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# A Summer of Maths

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Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
• View Poll Results: Have you studied any Group Theory already?
Yes, I did some Group Theory during/at my A-level.
37.84%
No, but I plan to study some before Uni.
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No, I haven't.
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1. I haven't done any maths for like 2 weeks and I already feel rusty :/
2. What is the derivative of y=x^x?

What is the derivatie and inverse of y=x^x^x^x^... (an infinite string of x's)

Here x^x^x=x^(x^x) not (x^x)^x

Also for what values of x does y exist?
3. (Original post by Brit_Miller)
I have a question but not an answer as I don't know how to do it. Hopefully someone can show how.

Let

I have never done anything like this so the following is possibly wrong. Even if it is correct I'm sure there's a much prettier way of doing it:

Spoiler:
Show

(for the bijection proof)

Suppose there exists such that

We would have:

Clearly

Furthermore:

Now consider the functions and

The above equality is satisfied when

Obviously is strictly increasing. Differentiating with respect to gives so decreases. Therefore has only one solution, namely . But this is defies our initial conditions.

Therefore is bijective.

4. (Original post by Brit_Miller)
I have a question but not an answer as I don't know how to do it. Hopefully someone can show how.

Let

Spoiler:
Show

Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:
Show

Suppose that, for some , we have

where is non-negative.

Now, if we take for granted that , we continue by

Therefore, since the result is trivially true for , we get

as desired.

Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
Spoiler:
Show

Existence of n-th roots:

If is positive, and , then there exists exactly one positive real number such that .

By Lemma 1, we can always write as

but then the above theorem says that they must be equal.

{**} A solution:

Spoiler:
Show

We want to show that is injective and surjective which would imply that it is bijective.

[1] Suppose that with and at least one of and holds.

Then, we obtain the following two equations.

Using Lemma 2, we see from that , and by substituting into we get

which contradicts our assumption. Therefore, is injective.

[2] We show surjectivity by existence.

Take any . Then, it suffices to show that such that and .

We find that and where by applying Lemmas 1 & 2.

Therefore, is bijective.

Now, its inverse is given by

Spoiler:
Show

Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:
Show

Suppose that, for some , we have

where is non-negative.

Now, if we take for granted that , we continue by

Therefore, since the result is trivially true for , we get

as desired.

Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
Spoiler:
Show

Existence of n-th roots:

If is positive, and , then there exists exactly one positive real number such that .

By Lemma 1, we can always write as

but then the above theorem says that they must be equal.

{**} A solution:

Spoiler:
Show

We want to show that is injective and surjective which would imply that it is bijective.

[1] Suppose that with and at least one of and holds.

Then, we obtain the following two equations.

Using Lemma 2, we see from that , and by substituting into we get

which contradicts our assumption. Therefore, is injective.

[2] We show surjectivity by existence.

Take any . Then, it suffices to show that such that and .

We find that and where by applying Lemmas 1 & 2.

Therefore, is bijective.

Now, its inverse is given by

Nice - and informative too! I guess I was completely off the map on this one.
6. (Original post by Lord of the Flies)
Nice - and informative too! I guess I was completely off the map on this one.
You can add some bits to it, and fix some, but in general I don't think you were off the map at all. (not that I am much on the map, but anyway)

For instance, , so that is a point in the plane.
The function is a mapping from points on the plane to points on the real plane; i.e. .

Spoiler:
Show

Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:
Show

Suppose that, for some , we have

where is non-negative.

Now, if we take for granted that , we continue by

Therefore, since the result is trivially true for , we get

as desired.

Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
Spoiler:
Show

Existence of n-th roots:

If is positive, and , then there exists exactly one positive real number such that .

By Lemma 1, we can always write as

but then the above theorem says that they must be equal.

{**} A solution:

Spoiler:
Show

We want to show that is injective and surjective which would imply that it is bijective.

[1] Suppose that with and at least one of and holds.

Then, we obtain the following two equations.

Using Lemma 2, we see from that , and by substituting into we get

which contradicts our assumption. Therefore, is injective.

[2] We show surjectivity by existence.

Take any . Then, it suffices to show that such that and .

We find that and where by applying Lemmas 1 & 2.

Therefore, is bijective.

