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phrontist
10-16-2005, 11:49 PM
Is there a function f(x) such that the derivative of f(x) is equal to the inverse of f(x)?

Prove it.

Rickertsen2
10-17-2005, 12:07 AM
this is kind of cheating but:
f(x)=0 where x=0

I assume you are asking this for a reason. Do you have a proof that no such function exists or something?

Interesting question.

---edit----
I was thinking he meant inverse function not 1/f(x)

David Guzman
10-17-2005, 12:22 AM
this is kind of cheating but:
f(x)=0 where x=0

Actually:

f(x)=0
f'(x)=0
f(x)^-1=1/0 which is a non-removable discontinuity. DNE

I dont think there is any, but then again I haven really taking time to look for a way.

Now you are gonna make me lose sleep on this. :p

Dave

Manoel
10-17-2005, 12:26 AM
this is kind of cheating but:
f(x)=0 where x=0

I assume you are asking this for a reason. Do you have a proof that no such function exists or something?

Interesting question.

The function you proposed (a constant) is not invertible, unfortunately.
I'm still giving this some thought, but I don't think there's such a function.
It is, indeed an interesting question. Does look like an assignment, though :p

More later...

[EDIT: David Guzman beat me to it...]

Rickertsen2
10-17-2005, 12:43 AM
wait, when you say inverse, are you refering to 1/f(x) or the inverse function of f(x)

I was assuming he meant inverse function, but now that i think about it, he probably would have inserted the word "function" if that is what he meant. Please clarify this.

Manoel
10-17-2005, 12:52 AM
wait, when you say inverse, are you refering to 1/f(x) or the inverse function of f(x)

I was assuming he meant inverse function, but now that i think about it, he probably would have inserted the word "function" if that is what he meant. Please clarify this.

Judging by the question, phrontist seems to know his math. As such, if he meant 1/f(x), I believe he'd have used "the reciprocal" of f(x).

David Guzman
10-17-2005, 12:59 AM
I might be wrong but:

f(x)=1
f'(x)=0
f(x)^-1=0/1=0

EDIT: OK, I am wrong :p

Manoel
10-17-2005, 01:07 AM
I might be wrong but:

f(x)=1
f'(x)=0
f(x)^-1=0/1=0

EDIT:
If you're considering the reciprocal, then 1/1 is not equal to 0. ;)
If you're considering the inverse function, then I didn't understand what you tried to do.

f(x)^-1 is a poor notation because it usually refers to the inverse of a function but can also be mistaken for the reciprocal of the function.

Rickertsen2
10-17-2005, 01:18 AM
grr math things like this frusterate my simple mind. I am no math person and i don't entirely understand why my solution is wrong. Maybie i am just stubborn. Lets define the inverse of a function as a being a function where f(x)=y and f^-1(y)=x for all x. By this logic if f(0)=0 then shouldn't f^-1(0) be 0 as well and shouldn't my solution work?

Manoel
10-17-2005, 01:38 AM
grr math things like this frusterate my simple mind. I am no math person and i don't entirely understand why my solution is wrong. Maybie i am just stubborn. Lets define the inverse of a function as a being a function where f(x)=y and f^-1(y)=x for all x. By this logic if f(0)=0 then shouldn't f^-1(0) be 0 as well and shouldn't my solution work?

Rickertsen2,

For a function to be invertible, it has to be bijective - "one on one" and "onto", that is, there's only one x value associated with a given y value, and that holds for every x. As an example, y = x^2 is not invertible, because for x = +- 2, y = 4. To think of it in another way, if you were to invert that function (rotate its graph about the line y = x), you'd get a function that, for one value of x, is defined for two y values. That is not a function.
For a quick test, draw an horizontal line in this graph and sweep it up and down. If it ever touches two points on the curve, then the function is not invertible.
You can see that your constant function does not meet the requisites for an invertible function.

