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Deeptrouble

Stupid MS Calculator

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I have already posted this on another forum, but anyway. MS Calculator does not perform one of the most elementary operations. To the calculator in Windows, 0^0=1. Everyone knows that 0^0 is an undetermined operation. How could they do not know. If they have a bug in such a small program like the calculator, I wonder what bugs lie in other programs.... e.g. Windows. Anyway, if I were in charge at Microsoft, I would fire anyone who worked at the calculator. Cya

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Guest Anonymous Poster
IS EVERYONE HERE A FUCKING IDIOT???????????
0^0 IS NOT 1 !!!!!

FUCK!!!
GO IN MATHEMATICA AND TRY IT!!! IT IS THE ONLY SOFTWARE THAT DOES IT CORRECT!!! GO ASK YOUR MATH TEACHER!!! TAKE ANY RELIABLE CALCULUS BOOK!!!!

"0 at any power equals 0"
"any number raised at 0 equals 1"

The two fucking rules are colliding when we try to do 0^0. IT IS AN UNDETERMINED CASE SUCKERS!!!!

MORONS!!! Doesn''t anyone know SHIT about math on this forum???????

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0^0!=1

For the ones that called me an idiot, read this:


If 0^0=1 than we can say that e^(0*ln[0])==1.
Therefore, we can say that e^(0*(-Infinity))==1
Which means that e^(0*-1*Infinity)==1
Which means that (e^0)*(e^Infinity)==1 (since 0*-1=0)
Which means that 1*(e^Infinity)==1
This means that 1*Infinity==1
This meanst that Infinity==1

Great job Einstein, you prooved that 1 is the biggest number in our Universe.

I hope you have the decency to admit you were wrong.

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Guest Anonymous Poster
quote:
Original post by Deeptrouble
If 0^0=1 than we can say that e^(0*ln[0])==1.

No, we can''t, since ln(0) is not defined. That we define 0^0 to be 1 does not magically make the the expression e^(0*ln[0]) valid.

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Deep Trouble: That doesn''t look right to me.
You''re multiplying by an infinite quantity, ln(0). You can''t do that, thus invalidating the very first step of your argument. It doesn''t really even have anything to do with the 0^0 case, it''s just wrong.


All of you should read the Dr. Math article (url posted earlier). It''s actually a reliable, sensible writing by college math professors, not retarded anonymous poster assholes.

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Saying that 0^0=1 is an error prone approach. I have already read the Dr. Math article. Let me just say that they use 0^0 there just as a shorthand.

There isn''t a single theorem in mathematics where the operation 0^0=1 cannot be prooved using 0^0=indeterminate and then using some other method to find its limit.

TO further sustain my point, let me say that I have also read Euler, Cauchy (obviously not their complete works, but a part of it) and the reason why Euler said 0^0=1 is because he didn''t know what a limit was (mathematically the concept was discovered by cauchy, weierstrass and others).

I have not multiplied by ln[0] in any way. I have merely used this formula: u^v==e^(v*ln[u]) (where ln[u]==logarithm in base e)

The statement that ln[0] is undefined can be easily removed if I used a string that went to 0 instead of the number 0 itself.

Another argument:

The array 1^0, (1/2)^0, (1/3)^0, (1/4)^0 obviously goes to 1. It also goes to 0^0.

On the other hand, the array 0^1, 0^(1/2),0^(1/3),0^(1/4), also goes to 0^0. ITS limit however is 0, not 1.

SO... here we are. 0==1. Then we can show that 1==2, etc.

0==1==2==3==4==5==...


The convention 0^0 is just that. A CONVENTION. Useful most of the time, but it should be taken carefully, as even Dr. Math suggests.

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check it out...

http://www.microsoft.com/windowsxp/pro/downloads/powertoys.asp

get MS Power Calculator, better than the standard calculator... and if u try 0^0 in it, it''ll give u an error...

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any power raised to 0 equals 1 and 0 is no exception. If you ever get to calculus and study l''hospitals theorem, you would realize that there as well. Though I imagine it doesn''t really take reaching to calculus to know that either.

I got a little chuckle out of it though..

