Question

how can I write 1*3*5*7...??

Answer 1

That could be interpreted two ways:

1. as the product of consecutive primes, or
2. as the product of consecutive odd integers.

Answer 2

;P

Answer 3

we could be fancy and write it with the \(\Pi\) notation :)

Answer 4

If it's the latter, then\[\prod_{i=0}^{\infty}2i+1\]

Answer 5

like that^

Answer 6

is as a product of odd numbers

Answer 7

Well, if it was the former, and you had an answer, then I would suggest that we get together. ;P

Answer 8

is just I have to write it in a general way is for a differential equation

Answer 9

but across, 1 is not prime and 2 is not on the list
so it must be option 2, nno?

Answer 10

I am just messing around. ^^

Answer 11

A minor problem is that 1*3*5*7*... is meaningless.

Answer 12

Are you wondering how to express the first n terms? That would be (2*1-1)(2*2-1)(2*3-1)...(2*n-1)

Answer 13

yes just like that the thing is that I'm wondering is possible that I could write that like a factorial?

Answer 14

Not one factorial. But compare just the format through 7, with 7! and you have 1*2*3*4*5*6*7 as the factorial and 1*3*5*7 as the number you want. How do you get from the first to the second?

Answer 15

What happens when you divide the factorial by 2*4*6?

Answer 16

my equation is (x-1)y''-xy'+y=0 in my solution I have to state 1*3*5*7...in the series

Answer 17

Notice that 1*3*5*7 is 1*2*3*4*5*6*7 divided by 2*4*6. Both of these can be expressed using factorials.

Answer 18

I don't like just giving answers, but gotta run. The top is (2n+1)! for and the bottom is 2*n! for n=3.

Answer 19

Oh! you're si right!!! but how can you state that just in factorials? sorry is just I can't see how U_U

Answer 20

The nth term in your sequence is (2n+1)! / (2*n!)