Case of the missing solution ... Sherlock?

[Dale10101]

I hope this isn't a math flub.:roll:




Tangential question: if the factor which has the independent variable x, or f(x) as an exponent contains a variable, is that expression still considered an exponential expression, is there a name for such an expression distinct from the usual b^x expressions where b is a constant?

P.S. I do realize that x =-2 does not work in the intermediate step of log (x+1), but that must mean there is an alternate solution path .... I am guessing.

 

[Quaid]

is [x^x] still considered an exponential expression

No, I don't think so. Can't remember for sure, but I think saw a professor at Penn State refer to this sort of thing as "super exponential".

(x + 1)^x is definitely not an exponential function.


Anyway, thanks for the "tricky" equation to solve. Gotta ponder awhile...

Cheers :cool:


PS: Wolfram Alpha also misses the solution -2 AND when I asked how it found x=0, it gives the following response. :lol:

 

[Dale10101]

Interesting

No, I don't think so. Can't remember for sure, but I think saw a professor at Penn State refer to this sort of thing as "super exponential".

(x + 1)^x is definitely not an exponential function.

It is interesting. I am making a collection of, and trying to classify different "types" of math problems that I run into and the different methods of solving them.

So the definition of an exponential equation stands as b^x where b is a number, and x is an independent variable. Follow up question, we usually see b as a positive real number, can it be a negative real number, can it be a complex number? Thanks for your response.

 

[lookagain]

I hope this isn't a math flub.:roll:


View attachment 3669
So how does one solve the original equation to find the x = -2 solution?

\(\displaystyle (-1)^{even \ integer} \ = \ 1 \ \ \)

so set \(\displaystyle \ \) x + 1 \(\displaystyle \ \) equal to -1:

x + 1 = -1 \(\displaystyle \ \implies \)

x = -2

That is an even integer, so \(\displaystyle \ \) x = -2 \(\displaystyle \ \) is one of the solutions.

 

[Dale10101]

Magnifying glass mode

\(\displaystyle (-1)^{even \ integer} \ = \ 1 \ \ \)

so set \(\displaystyle \ \) x + 1 \(\displaystyle \ \) equal to -1:

x + 1 = -1 \(\displaystyle \ \implies \)

x = -2

That is an even integer, so \(\displaystyle \ \) x = -2 \(\displaystyle \ \) is one of the solutions.

Sorry, but let me put it this way … huh? I am not sure whether you are pointing out something that enlarges the discussion or the clue that breaks the camel’s back …. call me Lestrade.

Does this observation provide a means to take the original problem and work it to find a (-2) solution?

I do see that x = -1 makes the equation undefined, and that x < -1 does mean that if we are working in the domain of Real numbers then x must be even to provide a solution since we have a positive number on the right side of the original equation, and that IS interesting because it narrows the possible solution set. I mean, where did (-2) come from … might there be other unfound solutions in the water, should we be swimming back towards the shore?

Anyway, I was able to pull this from Wolfram Alpha, maybe it was feeling more vigorous today. It provides (-2) as a solution but seems to have pulled it from its shorts.

 

[Deleted member 4993]

I hope this isn't a math flub.:roll:


View attachment 3669

Tangential question: if the factor which has the independent variable x, or f(x) as an exponent contains a variable, is that expression still considered an exponential expression, is there a name for such an expression distinct from the usual b^x expressions where b is a constant?

P.S. I do realize that x =-2 does not work in the intermediate step of log (x+1), but that must mean there is an alternate solution path .... I am guessing.

Roots of y = (x+1)^x-1 is in the complex domain when x<-1

The function does not exist in real domain when x<1 except at x = -2 (at this point y = 0)

So in the real domain - for x<-1, this function will exist "discretely" for "even integers" (x = -2, -4, -6...., -2n)

Maybe Daon or Pka can suggest way to catch this anomalous root of this discrete function!!!

 

[Quaid]

the definition of an exponential equation stands as b^x where b is a number, and x is an independent variable.

we usually see b as a positive real number, can it be a negative real number, can it be a complex number?

An exponential function cannot have a negative base.

If the base were negative, then even exponents would yield positive outputs and odd exponents would yield negative outputs, and in-between two adjacent Whole numbers (like 0 and 1, for example) there would be lots of undefined results.

EG:

(-2)^(1/2) is undefined



If the base were 0, then negative exponents would yield undefined results. (Powers of zero are not generally useful.)


