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[Christoff Yurievich]

I'm trying to find the inverse function ofthe equation f(x) = (2x)/(x-1). I know you start by putting y in the place of the f(x) and then exchanging the variables to get x= (2y)/(y-1). what now? Do I change the formula to read 2y * (y-1) to the -1 power and then multiple it out?

interesting note: my math professor said that the people who don't try to understand math usually do better at it. It's the ones who try to ask 'why' that end up not doing too well. It's a faith issue, he said. I have faith in my ability to do a problem, but I also have faith in my ability to screw it up. Give me the answer so I know if I did it correctly. Hell, most teachers only care about the work anyway so why not just give the answers and grade on the way the student got there?

Anyway, am I on the right track there? And no, it isn't for a test. Just an assignment that's due tomorrow.

 

[Dilbert]

I'm trying to find the inverse function ofthe equation f(x) = (2x)/(x-1). I know you start by putting y in the place of the f(x) and then exchanging the variables to get x= (2y)/(y-1). what now? Do I change the formula to read 2y * (y-1) to the -1 power and then multiple it out?

That's the right start. Now, you need to solve for y so that you have a new function of x.

I'll type it up right now...just wanted to give you a quick response.

 

[SD-Ness]

Not necessary. Solve for y.

Spoiler

x = (2y)/(y-1)

x(y-1) = 2y

[foil]

xy - x = 2y
xy - 2y - x = 0

[factor]

y(x - 2) = x
y = (x)/(x-2)

note: domain is x cannot equal 2

 

[Dilbert]

Spoiler

f(x) = (2x)/(x-1)
y = (2x)/(x - 1)

To find the inverse, switch x with y and then solve for y:

x = (2y)/(y - 1)
xy - x = 2y
-x = 2y - xy
-x = (2 - x)y
y = x/(x - 2)

(In the last step, I got rid of the -1 in the numerator by switching the terms in the denominator -- just makes it cleaner.)

INV f(x) = x/(x - 2)

Edit: Zero beat me to it...gotta learn to type faster. Also, good idea on the spoiler tags.)

 

[SD-Ness]

Goodie we got the same.

 

[Christoff Yurievich]

Thanks! So I was on the right track at least, that's always a good thing.

 

[SD-Ness]

What level math is this?

 

[belgurdo]

Why can't I be smart?

 

[LizardKing]

sweet, a math thread. maybe you guys can be kind enough to help me out also...

Prove the identity:

tan^2(theta) -1 / 1+tan^2 (theta) = 1-2cos^2(theta)

well i know there are some laws or properties of these but i don't know what they are. and i can't find my book. please help .

 

[fart]

1+1=????????????????

help me

 

[GG-Duo]

sweet, a math thread. maybe you guys can be kind enough to help me out also...

Prove the identity:

tan^2(theta) -1 / 1+tan^2 (theta) = 1-2cos^2(theta)

well i know there are some laws or properties of these but i don't know what they are. and i can't find my book. please help .

There are a lot of ways to do these, but here's a first crack.

Working with LHS

Apply Tan^2(x) + 1 = Sec^2(x)

Seperate the terms

Apply Tan^2(x) = sin^2(x) / cos^2(x), and apply sec = 1/cos

Apply Sin^2(x) = 1 - cos^2(x)

So you have 1 - cos^2(x) - cos^2(x)

=

For these, it's usually just
1. variations of Pythag.
2. variations of tan(x) = sin(x)/cos(x)
... unless you see cos(2x) or sin(2x). For those you need Double Angle identities

these should be in your book. and practice makes perfect.

 

[SD-Ness]

Meh.

