Why is Tomomi Okazaki chosen for the Trino Olympic athlete?

[asdf1]

why is 0*infinty=0?

 

[siddharth]

why is 0*infinty=0?

That is not true. It is indeterminate.In fact, infinity is not a real number, so how will you define multiplication? I think this question has come up and been addressed a lot of times before. You could search for the other similar threads.

 

[asdf1]

then if that's the case, then why's
-[e^(inifinty)][(infinity)^(v-1)] + -[e^(0)][(0)^(v-1)]=0?

 

[HallsofIvy]

It isn't. In fact, since infinity is not a number, it doesn't make any sense at all. Is it possible that you are thinking of a limit problem, as x-> \infty, where if you just replace x by \infty, you get that expression? You can't just replace x by a in lim_{x\rightarrow a} f(x) unless a is a number and f is continuous at a (which is a very special case).

Each of
lim_{x\rightarrow\infty}x^2\frac{1}{x}
lim_{x^2\rightarrow\infty}x\frac{1}{x}
lim_{ax\rightarrow\infty}x\frac{1}{x}
"evaluate" as \infty*0 but the first is 0, the second is \infty (in other words, the limit does not exist() and the third is a where a can be any number.

 

[asdf1]

the original question was the (intergral)[e^(t)][(t^v-1)]dt, range from 0 to infinity...
so if that integral is integrated, there'll be
-[e^(inifinty)][(infinity)^(v-1)] + -[e^(0)][(0)^(v-1)]=0 as part of the final answer?

 

[marytormey]

infinty=1 and 0*1 is still 0. You have 0 infinties which is nothing

 

[siddharth]

infinty=1 and 0*1 is still 0. You have 0 infinties which is nothing

That makes no sense whatsoever

 

[Hurkyl]

You don't integrate simply by plugging the endpoints into the integrand, so I don't understand why you're asking that question.

Secondly, recall that improper integrals (for example, integrals with -\infty or +\infty as a bound) are defined in terms of limits.

Finally, are you sure you aren't looking at \int_0^{+\infty} e^{-t} t^{v-1} \, dt?

 

[Morbid Steve]

I thought 0 times anything = 0? I have 0 apples, how many do I have?

 

[HallsofIvy]

the original question was the (intergral)[e^(t)][(t^v-1)]dt, range from 0 to infinity...
so if that integral is integrated, there'll be
-[e^(inifinty)][(infinity)^(v-1)] + -[e^(0)][(0)^(v-1)]=0 as part of the final answer?

One more time: infinity is a symbol, not a number. What you have written- plugging in infinity as if it were a number- is invalid.
\int_0^{\infty}e^t t^{v-1}dt
is defined as
lim_{A\rightarrow \infty}\int_0^Ae^t t^{v-1}dt

 

[Hurkyl]

I thought 0 times anything = 0? I have 0 apples, how many do I have?

When you say "anything", what you mean is something like "any real number" or "any integer". (Even if you don't know that's what you mean)

When you say "0 apples", that's not multiplication. (Although it can generally be manipulated as if it is a multiplication)

 

[ksinclair13]

I think this thread should be moved to General Math.

Last night, I figured out a way to show that 0*infinity = 1. I'll show you my thought process...

If .9999... = 1, then 1/infinity = 0. I said this because 1/infinity would look like .0000...n. However, n will never be placed at the end because there would be an infinite amount of zeroes. That means that 1/infinity = 0.

Using algebra, we could rearrange to say that 0*infinity = 1; and also, 1/0 = infinity.

If this were the case, then 0 and infinity would be multiplicative inverses of each other. This would make everything so much easier...

I know that none of this is true (infinity isn't an actual number for instance) - I was just having some fun playing around with these ideas while trying to get to sleep last night .

Edit:

Another way to show that 0*infinity = 1 :

log 0 = -infinity
10^-infinity = 0
1/(10^infinity) = 0
1/infinity = 0
0*infinity = 1

This assumes that log 0 actually does equal negative infinity.