If the base were 1, that would be an exponential function (again, not very useful).


Therefore, the base in an exponential function must be a positive number.


There are exponentiations involving complex numbers, but I don't know whether any of these are universally known as "an exponential function".

The following have definitions, for example.


e^(a + bi) = e^a*[cos(b) + i*sin(b)]

If the exponent is purely imaginary (i.e., a = 0), then we get Euler's formula:

e^(ib) = cos(b) + i*sin(b)


Stuff like (a + bi)^n and (a + bi)^(c + di) are discussed on the Internet.

---

Googling the topics above yields many results. Here are a few:

http://en.wikipedia.org/wiki/Exponential_function#Computation_of_ab_where_both_a_and_b_are_complex

http://mathforum.org/library/drmath/view/55604.html

http://en.wikipedia.org/wiki/Exponentiation#Powers_of_complex_numbers

http://www.math.wisc.edu/~angenent/Free-Lecture-Notes/freecomplexnumbers.pdf

 

[Quaid]

Does [lookagain's] observation provide a means to take the original problem and work it to find a (-2) solution?

I think so, but it's not an algebraic approach per se.

It's an analytical approach, and it can be used to find both solutions. (However, without first knowing -2 as a solution, I don't think that I would have come up with lookagain's approach.)

I was not able to come up with any algorithm, myself. I'll watch this space...

 

[JeffM]

I hope this isn't a math flub.:roll:


View attachment 3669

Tangential question: if the factor which has the independent variable x, or f(x) as an exponent contains a variable, is that expression still considered an exponential expression, is there a name for such an expression distinct from the usual b^x expressions where b is a constant?

P.S. I do realize that x =-2 does not work in the intermediate step of log (x+1), but that must mean there is an alternate solution path .... I am guessing.

There is an analytic solution to the problem of finding real values of x that satisfy \(\displaystyle (x + 1)^x = 1.\)

However, lookagain's solution is a lot more straightforward.

We start with a lemma: \(\displaystyle p,\ q \in \mathbb R\ and\ p = (- 1)^q \implies p = \pm\ 1.\)

Proof: \(\displaystyle q \in \mathbb R \implies \exists\ k \in \mathbb Z\ and\ r \in \mathbb R\ |\ k + r = q\ and\ \ 0 \le r < 1.\)

\(\displaystyle So\ p = (- 1)^q = (-1)^k * (-1)^r = \pm\ (-1)^r \implies \pm\ p = (-1)^r \implies |p| = |(-1)^r| = |-1|^r = 1^r = 1 \implies p = \pm\ 1.\)

Now as SK pointed out, the function \(\displaystyle (x + 1)^x\) is continuous on the real line if x > - 1,

is undefined if x = - 1, and is infinitely discontinuous on the real line if x < - 1. So there are two cases to deal with. One is if x > - 1, and you have dealt with that one analytically already. Now consider the case if x < - 1.

\(\displaystyle If\ \exists\ a\ |\ a \in \mathbb R,\ a < - 1,\ and\ (a + 1)^a = 1,\ then\)

\(\displaystyle (I):\ a + 1 < 0 \implies (- a - 1) > 0 \implies (-a - 1)^r > 0;\)

\(\displaystyle (II):\ 1 = (a + 1)^a = \{(-1) * (-a - 1)\}^a = (-1)^a * (- a - 1)^a = (\pm\ * 1)(-a - 1)^a = (-a - 1)^a;\ and\)

\(\displaystyle (III):\ log\left\{(- a - 1)^a\right\} = log(1) \implies log\left\{(- a - 1)^a\right\} = 0 \implies a * log(- a - 1) = 0 \implies log(-a - 1) = 0 \implies\)

\(\displaystyle (- a - 1) = 1 \implies - a = 1 + 1 = 2 \implies a = - 2.\)

Now \(\displaystyle (- 2 + 1)^{-2} = \dfrac{1}{(-1)^2} = \dfrac{1}{1} = 1\) so there is a real solution if x < - 1, and it is unique.

EDIT: Line 2 was missing a final exponent. That has now been corrected. Sorry for any confusion.

 

[Dale10101]

JeffM

Great and ausgezeichnet!

Comprehending your post is pushing my limits but, amazing to me, I am following. I even detected the glitch/typo that you edited and was going to ask about it. Am still reviewing to reflect on the many things it reveals. Thanks.