Identities here: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

The one you need: tan^2(theta) + 1 = sec^2(theta)
-----

This isn't right. I've confused myself too much...Maybe you can get something out of it. BLAH:

Spoiler

tan^2(theta) -1 / 1+tan^2 (theta) = 1-2cos^2(theta)

tan^2(theta) - 1 / (1 / cos(theta)) = 1 - 2cos^2(theta)(1 / cos

[divide with complex fractions; common denominator being cos(theta)]

cos(theta) * (tan^2(theta) - 1) = 1 - 2cos^2(theta)

cos(theta)(tan^2(theta) - cos(theta) = 1 - 2cos^2(theta)

(sin^2(theta)/cos(theta) - cos(theta) = 1 - 2cos^2(theta)

1- cos^2(theta)/cos(theta) - cos(theta) = 1 - 2cos^2(theta)

1/cos(theta) - cos(theta) - cos(theta) = 1 - 2cos^2(theta)

1/cos(theta) - 2cos(theta) = 1 - 2cos^2(theta)

 

[SD-Ness]

Seperate the terms

Apply Tan^2(x) = sin^2(x) / cos^2(x), and apply sec = 1/cos

Apply Sin^2(x) = 1 - cos^2(x)

So you have 1 - cos^2(x) - cos^2(x)

There we go.

 

[CavemanLawyer]

Yay for failing for Calc test tonight! I'm playng catch-up to make a B now

 

[fallout]

And while these trig identities seem pointless and stupid now, they become incredibly handy when you're dealing with ugly-as-hell integrals. By the time you start learning multivariate & vector calc, the common tricks should be second nature.

EDIT: I should probably point out that multivariate & vector calc is a second year university level course, heh, so uh, you won't have to worry for a little bit anyway (assuming you plan on going anywhere in physics/math/etc).

 

[Christoff Yurievich]

What level math is this?

ah...college algebra. Granted, it's "last chapter of the book" college algebra so it's more leaning towards trig.

I'm studying for my business degree so the most I'll have to do is economics, some finite math, and elements of calculus.

as to the why: I had an instance a week ago where I was in the turoting room and I was getting help from one of the professors. I had a problem and I had the answer. I knew my answer was right, I just didn't know why. He couldn't explain the why to me after 15 minutes of trying. It wasn't that I couldn't understand what he was saying, it was that he couldn't explain it. He just 'knew' that was the right answer because of years of teaching. :/

 

[fallout]

With algebra, being able to visualize things isn't necessary, but extremely helpful. Oh, and ask your econ prof how to find the area under the curve using integral calc. That always sends them into really long and confusing explanations (or so I hear).

Anyway, just because this is a math thread ...

Heh, I did that up in Maple during my comp. phys lecture. The syntax and actual equation can be found here. Sadly, I didn't actually come up with that equation, but I did play around with it a bit more to make it fit better.

 

[LizardKing]

Thank you guys. That table of identities will definately be a big help too as the test is tomorrow.

 

[Christoff Yurievich]

funnily enough, we just studied e yesterday so I actually got this one. yay, I'm one of the smart people.

 

[SD-Ness]

Here's something cool then:

e^(i * pi) = -1

[I'm guessing you've done "i" as well...]

 

[fallout]

Heh, below is a really lame math joke. Please don't kill me.

So this constant and e^x are walking along right? Then the constant notices a differential operator walking towards them and starts to freak out, saying, "Oh shit! That differential operator is going to annhilate me! I've gotta get out of here!", and runs off.

Meanwhile, e^x is laughing at the constant, watching him run away. So, feeling cocky, e^x walks up to the differential operator and says "You've got nothing on me man, I'm e^x." to which the differential operator replies "Oh yeah? Well I'm d/dy."

 

[SD-Ness]

:lol

 

[goodcow]

Heh, below is a really lame math joke. Please don't kill me.

So this constant and e^x are walking along right? Then the constant notices a differential operator walking towards them and starts to freak out, saying, "Oh shit! That differential operator is going to annhilate me! I've gotta get out of here!", and runs off.

Meanwhile, e^x is laughing at the constant, watching him run away. So, feeling cocky, e^x walks up to the differential operator and says "You've got nothing on me man, I'm e^x." to which the differential operator replies "Oh yeah? Well I'm d/dy."

