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What would be the simplest way to do this?

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    $\begingroup$ Go ahead, write out what you want, compute derivarives, the works. You are supposed to learn something by doing this, you know... $\endgroup$
    – vonbrand
    Apr 7, 2013 at 2:16

3 Answers 3

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First, let $z:=x-16$, then this is the same as $\sqrt{16+z}$ at $z=0$. Use the binomial series: $$(1+x)^\alpha=\sum_{n\ge 0}\binom\alpha n x^n$$ Now we have $$(16+z)^{1/2}=4\cdot\left(1+\frac z{16}\right)^{1/2}=\\ =4\cdot\sum_{n\ge 0} \binom{1/2}n\frac{z^n}{16^n}\,.$$

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  • $\begingroup$ Can you explain why z = x-16 is the same as sqrt(16 + z) at z=0 ? $\endgroup$ Apr 7, 2013 at 2:37
  • $\begingroup$ I mean, $f(x)=\sqrt x=\sqrt{16+z}$ as the function in $z=(x-16)$. $\endgroup$
    – Berci
    Apr 7, 2013 at 12:01
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    $\begingroup$ This is the way I would do it. (+1) If this were on a homework for Taylor series, however, I would do it the more formulaic way (taking derivatives at $a=16$, etc.) $\endgroup$
    – robjohn
    Apr 7, 2013 at 14:50
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The simplest way? Perhaps, Taylor expansion for f(x) = sqrt(x) centered at a=16.

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    $\begingroup$ This may be slightly misleading: this "simplest way", besides not contributing a lot to some newbie's understanding, would hardly be acceptable in any more or less decent college/university, so it may not be the wisest thing to do (for the student, of course). $\endgroup$
    – DonAntonio
    Apr 7, 2013 at 2:17
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    $\begingroup$ @DonAntonio Honestly, the question is so ridiculous, that it's hard to give a serious answer. As a college professor, I'd certainly expect even an average student to be able to do this with modest effort. Also as a college professor, I tell students to use all kinds of tools to understand things and I think that WolframAlpha is a bit under-utilized, as illustrated by the fact that I typed his exact title into WolframAlpha. $\endgroup$ Apr 7, 2013 at 2:35
  • $\begingroup$ Is this answer any worse than the fully worked out solutions to elementary HW questions that we see on this site all the time? This is a serious question. I honestly dont' know how to deal with this issue. $\endgroup$ Apr 7, 2013 at 2:35
  • $\begingroup$ Well @Mark: even full-worked solution leave a little window of hope for the student to understand a little: (s)he has to copy the question and, hopefully, understand it. WA doesn't care, and cannot either, about this. I also think this is a lame question from a college and up student, but... $\endgroup$
    – DonAntonio
    Apr 7, 2013 at 2:38
  • $\begingroup$ @Mark: I don’t downvote, but in my opinion it is indeed much worse: it teaches no mathematics at all, and I can’t see it doing anything to increase anyone’s understanding of Taylor series. Nor do I think it legitimate to justify a shoddy answer by ridiculing the question. Your answer would have been reasonable as a mildly facetious comment. $\endgroup$ Jun 2, 2013 at 11:10
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note that $f(x)=2^0x^{1/2}$, $f'(x)=-2^1 x^{-1/2}$, $f''(x)=\frac{2^2}{3} x^{-3/2}$,$f'''(x)=-\frac{2^3}{15}x^{-5/2}$ $\cdots$

Then just plug these into the normal formula for a taylor series expansion

formla for taylor series

here your $a=16$. Then see what patterns you can find to write these in summation notation!

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  • $\begingroup$ (+1) because this is the way it should be done on homework (which is my guess here). However, it is not what I would call "simpler" $\endgroup$
    – robjohn
    Apr 7, 2013 at 14:53

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