 

[Dale10101]

Summing up.

There is an analytic solution to the problem of finding real values of x that satisfy \(\displaystyle (x + 1)^x = 1.\)

However, lookagain's solution is a lot more straightforward.

We start with a lemma: \(\displaystyle p,\ q \in \mathbb R\ and\ p = (- 1)^q \implies p = \pm\ 1.\)

Proof: \(\displaystyle q \in \mathbb R \implies \exists\ k \in \mathbb Z\ and\ r \in \mathbb R\ |\ k + r = q\ and\ \ 0 \le r < 1.\)

\(\displaystyle So\ p = (- 1)^q = (-1)^k * (-1)^r = \pm\ (-1)^r \implies \pm\ p = (-1)^r \implies |p| = |(-1)^r| = |-1|^r = 1^r = 1 \implies p = \pm\ 1.\)

Now as SK pointed out, the function \(\displaystyle (x + 1)^x\) is continuous on the real line if x > - 1,

is undefined if x = - 1, and is infinitely discontinuous on the real line if x < - 1. So there are two cases to deal with. One is if x > - 1, and you have dealt with that one analytically already. Now consider the case if x < - 1.

\(\displaystyle If\ \exists\ a\ |\ a \in \mathbb R,\ a < - 1,\ and\ (a + 1)^a = 1,\ then\)

\(\displaystyle (I):\ a + 1 < 0 \implies (- a - 1) > 0 \implies (-a - 1)^r > 0;\)

\(\displaystyle (II):\ 1 = (a + 1)^a = \{(-1) * (-a - 1)\}^a = (-1)^a * (- a - 1)^a = (\pm\ * 1)(-a - 1)^a = (-a - 1)^a;\ and\)

\(\displaystyle (III):\ log\left\{(- a - 1)^a\right\} = log(1) \implies log\left\{(- a - 1)^a\right\} = 0 \implies a * log(- a - 1) = 0 \implies log(-a - 1) = 0 \implies\)

\(\displaystyle (- a - 1) = 1 \implies - a = 1 + 1 = 2 \implies a = - 2.\)

Now \(\displaystyle (- 2 + 1)^{-2} = \dfrac{1}{(-1)^2} = \dfrac{1}{1} = 1\) so there is a real solution if x < - 1, and it is unique.

EDIT: Line 2 was missing a final exponent. That has now been corrected. Sorry for any confusion.

I have learned a great deal from this problem including how an equation can have discrete real solutions beyond the domain of the real number system. That is, by saying x is an element of the real number set one tends to stop looking for equation solutions when the equation becomes undefined with respect to its domain , i.e. x < -1 in this case. Under those circumstances I am not sure that -2 IS a legitimate solution although it is a real solution that does fall within the set of real numbers. Perhaps one must just say that "it is what it is" and be cognizant of that fact. It seems important however to keep such a result in mind. In modeling a real life situation such an unrecognized solution could result in another Tacoma bridge collapse, no?

I see that the key to developing the analytic solution was the realization that expressing (x+1) as (-1)(-x-1) put the key factor (-x-1) back in the domain of the log function where the log function could be used again as it had been used when x >-1 => ( x+1) >0. The genius factor that goes beyond knowledge.

Proposition II concluded by stating that: \(\displaystyle (II):\ 1 = (a + 1)^a = (-a - 1)^a;\)

Should that have been \(\displaystyle (II):\ 1 = (a + 1)^a = (\pm\ )(-a - 1)^a ;\)

which would make no difference to the conclusion of Proposition III except to make it the pertinent rather than the only case.

Learned much, thanks for extending much effort.

 

[JeffM]

I have learned a great deal from this problem including how an equation can have discrete real solutions beyond the domain of the real number system. That is, by saying x is an element of the real number set one tends to stop looking for equation solutions when the equation becomes undefined with respect to its domain , i.e. x < -1 in this case. Under those circumstances I am not sure that -2 IS a legitimate solution although it is a real solution that does fall within the set of real numbers. Perhaps one must just say that "it is what it is" and be cognizant of that fact. It seems important however to keep such a result in mind. In modeling a real life situation such an unrecognized solution could result in another Tacoma bridge collapse, no?

I see that the key to developing the analytic solution was the realization that expressing (x+1) as (-1)(-x-1) put the key factor (-x-1) back in the domain of the log function where the log function could be used again as it had been used when x >-1 => ( x+1) >0. The genius factor that goes beyond knowledge.

