On 24 Jul 2009, at 14:20, m.a. wrote:
> Bruno,
> You overlooked the question at the bottom of the page
> that I tried unsuccessfully to work out. Brent supplied the answer
> but what I was looking for were the steps that lead up to the
> answer. marty
>
>
> ----- Original Message -----
> From: Bruno Marchal
> To: everything-list.domain.name.hidden
> Sent: Friday, July 24, 2009 4:48 AM
> Subject: Re: Seven Step Series
>
>>
>> a^n * a^(-m) = a^(n-m)
>>
>> Again, verify this on simple example of your own.
>>
>> OK: If a=10 and n=3 and m=4 Following the formula above
>> "a^n * a^(-m)" ,I get as the first half of the equation:
>>
>> 10^3 * 10/4 =1000 x 2.5= 2500 but for the second half "a^(n-
>> m) I get:
"first half":
It is 10^3 * 10^(-4) . OK?
Now, it looks like you are saying that 10^(-4) is 10/4. But we have
defined a^(-n) by 1/a^n. So 10^(-4) is 1/(10^4) = 1/10000
so it is 1000 * 1/10000 = 1000/10000 = 1/10 = 0.1
>>
>>
>> 10^(n-m)= 10^ -1= 10/1
>>
>> which of course makes no sense at all. Where did I go wrong?
"second half":
10^(n-m) = 10^(3-4) = 10^(-1) = 1/(10^1) = 1/10 = 0.1
(you found 10/1, which is 10).
My diagnostic: You have not integrate that a^(-n) is: ONE divided by
a^n. It is
1 / a^n.
So 10^(-4) = 1/10^4 = 1/10000 = 0.0001
and 10^(-1) = 1/10^1 = 1/10 = 0.1
Be careful with the little numbers 10^1 = 10, 10^0 = 1. (see below if
you have a problem)
1 divided by any number bigger than 01 is always a number little than
1. With decimal they begin by 0.<something>.
For example, when we write a rational number like 234.567
it is an abbreviation of 2*100 + 3*10 + 4*1 + 5*(1/10) + 6*(1/100) +
7*(1/1000),
which is the same as:
2*(10^2) + 3*(10^1) + 4*(10^0) + 5*10^(-1) + 6*10^(-2) + 7*10^(-3)
We say that 234.567 is written in base ten. All the digits are
coefficient of power of ten. (where a power of 10 is a number = 10^n,
n any integer).
Are you OK with 10^0 = 1? If not read below
1 = 100/100 OK?
100/100 = 10^2/10^2 OK?
10^2/10^2 = 10^(2-2) OK? (use of the formula above)
10^(2-2) = 10^0 OK?
So: 1 = 10^0 OK?
So 10^0 = 1 OK?
Question?
I think you could have found the mistakes by carefully reread the
definitions, in this case, of a^(-n), which is 1 / (a^n). Don't you
think so?
Bruno
http://iridia.ulb.ac.be/~marchal/
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Received on Fri Jul 24 2009 - 17:00:03 PDT