A Google search will provide thousands of good sources for information, this site is a nice one, with a simple explanation and pictures (who doesn't like pictures? :p)
http://archives.math.utk.edu/visual.calculus/0/inverse.6/

phrontist
10-17-2005, 07:29 PM
I meant (and said) inverse, not reciprocal. This isn't an assignment, just something that occured to me during my endless and tedious BC Caluclus (anyone else have Early Transcendentals?) homework.

I've been trying to approach the problem visually, the inverse of a function being that function "mirrored" about the line y=x. I think perhaps this is needlessly painful, I'm going to try some algerbraic manipulation of the definition of the derivative...

My guess is that no such function exists, but I wonder if there is an elegant proof of that.

sanddrag
10-17-2005, 08:37 PM
(anyone else have Early Transcendentals?)You mean the Stewart book? Yep, had that. I'd say it's not a bad book but there are some weird quirks every now and then where it won't explain some things very well. Mine came with CDs but I'd say they were pretty much useless.

Karthik
10-17-2005, 08:37 PM
I meant (and said) inverse, not reciprocal. This isn't an assignment, just something that occured to me during my endless and tedious BC Caluclus (anyone else have Early Transcendentals?) homework.

I've been trying to approach the problem visually, the inverse of a function being that function "mirrored" about the line y=x. I think perhaps this is needlessly painful, I'm going to try some algerbraic manipulation of the definition of the derivative...

My guess is that no such function exists, but I wonder if there is an elegant proof of that.

The following fact will lead you in the right direction.

Let f be a function that is differentialable on an interval I. Suppose that f has a defined inverse function, called g.

Using the defintion of an inverse, and the chain rule, it can be shown that

g'(x) = 1/f'(g(x)), where f'(g(x)) != 0

This should help with you algebraic manipulation.

sciguy125
10-17-2005, 08:56 PM
You know what? I was tempted to try this problem, but my hatred of calculus overcame me while I was working some stuff out. Calculus tends to have more letters than numbers. This is math, not grammar... These letters also tend to not be in the English alphabet. Again, this is math, not a foreign language.

Today's episode of Sesame Street was brought to you by... The number e ...And, the letter mu.

There also tends to be odd quirks. Take Gabriel's Horn. How can you possibly have something with an infinite surface area, infinite cross-section, but a finite volume!? And the volume isn't just finite, it happens to be a very specific number/non-English letter: pi.

Yes, I realize the power of calculus. I also respect it. Seeing it work in physics still amazes me. But, I just don't like it.

dubious elise
10-17-2005, 10:23 PM
This one is seriously making my head hurt. I recollect hearing a similar question proposed but my brain is too fuddled to remember the question in its entirety.

I say if 3rd semester Calc doesn't cover it, then it just isn't worth the headache.

...not yet, at least ;)

JamesBrown
10-17-2005, 11:36 PM
Ok well I cannot find a way to definitly prove it but I can prove that this is never the case in the following cases.

x^n where n is positive
x^n where n is negative
all linear functions,
All absolute value functions
x^n when x is not an integer
All trig functions
All step (integer) functions of x

Thats all I have time for now, Back to AP chem.

Denman
10-18-2005, 04:23 AM
lets call f^-1(x) the inverse and (f(x))^-1 the recipricol
anyway
Surely if you have f(x) which has powers in, when you differentiate it the power will decrease by one (unless your using e^x)
reflecting it in y=x inverts the power
(eg f(x)=2x^2+3 f^-1(x) = (x-3)^(1/2) / 2
as such differentiating it only lowers the power by 1
so you can't use integers as when you lower the power it won't be afraction
if you want to try fractions, a/b - 1= b/a which the only thing that that has an equal distance from 1 on either side is 1/2 which would give 3/2 -1 (not equals) 2/3 so that can't work
so not fractions

i'm wondering maybe something hyperbolic or complex?

EricH
10-20-2005, 03:31 PM
C'mon, guys, I haven't had calculus yet and I can tell you that there is a function like this. A function's inverse is its reflection in the line y=x, right? A function is any line or relation where any input has one and only one output, right? So a straight diagonal line is a function, correct? Now, the only function that will equal its own inverse has to have a slope of -1. So, any function whose equation reads something like f(x)=-x+b with any value for b will be its own inverse. Was that so hard?

