Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

Find the independent term of the expansion

Is it a must to expand the whole thing out and look at the independent term, or is there a shortcut??

i still can't figure out how to make the "0" smaller

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If you want a power to be more than one character, you need to put it in curly braces.

eg. x^{10} -->

There is a shortcut to this, because most of the expansion is irrelevant to you.

You're only interested in the term that doesn't contain any x. This is (x^2/2)^5 * (-1/x^2)^5 = -1/32.

The coefficient of this term is 10C5 = 252, which means the independent term of your expansion is -252/32 = -7.875.

Why did the vector cross the road?

It wanted to be normal.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi 1a2b3c2212;

Good answer mathysperson, that is the way. In case you can't follow it, I think this is his reasoning. Normally to find the independent term (constant term) you would just set x=0, but that is not possible here, because of the x in the denominator, Start with the binomial expansion of.

The general term is:

with

So:

Pull out of the brackets all the constants and get them on the left

Set the exponents of

and equal to 0Now just plug into (A).

*Last edited by bobbym (2009-05-05 01:02:33)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

mathsyperson, how did u factorise the original polynomial to (x^2/2)^5 * (-1/x^2)^5 = -1/32??

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

Offline

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

What you meant was

It doesn't make sense to me though.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi 1a2b3c2212;

That statement is not true. Mathsy did not mean that.

*Last edited by bobbym (2009-05-05 00:26:21)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

uh then where did he get that equation from?

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi 1a2b3c2212;

Look at my post, it is what he/she may have been thinking about when he/she solved the problem.

*Last edited by bobbym (2009-05-05 00:40:36)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

oh.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Did it make sense?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

i think i get the gist of it..'cause i haven't started on binomial theorem yet.

Offline

Pages: **1**