MoogleViper Posted June 4, 2009 Posted June 4, 2009 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 Posted June 4, 2009 Posted June 4, 2009 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 Posted June 4, 2009 Author Posted June 4, 2009 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 Posted June 4, 2009 Posted June 4, 2009 Yup. Zell is right. I had no idea how to explain it as I couldn't properly remember all the stages but yeah.
MoogleViper Posted June 4, 2009 Author Posted June 4, 2009 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 Posted June 4, 2009 Posted June 4, 2009 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 Posted June 4, 2009 Author Posted June 4, 2009 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 Posted June 4, 2009 Posted June 4, 2009 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 Posted June 4, 2009 Posted June 4, 2009 If you have (1 + bx)^a the expansion is valid for lbxl < 1 or lxl < 1/b I should be revising FP2 *leaves*
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