# 1 to the power of ∞ =e ?

1. Mar 8, 2012

1 to the power of ∞ =e ?????

Let there be function f(x):
f(x)=(b+1)^(b+1)/(b+1!)/[(b^b)/b!]
--an example of f(99): 100^100/100!/(99^99/99!)
-------------------------------------------------------------------------
and as b gets larger, f(x) converges to e.
so we have:

lim b→ ∞ (b+1)^(b+1)/(b+1!)/[(b^b)/b!]=e
(b+1)^(b+1)/(b^b)/(b+1)=e
(b+1)^(b+1-1)/(b^b)=e
(b+1)^b/(b^b)=e
[(b+1)/b]^b=e
(1+1/b)^b=e

and substitute lim b→∞:
(1+0)^∞=e
1^∞=e

I am happy to hear any disagreements.

2. Mar 8, 2012

### Office_Shredder

Staff Emeritus
Re: 1 to the power of ∞ =e ?????

You can't just substitute in b

and try to raise something to the infinity power sms pretend what you are doing is well defined. For example consider

lim (1+1/b)kb

taking the limit as b goes to infinity gives ek but trying to plug in infinity directly gives one to the infinity for every value of k

3. Jun 25, 2012

### Akshay_Anti

Re: 1 to the power of ∞ =e ?????

actually you are confused a bit! you might be familiar with the concept of limits. when we say limiting value, it means the value tangibly attained when the variable tends to i.e. approaches the value in a manner that it takes value that is not the exact value but a value that has a negligible error. the definition of e is given by a function which TENDS to 1 and is raised to the power of infinity. for a value of exact one raised to the power of infinity, the value shall be = 1 but the value e is attained when the function in the base has a limiting value of 1

4. Aug 18, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Well, n_kelthuzad's calculation is correct. You can substitute $b=\infty$ if You will, but the resulting equation makes no sense then. That's because the limiting process is lost then, and one starts to calculate with $\infty$ as if $\infty$ was an ordinary finite number. Why is that bad? Well, if one starts to perform alrithmetic operations with $\infty$ as if it was ordinary number, one can deduce, for example, $1-1+1-1+1-1+1-1+...... = 1/2$. Ramanujan and Hardy were into this kind of mathematics. Hardy wrote a book about it. This type of reasoning is called regularization and results are not unique. This clearly shows one cannot treat $\infty$ as an ordinary number. Instead, we have limits, and limits help us avoid errors. If You are into some mathematical calculations, then please do avoid such experimentations because results may begin to become mathematically strange indeed. That is, if You don't expect Your mathematical results to be strange. If You do want Your results to be strange, then welcome to a yet unexplored field of mathematics. Many mathematicians would agree that this new field is no mathematics at all, though... Well, many past mathematicians including Euler took a field trip into that fabled mathematics once in a while. So: be warned. Do not expect reasonable results from it.

Cheers.

5. Aug 19, 2012

### Anti-Crackpot

Last edited: Aug 19, 2012
6. Aug 19, 2012

### Hurkyl

Staff Emeritus
Re: 1 to the power of ∞ =e ?????

His argument coupled with your argument shows that it has to happen somewhere. Therefore, we conclude at least one of:

• n! can be negative for an integer n
• Your argument has an invalid step
• His argument has an invalid step
• The methods of calculus are inconsistent

but we cannot make any stronger logical conclusions without further work. We could apply the principle of Bayesian statistics and make an inference based on our relative confidence of the individual possibilities....

But in the end, all you've really done is just saying "you're wrong".

7. Aug 20, 2012

### Millennial

Re: 1 to the power of ∞ =e ?????

One to the power of infinity, in general, can be shown equal to $e^x$ for any x. This is the reason why we leave "raising 1 to the power of infinity" undefined, just like dividing by 0; because it has multiple values and not a single one.

8. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Well, $e^s,\; s \in \bf{C}$ is single valued in $\bf{C}$. But, yes, it has essential singularity at the point at infinity. And by the Weierstrass theorem, any function has all values from $\bf{C}$ in any neighborhood of its essential singularity.

Cheers.

9. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

It has all values except possibly one value. For example $e^x$ is nonzero, but it does have all other values. This is the big Picard theorem.

10. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

In C. It has all values in extended-C = C$\cup \{ \infty \}$.

Cheers.

11. Aug 20, 2012

### Millennial

Re: 1 to the power of ∞ =e ?????

The exponential function is nonzero everywhere along the complex plane. It takes negative values, positive values, but never a value of zero.

12. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

This is true in C. At the point at infinity it has essential singularity. So Your claim fails in extended-C.

Cheers.

13. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

OK, then please provide a point z in the neighborhood of $\infty$ such that $e^z=0$.

14. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

Or maybe just provide a reference for your claim that if z is an essential singularity of f, that in every neighborhood of z, f has every point??

(If you refer to the Casorati-Weierstrass theorem, then you should know that the theorem doesn't claim that, but rather says that the image of f is dense in $\mathbb{C}$)

15. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Check http://en.wikipedia.org/wiki/Weierstrass%E2%80%93Casorati_theorem, please. There's a sentence there: "In the case that f is an entire function and a=∞, the theorem says that the values f(z) approach every complex number and ∞, as z tends to infinity."

