feedback anyone??

dan · #1 $Old$ August 13th, 2006, 05:43 PM

Hope no one minds me asking this question...but I know that PH loves these

...
...and I’d like to see some augments for the other side of this.

A friend of mine once asked me if x^x=1 how many real values are there for x...
So if x^x=1
xLN(x)=LN(1)
xLN(x)=0 now divide by x
LN(x)=0 so we see that
x=e^0
=1 so x=1
but back to the equation...
xLN(x)=0 divide by LN(x)
x=0/LN(x)
=0 so x=0
so quite simply we see that if
x^x=1
x=1,x=0
so if that's true...why is 0^0 undefined it has to be defined as 1 if this is going to be true

.
any thoughts on why this is right of wrong?
Dan

Quick · #2 $Old$ August 13th, 2006, 06:01 PM

My response is located here

ThePerfectHacker · #3 $Old$ August 13th, 2006, 06:52 PM

It is wrong. The reason is rather complicated but it lies in the mathematically rigor definition of exponents.
---
What does $a^n$ mean, well, in school you are taught that,
$a^n=\underbrace{a\cdot a\cdot...\cdot a}_n$
Note, this definition only makes sense when $n$ is a positive integer

How would can you do,
$a^{-n}$ ? So mathematicians defined this to be $\frac{1}{a^n}$ . For a simple reason, it preserves the property that,
$a^x a^y=a^{x+y}$ along the other various properties of exponents.
Later, mathematicians defined exponents for rational number.
$x^{a/b}=\sqrt[b]{x^a}$ ---> you cannot prove it, it was defined like this. Why? Same reasons, it preserves the properties of exponents. But the problem is now you need to make a restriction that you did not have before. Namely, $a>0$ because otherwise you have negative roots.
Now, how can you define,
$a^r$ where $r$ is real? There are several methods which are all equivalent to each other. One method is like this, define,
$\ln x=\int_1^x \frac{dx}{x}, x>0$ (since it is countinous it is integratble). This function is called the natural logarithm function. We can then show it is a bijective function (meaning has an inverse). Furthermore, it satisfies the Intermediate value theorem which means that there is a unique number $e$ such as, $\ln e =1$ . From, here you define the function $\exp (x)=f^{-1}(x)$ . This function is called the natural exponential. It has the amazing property that, $\exp (a/b)=e^{a/b}$ . Thus, we define $e^r=\exp(r)$ , noting that the domain of $\exp$ is the range of $\ln$ . But the range of $\ln$ is the entire number line. Thus, $e^r$ is defined for any real number. Thus, $a^b=(e^{\ln a})^b=e^{b\ln a}=\exp (b\ln a)$ . But, the argument $e^{\ln a}=a$ is only valid for $a>0$ because the domain of the natural logarithmic is $a>0$ . Thus, from here we defined the operation $a^b=\exp(b\ ln a), a>0$ .
Thus, your argument cannot work because it does not follow the definition above. The necessary fact that the base must be a positive integer which you violated. Thus, it is not a supprise you got unusual results.

dan · #4 $Old$ August 14th, 2006, 06:43 AM

can't one difine lN(0) as -(infinity) (which is undifind)

but then 0ln0=0 would still be true
dan

Quick · #5 $Old$ August 14th, 2006, 08:00 AM

undefined does not mean infinity

dan · #6 $Old$ August 14th, 2006, 08:40 AM

yes i know, but is ln(0)undifined or -infinity or both?
dan

ThePerfectHacker · #7 $Old$ August 14th, 2006, 08:53 AM

Quote:

Originally Posted by dan

yes i know, but is ln(0)undifined or -infinity or both?

It is undefined.
Because, by defintion you are dealing with,
$\int^1_{0^+} \frac{dx}{x}$ which is an improper integral of the second type. Because the function $\frac{1}{x}$ is not countinous at $x=0$ and $x=0$ is an asymptote. It can be shown that this improper integral does not converge. Thus, there is no such value.

dan · #8 $Old$ August 18th, 2006, 06:48 AM

thanks for the feedback guys...I must say though...my opinion remains unchanged

dan

ThePerfectHacker · #9 $Old$ August 18th, 2006, 09:34 AM

Quote:

Originally Posted by dan

...my opinion remains unchanged

It is not a matter of opinion, it is fact that your reasoning was wrong.

Rebesques · #10 $Old$ August 18th, 2006, 12:51 PM

I think I saw the definition 0^0=1 in some textbook... Let me get back on this.

JakeD · #11 $Old$ August 18th, 2006, 02:11 PM

This page titled What is 0^0? gives a good, balanced discussion of how 0^0 could be defined. To quote one sentence: "depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent." The context is important because no one definition of 0^0 is useful in all contexts. But there are contexts--such as when using the binomial theorem--where defining 0^0 = 1 is a convenient, justifiable choice.

ThePerfectHacker · #12 $Old$ August 18th, 2006, 03:20 PM

Quote:

Originally Posted by JakeD

This page titled What is 0^0? gives a good, balanced discussion of how 0^0 could be defined. To quote one sentence: "depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent." The context is important because no one definition of 0^0 is useful in all contexts. But there are contexts--such as when using the binomial theorem--where defining 0^0 = 1 is a convenient, justifiable choice.

I would prefer the notion,
$0^0=1$ because it because extremely useful in infinite series.

dan · #13 $Old$ August 18th, 2006, 07:04 PM

Quote:

Originally Posted by Rebesques

I think I saw the definition 0^0=1 in some textbook... Let me get back on this.

oh yes, that is a valid definition. It's there and can be used legally.
it's the definition i perfer in most cases.

dan

dan · #14 $Old$ August 18th, 2006, 07:07 PM

Quote:

Originally Posted by JakeD

This page titled What is 0^0? gives a good, balanced discussion of how 0^0 could be defined. To quote one sentence: "depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent." The context is important because no one definition of 0^0 is useful in all contexts. But there are contexts--such as when using the binomial theorem--where defining 0^0 = 1 is a convenient, justifiable choice.

i've never seen it put better nice work!

the context is very inportant no question about it.
dan