Specialist's Specialist Thread

[Lasercookie]

What is a co-domain? I always see functions expressed something like this:
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?

The co-domain are the numbers that contain the range, while the range is the numbers that do come out of the function (in other words, the range is a subset of the co-domain). For the range [2, inf), the co-domain for that is Real (because the numbers between 2 and infinity are Real numbers).

This function would have a co-domain of integers:
f: J --> J, where f(x) = 5x

[Bhootnike]

What is a co-domain? I always see functions expressed something like this:
f : [5, inf) -> R, f(x) = 2 + sqrt(x - 5)
So you can see from this function that the domain is [5, inf) and the range is [2, inf). But the range isn't written down; only the domain and co-domain are.
What is the co-domain? Why is it just R? Can you please give me an example of a function where the co-domain isn't R?

i think the co-domain is the set of possible values which may be produced, when the specific domain is applied to the equation. the range  is the ACTUAL values which are produced. e.g.

if dom was [1,4], and the codomain was [1,10], and your equation is, say, y=x+1,
then your range is, [2,3,4,5].
so codomain is like what may possibly be obtained, whereas range is the actual/obtained values you get when you 'plug' your x-values into the equation.

the codomain thus wont = R when you have:
f:[1,2,3,4,5] --> [1,2,3,4,5,6,7,8,9,10], f(x) = x+1

[Special At Specialist]

So does that mean that the range is restricted by all 3 of these:
- The implied range
- The given range
- The codomain
And I have to find the values of the range which lie within all 3? And then to restrict it even further is the actual domain which will only produce certain values to begin with.

[TrueTears]

just to re-emphasize what has been said, but diagrammatically, i find the diagram on the wiki page to be very helpful, check it out: http://en.wikipedia.org/wiki/File:Codomain2.SVG

[Bhootnike]

So does that mean that the range is restricted by all 3 of these:
- The implied range
- The given range
- The codomain
And I have to find the values of the range which lie within all 3? And then to restrict it even further is the actual domain which will only produce certain values to begin with.

the range will definitely be restricted by codomain, and the actual domain as you've said, and i reckon your other points would be other restrictions too, (by 'given range' i guess you mean that they have specifically stated that the range is [a,b]) . implied domain - yeah makes sense, if you had, , x cannot equal 1 thus the implied domain , if not already explicitly stated, would have to be implied to be [1,inf), or R+,  and thus implied range is also restricted.

 just to be sure, someone more mathematical than me will have to confirm that !

[Special At Specialist]

Okay thanks for clearing that up.

You know how:
sin^2(x) = (sin(x))^2
sin^3(x) = (sin(x))^3
sin^-1(x) = arcsin(x)

Well what does sin^-2(x) equal?
Is it (sin(x))^-2 or is it (arcsin(x))^2 ?

[brightsky]

always cosec^2(x).

[pi]

You won't see that notation on exams, but brightsky is right

[Special At Specialist]

Please help me with this complex numbers question:

w = (4z + 9) / (z - 4) and z = w bar
If z = x + yi, show that (x - 4)^2 + y^2 = 25

[Mao]

Well, here is a not-so-usual approach, I'm sure the more typical algebra-hack approach will be posted by someone else.





So, a circle centred at with a radius of 5. Which is the cartesian equation you have given.

[Special At Specialist]

Thanks Mao I have another question:

If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?

[Mao]

Thanks Mao I have another question:

If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?

The answer is undefined. X=5 is a vertical line. It's gradient would be +/- inifinity.

[Special At Specialist]

Thanks Mao I have another question:

If you have the equation x = 5, how do you find d(5) / dx ?
Would the answer be 1, since you treat 5 as the function of x and you take the derivative with respect to x, or would it be 0, since there is no change in the value of 5?

The answer is undefined. X=5 is a vertical line. It's gradient would be +/- inifinity.

Nooooo I'm not asking for the gradient. The gradient would be dy/dx.
I'm asking for d(5) / dx.
If you look at it from an algebraic perspective, you have 5 = x, therefore d(5) / dx = 1. But I'm not sure if it works that way which is what I'm asking.
But thanks for trying anyway.

[kamil9876]

You differentiate functions, not equations. (in the context of specialist maths I presume)

[#1procrastinator]

Hope I'm not hijacking the thread too much but what's the difference between differentiating functions and differentiating equations?