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Validity interval Maclaurin series

How do we work out a validity interval for a Maclaurin series? I've seen that with something such as ln(1 - 3x), the interval is -1 < -3x < 1, so -(1/3) < x < (1/3).

My intuition says that we take the x multiple, slap it in the middle and take the constants and put them at either side, making one positive and one negative.
But I doubt this rule follows for everything.

So what do we generally do to find the validity interval for a Maclaurin series?
Original post by Heirio
How do we work out a validity interval for a Maclaurin series? I've seen that with something such as ln(1 - 3x), the interval is -1 < -3x < 1, so -(1/3) < x < (1/3).

My intuition says that we take the x multiple, slap it in the middle and take the constants and put them at either side, making one positive and one negative.
But I doubt this rule follows for everything.

So what do we generally do to find the validity interval for a Maclaurin series?


When you have something like ln⁡(2−5x)\ln(2-5x) or something, get in the form that is in the formula booklet. That is ln⁡(1+x)\ln(1+x). So you want to get something like ln⁡(1+[something that is between -1 and 1])\ln(1+[\text{something that is between -1 and 1}])

This is simple as ln⁡(2−5x)=ln⁡(2)+ln⁡(1−52x)\ln(2-5x)=\ln(2)+\ln(1-\frac{5}{2}x) which for easy comparison can be written as ln⁡(2)+ln⁡(1+(−52x))\ln(2)+\ln(1+(-\frac{5}{2}x))

Now here that 'something' is clearly −52x-\frac{5}{2}x, so say −1<−52x<1-1<-\frac{5}{2}x<1 and so get the answers of −25<x<25-\frac{2}{5}<x<\frac{2}{5} by some basic rearranging.
(edited 6 years ago)
Reply 2
Avatar for Heirio
Heirio
OP
17
I see. I saw on the powerpoint I'm using to revise that another expression that requires a validity interval is (1 + x)^n. Are there any more kinds of expressions apart from the two I've mentioned in this thread and do we work them out in the same way?
Original post by Heirio
I see. I saw on the powerpoint I'm using to revise that another expression that requires a validity interval is (1 + x)^n. Are there any more kinds of expressions apart from the two I've mentioned in this thread and do we work them out in the same way?


Yes the principle carries over, and yes there are other expressions but at A-Level it's just those 2 I think.
Reply 4
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Heirio
OP
17
I see, so the rule you described is the only one I need for my studies atm? That's good.

A small question though, you said we put it into a different form, so ln(2 - 5x) becomes ln2 + ln(1 - 5x/2). If we're using the right hand side of the expression, what happens to the ln2? Do we simply disregard it?
Original post by Heirio
I see, so the rule you described is the only one I need for my studies atm? That's good.

A small question though, you said we put it into a different form, so ln(2 - 5x) becomes ln2 + ln(1 - 5x/2). If we're using the right hand side of the expression, what happens to the ln2? Do we simply disregard it?


It's a constant outside the variable function. It doesn't affect the interval of validity, so we don't need to touch it as far as the validity is concerned.

And it's not so much of a 'rule' - it's just how I remember it. You may find an easier way and stick with that one, whatever gets the job done.
Reply 6
Avatar for Heirio
Heirio
OP
17
Original post by RDKGames
It's a constant outside the variable function. It doesn't affect the interval of validity, so we don't need to touch it as far as the validity is concerned.

And it's not so much of a 'rule' - it's just how I remember it. You may find an easier way and stick with that one, whatever gets the job done.


yeah that all sounds good
thanks for the help!

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