Help with binomial expansion!

nevinsmith · 2011-04-12 19:51:41

so the question is expand 1 / (x^2+3)

so i take 1st term as x^2, 2nd term as 3 and power as -1

using binomial series i got-

1/3-(x^2)/9+(x^4)/27-(x^6)/81..... true for all values of x???

but the answer says-

3-9x^2+27x^4-81x^6.... true for -sqrt3>x>sqrt3

??? why does that happen?
is it to do with requirements of convergence or am i just wrong?

so how do i calculate the valid values of x for convergence?

Bob · 2011-04-12 19:58:01

hi nevinsmith

stay on-line answer coming up

Bob

Bob · 2011-04-12 20:04:13

hi nevinsmith

You have

You can only use the general form of the binomial theorem for n = -1 and this form also requires a 1 not a 3 in the bracket.

So do this:

Now you can proceed with the expansion.

Post again if you need more help

Bob

Last edited by Bob (2011-04-12 20:07:49)

nevinsmith · 2011-04-12 20:11:49

hi bob bundy, thanks for helping

so i expand 1/3(1/3x2+1)^-1

i still get the terms in fractions with convergence of every value of x

1/3-(x^2)/9+(x^4)/27-(x^6)/81.....

can you please explain how the answer is different?

which is 3-9x^2+27x^4-81x^6.... true for -sqrt3>x>sqrt3

how did the terms not have fractions?

thanks for your reply

Bob · 2011-04-12 20:35:16

hi again,

n = -1 'x' = x^2 / 3

I think I've sorted it now :

Sorry. I'm not getting a fractionless result. I've got to go out now but I'll look again in about 3 hours time (1pm BST)

Bob

Last edited by Bob (2011-04-12 20:51:59)

bobbym · 2011-04-12 21:24:17

Hi nevinsmith;

Did you try a value for x in that supposed series without denominators? You know, between -√3 and √3. And then try that value in 1 / (x^2 + 3)?

It certainly is not the series expansion for 1 / (x^2 + 3).

Actually:

Bob · 2011-04-13 00:16:59

hi nevensmith

I stand by my answer:

Is the answer you have from a book. I would suspect a printing error.

The test is "Does either formula actually work"?

Here's my figures

Pick a value for x in the range let's say x = 0.5

value of original expression = 0.307692307692308

Target answer: the values for the first five terms 3 - 2.25 + 1.6875 -1.265625 +0.94921875 = 2.12109375

not even close!

My answer = 0.333333333333333 -0.0277777777777778 + 0.00231481481481481 -0.000192901234567901 + 0.0000160751028806584 = 0.307693544238683

That's close enough for me. Take more terms if you want to get closer

this is the ultimate test. Does the expansion actually work?

The range is found by having -1 < (x^2)/3 < 1 and re-arranging to get -root 3 < x < root 3

Hope that helps,

Bob

gAr · 2011-04-13 00:44:46

Hi all,

Looking at the answer which the source says,
there's probably a mistake with the question.
It may be

Bob · 2011-04-13 00:58:25

hi gAr

You may be right but then

true for -sqrt3>x>sqrt3

isn't right.

edit: In fact, now I look at it again, it isn't right anyway. The signs are back to front!

Bob

Last edited by Bob (2011-04-13 00:59:26)

gAr · 2011-04-13 01:14:27

Hi Bob,

Oh, didn't check that condition when posting.

edit: In fact, now I look at it again, it isn't right anyway. The signs are back to front!

Yes, I observed it as well!

Looks like multiple mistakes then.

bobbym · 2011-04-13 07:27:13

Hi nevinsmith, Bob, gAr, bobbym;

Bob Bundy wrote:

I stand by my answer:

I did not explain myself right. I was not questioning your answer. That is what I got, so it has to be right. Why would I get a wrong answer? What I do not like is 3-9x^2+27x^4-81x^6.... as the answer . As I showed in post #6 that is not the series for 1 / (x^2+3)

Hi nevinsmith;

Math results have to withstand the test of numbers. Numerics is part of math too. Whenever you get a series representation for some expression always check to see that it is at least close. Then produce a few more terms of your series and convince yourself that it is converging to the correct answer. There are more precise means but this type of testing will weed out most of the errors quickly.

Bob · 2011-04-13 20:44:55

Good morning bobbym,

I did not explain myself right. I was not questioning your answer.

I wasn't upset, thinking you were criticising me. I had already made up that post in my head while I was out of the house. Oh yes, and I knew I was right too! (confidence = 90%)

That is what I got, so it has to be right. Why would I get a wrong answer?

Well now I'm totally confident! (confidence = 99.99%) Together, we must be on top of the Pyramid of Correctness. You can shapen razor blades just by holding them close.

Best wishes.

Bob

bobbym · 2011-04-13 20:56:18

Yes! Pyramid power!

nevinsmith · 2011-04-14 00:22:05

yea i would think there is a print error

the origional problem was probably what gAr said it was

the book is pretty darn useless at explaining stuff
thanks for the help guys!

bobbym · 2011-04-14 06:03:03

Hi nevinsmith;

What is the name of that book?

Math Is Fun Forum

#1 2011-04-12 19:51:41

Help with binomial expansion!

#2 2011-04-12 19:58:01

Re: Help with binomial expansion!

#3 2011-04-12 20:04:13

Re: Help with binomial expansion!

#4 2011-04-12 20:11:49

Re: Help with binomial expansion!

#5 2011-04-12 20:35:16

Re: Help with binomial expansion!

#6 2011-04-12 21:24:17

Re: Help with binomial expansion!

#7 2011-04-13 00:16:59

Re: Help with binomial expansion!

#8 2011-04-13 00:44:46

Re: Help with binomial expansion!

#9 2011-04-13 00:58:25

Re: Help with binomial expansion!

#10 2011-04-13 01:14:27

Re: Help with binomial expansion!

#11 2011-04-13 07:27:13

Re: Help with binomial expansion!

#12 2011-04-13 20:44:55

Re: Help with binomial expansion!

#13 2011-04-13 20:56:18

Re: Help with binomial expansion!

#14 2011-04-14 00:22:05

Re: Help with binomial expansion!

#15 2011-04-14 06:03:03

Re: Help with binomial expansion!

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