Differentiate y = arccosec(x) Watch

Zii
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#41
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#41
(Original post by Speleo)
It's because d/dx (x+C) = 1. You don't know any properties of exp(x) at this point other than that it is the inverse of log(x) and is its own derivative. i.e. you don't know that e^(a+b) = e^a.e^b
You could try deriving this from log(ab) = log(a) + log(b) though.
Oh, so by the chain rule:

y = \mathrm{exp}(x + C)

\frac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{exp}(x + C) \cdot \frac{\mathrm{d}}{\mathrm{d}x}(x + C) = \mathrm{exp}(x + C)

Cool. I've learnt loads through this thread.
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Speleo
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#42
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I don't think A-level handles exp and log very well.
It's nice to see how the various properties of exp and log are all connected, i.e. the facts that log(a) + log(b) = log(ab) and d/dx (logx) = 1/x aren't just coincidental.

If you are still interested, starting with the definition:
exp(x) = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...
and the fact that exp(1) is a finite number, e say, prove that:

exp(a+b) = exp(a)exp(b)
then
exp(n) = e^n for n = 0, 1, 2, ... (you can assume that exp(0) = 1)
then
exp(z) = e^z for z = ..., -2, -1, 0, 1, 2, ...

What is e^[a*log(b)]?
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Zii
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#43
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#43
(Original post by Speleo)
If you are still interested, starting with the definition:
exp(x) = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...
and the fact that exp(1) is a finite number, e say, prove that:

exp(a+b) = exp(a)exp(b)
then
exp(n) = e^n for n = 0, 1, 2, ... (you can assume that exp(0) = 1)
then
exp(z) = e^z for z = ..., -2, -1, 0, 1, 2, ...

What is e^[a*log(b)]?
I have tried to do that before but I didn't really know what route to take to do it. I ended up expanding things binomially and it got very messy - but should I have kept on going?

Can I assume you're studying maths at university? Or have some higher level qualification in something maths-related?
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Speleo
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#44
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Yeah, start with (1 + a + a^2/2! + ...)(1 + b + b^2/2! + ...) then consider which terms when multiplied out have a total exponent of n, i.e. the sum of the exponent of a and the exponent of b is n, then use the binomial theorem and sum over n.

I am a year into maths at university.
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Zii
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#45
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#45
(Original post by Speleo)
Yeah, start with (1 + a + a^2/2! + ...)(1 + b + b^2/2! + ...) then consider which terms when multiplied out have a total exponent of n, i.e. the sum of the exponent of a and the exponent of b is n, then use the binomial theorem and sum over n.

I am a year into maths at university.
Is it good? Do you do a lot of stuff like this? I'm really looking forward to it myself. I'm trying to do lots of maths over the summer to (a) keep myself brushed up and (b) because hell, I like it.

And I've tried doing what you suggested and it's working a treat. Things keep factorising nicely to give things like \frac{(a+b)^2}{2!} which is what I'm after!
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Speleo
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#46
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I enjoy it. This stuff is perhaps more university-style than A-levels are, i.e. "here is a definition of an object, what properties of this object can we deduce from this definition? what theorems can we prove about this object?" rather than "here is an object and a list of properties and applications of this object".

Well you know that a^n/n!.1 and a^(n-1)/(n-1)!.b are in this sum to start with...
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Zii
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#47
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#47
(Original post by Speleo)
I enjoy it. This stuff is perhaps more university-style than A-levels are, i.e. "here is a definition of an object, what properties of this object can we deduce from this definition? what theorems can we prove about this object?" rather than "here is an object and a list of properties and applications of this object".

Well you know that a^n/n!.1 and a^(n-1)/(n-1)!.b are in this sum to start with...
Sounds a lot more interesting than A-levels to be honest! Also, I managed to show \mathrm{exp}(A+B) = \mathrm{exp}(A)\mathrm{exp}(B) and given that \mathrm{exp}(0) = 1 is this enough to conclude that the \mathrm{exp}(x) function is an exponential one, e^x?
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Speleo
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#48
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Not immediately.
Try proving it for the natural numbers first.

We know what a^n is when n is an integer. We know what it is for a rational a^(p/q), but what is the definition of a^x when x is irrational?
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Zii
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#49
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Well, for a^n when n is an integer, it's simply a \times a \times a \times a \times a \times a \times a, n times.

For when n is a plain old regular fraction with p, q natural numbers, It's simply the same as before except p times, and then you have to q-root it. I must say actually defining this is a bit hard to do to be honest. As for a^x when x is irrational, well, I don't actually know how to define it at all.

It's surely easy to do for \mathrm{exp}(n) where n is an integer, as I've shown \mathrm{exp}(A+B) = \mathrm{exp}(A)\mathrm{exp}(B), essentially I can just let A + B = n, or worse comes to worse, break it down into (\mathrm{exp}1)(\mathrm{exp}1)(\  mathrm{exp}1) , n times.
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Speleo
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a^b where b is irrational is defined as exp[b*log(a)], which you can see is defined in terms of positive integer powers.

Basically we are trying to show that this is consistent with our definition of a^b for rational b.


EDIT: use induction for integer n
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Zii
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#51
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Oh of course! Induction!

I'm still a little confused... how do we actually define x^{\frac{p}{q}} ?
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Speleo
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q>0
x^(1/q) is the unique positive number with x as its qth power (q is an integer so this is defined).
Suppose a^q = b^q. Then (a/b)^q = 1 => a/b = +-1 since a, b are real. Therefore only one of a, b can be positive if a and b are distinct.

q<0
x^(1/q) = 1/[x^(1/-q)]

Then take the pth power, which is defined as p is an integer.
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