Manoel
10-20-2005, 04:24 PM
C'mon, guys, I haven't had calculus yet and I can tell you that there is a function like this.

Well, you didn't read the question with enough attention. Phrontist wants a function such that its derivative is equal to its inverse, not the function itself.

A function is any line or relation where any input has one and only one output, right?

No, not at all. My previous example of y = x^2 would not be a function if you were correct. You described a bijective (http://mathworld.wolfram.com/Bijection.html) function, but there are also injective (http://mathworld.wolfram.com/Injection.html) and, the ones that disproves you, surjective (http://mathworld.wolfram.com/Surjection.html) functions.

So a straight diagonal line is a function, correct? Now, the only function that will equal its own inverse has to have a slope of -1. So, any function whose equation reads something like f(x)=-x+b with any value for b will be its own inverse. Was that so hard?

Yeah, that is correct if you disregard the derivative part of the problem.

EricH
10-20-2005, 04:45 PM
I stand corrected.

Denman
10-20-2005, 07:42 PM
Well, you didn't read the question with enough attention. Phrontist wants a function such that its derivative is equal to its inverse, not the function itself.

No, not at all. My previous example of y = x^2 would not be a function if you were correct. You described a bijective (http://mathworld.wolfram.com/Bijection.html) function, but there are also injective (http://mathworld.wolfram.com/Injection.html) and, the ones that disproves you, surjective (http://mathworld.wolfram.com/Surjection.html) functions.

Yeah, that is correct if you disregard the derivative part of the problem.
but you can only actually call it a function if its 1-1 in theory...

Manoel
10-20-2005, 08:49 PM
but you can only actually call it a function if its 1-1 in theory...

Care to elaborate? Not sure if I understand what you wrote...

EricH
10-20-2005, 11:18 PM
A 1-1 function (also written as one-to-one) is one that only goes down or up, not both, over its entire length. I don't understand why Denman says you can only call a function if it's 1-1 in theory, but he may know more than I do, which is quite likely. However, y=x^2 is a function, and it's not 1-1.

Manoel
10-21-2005, 12:18 AM
A 1-1 function (also written as one-to-one) is one that only goes down or up, not both, over its entire length. I don't understand why Denman says you can only call a function if it's 1-1 in theory, but he may know more than I do, which is quite likely. However, y=x^2 is a function, and it's not 1-1.

Well, that's not the definition of a one-to-one function; in general it is, indeed, a consequence of the function being one-to-one. However, it does not apply to every function. For example: y = x^3 would be one-to-one by your definition, but, if you consider its complex roots*, then it is not.

* - For the record, I never had much of a formal, theoretical education on complex numbers, just its practical applications. Therefore, I may be wrong on the above statement. ;)

Denman
10-21-2005, 04:42 AM
sorry my bad i meant one-many, as long as its only one on the start so you cant get 2 values of f(x) for the same number.... however when you take the inverse (eg y=x^1/2) you need to limit your domain and so would need to limit the domain/range of the original function anyway

jdiwnab
10-21-2005, 11:02 AM
A one-to-one function is a function that has only one value of y (f(x)) for every value of x. You could also say that it passes the "horazantal line test" where if you were to draw horizantal lines, each would only pass though the graph once. One-to-one functions are important becuase they are functions that, when inversed, are still functions. y=x^2 is a function, but not one-to-one becuase f(1) and f(-1) have the same value. y=x^3 is one-to-one becuase no values of y repeat themselves. You can inverse y=x^3 and still have a function.
y=x^3 becomes x=y^3 when inversed.
x^(1/3)=y thus is also a function.
"One-to-one" comes from the fact that there is one value of x for every value y and one value y for every value x.
Nothing says that a functions has to be one-to-one. To inverse it and still have a function, yes. But a functions has to pass the virtical line test: if virtical lines were drawn, they would pass though the graph no more than once.
y^2=x passes the horizantal line test, so you can inverse it and have a function, but doesn't pass the virtical line test so it, in itself, is not a function.
The opposite is true of y=x^2. Is a function, but can't have the inverse be a function. Infact, it is the inverse of the above, but it isn't a function on the traditional x-y axis.

