Chickenslayer69
Badges: 7
Rep:
?
#1
Report Thread starter 4 years ago
#1
Probably a stupid question, but why is the range of values of x for which the expansion is valid -1/2 < x < 1/2?

http://prntscr.com/dknt6h

Thanks
0
reply
Zacken
Badges: 22
Rep:
?
#2
Report 4 years ago
#2
(Original post by Chickenslayer69)
Probably a stupid question, but why is the range of values of x for which the expansion is valid -1/2 < x < 1/2?

http://prntscr.com/dknt6h

Thanks
Do you know that the expansion of \ln(1+x) is valid for  -1 &lt; x \leq 1? If not: the proof is likely out of your grasp at the moment, but it should be a given result.

Any, then the expansion of \ln(1+ 2x) is valid for -1 &lt; 2x \leq 1 and the expansion of \ln(1-2x) valid for -1 &lt; -2x \leq 1.

Since you're taking the product of those two series, you need the interval of validity (known as the radius of convergence) to satisfy both of the above inequalities, i.e -\frac{1}{2} &lt; x \leq \frac{1}{2} and -\frac{1}{2} \leq x &lt; \frac{1}{2}. From which you get the required result.
0
reply
Chickenslayer69
Badges: 7
Rep:
?
#3
Report Thread starter 4 years ago
#3
(Original post by Zacken)
Do you know that the expansion of \ln(1+x) is valid for  -1 &lt; x \leq 1? If not: the proof is likely out of your grasp at the moment, but it should be a given result.

Any, then the expansion of \ln(1+ 2x) is valid for -1 &lt; 2x \leq 1 and the expansion of \ln(1-2x) valid for -1 &lt; -2x \leq 1.

Since you're taking the product of those two series, you need the interval of validity (known as the radius of convergence) to satisfy both of the above inequalities, i.e -\frac{1}{2} &lt; x \leq \frac{1}{2} and -\frac{1}{2} \leq x &lt; \frac{1}{2}. From which you get the required result.
Yeah I knew it was something to do with that but I got confused with why less than was used and not less than or equal to. I know nothing about the last paragraph you wrote so I guess I should look that up, haha. Thanks.

Edit: Looking at it now it's easy to understand, not sure why I was confused, haha. Thanks
0
reply
Zacken
Badges: 22
Rep:
?
#4
Report 4 years ago
#4
(Original post by Chickenslayer69)
Edit: Looking at it now it's easy to understand, not sure why I was confused, haha. Thanks
Glad you understand.
0
reply
atsruser
Badges: 11
Rep:
?
#5
Report 4 years ago
#5
(Original post by Zacken)
Since you're taking the product of those two series, you need the interval of validity (known as the radius of convergence) to satisfy both of the above inequalities,
It's the sum of the series, no?
0
reply
Zacken
Badges: 22
Rep:
?
#6
Report 4 years ago
#6
(Original post by atsruser)
It's the sum of the series, no?
Oh yes, my bad. I misread it as two seperate logarithms, didn't see it was all encapsulated in one log.

Thanks.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Which of these would you use to help with making uni decisions?

Webinars (54)
13.85%
Virtual campus tours/open days (89)
22.82%
Live streaming events (37)
9.49%
Online AMAs/guest lectures (38)
9.74%
A uni comparison tool (88)
22.56%
An in-person event when available (84)
21.54%

Watched Threads

View All