Taylor series and Taylor's inequality

[rotund wagon]

So I have the Maclaurin series $$T_5 (x)=\sqrt{x+1}\approx1+\frac1{2}x-\frac1{8}x^2+\frac1{16}x^3-\frac{5}{128}x^4+\frac{7}{256}x^5$$ where $$f^{(6)}(x)=-\frac{945}{64(x+1)^{\frac{11}{2}}}.$$ I'm trying to find the error between $x\in[0,0.5]$ and $x\in[-0.5,0]$. \ So for $x\in[0,0,5]$ it's an alternating series so I used the error estimate for that, $$\abs{s-s_n}\le a_{n+1}=\abs{-\frac{21}{1024}(0.5)^6}\approx 3.2\times 10^{-4}.$$ While for $x\in[-0.5,0]$ you'd use Taylor's inequality, so, $$\abs{E_5 (x)}\le \abs{\frac{f^{n+1}(z)}{(n+1)!}x^{n+1}}=\abs{\frac{f^6(-0.5)}{6!}(-0.5)^6}\approx 0.015$$

[slim rivet]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[lyric trellis]

Anonymous

[rotund wagon]

Does my working look right here?

[vagrant socket]

yeah pretty much

[rotund wagon]

Cool ty!

[tacit sentinel]

@rotund wagon has given 1 rep to @vagrant socket