Things have been drifting, but I felt the need to clear the air about what is a function and a one-to-one function.

Manoel
10-24-2005, 01:34 AM
Well, turns out there is such a function.

y(x) = 0.743*x^(1.618)

;)

I took a more formal approach to the problem that didn't get me anywhere, so I talked to a buddy and here's what we found.
Explanation (http://www.colegio-provincia.com.br/calc_query.pdf).
As you see, nothing formal. A mathematician will say it isn't a proof at all, but it does show that there's at least one said function.

PS - My statement on a previous post that y=x^3 was not bijective is wrong. :p

Denman
10-24-2005, 07:33 AM
thats scary
do you know what 1/2(1+sqrt5) is!?
shudders
rep point for that

Manoel
10-24-2005, 03:01 PM
thats scary
do you know what 1/2(1+sqrt5) is!?
shudders
rep point for that

Amusing, isn't it? ;)
Of course, it isn't just a coincidence, it's a consequence of what we tried to do. We wanted an exponent n with the same characteristics as the Greeks wanted, just for different purposes. :D

Leo M
10-26-2005, 02:39 PM
Absolutely brilliant, Manoel! This is what I love about First. When phrontist in Virginia can put out a terrifically interesting question from Virginia, and have answers come in right on point from the UK and Brazil – well, we didn’t have anything like that going for us when I was in high school.

Manoel, however did you think of approaching the problem this way? You show extraordinary mathematical insight in choosing the right form for f(x). And phrontist – what made you think up this question? It’s absolutely first rate – you’ll do just fine at MIT when you get there.

For those who struggle with these concepts like I do, the inverse of a function exactly “undoes” what the original function did to the variable. If you perform a ‘function operation’ on a variable x, then follow up with the ‘inverse function operation’, you get the original x back.

So, if f(x) = 0.743 * x^1.618, and we take the first derivative,

then f’(x) = 0.743 * 1.618 * x^0.618 = 1.202 * x^0.618.

But, if f’(x) = Inv-f(x), we can substitute f(x) into the inverse function and get the original x back :

Inv-f(x) = 1.202 * ( f(x) )^0.618 =? x

Inv-f(x) = 1.202 * [ 0.743 * x^1.618] ^ 0.618 = 1.202 * 0.743^0.618 * (x^1.618)^0.618
= 1.202 * 0.832 * x = x, just what we started with.

So, the derivative of f(x) is the same as the inverse of f(x), just as Manoel showed us.

If we keep differentiating – which we can do as long as x >= 0, we can learn some interesting things about these functions, slopes, inflection points, concavity, etc.

f(x) = 0.743 * x^1.618
f’(x) = 1.202 * x^0.618
f’’(x) = 0.743 * x^-0.382
f’’’(x) = -0.284 * x^-1.382

From x=0 to infinity, the function f(x) increases continuously, as does the first derivative/inverse. But f(x) approaches infinity much faster than the derivative.

From f’(x) we can see that the slope of the function f(x) approaches infinity as x gets larger and larger. Eventually, f(x) is heading almost straight up.

From f ’’(x) we can see that the function is everywhere concave upwards, as f’’(x) is positive for all x>0. But f ’’(x) is the slope of the first derivative/inverse, so we can see that the inverse always has a positive slope. But f ’’(x) goes towards zero as x gets larger and larger, so the slope of the inverse goes to zero, i.e., the inverse function approaches a horizontal line. Not an asymptote, however – the inverse function can be made as large as desired by taking large enough values of x. There’s no upper limit for it, so no asymptote.

