The magical thread of C4 revision

[MoogleViper]

Right I have an exam tomorrow so I want a few things clearing up for me.

 

First thing:

 

Q (i): Expand (2+x)^(-2) in ascending powers of x up to and including the term in x³, and state the set of values for which the expansion is valid.

 

I did the expansion and got

0.25 - 0.25x + 0.1875x² - 0.125x³ which is correct.

 

But I don't get the second part. The answer is -2 < x < 2 but I don't know why.

[Zell]

Not 100% on this but I think you have to use the general result that the series (1 + x)^a converges for all |x|< 1.

 

(x + 2)^-2 = (2(0.5x + 1))^-2

 

Hence |0.5x|< 1 for validity ==> -1 < 0.5x < 1 ==> -2 < x < 2

[MoogleViper]

Not 100% on this but I think you have to use the general result that the series (1 + x)^a converges for all |x|< 1.

 

But why is that?

[Ellmeister]

Yup. Zell is right. I had no idea how to explain it as I couldn't properly remember all the stages but yeah.

[MoogleViper]

Another question.

 

I had to integrate;

 

2x + 3 + x/(x²+4)

 

Why is the integral of the fraction 0.5ln(x²+4)? I understand the ln but why does the x disappear?

[Zell]

But why is that?

 

Ahh, that would involve a bit of analysis, something you do at university level. There's probably some kind of justification in the C4 textbook, but I doubt you have to know why, just use the result.

 

Basically, (1 + x)^a is equal to the power series:

 

SUMn=0 up to positive infinity [(a n) * x^n]

 

where (a n) is aCn, ie the binomial coefficient.

 

From here, we can use many analytical techniques to determine for what values of x will make the series converge to a real number, such as the ratio test and the Cauchy root test. The end result is that the series converges for all |x|< 1 (we'll ignore the case x=1).

 

I can into it a bit more if you want.

 

Another question.

 

I had to integrate;

 

2x + 3 + x/(x²+4)

 

Why is the integral of the fraction 0.5ln(x²+4)? I understand the ln but why does the x disappear?

 

 

Suppose the integral was ln(x^2 + 4). Differentiating gives:

 

2x/(x^2 + 4)

 

However we require x/(x^2 + 4). Hence halving the proposed intergral gives the correct answer. To check we differentiate (using the chain rule). Note the "x" term is obtained after differentiating x^2 +4 inside the logarithm.

[MoogleViper]

Ahh, that would involve a bit of analysis, something you do at university level. There's probably some kind of justification in the C4 textbook, but I doubt you have to know why, just use the result.

 

Basically, (1 + x)^a is equal to the power series:

 

SUMn=0 up to positive infinity [(a n) * x^n]

 

where (a n) is aCn, ie the binomial coefficient.

 

From here, we can use many analytical techniques to determine for what values of x will make the series converge to a real number, such as the ratio test and the Cauchy root test. The end result is that the series converges for all |x|< 1 (we'll ignore the case x=1).

 

I can into it a bit more if you want.

 

So at A-level it's more of a case of just accepting that it is?

 

Suppose the integral was ln(x^2 + 4). Differentiating gives:

 

2x/(x^2 + 4)

 

However we require x/(x^2 + 4). Hence halving the proposed intergral gives the correct answer. To check we differentiate (using the chain rule). Note the "x" term is obtained after differentiating x^2 +4 inside the logarithm.

 

 

Oh right I get it now. Thank you very muchly.

[Zell]

So at A-level it's more of a case of just accepting that it is?

 

 

 

 

Oh right I get it now. Thank you very muchly.

 

Pretty much, you have to accept a lot of things at A level without even realising it, like integration being the reverse of differentiation.

 

Anywhos, I'm now going to do some revision of my own. Toodles.

[chairdriver]

If you have (1 + bx)^a

 

the expansion is valid for lbxl < 1

 

or lxl < 1/b

 

 

I should be revising FP2 *leaves*