 

[Hurkyl]

...
1/infinity = 0
0*infinity = 1

And also

2 / infinity = 1/infinity + 1/infinity = 0 + 0 = 0
0*infinity = 2
1 = 2

 

[maverick6664]

In many cases, you must not treat infinity as a norml number. Infinity is a kind of state "when one or more parameters are getting close to something or getting greater/smaller, the "value" is getting greater than anything". So it's written as lim_{x \rightarrow \infty} x or many other ways. So if you try to calculate 0*inf, it means nothing, but if you define "infinity" in each case properly, it can be "calculated" as such.

To prove infinity is not a number, you can tell this:

Assume infinity is a number, it must be greater than any number. Then how about x = infinity + 1? Clearly x > infinity, and this contradicts our assumption of infinity, because x (number) is greater than infinity. So our assumption "infinity is a number" is wrong; i.e. infinity is not a number. And calculation cannot be done in the same way as with usual numbers.

 

[arildno]

To prove infinity is not a number, you can tell this:

Assume infinity is a number, it must be greater than any number. Then how about x = infinity + 1? Clearly x > infinity, and this contradicts our assumption of infinity, because x (number) is greater than infinity. So our assumption "infinity is a number" is wrong; i.e. infinity is not a number. And calculation cannot be done in the same way as with usual numbers.

That assumes that the operation of addition works in exactly the same manner when one of the addends is the number infinity.
If, however, you say that inf+a=inf, where a is any number, then your contradiction won't appear, and you may treat infinity as a number.
(Of course, then you've basically changed the meaning of addition, and you are no longer talking about the "normal" numbers..)

 

[HallsofIvy]

"Last night, I figured out a way to show that 0*infinity = 1. I'll show you my thought process..."
Have you paid no attention to what has been said previously? Infinity is not a number and "0*infinity" makes sense only as shorthand for a limit.

"If .9999... = 1, then 1/infinity = 0. I said this because 1/infinity would look like .0000...n. "

?? What possible justification do you have for saying what 1/infinity would "look like"?

 

[maverick6664]

That assumes that the operation of addition works in exactly the same manner when one of the addends is the number infinity.
If, however, you say that inf+a=inf, where a is any number, then your contradiction won't appear, and you may treat infinity as a number.
(Of course, then you've basically changed the meaning of addition, and you are no longer talking about the "normal" numbers..)

I used a simple reduction to absurdity to prove "infinity" isn't a number. I just say "If infinity were assumed to be a number, our traditional calculation would collapse. So infinity isn't a number." That's why we use lim sign to "calculate" infinity especially in \frac \infty \infty and in 0 * \infty. Addition seems to work, but even after addition, comparision won't work: i.e. \infty + a > \infty (a>0) doesn't hold (cannot be defined.) And when we have -\infty addition is no longer valid. I just wanted to say "infinity isn't a number." As in the title of this thread, 0*\infty isn't defined, and we need lim sign to calculate infinity-like numbers.

In order to treat infinity like a number, "non-standard" analysis was created (but I forgot the details because it's unnecessarily complicated...)

 

[arildno]

And calculation cannot be done in the same way as with usual numbers.

I overlooked this original statement of yours.
Sorry, won't happen again. I'll put my head in a bucket of water, if that will temper your just wrath over me.

 

[ksinclair13]

Please re-read my first post, HallsofIvy, before making me look stupid. I mentioned in there that I know that none of that is true and that I was just having a little fun in my head. I also mentioned that I know that infinity is not a number.

 

[maverick6664]

I overlooked this original statement of yours.
Sorry, won't happen again. I'll put my head in a bucket of water, if that will temper your just wrath over me.

No problem. Today my favorite female speed skater, Tomomi Okazaki was chosen for Trino olympic athlete and I am happy :!) She will win gold medal. :)