trick question?

[Sarahisme]

hey is this like a trick question or something?

 

because i just get 0

[Sarahisme]

oops, just ignore that comment, but hang on i'll post my updated answer, and if anyone has the time, could they please tell me if its correct?

 

Thanks

 

Sarah

[Sarahisme]

actually, one second thoughts, could i please get a bit of help (hints or whatever) with this?

[Sarahisme]

what does a polynomial of degree 2n+1 mean?

[Sarahisme]

ok is this correct?

[matt grime]

A poly of degree k means that the largest power of x is x^k, eg x^2+3x+1 is degree 2, and so on.

 

log((1+x)/(1-x)) = log(1+x) - log(1-x)

 

so subtract the two taylor series (up to the 2n+1'th term) and divide by two.

[Sarahisme]

yeah, i think thats what i did, and then answer in my last post is what i got

 

is that the right kind of thing?

[matt grime]

The answer you got in the last post is not a polynomial of degree 2n+1. Write out the talyor polys of log(1+x) and log(1-x) and add them together and divide by 2, what is the resulting polynomial?

 

Note your last thunmbnail has a nicer form as

 

0 if n is even and x^n/n if n is odd.

[Sarahisme]

that thumbnail is the 2 things put together, ie. if n=1, that is a polynomial of degree one and if you add that same thing (with n=2) that would give a 2nd degree polynomial (which in this case doesnt exist? is that really true) and then if you add that thing with n=3, you would get a 3rd order polynomail..... and so on until you got to the addtion where n = 2n+1 i guess

 

???

[Sarahisme]

oh, so if its of degree 2n+1 then it is an "odd degree" so to speak.

 

so then the answer would be:

P= x^n/n

 

??

[matt grime]

Eh? n is what? Where are the smaller terms. I see nothing in your thumbnail that indicates adding terms together - that would require a Sigma sign, or soemthing.

 

Ok, let's start more simply. What is the taylor polynomial of log(1+x) of degree 2n+1?

 

Hint, the talyor poly for sin of degree 2n+1is:

 

x+x^3/3! + x^5/5! + ... + x^{2n+1}/(2n+1)! =

[Sarahisme]

hows this?...this is what i have been trying to say...

[matt grime]

you're raising -1 to an even power, so you can simplify it a hell of a lot.

[Sarahisme]

lol oops, sorry, the last part of that equation should be:

[Sarahisme]

and then thats the correct answer isnt it?!?

[matt grime]

Have you remembered to divide by two? it's one half of log(1+x)-log(1-x) we're after.

[Sarahisme]

oops yeah half of that whole expression then (i.e. the real expression (lets call it P(x)) is :

P(x) = .5p(x)

 

assuming the expression before was p(x)

 

and then that is defintely correct??, right??

[matt grime]

Stop being panicky about the result. Trust yourself to add together two things and divide by 2.

[Sarahisme]

lol, i'll take that as a "yes its correct" then

[matt grime]

Why would you trust my opinion? I can't even remember what the taylor series of log(1+x) is.

[Johnny5]

Why would you trust my opinion? I can't even remember what the taylor series of log(1+x) is.

 

That one's easy...

 

Start off with this...

 

Integral 1/U du = ln U + C

 

Now, do what you have to to make the C=0

 

So you are going to write an indefinite integral instead...

 

ln 1 = 0

 

 

Integral_U=1^U=U(x) 1/U du = ln U(x)

 

now make the following substitution:

 

U(x) = 1+x

 

dU=dx

 

Thus you have:

 

Integral_x=0^U=1+x 1/(1+x) dx = ln (1+x)

 

So the RHS is what you want explicitely, and you can use the LHS to figure out the series for the expression for the RHS.

 

Write the integrand so that you can use Newton's binomial series expansion formula:

 

Integral_x=0^U=1+x (1+x)^-1 dx = ln (1+x)

 

 

Now, you can rewrite the integrand as a series switch the summation sign and the integral sign, and then integrate term by term.

[matt grime]

Johnny, I know what a Talyor series is, and how to find them (and I certianly wouldn't sue your method since you would need to prove that sum and integral were allowed to be intechanged, and that is difficult usually) and how to find its remainder in one of three forms, but I do not memorize them any more, nor do I have any need to, nor is it important what the Taylor series is. You have missed the point that I was trying to make to Sarah that as long as she has the correct idea as to how to solve it that is the main thing. The actual answer is pretty immaterial. Please don't hijack this thread.

[Johnny5]

Johnny, I know what a Talyor series is, and how to find them (and I certianly wouldn't sue your method since you would need to prove that sum and integral were allowed to be intechanged, and that is difficult usually) and how to find its remainder in one of three forms, but I do not memorize them any more, nor do I have any need to, nor is it important what the Taylor series is. You have missed the point that I was trying to make to Sarah that as long as she has the correct idea as to how to solve it that is the main thing. The actual answer is pretty immaterial. Please don't hijack this thread.

 

Well it is an alternative approach to finding the answer. I didn't intend to hijack the thread, just offer an alternative, since you can find the expansion of ln(1-x) identically. Anyways, I just happened to know the answer to this off the top of my head, so i thought i would share.

 

Regards

[Sarahisme]

lol, i trust you matt

 

thanks for all your help

 

-Sarah

[Sarahisme]

and you too Johnny, thanks very much for your help