- \frac{2}{k}(1-kx)^{1/2} = x+ \displaystyle\sum_{n=0}^ \infty \dfrac{(2n)!}{4^{n}n!(n+1)!} k^{n}x^{n+1} +C

where C is the constant of integration. However, you may ask (you as in me earlier today) why we dont get an identical +C on either side of the expansion which cancels? Well we can clearly see that the result would not make sense, e.g. without a +C, when subbing in x=0 we get an immediate contradiction. But we could claim they are identical expressions to begin with so why would the +C be different? I believe this is just due to the nature of being an indefinite integral (think about a differential equation in mechanics with t=0 when v=k, k is not 0) ( @Zacken any nice way to explain this? or is it just trivially true?) .

Upon subbing in a value within the criteria for convergence (

|x| \leq \frac{1}{k}

) we see

C= - \frac{2}{k}

and we have come up with a new expression to use.

Reply 129

Zacken

Original post by EnglishMuon

( @Zacken any nice way to explain this? or is it just trivially true?) .

I guess it's just the "family of solutions" thing all over again. Just because two things have the same derivative doesn't mean that when you integrate them they're still identical. Basically, integrating kind of removes the "equality' sign and to put it back in, you need to add the +c.

I've been sitting here trying to think of an example where we could have a non-differential-equation example where

f(x) = g(x) but int f(x) =/= int g(x).

But I can't come up with anything interesting right now that motivates the use of the + c.

A trivial one is:

\frac{dy}{dx} = x

, i.e: you're saying the gradient is identically equal to x, but integrating both sides gets you y = x + c and not y = x. This just feels a little tautological though, in this context. Let me know if you can come up with an interesting example! I've tried thinking about integrating the geometric series term by term and integrating both sides of that inequality, but no fruit (I think?) .

Reply 130

Insight314

How is STEP Prep going? I see you are getting them S grades, is that correct? When did you start preparing for STEP and how much time do you spend on STEP Prep? :smile:

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Reply 131

Insight314

Original post by EnglishMuon

Day 25-28 Summary
One point that I seem to have overlooked (or at least not appreciated enough) a few times now is the constant of integration when integrating series.
Consider the expansion of

(1-kx)^{-1/2}= 1+ \displaystyle\sum_{n=0}^ \infty \dfrac{(2n)!}{4^{n}(n!)^2} k^{n}x^{n}

(which may be derived via binomial theorem).
A common way of then finding new expressions for infinite series can come from integrating either side.
So we end up with

- \frac{2}{k}(1-kx)^{1/2} = x+ \displaystyle\sum_{n=0}^ \infty \dfrac{(2n)!}{4^{n}n!(n+1)!} k^{n}x^{n+1} +C

|x| \leq \frac{1}{k}

) we see

C= - \frac{2}{k}

and we have come up with a new expression to use.

I loved doing that question. :smile:

STEP III 2007, I think?

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(edited 7 years ago)

Reply 132

EnglishMuon

Original post by Insight314

I loved doing that question. :smile:

STEP III 2007, I think?

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haha yep

It was quite nice.

Reply 133

EnglishMuon

Day 29 Summary
I kept flicking by the formula for 'radius of curvature' in the formula book, but never really thought about what it was about. So after looking up the basic definition that it was "the radius of the circular arc which best approximates the curve at that point", I thought the best way to understand it is to true derive it myself! Here is my attempt below- it all seems to work out, but the last line where somehow

( \dfrac{dx}{dt})^{3} \dfrac{d^{2}t}{dx^{2}} = -\dfrac{d^{2}x}{dt^{2}}

if my workings actually hold...

ImageUploadedByStudent Room1462100269.169224.jpg

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(edited 7 years ago)

Reply 134

Marxist

Original post by EnglishMuon

( \dfrac{dx}{dt})^{3} \dfrac{d^{2}t}{dx^{2}} = -\dfrac{d^{2}x}{dt^{2}}

if my workings actually hold...

ImageUploadedByStudent Room1462100269.169224.jpg

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Reply 135

username1162972

Original post by EnglishMuon

( \dfrac{dx}{dt})^{3} \dfrac{d^{2}t}{dx^{2}} = -\dfrac{d^{2}x}{dt^{2}}

if my workings actually hold...

ImageUploadedByStudent Room1462100269.169224.jpg

Posted from TSR Mobile

III 2007 Q4 this pops up on 'I think' not 100% sure.

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Reply 136

EnglishMuon

Original post by physicsmaths

III 2007 Q4 this pops up on 'I think' not 100% sure.

Posted from TSR Mobile

oh nice. I did this paper the other day but skipped over that q. It doesnt look like you need to know what its about in the q though but nice to see it used anyway!

Reply 137

username1162972

Original post by EnglishMuon

oh nice. I did this paper the other day but skipped over that q. It doesnt look like you need to know what its about in the q though but nice to see it used anyway!

Na you don't. It is just nice haha.

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Reply 138

EnglishMuon

Day 30 Summary
Although Im coming up with mostly full solutions to most STEP qs now, I still make the odd mistake and can look in to the interesting topics further!
Firstly, one very important and standard fact that I almost missed earlier:
Suppose we have two lines in the complex plane. From our knowledge of De Moivre's theorem, we know multiplying by a complex number can be seen by a rotation by addition of arguments (as well as a 'stretch' ) hence the transformation of one line to its perpendicular can be done by multiplying by

i

. So if we have 4 complex numbers

a,b,c,d

and the line segment

\vec{AB} = b-a

is perpendicular to

\vec{CD} = d-c

, we can say

b-a= \lambda i (d-c)

( lambda is a real scalar).
Furthermore, we know complex conjugates have the following properties:

\bar{z+w}= \bar{z} + \bar{w}

and

\bar{zw} = \bar{z} \bar{w}

b-a= \lambda i (d-c) \Rightarrow \bar{b}- \bar{a} = -\lambda i ( \bar{d}- \bar{c})

.

Here are also some interesting features of Fibonacci numbers that I have looked into:
From Pascal's triangle, it can be seen the sums of the

nth

'slanted' diagonals equals the

nth

Fibonacci number, i.e.

Unparseable latex formula:

\displaystyle\sum_{r=0}^{ \left[ \frac{n-1}{2} \right] } ^{n-r-1}\mathrm{C}_r = F_{n}

Also,

F_{n} = \dfrac{ \phi ^{n} - \psi ^{n}}{ \sqrt{5}}

where

\phi

is the golden ratio, and

\psi = \dfrac{1- \sqrt{5}}{2}

\displaystyle\lim_{x \to \infty} \dfrac{F_{n+1}}{F_{n}} = \phi

Reply 139

EnglishMuon

@Insight314 Would you be able help with a probable silly slip on an M4 q? On June 2011 q2, when applying Newton's Law of restitution to the wall I got

- \frac{3}{4} u \sin \theta = v \sin \phi

(where

\theta , \phi

are the angles between the wall and initial velocity and wall and final velocity respectively, but I shouldnt have a -ve there apparently. Why is this the case?