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# A Summer of Maths Tweet

Maths and statistics discussion, revision, exam and homework help.

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1. Re: A Summer of Maths
(Original post by Lord of the Flies)
Having , does not yield

Also, you stated which negates your conclusion.
Spoiler:
Show

If , then what value should take?
Typo, that k was meant to be a 1. This then gives f(k)=1 for all k.
2. 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).pdf
Last edited by jj193; 01-07-2012 at 22:55.
3. Re: A Summer of Maths
What are people going to read up on over the summer?
4. Re: A Summer of Maths
{*} Question:

Find such that for all and the inequality

is satisfied.

{**} Required:

Spoiler:
Show

The notation explicitly specifies the set of all natural numbers together with zero; i.e. the union of the natural numbers with the singleton that contains zero.
Does this work?

Spoiler:
Show

Therefore
Last edited by Lord of the Flies; 01-07-2012 at 23:19.
5. Re: A Summer of Maths
(Original post by james22)
Typo, that k was meant to be a 1. This then gives f(k)=1 for all k.
Oh 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! Sorry!
Last edited by Lord of the Flies; 01-07-2012 at 23:09.
6. Re: A Summer of Maths
(Original post by ben-smith)
What are people going to read up on over the summer?
Stats
7. Re: A Summer of Maths
(Original post by Lord of the Flies)
Does this work?
It does. Can you please use a
HTML Code:
[spoiler]blah[/spoiler]
to add solutions or hints. Thanks!
8. Re: A Summer of Maths
(Original post by ben-smith)
What are people going to read up on over the summer?
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

What do people with experience think about Vector Analysis and Cartesian Tensors by Bourne?

(Original post by Farhan.Hanif93)
...
9. Re: A Summer of Maths
It does. Can you please use a
HTML Code:
[spoiler]blah[/spoiler]
to add solutions or hints. Thanks!
Done!
10. Re: A Summer of Maths
(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

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.
Have you got any more questions like this?
11. Re: A Summer of Maths
(Original post by Tobedotty)
Stats
errr, really?
OK
12. 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.
13. Re: A Summer of Maths
(Original post by ben-smith)
errr, really?
OK
yeah! I've been finding it interesting for some reason. I'll probably get bored of it soonish and turn back to pure though and gorge on Olympiad problems I imagine.

What are you doing?
14. Re: A Summer of Maths
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

What do people with experience think about Vector Analysis and Cartesian Tensors by Bourne?
It 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.
15. Re: A Summer of Maths
(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

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.
Love these kind of problems 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?
16. Re: A Summer of Maths
{*} 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.

Last edited by jack.hadamard; 02-07-2012 at 12:34. Reason: Forgot about something. :p
17. Re: A Summer of Maths
(Original post by Farhan.Hanif93)
.. this does a sufficiently good job..
It is one of the books suggested for the first-year course; it looks good, I'll check it.
18. Re: A Summer of Maths
I have a question but not an answer as I don't know how to do it. Hopefully someone can show how.

Let

19. Re: A Summer of Maths
{*} 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.

1.

Spoiler:
Show
(n-1)/(a_1*a_(n-1))
Partial fractions did the trick here.
20. Re: A Summer of Maths
{*} 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.

ii)
Spoiler:
Show
use partial fractions: