I haven't done any maths for like 2 weeks and I already feel rusty :/

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 02072012 13:54

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 02072012 14:36
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?Last edited by james22; 02072012 at 19:13. 
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 02072012 14:58
(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
(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.
Last edited by Lord of the Flies; 02072012 at 15:00. 
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 44
 02072012 15:07
(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
Lemma 2 (Corollary of Existence of nth roots). Two cubes are equal if and only if the are the same.
Spoiler:Show
{**} 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
Last edited by jack.hadamard; 02072012 at 15:10. 
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 02072012 15:31
(Original post by jack.hadamard)
{*} Additional:
Spoiler:Show
Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:Show
Lemma 2 (Corollary of Existence of nth roots). Two cubes are equal if and only if the are the same.
Spoiler:Show
{**} 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

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 46
 02072012 15:46
(Original post by Lord of the Flies)
Nice  and informative too! I guess I was completely off the map on this one.
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. . 
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 02072012 16:04
(Original post by jack.hadamard)
{*} Additional:
Spoiler:Show
Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:Show
Lemma 2 (Corollary of Existence of nth roots). Two cubes are equal if and only if the are the same.
Spoiler:Show
{**} 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

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 02072012 18:33

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 02072012 18:52
(Original post by jack.hadamard)
Is anyone interested in summing up
where is a positive integer? Ideas how we can do this?
EDIT: After doing some maths, it seems that the above reasoning is flawed, the terms do indeed tend to 0 so it may converge.Last edited by james22; 02072012 at 18:55. 
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 02072012 18:54
(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. 
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 51
 02072012 19:01
(Original post by jack.hadamard)
Is anyone interested in summing up
where is a positive integer? Ideas how we can do this? 
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 52
 02072012 19:11
subscribing.....around 2 in the night I tend to get bored

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 02072012 19:13
(Original post by Lord of the Flies)
As a side question to it, for what values of x is f(x) defined? 
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 02072012 19:27
(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?
... which makes the question more difficult, but more interesting! Hm...Last edited by Lord of the Flies; 02072012 at 19:43. 
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 55
 02072012 19:31
(Original post by jack.hadamard)
{*} Additional:
Spoiler:Show
Lemma 1. The cube of a negative number is the negative of its magnitude cubed.
Spoiler:Show
Lemma 2 (Corollary of Existence of nth roots). Two cubes are equal if and only if the are the same.
Spoiler:Show
{**} 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
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 .Last edited by jj193; 02072012 at 19:33. 
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 02072012 20:04
{*} 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 lefthand side.
ii) Evaluate
{**} Required:

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 02072012 20:04
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. 
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 02072012 20:12
(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.
Also quite easy: How many (real) solutions does have? Knowledge required: GCSE & below.Last edited by electriic_ink; 02072012 at 20:19. 
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 02072012 20:16
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 afterlunch typing orders are possible? (That there are no letters to be typed is on of the possibilities).
No ugrad knowledge required beyond combinations/ permutations.Last edited by Blutooth; 02072012 at 20:34. 
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 02072012 20:30
(Original post by electriic_ink)
Let z=y'.
Also quite easy: How many (real) solutions does have? Knowledge required: GCSE & below.
(and none surely?)
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