I don't get it.

 

[SD-Ness]

Then you probably haven't taken pre-calculus or calculus...

 

[TheQueen'sOwn]

That reminds me... I have algeo homework .

 

[goodcow]

Then you probably haven't taken pre-calculus or calculus...

You're right, I haven't.

 

[fallout]

I don't get it.

Time for some basic calc. I don't know how much detail I need to get into here, so I'll explain what I was talking about then throw in a few proofs maybe.

A pretty common notation for the most basic differential operator is d / dx, where you're taking the derivative of a function with respect to x (which can be any variable). Taking the derivative of a function is pretty basic and if you don't care about the essence of why it works, you can just use general equations to solve any function.

So for anything of the form x^n (where n is a constant), d / dx of that function equal nx^(n - 1). Don't really need to get into it any more than that. As an example,

d / dx of x^3 = 3x^2
d / dx of x = 1
d / dx of 3 = 0

I could get into stuff like the chain rule, product rule, blah blah, but that'd require me being here all night and hardly has anything to do with the joke, heh.

Now, let's talk about natural logarithms (ln) and e. Here are just some basic properties.

e^(ln(x)) = x
ln(e^) = y
ln(e) = 1
d / dx of ln(x) = 1 / x

If we have a function such as,

y = e^x

However, that might not be the most useful form of it. Alternatively, we can write that as,

ln = x

So, if we take d / dx of both sides, we get,

d / dx (ln) = 1

By the Chain Rule (hmm ... maybe I should've talked about that afterall ... whatever),

(d / dy) (ln) (dy / dx) = 1 and hence, (1 / y) (dy / dx) = 1

Multiply both sides by y and you get dy / dx = y, and since we already stated that y = e^x, dy / dx = e^x.

Therefore, d / dx of e^x = e^x

Really, the only thing you need to know here is that d / dx of e^x = e^x. So basically, the joke is that a constant will be turned to 0 by any differential operator, but d / dx will have no effect on e^x. However, if it's differentiated with respect to y (d / dy), then "x" is just another constant to it and will therefore become 0.

 

[bachikarn]

Really, the only thing you need to know here is that d / dx of e^x = e^x. So basically, the joke is that a constant will be turned to 0 by any differential operator, but d / dx will have no effect on e^x. However, if it's differentiated with respect to y (d / dy), then "x" is just another constant to it and will therefore become 0.

What? Wouldnt d/dy e^x = (e^x) * dx/dy ?

 

[SD-Ness]

What? Wouldnt d/dy e^x = (e^x) * dx/dy ?

Well (e^x) * dx/dy = (e^x)

So it's not necessary.

 

[fallout]

What? Wouldnt d/dy e^x = (e^x) * dx/dy ?

Haha, shit, I'm retarded. Didn't even think about that. Although, I know what would fix it, but there's no in hell I'm explaining it.

∂ / ∂y of e^x = 0

Huzzah, partial derivatives. Damn, sorry bout that.

 

[boredJonathan]

I feel geekier for just having opened this thread! w00t!

*ahem*

On a related note.. I now feel really dumb because of this thread.

 

[fallout]

I feel geekier for just having opened this thread! w00t!

*ahem*

On a related note.. I now feel really dumb because of this thread.

You feel dorkier? I've been sitting here trying to explain a calculus joke on a saturday night. The very words "calculus joke" are bad enough, even if you don't take into account what day of the week it is.

Although, in my own defense, finals are coming up and everyone's studying.

 

[fart]

what you want is del over dely

 

[impirius]

Hey guys, why is 6 afraid of 7

 

[B'z-chan]

Hey guys, why is 6 afraid of 7

Cause 7 8 9

 

[impirius]

Cause 7 8 9

Bingo! Wooooooooooo! I also would have accepted "because Dustin Runnels is roughly twice the size of Sean Waltman"

 

[Mumbles]

Heh, below is a really lame math joke. Please don't kill me.

