Express (3+4x)^0.5 as a series of descending powers of x, up to and including the third non-zero term, simplifying the coefficient. State the set of values of x for which the series expansion is valid.
I got the expression but I don't know how to find the validity. Is it |3x^-1|<1? Anyone can help?
Thanks =)
|x|>3/4, i.e. x>3/4 or x<-3/4
By the way, TJC revision set A maths question?
Wow how you know?
Because we are all from TJC, my dear jianliang (google rocks ok)
And you might wanna help me expose who is the intellectual one behind the Occam's Razor ign, and that avatar.
And ppl at sgf know that I won't post for such a mindless reponse, so I'll get back to topic as I intended to.
Where did I make a mistake? Isn't it:
(3+4x)^0.5 --> 3^0.5 (1+4/3x)^0.5
so valid when |4/3x| < 1 so |x| < 3/4
Blast it, since when did we ever had a chance to write validity as |x| > c ? Is this a trick of some sort?
Originally posted by Garrick_3658:Where did I make a mistake? Isn't it:
(3+4x)^0.5 --> 3^0.5 (1+4/3x)^0.5
so valid when |4/3x| < 1 so |x| < 3/4
Hi,
(1 + 4/3 x)^0.5 will still give terms in ascending powers of x, which is not meeting the requirement of the question.
The correct approach should instead be:
(3+4x)^0.5 = (4x)^0.5 [1 + 3/(4x)]^0.5
and the validity range is then |3/(4x)| < 1, i.e. |x| > 3/4.
Thanks!
Cheers,
Wen Shih
Originally posted by Garrick_3658:Because we are all from TJC, my dear jianliang (google rocks ok)
And you might wanna help me expose who is the intellectual one behind the Occam's Razor ign, and that avatar.
And ppl at sgf know that I won't post for such a mindless reponse, so I'll get back to topic as I intended to.
Where did I make a mistake? Isn't it:
(3+4x)^0.5 --> 3^0.5 (1+4/3x)^0.5
so valid when |4/3x| < 1 so |x| < 3/4
Blast it, since when did we ever had a chance to write validity as |x| > c ? Is this a trick of some sort?
Nahhhh google doesn't rock, my blog does =)
Anyway, it seems that I made a stupid mistake that 3/4 = 3 XD