Thread: Maths problem 9999

i'll take your word for it

Brilliant! Still illegal IMHO, though, for the same reason as before: the operation is not performed on the nines, in this case the nines only identify a subset of an unlimited, and therefore infinite, set of functions. If that were allowed, any and all numbers can be produced.

If, however, you have a solution that accepts the way I suggested of producing 1, I would still be keen to see it. We would then let the professors fight over it. (If you can still find any who champions your cause...) I have to admit that I do not have a great opinion of Adelaide academics, though, you might end up being right...

Well it is as illegal as the factorial. Both are mathematics notations not operators. We accepted factorial so we should accept summation too.

I think it is ok. Trust me.


I will let you know the other 2 ways when I remember them.

Originally Posted by ss_11000

i'll take your word for it

C’mon grasshopper. That is not the right attitude.

You need to work it out.

Originally Posted by Wongo

Well it is as illegal as the factorial. Both are mathematics notations not operators. We accepted factorial so we should accept summation too.

I think it is ok. Trust me.


I will let you know the other 2 ways when I remember them.

Hmmm.... As I always say to my daughter, never trust anybody who says "trust me"...

I still think that not being anally retentive about notations v. operators is OK (it reinforces my acceptance of lim x) , but does not change my point that you are working on infinite functions, not on the four nines, which denies the intention.

No worries mate. Take away both the ! notation and E notation. (sorry Greg, your solutions for 1 to 20 were wrong )

Re the acceptance of lim x, mate let me say it one more time 0.99999... is not 1. It's not even close.

Originally Posted by Wongo

Re the acceptance of lim x, mate let me say it one more time 0.99999... is not 1. It's not even close.

Well, tell that to my son's calculus professor... I'm too old for this...
You wouldn't still be stuck in Euclidean space, perchance?

It doesn’t matter if it is for a professor in calculus. 0.999999… is not equal to 1. It is a fact. You can round it up to 1 but it is not 1.

Originally Posted by Wongo

C’mon grasshopper. That is not the right attitude.

You need to work it out.

yeah sure.

i'll let you know in a couple of months

Originally Posted by Wongo

It doesn’t matter if it is for a professor in calculus. 0.999999… is not equal to 1. It is a fact. You can round it up to 1 but it is not 1.

Do you accept that for any two numbers a and b there is an average number in between c = (a+b)/2?

No I don't

f

Originally Posted by Wongo

Anyone wants to explain it?

Quote:
Originally Posted by Wongo
C’mon grasshopper. That is not the right attitude.

You need to work it out.

yeah sure.

i'll let you know in a couple of months
__________________
S T I R L O

gday wongo, i just came across this thread again.

dunno if i'm right but does that (^^^) mean that the sum of the 3rd through to 6th terms of that series is 68, and minus the 1 it equals 67?


btw wongo, i got a headband that looks just like your avatar

Originally Posted by ss_11000

dunno if i'm right but does that (^^^) mean that the sum of the 3rd through to 6th terms of that series is 68, and minus the 1 it equals 67?

Correct