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The first part of my question (for my homework) states "Find the Taylor series about $x=0$ of the function $f(x)=\frac{1}{(1-x)^2}$ "...

I have found the taylor series of $\frac{1}{(1-x)^2}$ to be $$\sum_{n=0}^\infty\dfrac{\dfrac{1}{(1-a)^2}^n{(x-a)^n}}{n!}$$ and the Maclaurin series to be $$\sum_{n=0}^\infty(n+1)x^n$$

I have a couple of questions...

  1. Am I correct in starting my bound as $n=0$? I wasn't sure because some examples have $n=1$. Maybe that's just if the function is undefined for $n=0$?
  2. I have found both the taylor series and the maclaurin series as the question asks for the "Taylor series" about $x=0$ which i thought was specifically called a Maclaurin series? But they are different series, so... I am asked to then find the radius and interval of convergence of the answer but I am really confused as to whether I am meant to use the Taylor series version or the Maclaurin series version to do this. Can someone clarify which series out of the ones I found to use for this part?
  3. The last half of the question states "Hence or otherwise determine the sum of the series summed from $n=0$ to infinity of $\frac{(n+1)}{2^n}$... Does that part have anything to do with either of the series I found for the first half? Or is it just a new part of the question and I should apply the same techniques to find the sum of that series as any other and just ignore the first half of the question to complete this part?
User1997
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  • Learn how to format @ https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – For the love of maths Oct 27 '18 at 04:06
  • Okay the editing is fine now, can anyone please help with my questions? I'm really stuck. – User1997 Oct 27 '18 at 04:34
  • Your "Taylor series" adds to $$\frac{\exp(x-a)}{(1-a)^2}$$ so cannot be correct. In any case a Taylor series about $0$ is a Maclaurin series. – Angina Seng Oct 27 '18 at 04:38
  • @LordSharktheUnknown

    How does it add to that?

     – User1997 Oct 27 '18 at 04:44
  • I plugged f(x) into the taylor polynomial equation. – User1997 Oct 27 '18 at 04:45
  • $$\frac{1}{(1-a)^2}+\frac{1}{(1-a)^2}{x-a}+\dfrac{\dfrac{1}{(1-a)^2}{(x-a)^2}}{2!}+\dfrac{\dfrac{1}{(1-a)^2}{(x-a)^3}}{3!}+\dfrac{\dfrac{1}{(1-a)^2}{(x-a)^4}}{4!}+...$$

    So the general taylor series for this would be $$\dfrac{\dfrac{1}{(1-a)^2}{(x-a)^n}}{n!}$$ wouldnt it?

     – User1997 Oct 27 '18 at 04:51
  • When asked to find the value of a specific numerical infinite series, the thing to look for is how it's a special value (at some specific $x$) of your power series. For example, if a series contains $1/2^n = (1/2)^n$ it might be a power series evaluated at $1/2$. Concerning the Maclaurin and Taylor terminology, "Maclaurin" just means "Taylor series at 0". Outside of calculus courses nobody who works in math ever uses the phrase "Maclaurin series". We just talk about the power series at 0. – KCd Oct 27 '18 at 04:52
  • The factorials in the denominator are wrong, as at $a = 0$ you can see it's not the answer that you already know ($\sum_{n \geq 0} (n+1)x^n$). You should get out a big piece of paper and carefully differentiate $1/(1-x)^2$ six times so you see what the pattern really is. Don't stop after differentiating once or twice. – KCd Oct 27 '18 at 04:54
  • The answer i got after differentiating etc is =$$1+2x+3x^2+4x^3+5x^4+...+(n+1)x^n$$ – User1997 Oct 27 '18 at 04:58
  • I can't edit it now in the comments but ive edited it in the main body but the f(a) is meant to be raised to the nth derivative – User1997 Oct 27 '18 at 05:04
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    The Taylor series at $0$ is the Maclaurin series, which you have found. All that stuff you're writing with "general" $a$ is irrelevant (and wrong) and should just be deleted. The power series representation equals the function where the power series converges. This stuff all has real numerical meaning. Look at the graphs of $y = 1/(1-x)^2$ together with the graphs of the partial sums of the Taylor series at $0$: $y = 1$, $y = 1 + 2x$, $y = 1 + 2x + 3x^2$, and so on. Around $x = 0$ they should look quite close to each other. – KCd Oct 27 '18 at 05:26
  • Okay, so for my first question in the Maclaurin series, am I correct in staring at n=0 and not at n=1? You have answered my second question. For my third question, could you explain how i am meant to use the Maclaurin series to find the sum of the series they provided? I can see how they are similar, but im not too sure how to go about this correctly... I dont have much experience with these yet. – User1997 Oct 27 '18 at 05:31
  • Regarding “undefined for $n=0$”: it's not, since $0!=1$ (see here). And regarding your third question, see the most frequently asked question on this site: here. – Hans Lundmark Oct 27 '18 at 06:21

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Your last question: on your second attempt you successfully computed the Maclaurin series as $$\frac1{(1-x)^2}=\sum_{n=0}^\infty(n+1)x^n.\tag{$*$}$$ You are enjoined to calculate $$\sum_{n=0}^\infty\frac{n+1}{2^n}.$$ This looks very similar to the RHS of $(*)$. Can you obtain it from the RHS of $(*)$ by making a substitution?

Angina Seng
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  • Do you mean that i substitute something into $$(*)$$ to make it like the other series? x would have to equal 2 and n would have to be -n – User1997 Oct 27 '18 at 05:17
  • @KaylaMartin Very big hint: $(\frac{1}{a})^n = \frac 1{a^n}$. – Deepak Oct 27 '18 at 05:18
  • Yeah i can see thats what i need to do but i cant really tell how to do that by substition? What am i allowed to do to it? Can i let $$x^n$$ = $$\frac{1}{2^{n}}$$ ? Thats all i can think of. – User1997 Oct 27 '18 at 05:20
  • @KaylaMartin What happens when $x= \frac 12$? – Deepak Oct 28 '18 at 03:16