Math Is Fun Forum

gnits

Member

Registered: 2006-02-27

Posts: 3

Maclaurin\Taylor Series

Hi,

Can anyone give me some hints as to how to go about solving the following:

Prove that if m and n are of the same sign then:

Thanks for any help. I know the method involves a Taylor expansion but of what function and about which point I can't find.

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: Maclaurin\Taylor Series

Does it help if you know that, according to Taylor\Maclaurin:

In cases where |x-1| ≤ 1?

As long as you assume that m and n are the same sign and m ≤ 2n, then you could substitute in m/n for x in the above formula. You might need to convert the 1s in the equation to "fancy" 1s (ie n/n, m/m, or nm/nm)

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

gnits

Member

Registered: 2006-02-27

Posts: 3

Re: Maclaurin\Taylor Series

Hi,

Thanks for your reply.

This is the method I have found:

Consider the Maclaurin expansion of:

this is the same as:

Which are both standard series and so this is equal to:

which is equal to:

Now substitute into this the value of:

With this value of x we have:

and so overall we arrive at:

as required.

The series is only convergent for:

which implies that m and n must be of the same sign - as expected to prevent the ln function from having a negative argument.