From f ’’’(x), which is always negative for all x>0, we can see that the inverse is always concave downwards.

(Manoel – you might want to re-check your graph of these functions. It’s always a good idea to double-check computer generated graphs.

Note that the function is numerically equal to its first derivative/inverse when x = 1.618.

A magical number, to be sure. More interesting properties than can be addressed here.

I haven't had this much fun since the last First competition.

phrontist
10-26-2005, 03:50 PM
Well, turns out there is such a function.

Manoel, you are brilliant. I just went through this with a teammate, and while your explanation was a bit hard to follow (not your fault, ours) we eventually got it (and checked it with my brick of a TI-92). Highest compliments man! How old are you? (EDIT: Ah, I didn't see you were a mentor, I thought you were a student) This demonstrates rather impressive intuition.

An alumni emailed me a 4 page proof that there wasn't such a function, thanks for putting him in his place. :D

What regional are you attending this year?

Manoel
10-28-2005, 03:22 PM
Manoel, however did you think of approaching the problem this way? You show extraordinary mathematical insight in choosing the right form for f(x). And phrontist – what made you think up this question? It’s absolutely first rate – you’ll do just fine at MIT when you get there.

Leo M and phrontist,

Thanks for the compliments! As I told you, it was a joint effort - a buddy helped me with that one.
At first I thought (like everyone) that there wasn't such a function, and spent some time trying to prove that. Since I'm studying to become an Engineer, and not a Mathematician, that formal, axiom driven kind of math doesn't go well with me, so I gave up.
Better than proving that it didn't exist would be finding such a function, so that's what I tried to do next.
I looked at the common forms of functions - exponential, trigonometric, etc. - and did a quick check. That quick check threw me off, because I was considering only integer exponents and that didn't cut it.
I had to try functions of the form e^x, because that's the answer to pretty much everything ;), and then came the idea to try an exponent that when inverted would be itself minus one. If I had remembered the golden ratio in the first place, I'd have arrived at the answer quicker. :D

Manoel, you are brilliant. I just went through this with a teammate, and while your explanation was a bit hard to follow (not your fault, ours) we eventually got it (and checked it with my brick of a TI-92). Highest compliments man! How old are you? (EDIT: Ah, I didn't see you were a mentor, I thought you were a student) This demonstrates rather impressive intuition.

An alumni emailed me a 4 page proof that there wasn't such a function, thanks for putting him in his place.

What regional are you attending this year?

Phrontist,
I agree my explanation wasn't very easy to follow; I should have explained some passages a little better, but I was trying to do a one-pager. :p
Add to that the fact that I lack the proper technical English knowledge and you got what you got. :ahh:
The only thing I hate more than a bad explained deduction is one that is "left as an exercise to the student". :rolleyes:
I'm a mentor but a young one, I'm just 20 and on my way to receiving a BSEE, somewhere around 2007 (Engineering takes five years in Brazil).
We'll probably be going to Boston this year, if you make it to the MIT (and I sure hope you do!) make sure to stop by!
Care to share that 4-page proof? I'd like to take a look at it, if possible.

(Manoel – you might want to re-check your graph of these functions. It’s always a good idea to double-check computer generated graphs.

Yeah, I just threw the graphic to show its reflection around the y=x line. Computer graphs are bad, but one that goes from -infinity to infinity is even worse!
By the way, I'm a big fan of yours for that sine/cosine approximation you shared with us back in the PBasic days. Good stuff, man, good stuff!

Travis Hoffman
10-28-2005, 04:06 PM
Somewhere, Leonardo da Vinci is smiling:

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibInArt.html

And a mathematician's perspective -

http://mathworld.wolfram.com/GoldenRatio.html

My brain hurts after reading through that.

Leo M
10-29-2005, 08:48 AM
Like the Fibonacci series itself, it just goes on and on :

http://www.mscs.dal.ca/Fibonacci/

Fibonacci Association, Fibonacci Quarterly..................