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Guest Anonymous Poster
quote:
Original post by Deeptrouble
I have not multiplied by ln[0] in any way. I have merely used this formula: u^v==e^(v*ln[u]) (where ln[u]==logarithm in base e)

That formula is only valid for u!=0, and that''s the whole reason you DEFINE 0^0 to be 1. It''s a special case. If it followed from u^v==e^(v*ln[u]) you wouldn''t have to define it specifically. Obviously, defining 0^0=1 doesn''t imply that e^(0*ln[0])=1.

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Thanks, I didn''t know about the Power MS Calculator.

I am awed that people do not recognize this as being plain wrong.

Nervo, and all the anonymous posters... It is amazing. Sorry, for restating what I have already stated before, but if you really want to be rigurous about it 0^0 is NOT 0. Mathematica shows you that it is a mistake.

ALSO, nobody said ANYTHING about my array example. Those are two arrays, with two different limits that reach 0^0.

This is truly amazing as to how I am considered an idiot for stating something that is true. The fact is that this has been under some debate, BUT IN THE 19 th century.

Anyway, do not try to convince me that 0^0 is defined as being 1. IT IS NOT. I have those two arrays that show it

a(n)= 0 ^ (1/n) --> 0
and
b (n) = (1/n)^0 --> 1

Also, Nervo, let me say once again that the entire statement can be reformulated without using ln[0] but just using a string that goes towards 0 (never reaching it). In this case, ln out of this array is always defined and it''s limit would be actually -Infinity. I thought it was understood.

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Guest Anonymous Poster
I suppose the Power Calculator of Microsoft is buggy since it outputs an error for 0^0, and the NORMAL calculator from Microsoft is not buggy.

Also, I suppose that Mathematica is buggy and Maple (who doesn''t see the error and outputs 1) is not...

Scary

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Let me add another trick here.

Suppose that 0^0=1
This means that 0^(3-3)=1
This means that (0^3) / (0^3) =1
And since 0^3 == 0,
This means that 0/0 = 1

Again, this is WRONG!!! 0/0 is indeterminate.

If you consider 0/0=1 I can prove some other interesting things.

Consider this (contradiction that 0/0=1):

4^2-4^2 == 4^2-4^2
4(4-4) == (4+4)(4-4)
// Here follows the mistake
// Because 4-4=0 I cannot do (4-4)/(4-4)=1
// If I do that, I get:
4==2*4

=> 1==2
And after this, I can get again, 1==2==3==4==5==...

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Looks like someone, namely the above poster, is in need of some math lessons.

Wow. I''m just awed that something as ridiculous as this spawned 2 pages already. Think about it: it''s a problem with no universal solution. What do you expect MS, or anyone for that matter, to do? Euler argues that it''s 1, others argue it''s undefined and yet others argue that it''s an impossible result. So now we''re expecting MS to solve it all for us.

Cool, I can live with that. Eager to see their take on sqrt(-1)... Wait, it''s not ''i''? sqrt(-4) isn''t ''2i''? What the!! Oh no! MS is so wrong!! o.o

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RuneLancer.

Obviously, what I argued above is wrong. The POINT OF IT was to be wrong. I tried to show that if you consider 0^0==1 than you are WRONG. I tried to show why.

The calculator shows that 0/0 and 1/0 are errors. However, it does not say anything about 0^0 (which IS an error).

Power Calculator shows that it is an error. Euler said it was 1 but Euler was wrong. The reason for this is because he didn''t knew about limits. He was wrong on other occasions too (most notably when saying that a number that is less than 1/n for all n, then this number is 0. We know from nonstandard analysis that this is not true).

One more thing sqrt(-1) is i only if you are working with complex numbers. The MS Calculator seems to be a real calculator (it has not support for introducing complex, or working with complex numbers).

I hope that you didn''t mean to ridiculize me when you said that I need some math lessons. I really think I DO need some math lessons. But at least in some more advanced topics than these discussed here.

This is my last post.

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Guest Anonymous Poster
quote:
Original post by Deeptrouble
ALSO, nobody said ANYTHING about my array example. Those are two arrays, with two different limits that reach 0^0.

They probably didn''t say anything because there is no relevance to it. Everyone knows x^0 -> 1 and 0^x -> 0 when x -> 0, but that is no argument against 0^0 being defined as 1.

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