Question

Calculus help..an someone explain why the following equals 1

Answer 1

Why what equals 1 ?

Answer 2

d/dx x

Answer 3

\[\LARGE the \ following = 1\] this is true because "ollo" is an even function so it disappears in the limit and "wing" flies away. "the f" is left and that's just the function, which equals 1 of course.

Answer 4

\[\sum_{n=0}^{\infty} n!0^{n}=1+0+0+0....=1\]

Answer 5

They've decided that \[\LARGE 0!=1\] and \[\LARGE 0^0=1\] but of course 0 to any power other than 0 is not longer an indeterminant form so it's just 0. Adding up an infinite number of 0s still gives you just zero.

Answer 6

why is 0^0 defined by 1 as some teachers I don't know

Answer 7

\(0^0\) is indeterminate.

Answer 8

I am giving zero pizza slices to zero friends. So each friend gets 1 slice.
Makes perfect sense to me....

Answer 9

0^0 is just same as 0/0, in THAT sense

Answer 10

Well one good reason to define 0^0=1 is in the power series of e^x. \[\LARGE e^x= \sum_{n=0}^\infty \frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\] So any time you plug in x=0 to get e^0 you are sort of implicitly saying in the power series that you've decided 0^0=1 for this instance.

Answer 11

Well if 0^0 = 1 then you would have 1+2+6+18....as your answers

Answer 12

I don't get how e^0=1 implies 0^0=1

Answer 13

@nelsonjedi How do you suppose that? Where is the 2 and 6 and 18 coming from?

Answer 14

@myininaya write out the power series

Answer 15

n!

Answer 16

The power series for e^x?

Answer 17

Yes, but look at just the first term. Any power of 0 is 0.\[\LARGE 1!*0^1=0\] For example, \[\LARGE 0^3=0\]

Answer 18

Agree..so every term is n! multiplied by 0. So how is it you get 1 for the first factorial

Answer 19

\[\LARGE e^x= \sum_{n=0}^\infty \frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+...\]

So plugging in x=0 we have: \[\LARGE e^0=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+...\]

Answer 20

So what you are telling me is 0/0! = 1.

Answer 21

well no he defining 0^0 as 1 not as 0

Answer 22

\[\text{ he is saying } \frac{0^0}{0!}=1 \]

Answer 23

If for this instance 0^0 does not =1 then e^0 will not =1 since it's the only term that saves us.

@nelsonjedi That's not what I'm saying. I'm saying \[\LARGE 0^0=1\] We already know that \[\LARGE 0!=1\] so they're both 1, kind of confusing.

0!=1 as sort of a rationale for why this is true is because there is only 1 way to arrange 0 things.

Answer 24

Thank you

Answer 25

I guess people rather say x^0=1 for all x instead of 0^x=0 for all x?

Answer 26

Even so, it's not necessarily true. 0^0 is still an indeterminate form.

Answer 27

And yes I know 0^0 is an indeterminate form

Answer 28

In general it is indeterminate as it can be either 0 or 1 depending on how you take the limit of x^y. But in specific situations it can be assigned one of the two possibilities.

Answer 29

And it is so weird though we define it as 1 instead of 0 sometimes

Answer 30

Is there a situation where 0^0 is assigned a value of 0?

Answer 31

The problem with doing pure math lies in the fact that you can come to a lot of different results and they depend on the context in which the math describes. Otherwise you're essentially doing meaningless playing around.

For example we can take the average of 2 and 3 and get 2.5. But this is actually a fairly meaningless result if I am saying the average number of children that two families have is 2.5 because one family has 2 children and the other has 3 since you can't have half a child last time I checked.

Derivatives are the same way, their definition is all about the ratio of two infinitesimals. If you just plug it in without doing some algebra you will always get 0/0. So it's undefined and calculus allows us to look at 0/0 in different contexts to define it depending on what we are looking at.

Answer 32

Can I have example of when 0^0 is defined as 0
Like you provided an example of why we needed 0^0 be 1 in that one particular case your brought up with the power series of e^x and replacing x with 0 showing 0^0=1

Answer 33

None really come to mind. Honestly when I think of 0 to any positive power I think it must be 0. When 0 is to any negative power then we easily have positive or negative infinity. (Just think of the graph y=1/x) When 0 is raised to 0 it is sort of meaningless by itself. The only context in which I have ever seen it is when 0^0=1.

Just keep it in mind, maybe this should be an axiom? I don't know.

Answer 34

agreed
I know 0^a=0 for when a>0
I was just trying to find if 0^0 being defined as 0 would ever be useful

Answer 35

Thanks @kainui for showing the power series thing. It is kinda convincing for the argument that 0^0 can be defined as 1