I was given a function and told to use the geometric series to find the Taylor series. I was then asked where the series converges. What does this mean? How do I go about solving a problem like this?
The function I was given was:
$$\frac{1}{2-x}$$ However, in general, how should I solve such a problem?

Also, does the Taylor series only help for inputs near zero?
Is that what convergence means - that the Taylor Series works for those numbers? If yes, then does the sine function converge for all real numbers?

  • 2
    $\begingroup$ How do you want us to answer if you don't tell us which function you are talking about? Concerning your final question, yes, it converges for all real numbers. $\endgroup$ – José Carlos Santos Aug 28 '19 at 21:03
  • $\begingroup$ I put in the function $\endgroup$ – Burt Aug 28 '19 at 21:20

As examples which might help :

  • $1+x+x^2+x^3+\cdots$ only converges for $-1 \lt x \lt 1$ and then (as a geometric series) is equal to $\frac1{1-x}$ and is its Taylor series around $x=0$

  • $1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ converges for all real $x$ and is equal to $e^x$ and is its Taylor series around $x=0$

Your case of $\sin(x) = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ is related in many ways to the second example and also converges for all real $x$

Added: Your later addition of $\frac{1}{2-x}$ which you might expand to $\frac12 +\frac{x}{2^2} +\frac{x^2}{2^3} +\cdots$ is similar to the first example and only converges when $-2 \lt x \lt 2$

  • $\begingroup$ Since sin(x) and $e^x$ both converge for all real x, that means that the Taylor series is a good approximation for all real x, right? $\endgroup$ – Burt Aug 28 '19 at 21:18
  • $\begingroup$ @burt - If $x$ is large than you will need a lot of terms for a good approximation. For example $\sin(10) \approx -0.544$ but you need to add up about sixteen terms in the Taylor series to get this to three decimal places. Meanwhile $\sin(100) \approx -0.506$ but you probably cannot approach this through the Taylor series without much more precision than usual $\endgroup$ – Henry Aug 28 '19 at 21:33
  • $\begingroup$ I see. But technically it would converge for all x. $\endgroup$ – Burt Aug 28 '19 at 22:55
  • $\begingroup$ And, if I use the Taylor series centered around 100 I'll get a much quicker approximation? $\endgroup$ – Burt Aug 28 '19 at 22:56
  • $\begingroup$ @burt: The Taylor expansion of $\sin(x)$ about $x=100$ involves $\sin(100)$ and $\cos(100)$. If you want to find an approximation to $\sin(100)$ then it is perhaps better to do the expansion about $x=32\pi \approx 100.53$ $\endgroup$ – Henry Aug 28 '19 at 23:02

Let's use your example: $\dfrac1{2-x}=\dfrac 12\dfrac 1{(1-\frac x2)}=\dfrac 12\sum_{n\ge0}(\dfrac x2)^n$, and converges when $\mid\dfrac x2\mid\lt1$, or $\mid x\mid\lt2$. This is the Taylor series at $0$.

You can use the same sort of procedure to get the Taylor series at $1$, say: $\dfrac 1{2-x}=\dfrac 1{1-(x-1)}=\sum_{n\ge0}(x-1)^n$, which converges when $\mid x-1\mid\lt1$.

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    $\begingroup$ But how do you know when it converges? $\endgroup$ – Burt Aug 28 '19 at 23:16
  • $\begingroup$ The geometric series has partial sum equal to $\dfrac {1-x^{n+1}}{1-x}$, which converges to $\dfrac 1{1-x}$, precisely when $\mid x\mid\lt 1$. Look at the numerator, and note that $x^{n+1}\to0$. $\endgroup$ – Chris Custer Aug 28 '19 at 23:21

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