
#1
Mar812, 08:40 PM

P: 26

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+11)/(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
Mar812, 08:47 PM

Mentor
P: 4,499

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 e^{k} but trying to plug in infinity directly gives one to the infinity for every value of k 



#3
Jun2512, 11:48 AM

P: 62

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
Aug1812, 06:41 AM

P: 77

1 to the power of ∞ =e ?????
Hi.
Well, n_kelthuzad's calculation is correct. You can substitute [itex]b=\infty[/itex] 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 [itex]\infty[/itex] as if [itex]\infty[/itex] was an ordinary finite number. Why is that bad? Well, if one starts to perform alrithmetic operations with [itex]\infty[/itex] as if it was ordinary number, one can deduce, for example, [itex]11+11+11+11+...... = 1/2[/itex]. 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 [itex]\infty[/itex] 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
Aug1912, 07:00 PM

P: 75

Hmmm... 1/1! + 1/2! + 1/3!... = 1 at the limit? I don't think so lessun' you can show me where 1/n! starts to become a negative number...
RELATED LINK: Taylor Series  List of Maclaurin series of some common functions http://en.wikipedia.org/wiki/Taylor_...mmon_functions 



#6
Aug1912, 09:39 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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
Aug2012, 03:40 AM

P: 295

One to the power of infinity, in general, can be shown equal to [itex]e^x[/itex] 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
Aug2012, 04:20 AM

P: 77

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



#9
Aug2012, 11:29 AM

Mentor
P: 16,561





#10
Aug2012, 11:39 AM

P: 77

Hi.
In C. It has all values in extendedC = C[itex]\cup \{ \infty \}[/itex]. Cheers. 



#11
Aug2012, 11:41 AM

P: 295

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




#12
Aug2012, 06:37 PM

P: 77

Hi.
This is true in C. At the point at infinity it has essential singularity. So Your claim fails in extendedC. Cheers. 



#13
Aug2012, 06:54 PM

Mentor
P: 16,561





#14
Aug2012, 07:24 PM

Mentor
P: 16,561

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 CasoratiWeierstrass theorem, then you should know that the theorem doesn't claim that, but rather says that the image of f is dense in [itex]\mathbb{C}[/itex]) 



#15
Aug2012, 09:16 PM

P: 77

Hi.
Check http://en.wikipedia.org/wiki/Weierst...sorati_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 [itex]e^s[/itex] near essential singularity [itex]\infty[/itex]. To make life easier, let us first shift essential singularity to origin by substituting [itex]s \to 1/z[/itex]. So let us consider function [itex]e^{1/z}[/itex]. Essential singularity is now shifted to origin [itex]z=0[/itex]. Rewrite [itex]z[/itex] as [itex]z=r e^{\rm{i} \phi}[/itex]. Hence [itex] e^{1/z} [/itex] is now rewritten in form [itex]e^{1/z}=e^{e^{\rm{i} \phi}/r}=e^{\cos \phi /r} e^{ \rm{i} \sin \phi /r}[/itex]. Let [itex]z[/itex] approach origin from the left, [itex]z \to 0[/itex]. This way [itex] \phi = \pi [/itex], [itex] \cos \phi = 1[/itex], and [itex]e^{\cos \phi /r}=e^{1/r}[/itex]. Now let [itex]r \to 0+[/itex]. Obviously [itex]e^{1/r} \to e^{\infty} =0[/itex]. So [itex]e^{1/z}[/itex] is zero as [itex]z[/itex] approaches origin from the left. In terms of [itex]s[/itex], we have [itex]s=1/z=e^{\rm{i} \phi }/r[/itex]. Rewrite [itex]s[/itex] as [itex]s=Re^{\rm{i} \theta}[/itex]. Obviously [itex]R=1/r[/itex] and [itex]\phi = \theta[/itex]. Hence, as [itex]s[/itex] approaches point at infinity along negative real axis [itex]\theta=\pi[/itex], exponential function approaches zero. Well, we knew that since high school, I guess... I hope this helped a bit. Cheers. 



#16
Aug2012, 09:17 PM

P: 77

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



#17
Aug2012, 09:35 PM

Mentor
P: 16,561

If you say that we can approach every value of [itex]\mathbb{C}[/itex], then it is true of course, by CasoratiWeierstrass. 


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