Limiting Sums Help! (1 Viewer)

SGSII · Feb 23, 2013

Hi Guys

Could someone please help me convert these reoccurring decimals to fractions using limiting sums from f) onwards showing working out:

I can do the previous ones before f) but have no idea how to do the rest :S
Please Help!
Thanks

Sy123 · Feb 23, 2013

SGSII said:
Hi Guys
Could someone please help me convert these reoccurring decimals to fractions using limiting sums from f) onwards showing working out:

View attachment 27738

I can do the previous ones before f) but have no idea how to do the rest :S
Please Help!
Thanks

So by limiting sum I am assuming you are allowed to use the formula:

$\frac{a}{1-\alpha} = a(1 + \alpha + \alpha^2 + \dots + \alpha^n + \dots) \ \ \ \ |\alpha| < 1$

We can rewrite the decimal in f:

$$f)$ \ \ 0.2 \dot{3} = 0.2 + 0.0\dot{3} = 0.2 + (0.033333\dots)$

$=0.2 + (0.03 + 0.003 + 0.0003+ \dots)$

$=0.2 + (\frac{3}{100}+ \frac{3}{1000}+ \frac{3}{10000} + \dots)$

$=\frac{1}{5} + \frac{\frac{3}{100}}{1-\frac{1}{10}}$

$= \frac{1}{5} + \frac{1}{30}$

$=\frac{6}{30} + \frac{1}{30} = \frac{7}{30}$

So basically you split up the decimal, do this VERY similarly for the next parts.

SGSII · Feb 23, 2013

Sy123 said:
So by limiting sum I am assuming you are allowed to use the formula:

$\frac{a}{1-\alpha} = a(1 + \alpha + \alpha^2 + \dots + \alpha^n + \dots) \ \ \ \ |\alpha| < 1$

We can rewrite the decimal in f:

$$f)$ \ \ 0.2 \dot{3} = 0.2 + 0.0\dot{3} = 0.2 + (0.033333\dots)$

$=0.2 + (0.03 + 0.003 + 0.0003+ \dots)$

$=0.2 + (\frac{3}{100}+ \frac{3}{1000}+ \frac{3}{10000} + \dots)$

$=\frac{1}{5} + \frac{\frac{3}{100}}{1-\frac{1}{10}}$

$= \frac{1}{5} + \frac{1}{30}$

$=\frac{6}{30} + \frac{1}{30} = \frac{7}{30}$

So basically you split up the decimal, do this VERY similarly for the next parts.

Okays Thank youu so much!

SGSII · Feb 23, 2013

Another quick question that i happened to get stuck on:

Its still limiting sums
Please help!

Uniguy · Feb 23, 2013

SGSII said:
Another quick question that i happened to get stuck on:
View attachment 27739
Its still limiting sums
Please help!

Well, you have a= 1.5, r= 2/3

therefore.....a/(1-r)
=

1.5/(1-2/3)= 4.5m

SGSII · Feb 24, 2013

Uniguy said:
Well, you have a= 1.5, r= 2/3

therefore.....a/(1-r)
=

1.5/(1-2/3)= 4.5m

But the answer says 3.5m

Sy123 · Feb 24, 2013

SGSII said:
Another quick question that i happened to get stuck on:
View attachment 27739
Its still limiting sums
Please help!

Ok, so when she drops the ball, it travels 1.5m

Then when it bounces it bounces back up at $1.5 \times 2/5$ m
Then it bounces back down at $1.5 \times 2/5$ m

Then it bounces back up at: $1.5 \times (2/5)^2$
Then it bounces back down at: $1.5 \times (2/5)^2$

So we can establish the pattern:

$S= 1.5+ 2\times \frac{2}{5} \times 1.5 + 2\times \left(\frac{2}{5} \right)^2 \times 1.5 + \dots$

Lets manipulate so we can compute the Sum more easily

$S=1.5 (1+2(\frac{2}{5} + \frac{2^2}{5^2}+\frac{2^3}{5^3}+ \dots))$

So we can calculate the sum in the middle using our formula:

$S=1.5(1+2\frac{2/5}{1-2/5})$

$S=1.5(1+2\frac{2}{3}) = 3.5 m$

asianese · Feb 24, 2013

You need to visualise such physical problems before jumping into them!

SGSII · Feb 24, 2013

Ohhh! That makes sense! AHAHA
Thank you so much!

Limiting Sums Help! (1 Viewer)

SGSII

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Sy123

This too shall pass

SGSII

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SGSII

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Uniguy

Banned

SGSII

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Sy123

This too shall pass

asianese

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