Let's demonstrate the behavior of $e^s$ near essential singularity $\infty$. To make life easier, let us first shift essential singularity to origin by substituting $s \to 1/z$. So let us consider function $e^{1/z}$. Essential singularity is now shifted to origin $z=0$. Rewrite $z$ as $z=r e^{\rm{i} \phi}$. Hence $e^{1/z}$ is now rewritten in form $e^{1/z}=e^{e^{-\rm{i} \phi}/r}=e^{\cos \phi /r} e^{- \rm{i} \sin \phi /r}$. Let $z$ approach origin from the left, $z \to 0-$. This way $\phi = \pi$, $\cos \phi = -1$, and $e^{\cos \phi /r}=e^{-1/r}$. Now let $r \to 0+$. Obviously $e^{-1/r} \to e^{-\infty} =0$. So $e^{1/z}$ is zero as $z$ approaches origin from the left. In terms of $s$, we have $s=1/z=e^{-\rm{i} \phi }/r$. Rewrite $s$ as $s=Re^{\rm{i} \theta}$. Obviously $R=1/r$ and $-\phi = \theta$. Hence, as $s$ approaches point at infinity along negative real axis $\theta=-\pi$, exponential function approaches zero. Well, we knew that since high school, I guess...

I hope this helped a bit.

Cheers.

Last edited: Aug 20, 2012
16. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

How do You type C in latex here for set of complex numbers? \C does not work for me... Thanks.

Cheers.

17. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

This is something completely different from what you said!!! You said originally that f has all values of $\mathbb{C}$ in a neighborhood of the essential singularity. I hope you realize that this statement was false.
If you say that we can approach every value of $\mathbb{C}$, then it is true of course, by Casorati-Weierstrass.

18. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

Use [NOPARSE]\mathbb{C}[/NOPARSE].

19. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

I only use word "approach" because essential singular point is not a limit point, it is singularity. If You realize that $e^{-\infty}=0$ translates to $e^{1/0-}=0$ then we are done. If not: think about it... I don't see any problem at origin with graph of $e^{1/x}$ in interval $\left[-1,0 \right]$.

Cheers.

20. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Just to make it a bit more clear: function fails to be analytic at essential singularity. All our analytic notions and phrases are worthless there. However, it is easy to see that if we fix angle of approaching essential singularity and consider real functions exactly at essential singularity, the extended function well defined at essential singularity does exist. It has 2 values at essential singularity though. Point being: extension is obvious.

Cheers.

21. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Shouldn't "I am become the supreme onion, the saddener of worlds" be "I have become the supreme onion, the saddener of worlds"?

Cheers.

22. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Thanks for $\mathbb{C}$.

Cheers.

23. Aug 20, 2012

### micromass

Re: 1 to the power of ∞ =e ?????

OK, there is no problem with the function in [-1,0]. But this is a real function, not a complex function. If you look at $e^{1/x}$ on [-1,0], then 0 is a removable singularity there, not an essential singularity. The essential singularity arises when looking in the complex plane.

I don't see what the point is of considering real functions at essential singularities.
But anyway, let's look at $e^{1/z}$ when approaching 0 along the imaginary axis. So we're looking at

$$e^{1/(ix)}=e^{-i/x}=\cos(1/x)-i\sin(1/x)$$

This is a whole lot more problematic. Now there is no well-defined value that the essential singularity can take on. Indeed, in any neighborhood U around 0, we see that $f(U)$ is dense in [-1,1] in this case. Again, this is for the real function.

When dealing with the complex function $f:\mathbb{C}\setminus \{0\}\rightarrow \mathbb{C}:z\rightarrow e^{1/z}$, we see that for any neighborhood U around 0, the image f(U) attains every value of $\mathbb{C}$ except 0. However, f(U) is dense in $\mathbb{C}$.

24. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hello.

Ha ha, kk, no harm done.

Yes, in complex analysis one cannot analyze the behavior of function at the essential singularity itself, because function fails at being analytic there horribly. We all know that and agree on that, of course.

Well, since high school, when I first encountered Weierstrass theorem on essential singularities, I also encountered the Riemann theorem on removable singularities. Riemann's theorem is fairly simple: if analytic function is analytic in neighborhood of a and if limit at a exists, then f is obviously extended to a this way. There exits a limit: hence, there exists extension.

Thus, when speaking about essential singularities, one can tackle values of function f at essential singularity in some limited way too. Function f fails to be analytic at essential singularity horribly. However, some real non-analytic limits do exist. For instance, $\lim_{x \to 0-} e^{1/x}$ exists, obviously, and is equal to $0$. It is easily checked by just observing the graph of $e^{1/x}$ at origin and to the left. So, there are continuous extensions of f at essential singularity along some paths.

Let me get this straight: this is not a well established theorem. No madam, no sir, not established at all. Actually, this observation of mine is obvious, and I therefore put it here for sake of discussion. Who knows: maybe something good comes out of it in the end... I certainly hope we can get a good chatter from it!

Cheers.

25. Aug 20, 2012

### Kraflyn

Re: 1 to the power of ∞ =e ?????

Hi.

Ha ha, yes, we are obviously at the same track.

Yes, I agree. The point of departure is: the very first original observation done by author who originated this thread, considered only reals. So this little detail might have produced a bit of a confusion, since we all departed a bit into complexes as thread moves on and develops.

Well, good conversation is always welcome.

Cheers.