A Summer of Maths
Maths and statistics discussion, revision, exam and homework help.
-
View Poll Results: Have you studied any Group Theory already?
Yes, I did some Group Theory during/at my A-level. 24 34.78% No, but I plan to study some before Uni. 15 21.74% No, I haven't. 30 43.48%
-
Re: A Summer of Maths(Original post by Lord of the Flies)
Having
,
does not yield 
Also, you stated
which negates your conclusion.
Typo, that k was meant to be a 1. This then gives f(k)=1 for all k. -
Re: A Summer of Maths
Q1
Can you spot the mistake?
We use induction to prove that n horses are the same colour:
Base case, n=1, true.
assume true for n=k \geq 1
If we have k+1 horses. Consider the first k horses. By assumption, they are all the same colour. Take one of these k horses, and the (k+1)th horse - again, by assumption, they are the same colour.
Hence all k+1 are the same colour (here we use 'transitivity of colour' - to be formalised below)
------------------
Q2
Above i mentioned 'transitivity of colour'.
Let S be a set, and ~ be a 'binary relation' on S. (this means that it represents true or false. It's like a magic ball, you can give it two elements a,b of S and ask it the question, are they related? and it will return true or false (This is notated as a~b)
In maths this is often stated as ~
where
denotes the Cartesian product (look it up!)
We say that '~' is:
transitive if and only if (a~b and b~c implies a~c) for any a,b,c in S
symmetric if and only if (a~b implies b~a) for any a,b in S
reflexive if and only if (a~a) for any a in S
If all three of the above are satisfied, then ~ is called an equivalence relation.
Convince yourself that "having the same colour" is an equivalence relation on any set of elements with a colour.
Question:
Is
is an equivalence relation? Is > an equivalence relation?
----------------
Q3
Let
be finite.
Use the principle of induction to show that it has a least element.
------------------
Really tough question:
The guests at Hilbert's hotel keep asking the receptionist if they can use the phone to make a call.
The female receptionist at Hilbert's hotel wishes to keep track of how many times each room number has used the phone.
She wishes to use her typewriter because he pencil broke and Mr Hilbert chewed her pen so that it no longer works (he was hungry). Again, Mr Hilbert has been silly and he chewed all but 2 of the keys off the receptionist's typewriter,
Q4 i) how can she keep her record?
Later that evening, after an unsatisfactory meal produced by Mrs Hilbert, Mr Hilbert is hungry again. Mr Hilbert eats another key from the typewriter, leaving just one key!
ii)It is possible for the receptionist to keep track of the room numbers, but how?
Spoiler:Show
hint:
ii) there are infinitely many primes.
any natural number has a unique prime factorisation. (Fundamental Theorem of Arithmetic)
------------
The first few pages of chapter 8 in Korner's "A Companion to Analysis" are readable and somewhat surprising.
it is found here: ftp://195.214.211.1/books/DVD-002/Ko...on_to_Analysis[c]_A_Second_First_and_First_Second _Course_in_Analysis_(2004)(en)(6 13s).pdfLast edited by jj193; 01-07-2012 at 22:55. -
Re: A Summer of MathsDoes this work?(Original post by jack.hadamard)
{*} Question:
Find
such that for all
and
the inequality

is satisfied.
{**} Required:
Last edited by Lord of the Flies; 01-07-2012 at 23:19. -
Re: A Summer of MathsOh right! I couldn't see what you where getting at with the mistake but by having a go at it myself I can see it now! Hence I kind of unknowingly rephrased your 'corrected' solution above!(Original post by james22)
Typo, that k was meant to be a 1. This then gives f(k)=1 for all k.
Sorry!
Last edited by Lord of the Flies; 01-07-2012 at 23:09. -
Re: A Summer of MathsStats(Original post by ben-smith)
What are people going to read up on over the summer? -
Re: A Summer of MathsIt does. Can you please use a(Original post by Lord of the Flies)
Does this work?
to add solutions or hints. Thanks!HTML Code:[spoiler]blah[/spoiler]
-
Re: A Summer of MathsI have put one of the books on which I'll spend some time over the summer.(Original post by ben-smith)
What are people going to read up on over the summer?
The topics that interest me are Group Theory, Analysis and some Vector Calculus.
Spoiler:Show
(Original post by Farhan.Hanif93)
... -
Re: A Summer of MathsDone!(Original post by jack.hadamard)
It does. Can you please use a
to add solutions or hints. Thanks!HTML Code:[spoiler]blah[/spoiler]
-
Re: A Summer of MathsHave you got any more questions like this?(Original post by DJMayes)
At the Nottingham University Open Day there was a "Maths Trail" with several interesting questions on it. The questions ranged from requiring a working knowledge of arithmetic progressions and combinations to lowest common multiples and counting squares; and more emphasis was put on thinking about them than slogging through endless manipulation. I thought I'd share one with you. The question is of the kind that could be set in C1, but is an interesting one:
A rectangle is inscribed inside a circle of radius 6 units such that each of the vertices of the rectangle lie on the circumference of the circle. Given that the perimeter of the rectangle is 28 units, what is the area?
Required Knowledge:
Spoiler:Show
- Pythagoras' Theorem
- Simultaneous Quadratic Equations
Hints:
Spoiler:Show
What information are you given? You have both the perimeter of the rectangle, and the radius of the circle it is inscribed inside. How can you relate these in terms of the length and width of the rectangle?
