Question

can you help me on
o to the zero power

Answer 1

thanks

Answer 2

in general x^0 = 1 , unless x = 0

Answer 3

there might be a good reason why 0^0 is undefined.
I graphed y = x^x , and it is not defined when x = 0

Answer 4

define it as 1 if you are using binomial theorem or sequences

Answer 5

ok, since the limit of x^x as x->0+ is 1 ?

Answer 6

limit has got nothing to do with the function being defined right ?

Answer 7

the right side limit is 1 irrespective of whether x^x is defined at 0 or not

Answer 8

why would you define it as 1? seems kind of arbitrary

Answer 9

binomial theorem is not arbitrary

Answer 10

that should be (a+b)^n

Answer 11

also in calcII, while working with sequences we define it 0^0 as 1

Answer 12

yes because the limit of x^x is 1 as x->0+ , it seems natural to define it as 1

Answer 13

\[(a+b)^n = \sum\limits_{k=0}^n a^{n-k}b^k\]

you want binomial theorem to be general and work for a=-b and n=0

Answer 14

just as you can define f(x)= sinx / x as 1 , when x = 0

Answer 15

sinx/x is not defined at x = 0

Answer 16

plugging up the removable discontinuity

Answer 17

you can read this
http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

Answer 18

im not sure where you would see this turn up in sequences

(-b + b )^0 = sum k=0 ..0 , and gets kind of gnarly

Answer 19

(-b + b )^0 = sum {k=0 ..0} [(-b)^0 * b^0] = 1*1

Answer 20

\[e^x = \sum\limits_{n=0}^{\infty} \dfrac{x^n}{n!}\]

what can you say about the value of e^x at x = 0?

Answer 21

most power series representation require a piecewise definition if you dont define 0^0 = 1 when you're working with sequences

Answer 22

e^x = 1 + x + x^2/2 + x^3/ 6 + ...

Answer 23

e^0 = 1 + 0 + 0 + ...

Answer 24

you're just masking the actual problem by expanding like that

Answer 25

that expansion is a result of power series representation of e^x, not an alternative