marcus888
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Hi

I am looking at convergent geometric series and did not know why the common ratio 'r' is in those lines.

My book originally states -1 < r < 1

So does |r| < 1 just mean that?

Could someone explain it please?

Thanks
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poorform
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Yes it does. It is the modulus function. http://www.examsolutions.net/maths-r...troduction.php

|a| < b then -b < a < b,
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dumber student
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(Original post by marcus888)
Hi

I am looking at convergent geometric series and did not know why the common ratio 'r' is in those lines.

My book originally states -1 < r < 1

So does |r| < 1 just mean that?

Could someone explain it please?

Thanks
I think it comes from the fact that lim r^(n+1), which you evaluate when deriving the geometric series formula, is 0 if |r|<1.
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marcus888
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(Original post by poorform)
Yes it does. It is the modulus function. http://www.examsolutions.net/maths-r...troduction.php

|a| < b then -b < a < b,
Thanks, very helpful.
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atsruser
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(Original post by marcus888)
Hi

I am looking at convergent geometric series and did not know why the common ratio 'r' is in those lines.

My book originally states -1 < r < 1

So does |r| < 1 just mean that?

Could someone explain it please?

Thanks
It's useful to think of the modulus function as measuring the distance between two points on the number line. So, if we call d(a,b) the distance function between a and b then, e.g:

d(5,2) = 3
d(2,5) = 3
d(4,-3) = 7

etc. In general d(a,b)=d(b,a) since the distance from a to b equals the distance from b to a.

To get a formula for d(a,b) note that the distance between two numbers is always +ve or 0. So if we think in terms of directed numbers on the number line (or more graphically, arrows pointing in the +ve or -ve direction), then consider:

5-2 = +3 - this is the distance from 2 to 5, or from 5 to 2
2-5 = -3 - this is the -ve of the distance from 2 to 5, or from 5 to 2

From this we can see that -(2-5)=5-2=+3 is the distance between 2 and 5. In general we have:

if a &gt; b \Rightarrow a-b &gt;0 then d(a,b) = a-b
if a &lt; b \Rightarrow a-b &lt; 0 then d(a,b) = -(a-b)
if a = b \Rightarrow a-b =0 then d(a,b) = a-a=0

So we define a function, which we write |a-b|:

|a-b| = \begin{cases}

a-b & \text{ if } a-b &gt;0 \\

-(a-b) & \text{ if } a-b &lt; 0 \\

0 & \text{ if } a=b \text{ (or } a-b=0 \text{)}

\end{cases}

We call this the modulus function.

Note that since a=a-0 then |a| = |a-0| which is the distance from a to 0.

So the equality |r| &lt; 1 \Rightarrow |r-0| &lt;1 asks you to find all of the numbers which are less than 1 unit away from 0 on the number line. Of course, this means you can go almost from 0 to 1 in the +ve direction, or almost from 0 to -1 in the -ve direction. So we can see that |r| &lt; 1 \Rightarrow -1 &lt; r &lt; 1.

We can also use the definition above to write down a definition for, say, |x| directly, by making the substitution a-b \to x, which gives:

|x| = \begin{cases}

x & \text{ if } x &gt;0 \\

-x & \text{ if } x &lt; 0 \\

0 & \text{ if } x=0 

\end{cases}

We can now say that |r| &lt; 1 gives us:

if r &gt;0 then r &lt; 1
if r &lt; 0 then -r &lt; 1 \Rightarrow r &gt; -1

and combining those two inequalities we get -1 &lt; r, r &lt; 1 \Rightarrow -1 &lt; r &lt; 1
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Louisb19
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(Original post by atsruser)
It's useful to think of the modulus function as measuring the distance between two points on the number line. So, if we call d(a,b) the distance function between a and b then, e.g:

d(5,2) = 3
d(2,5) = 3
d(4,-3) = 7

etc. In general d(a,b)=d(b,a) since the distance from a to b equals the distance from b to a.

To get a formula for d(a,b) note that the distance between two numbers is always +ve or 0. So if we think in terms of directed numbers on the number line (or more graphically, arrows pointing in the +ve or -ve direction), then consider:

5-2 = +3 - this is the distance from 2 to 5, or from 5 to 2
2-5 = -3 - this is the -ve of the distance from 2 to 5, or from 5 to 2

From this we can see that -(2-5)=5-2=+3 is the distance between 2 and 5. In general we have:

if a &gt; b \Rightarrow a-b &gt;0 then d(a,b) = a-b
if a &lt; b \Rightarrow a-b &lt; 0 then d(a,b) = -(a-b)
if a = b \Rightarrow a-b =0 then d(a,b) = a-a=0

So we define a function, which we write |a-b|:

|a-b| = \begin{cases}

a-b & \text{ if } a-b &gt;0 \\

-(a-b) & \text{ if } a-b &lt; 0 \\

0 & \text{ if } a=b \text{ (or } a-b=0 \text{)}

\end{cases}

We call this the modulus function.

Note that since a=a-0 then |a| = |a-0| which is the distance from a to 0.

So the equality |r| &lt; 1 \Rightarrow |r-0| &lt;1 asks you to find all of the numbers which are less than 1 unit away from 0 on the number line. Of course, this means you can go almost from 0 to 1 in the +ve direction, or almost from 0 to -1 in the -ve direction. So we can see that |r| &lt; 1 \Rightarrow -1 &lt; r &lt; 1.

We can also use the definition above to write down a definition for, say, |x| directly, by making the substitution a-b \to x, which gives:

|x| = \begin{cases}

x & \text{ if } x &gt;0 \\

-x & \text{ if } x &lt; 0 \\

0 & \text{ if } x=0 

\end{cases}

We can now say that |r| &lt; 1 gives us:

if r &gt;0 then r &lt; 1
if r &lt; 0 then -r &lt; 1 \Rightarrow r &gt; -1

and combining those two inequalities we get -1 &lt; r, r &lt; 1 \Rightarrow -1 &lt; r &lt; 1
I wonder how long that took to LaTeX?
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