Spoiler:Show
Consider the diagonal of the rectangle
Full Solution:
Spoiler:Show
Let X and Y represent the length and width of the rectangle. Using this and the perimeter we can automatically deduce an equation:
2X + 2Y = 28
Which simplifies to X + Y = 14
As we are told that the radius of the circle is 6, we also know that its diameter must be 12. This diameter forms the diagonal of the rectangle. From this, another equation can be deduced using Pythagoras' Theorem:
X^2 + Y^2 = 144
We now have a set of simultaneous equations to solve. Re-arranging the first to leave Y in terms of X and substituting in the second equation leaves you with this:
X^2 + (14 - X)^2 = 144
This expands to:
X^2 + X^2 - 28X + 196 = 144
Which then simplifies to:
2X^2 - 28X + 52 = 0
or:
X^2 - 14X + 26 = 0
You can then complete the square or use the quadratic formula to get the result:
X = 7 + rt(23) or X = 7 - rt(23)
Using this you can then confirm that Y has the same values (i.e. if X = 7 + rt(23) then Y = 7 + rt(23) and vice-versa)
Now you know both side lengths, you simply multiply them together to get the area. The final answer is 26.
-
Re: A Summer of Maths
Saw something like this on TSR a while ago:
Question
Let the set
We randomly and independently generate elements of
times. If the sum of the results is
we stop. If the sum of the results is
, we randomly and independently generate elements of
times etc.
Given that
, what is the probability of this process ending?
Required
Spoiler:Show
Basic laws of AND/OR statements, conditional probabilities, total probabilities, and knowing what an independent event is.
-
Re: A Summer of MathsIt looks good but it's a little on the expensive side given that this does a sufficiently good job, in my opinion (and is mentioned on one of the Colleges' online reading list, I forget which). If you want more stuff on Tensors, there's another book from the same series (although I haven't personally tried it); and you can buy both for a fraction of the price of Bourne's book.(Original post by jack.hadamard)
I have put one of the books on which I'll spend some time over the summer.
The topics that interest me are Group Theory, Analysis and some Vector Calculus.
Spoiler:Show -
Re: A Summer of MathsLove these kind of problems(Original post by DJMayes)
At the Nottingham University Open Day there was a "Maths Trail" with several interesting questions on it. The questions ranged from requiring a working knowledge of arithmetic progressions and combinations to lowest common multiples and counting squares; and more emphasis was put on thinking about them than slogging through endless manipulation. I thought I'd share one with you. The question is of the kind that could be set in C1, but is an interesting one:
A rectangle is inscribed inside a circle of radius 6 units such that each of the vertices of the rectangle lie on the circumference of the circle. Given that the perimeter of the rectangle is 28 units, what is the area?
Required Knowledge:
Spoiler:Show
- Pythagoras' Theorem
- Simultaneous Quadratic Equations
Hints:
Spoiler:Show
What information are you given? You have both the perimeter of the rectangle, and the radius of the circle it is inscribed inside. How can you relate these in terms of the length and width of the rectangle?
Spoiler:Show
Consider the diagonal of the rectangle
Full Solution:
Spoiler:Show
Let X and Y represent the length and width of the rectangle. Using this and the perimeter we can automatically deduce an equation:
2X + 2Y = 28
Which simplifies to X + Y = 14
As we are told that the radius of the circle is 6, we also know that its diameter must be 12. This diameter forms the diagonal of the rectangle. From this, another equation can be deduced using Pythagoras' Theorem:
X^2 + Y^2 = 144
We now have a set of simultaneous equations to solve. Re-arranging the first to leave Y in terms of X and substituting in the second equation leaves you with this:
X^2 + (14 - X)^2 = 144
This expands to:
X^2 + X^2 - 28X + 196 = 144
Which then simplifies to:
2X^2 - 28X + 52 = 0
or:
X^2 - 14X + 26 = 0
You can then complete the square or use the quadratic formula to get the result:
X = 7 + rt(23) or X = 7 - rt(23)
Using this you can then confirm that Y has the same values (i.e. if X = 7 + rt(23) then Y = 7 + rt(23) and vice-versa)
Now you know both side lengths, you simply multiply them together to get the area. The final answer is 26.
Just a note though - could you not observe that you need the product of the roots of the quadratic equation, which is given by c/a in 0=ax^2+bx+c, so you know the answer is 26 without having to find the roots then multiply them together?
-
Re: A Summer of Maths1.(Original post by jack.hadamard)
{*} Question:
i) Let
be an arithmetic sequence with non-zero terms.
Obtain an expression for
.
ii) Find the sum to infinity of the following series.
Spoiler:Show(n-1)/(a_1*a_(n-1))
Partial fractions did the trick here. -
Re: A Summer of Mathsii)(Original post by jack.hadamard)
{*} Question:
i) Let
be an arithmetic sequence with non-zero terms.
Obtain an expression for
.
ii) Find the sum to infinity of the following series.
Sorry!
Just a note though - could you not observe that you need the product of the roots of the quadratic equation, which is given by c/a in 0=ax^2+bx+c, so you know the answer is 26 without having to find the roots then multiply them together?
