Sums of Infinate Series...HELP! I need somebody! HELP!

S

skyd92

New member
Joined
Jun 29, 2009
Messages
16
The book can't explain how to solve this problem worth anything and I can't figure out anything about this problem.

The problem: "Express each decimal as a ratio of two integers: .45454545..."

I really need help understanding how to figure out this problem. Thanks a bunch! :)
 
.454545454545... = .45+.0045+.000045+.00000045+.0000000045+.....

This is a geometric progression with the first term being a = .45 and r being .0045/.45 = .01.

\(\displaystyle Ergo, \ \sum_{n=0}^{\infty} ar^{n} \ = \ \frac{a}{1-r} \ = \ \frac{.45}{1-.01} \ = \ \frac{.45}{.99} \ = \ \frac{5}{11}.\)

Note: 0<|r|<1
 
FWIW, any repeating decimal may be easily turned into a rational number:

If we want to express \(\displaystyle x = 0. \overline{d_1d_2...d_n} \,\,\) as \(\displaystyle \,\, \frac{a}{b}\)

Then we can write

\(\displaystyle x = \frac{d_1d_2...d_n}{\underbrace{9999...9}_{\text{n times}}}\)
 
Good show daon, but that formula only works if the first decimal (tenths) is part of the repeatment.

\(\displaystyle For \ example, \ .00\overline{453} \ = \ \frac{.00453}{1-.001} \ = \ \frac{.00453}{.999} \ = \ \frac{453}{99900} \ = \ \frac{151}{33300}.\)

Hence, one should stick to the original formula [a/(1-r)] to avoid confusion.
 
daon said:
FWIW, any repeating decimal may be easily turned into a rational number:

If we want to express \(\displaystyle x = 0. \overline{d_1d_2...d_n} \,\,\) as \(\displaystyle \,\, \frac{a}{b}\)

Then we can write

\(\displaystyle x = \frac{d_1d_2...d_n}{\underbrace{9999...9}_{\text{n times}}}\)
Click to expand...

Nit picking, I know, but you meant: any repeating decimal IS a rational number, which can be turned into the form a/b.

Now it is your turn to complain about how many words "nit picking" should be.
 
PAULK said:
daon said:
FWIW, any repeating decimal may be easily turned into a rational number:

If we want to express \(\displaystyle x = 0. \overline{d_1d_2...d_n} \,\,\) as \(\displaystyle \,\, \frac{a}{b}\)

Then we can write

\(\displaystyle x = \frac{d_1d_2...d_n}{\underbrace{9999...9}_{\text{n times}}}\)
Click to expand...

Nit picking, I know, but you meant: any repeating decimal IS a rational number, which can be turned into the form a/b (where a and b are rational numbers).

Now it is your turn to complain about how many words "nit picking" should be.
Click to expand...
 
Yes, yes, I failed in more than one way on that post. Time to pick up the pieces of my dignity and move on now :wink:
 
Re: Sums of InfinIte Series...HELP! I need somebody! HELP!

quote="skyd92"]... I am taking a distance learning course, I just want someone to explain this problem a bit clearer than my book did, that is all I ask.[/quote]

As DrMike showed, you should be able to look up 'infinite geometric series' in your text or on the internet. (And, of course, geometric series is a relatively new thing; if your book is more than 1000 years old, it won't have it.)

And, as to your 'distance learning' course -- do you actually pay someone for it? If so, someone there should earn their money and help you.
 
You can do them like this as well, but the geometric formula is easier.

\(\displaystyle x=.45454545....\)............[1]

Multiply by 100:

\(\displaystyle 100x=45.45454545......\)...........[2]

Subtract [1] from [2]:

\(\displaystyle 99x=45\)

\(\displaystyle x=\frac{45}{99}=\frac{5}{11}\)
 
You must log in or register to reply here.
Top