1

$$f(x) = \frac{\sin x}{1-x}$$

I tried to differentiate it but it is very difficult, what is the trick?

Michael Hardy
  • 1
Paul the Pirate
  • 1,566

1 Answers1

4

I agree with David. Find the series for $\sin(x)$ and $\dfrac1{1-x}$ (a much simpler task), then find the first several terms of the product.

Cameron Buie
  • 102,994
  • What do you mean? I don't follow what you are trying to say. – Paul the Pirate Jul 30 '13 at 00:26
  • @PaulthePirate: the product of the series will give you the series of the product. – tomasz Jul 30 '13 at 00:27
  • So I need to find the approximation for the sum and then multiply them? – Paul the Pirate Jul 30 '13 at 00:29
  • @PaulthePirate $\sin x = x-x^3/3!+\cdots$ and $1/(1-x)=1+x+x^2+x^3+\cdots$. Just find the terms up to order $4$ of $( x-x^3/3!+\cdots)(1+x+x^2+x^3+\cdots )$. Take care to note that in the product the "dots" give terms $c_k x^k$ with $k>4$, so you can ignore them. – David Mitra Jul 30 '13 at 00:33
  • @PaulthePirate: As tomasz points out, the product of the two Maclaurin series for $\sin(x)$ and $\dfrac1{1-x}$ will be precisely the Maclaurin series for $f(x)$. David has been gracious enough to narrow your parameters further, but you should be certain that you understand why you don't actually need the whole Maclaurin series for $\sin(x)$ or $\dfrac1{1-x},$ so that you can draw conclusions appropriately for future problems. – Cameron Buie Jul 30 '13 at 00:40
  • I don't understand. Is it somewhat the idea that $x^2 * x^2 = x^4$? – Paul the Pirate Jul 30 '13 at 00:50
  • @PaulthePirate: Note that all of the rest of the (non-zero) terms of the expansion of $\sin(x)$ are of degree strictly higher than $4$, so multiplication by any of the terms of the expansion of $\dfrac1{1-x}$ will again yield a term of degree strictly higher than $4$, which we don't need. Likewise, the rest of the terms of the expansion of $\dfrac1{1-x}$ are of degree strictly higher than $3$, so multiplication by any of the (nonzero) terms of the expansion of $\sin(x)$ will yield a term of degree strictly higher than $4$, which we don't need. – Cameron Buie Jul 30 '13 at 00:55