I have been a plant engineer for fifteen years +, and sometimes I
really need something more than pipe leaks, burned-out motors,
and managers running around 90 miles an hour with their hair on fire.

It regenerates something deep inside me to work on the clever problems and talk to the ingenious people.

Here's a question I've been wondering about.

Who do you think will be the first First person to win a Nobel Prize?

Maybe that's poorly phrased and a separate thread besides, but I wonder sometimes whether I am communicating with a future prize winner. I hope I live long enough to see it (I am a year older than Dean).

I'd sure love to see one of you guys in Stockholm, with dean and Woodie in the audience.

phrontist
10-31-2005, 09:48 PM
Bad news! My teammate has carefully explained to me that we have all been deceiving ourselves :(

Quoth the amazing Dustin:

Bjorn I looked back to the calculus post on CD, this person is wrong:

f^-1(x)=1.202x^0.618 is not the inverse of f(x)=0.743x^1.618,

inv f(x) is a function of y, not x, it would have to be expressed as + and – 1.202x^0.618 for graphing on your calculator (on the interval of 0 to infinity)(considering the x axis to be the domain, which in reality it is not), which is not the same as the derivative operation on f. Look at the graph!!!!!!!! If you would of said of such a constraint, that the function didn’t have to be on its full sets which is in R^2 for the domain and range of said function, then yes this is true, but that’s not the real definition of a function!

f^-1(f(z))=f(f^-1(z))=z
z=-1.0

f(z)=0.743*(-1.0)^(1618/1000)
=0.743
f^-1(0.743)=1.202(0.743)^(618/1000)
=1.0004 Not –1.0
really =1.0
and my proof as it stands with the true definition of a function is still true.

Dustin

You wonted a proof and that’s what I gave you, you have to be very specific on your definitions when considering such a problem, next time be true to the language or include exceptions or I wont bother with such problems.

Leo M
11-01-2005, 08:43 AM
Maybe I am extraordinarily dense this morning, but I cannot follow this argument at all.

"f^-1(f(z))=f(f^-1(z))=z
z=-1.0"

The first equation is just a way of defining the inverse of a function. The second (z=-1) is just a statement assigning a value to z, and does not follow from the previous statement at all. In the words of Wolfgang Pauli, "That isn't right. That isn't even wrong."

"inv f(x) is a function of y, not x, it would have to be expressed as + and – 1.202x^0.618 for graphing on your calculator (on the interval of 0 to infinity)(considering the x axis to be the domain, which in reality it is not), which is not the same as the derivative operation on f. Look at the graph!!!!!!!! If you would of said of such a constraint, that the function didn’t have to be on its full sets which is in R^2 for the domain and range of said function, then yes this is true, but that’s not the real definition of a function!"

If anyone can explain what the above statements mean, I'd be happy to listen, but as they stand they do not appear to be anything but randomly arranged pseudo-mathematical jargon with lots of !!! thrown in.

"You wonted a proof and that’s what I gave you, you have to be very specific on your definitions when considering such a problem, next time be true to the language or include exceptions or I wont bother with such problems. "

Concerning the above statement, in legal terms: re ipso loquitor - the thing speaks for itself.

Manoel is still correct.

Manoel
11-01-2005, 08:24 PM
Bad news! My teammate has carefully explained to me that we have all been deceiving ourselves :(

Quoth the amazing Dustin:

As Leo M, I'm not sure I got it right.

Anyway, here are two possible readings:
- The aforementioned function is not a function because a function's domain and range has got to be (-infinity, infinity). Well, that's certainly not the definition of a function (neither a requirement) and many "well established" functions do not fulfill those requirements. Ex: y=sqrt(x), y = tan x, ...
- The function is no good because it is not defined for "z"=-1 (not sure why that came up). Again, that's wrong, because the function's domain is [0, infinity) and we're all OK with it.

As those possibilities were easily refuted, I'm pretty sure I just didn't understand what Dustin was trying to say. Maybe some punctuation (and please, don't take it the wrong way) or the 4-page proof could help us here.

Regards,