Proposition II concluded by stating that: \(\displaystyle (II):\ 1 = (a + 1)^a = (-a - 1)^a;\)

Should that have been \(\displaystyle (II):\ 1 = (a + 1)^a = (\pm\ )(-a - 1)^a ;\)

No. \(\displaystyle (- a - 1)^a > 0\ so\ (- 1) * (- a - 1)^a = 1 > 0\ is\ impossible,\ meaning\ (1) * (-a - 1)^a = 1 \implies (-a - 1)^a = 1.\)

What is tricky about this EQUATION is that you can use the log function only for values in the domain of the log, but any real solution (if it exists) will be in that domain. In other words, if the solution exists, it will have certain properties, which you can use to determine what the solutions are.

which would make no difference to the conclusion of Proposition III except to make it the pertinent rather than the only case.

Learned much, thanks for extending much effort.

Dale

Much of your last post is about the meaning of mathematics. Epistemology is not one of my fields so you need to take what I say with a huge bushel of salt.

First of all, in many practical problems to which math is applied, there are explicit or implicit restrictions on the domain. For example, in economics, the variables are frequently constrained to be non-negative. For some economists, quantities produced and consumed must be non-negative integers, and prices must be non-negative rational numbers. (The latter position is realistic but leads to ugly math so most economists ignore it.) So if I got the equation (x + 1)^x = 1 in the context of certain kinds of practical problem, it might be quite appropriate to say that x = - 2 is not a relevant answer.

However, the fact that a function is not defined for certain values seems to me completely irrelevant to the solution set of an equation that incorporates that function. We are looking for the set of solutions that satisfy the equation; we have no interest in values that do not satisfy the equation.

Finally, I have an opinion about real and complex numbers: they are figments of the human imagination that can never be observed in the physical world but that (amazingly) are extremely useful in analyzing the real world. If you think about the function (x + 1)^x in terms of complex numbers, the domain is all complex numbers except -1 + 0 * i. And the equation (x + 1)^x = 1 has solutions among the complex numbers and can presumably be solved analytically there (don't ask me how: I avoid complex numbers). The fact that only two of the complex solutions are of the form r + 0 * i does not strike me as unusual or disturbing in the slightest. That sort of thing happens in the complex numbers (where I seldom venture). For example,

\(\displaystyle c \in \mathbb C\ and\ c^3 = - 1 \implies c^3 + 1 = 0 \implies (c + 1)(c^2 - c + 1) = 0 \implies c + 1 = 0\ or\ c^2 - c + 1 = 0.\)

\(\displaystyle But\ c^2 - c + 1 = 0 \implies c = \dfrac{-(-1) \pm \sqrt{(- 1)^2 - 4(1)(1)}}{2 * 1} \implies c = \dfrac{1 \pm i\sqrt{3}}{2}.\)

So there are three solutions to c^3 = 1 in the complex numbers:

\(\displaystyle c = - 1 + 0i,\ or\ c = \dfrac{1}{2} + i\left(\dfrac{\sqrt{3}}{2}\right),\ or\ c = \dfrac{1}{2} - i\left(- \dfrac{\sqrt{3}}{2}\right).\) Only the first is a real number.

Check if you do not believe me.

 

[Dale10101]

OK

Thank you JeffM, and the aspirin portion of the pharmaceutical thanks you too ... but that is ok, I am a math masochist for better or worse, and get pleasure from the pain. I shall endeavor to pull my brain up by its ear straps and comprehend what you are saying ... a bit at a time.

again, thanks for the follow up

 

[Dale10101]

Ding

Originally Posted by Dale10101


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What is tricky about this EQUATION is that you can use the log function only for values in the domain of the log, but any real solution (if it exists) will be in that domain. In other words, if the solution exists, it will have certain properties, which you can use to determine what the solutions are.

which would make no difference to the conclusion of Proposition III except to make it the pertinent rather than the only case.
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Well nuts, I might have realized that had I been in the ball park instead of the parking lot. Running on two ingrained habits I was thinking you just willy-nilly disappeared part of a factored quantity; but of course (+ or -) is not a factor it is short hand for a case statement just as in the quadratic equation.

Also ingrained was the notion that one seeks a solution set and then goes back and checks the cases to see if they are viable, so again I did not even think about the possibility of eliminating a solution path before concluding it.

In retrospect, like ... duh !!! but anyway another inch forward.