Now, its inverse is given by

Nicely done!
8. Is anyone interested in summing up

where is a positive integer? Ideas how we can do this?
Is anyone interested in summing up

where is a positive integer? Ideas how we can do this?
Without doing any mathematics, I think that should diverge. For large n the value of x can be ignored so n!/(n+x)! is approximately 1 so it must diverge.

EDIT: After doing some maths, it seems that the above reasoning is flawed, the terms do indeed tend to 0 so it may converge.
10. (Original post by james22)
Without doing any mathematics, I think that should diverge. For large n the value of x can be ignored so n!/(n+x)! is approximately 1 so it must diverge.
Take x = 2, then n!/(n+2)! = 1/(n+1)(n+2) which converges.
Is anyone interested in summing up

where is a positive integer? Ideas how we can do this?
Spoiler:
Show
I'd start off by noting that , which should converge for .

I'll finish it off when I get back if someone hasn't already.
12. subscribing.....around 2 in the night I tend to get bored
13. (Original post by Lord of the Flies)
Spoiler:
Show

If is the number of powers of we can define:

In particular

I don't believe it is possible to express the inverse in the form though.

Sorry I wasn't clear, there are meant to be infinite x's.

As a side question to it, for what values of x is f(x) defined?
14. (Original post by james22)
Sorry I wasn't clear, there are meant to be infinite x's.

As a side question to it, for what values of x is f(x) defined?
Ah. In any case my working is wrong, I stupidly misread the question!...

... which makes the question more difficult, but more interesting! Hm...

Spoiler:
Show

Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:
Show

Suppose that, for some , we have

where is non-negative.

Now, if we take for granted that , we continue by

Therefore, since the result is trivially true for , we get

as desired.

Lemma 2 (Corollary of Existence of n-th roots). Two cubes are equal if and only if the are the same.
Spoiler:
Show

Existence of n-th roots:

If is positive, and , then there exists exactly one positive real number such that .

By Lemma 1, we can always write as

but then the above theorem says that they must be equal.

{**} A solution:

Spoiler:
Show

We want to show that is injective and surjective which would imply that it is bijective.

[1] Suppose that with and at least one of and holds.

Then, we obtain the following two equations.

Using Lemma 2, we see from that , and by substituting into we get

which contradicts our assumption. Therefore, is injective.

[2] We show surjectivity by existence.

Take any . Then, it suffices to show that such that and .

We find that and where by applying Lemmas 1 & 2.

Therefore, is bijective.

Now, its inverse is given by

Nice

The following result could of been quoted to make your answer alot shorter (though it's always good practice to do things straight from the definitions)
Related exercise:
Let
g is said to be a left [or right] inverse of f if respectively

Show that f is surjective iff it has a right inverse
Show that f is injective iff it has a left inverse

Hence a map is bijective iff it has an (left and right) inverse. (an g is said to be an inverse iff g is both a left and right inverse.)

Using this, you only need to verify that your inverse is an inverse .
16. {*} Question:

The polynomial is irreducible over .

i) By completing the square, show that is not irreducible over the set of real numbers.

Hence, derive the Sophie Germain algebraic identity

by starting from the left-hand side.

ii) Evaluate

{**} Required:

Spoiler:
Show

A polynomial is said to be irreducible over a set if it cannot be factored into polynomials with coefficients from the given set.

As an example,

is irreducible over the set of rational numbers denoted by .

17. Okay, I have another question which I'm sure is relatively simple and someone will know (sorry for using the thread without answers, but it's a nice place to ask questions!)

Consider this 2nd order differential equation:

Write this as a system of 1st order equations with appropriate initial conditions.
18. (Original post by Brit_Miller)
Okay, I have another question which I'm sure is relatively simple and someone will know (sorry for using the thread without answers, but it's a nice place to ask questions!)

Consider this 2nd order differential equation:

Write this as a system of 1st order equations with appropriate initial conditions.
Let z=y'.

Also quite easy: How many (real) solutions does have? Knowledge required: GCSE & below.
19. In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order 1,2,3,4,5,6,7,8,9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based on the above information, how many such after-lunch typing orders are possible? (That there are no letters to be typed is on of the possibilities).

No ugrad knowledge required beyond combinations/ permutations.
20. (Original post by electriic_ink)
Let z=y'.

Also quite easy: How many (real) solutions does have? Knowledge required: GCSE & below.
Thanks

(and none surely?)

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