So this constant and e^x are walking along right? Then the constant notices a differential operator walking towards them and starts to freak out, saying, "Oh shit! That differential operator is going to annhilate me! I've gotta get out of here!", and runs off.

Meanwhile, e^x is laughing at the constant, watching him run away. So, feeling cocky, e^x walks up to the differential operator and says "You've got nothing on me man, I'm e^x." to which the differential operator replies "Oh yeah? Well I'm d/dy."

I feel bad for laughing at that. I think I'll just blame it on my drunkenness.

I don't think that's true. You can understand math far better if you try to find the why behind the answers. In math I always found that if I understood a "why," less memorizing was required for tests.

It depends on the class, actually. For the lower classes (and let's be honest, college algebra is the lowest), trying to understand the why of the ideas will really slow most people down, and so they'll be stuck while the rest of the class is being taught the equations they need for the next test. OTOH, for even the mid-level stuff, it's *far* easier to figure out the why, because the how - the equations - flow naturally from them, and the professor will expect you to have that level of understanding anyway.

I'm having Fourier Transform flashbacks, so I'll stop talking now.

 

[boredJonathan]

You feel dorkier? I've been sitting here trying to explain a calculus joke on a saturday night.

Point taken. Although, who was reading a thread about math on a saturday night, trying to follow the calculus joke?

Hmm.. I think we shall have to call this one a draw good sir.

 

[GG-Duo]

I'm having Fourier Transform flashbacks, so I'll stop talking now.

You understand Fourier Transform?
Could you talk a little bit about that?

I've taken two classes that mentioned FT, and FFT, and I still don't understand how the equation works, how it's derived, etc etc etc. Or, at least, my knowledge of it is very basic.

 

[nitewulf]

You understand Fourier Transform?
Could you talk a little bit about that?

I've taken two classes that mentioned FT, and FFT, and I still don't understand how the equation works, how it's derived, etc etc etc. Or, at least, my knowledge of it is very basic.

I know you didn't ask me, but:
Fourier theorized that all periodic signals can be broken down into infinite sums of sines and cosines. And in case a signal isnt periodic, he postulated that the period can be considered infinite, and the signal can be described by integrals of sinusoids.
So breaking down a continuous function, f(x) into infinite sums of sines and cosines is called Fourier Series Analysis.
And breaking down an aperiodic function like the exponential function, into integrals of sinusoids is known as Fourier Transform, in honor of Fourier.
Just like Laplace Transforms, FT is an integral transform that's used to convert extremely complex differential equations into algebraic functions, whereupon one could simply seperate the required variable and solve for it.
All this assumes that there exists an Inverse Transform to invert the result into its original form.
By example,
A time domain function f(t) is transformed into the frequency domain w (omega) by the Fourier Transform, like so:

So let's say we want to find the FT of the time domain exponential function f(t)=Ae^(-bt) where A is an arbitrary amplitude.
I'll use "int" for the integral symbol.
By definition,
F(w) = int(0 to infinity)Ae^(-bt)*e^(-jwt)dt
F(w) = A*int(0 to infinity)e^(-(b+jw)t)dt
F(w) = [-A(e^(-(b+jw)t)/b+jw)] (infinity to 0)
F(w) = A/(b + jw) for b > 0
The FT does not exist for b < 0 as the exponential becomes unbounded and the series diverges.
Now for purposes involving DSP (digital signal processing) and solutions of higher order circuits (RLC) with lots of components, it is not alwyas practical to solve these problems by hand. Therefore we need to insert the data into a computer which can perform numerical analyses and churn out results quickly.
This involves discretizing the continuous Fourier Series and writing algorithms that may be turned into computer code. FFT, or the Fast Fourier Transform is an efficient algorithm that can compute the FT and its inverse quickly.