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SweatingBear
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Continuing from http://www.mathhelpboards.com/f10/taylor-series-x-%3D-1-arctan-x-5056/:
The discussion in that thread gave rise to a general question to me: Does not the point of differentiation change when one makes the substitution \(\displaystyle h = x -a\)? I like Serena affirmed this "conjecture but ZaidAlyafey seemingly disagreed.
Let me illustrate my argument by an example: Suppose we wish to find the Taylor series of \(\displaystyle f(x) := \sqrt{x+2}\) about \(\displaystyle x = 2\). Instead of tediously computing a plethora of derivatives, let us somehow take advantage of any Maclaurin series we know of. Judiciously so, we can attempt to use the binomial series. As a general statement, we can write
\(\displaystyle \Large f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2} }{n!} (x-2)^n \, .\)
Now, we let \(\displaystyle h = x - 2\). This tells us that when \(\displaystyle x\) is \(\displaystyle 2\), \(\displaystyle h\) is \(\displaystyle 0\). Thus derivatives will be taken at \(\displaystyle h = 0\) post-substitution (which is after all what allows us to take advantage of Maclaurin series!). Moreover, we can infer \(\displaystyle h + 4 = x + 2\). So, let us replace all instances of \(\displaystyle x\) with instances of \(\displaystyle h\).
\(\displaystyle \Large f(2 + h) = \sqrt{h+4} = \underbrace{2\sqrt{1 + \frac{h}{4}} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0} }{n!} h^n}_{\text{We can use Maclaurin series!}} \, .\)
Now we see from the sum in the right-hand side that \(\displaystyle 2\sqrt{1 + \frac {h}{4} }\) has a power series about \(\displaystyle h = 0\) i.e. a Maclaurin series! Great, we can now use the binomial series and re-substitute for \(\displaystyle x\), thusly arriving at the Taylor series for \(\displaystyle \sqrt{x+2}\) about \(\displaystyle x=2\). Awesome!
Is there something erroneous in my line of reasoning? I think "changing the point of differenation" from \(\displaystyle x=2\) from \(\displaystyle h=0\) makes perfect sense but maybe there is something I am not seeing.
Addition: I suspect there is some kind of mathematical error when I go from \(\displaystyle \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2}\) to \(\displaystyle \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0}\) (i.e. taking the liberty to shift the functions dependence from \(\displaystyle x\) to \(\displaystyle h\)), but I am not able to pinpoint the error. Perhaps because \(\displaystyle f\) depends on \(\displaystyle x\) and not on \(\displaystyle h\)? Or maybe it depends on both? Hm, much appreciated if somebody helps me see things clearer!
The discussion in that thread gave rise to a general question to me: Does not the point of differentiation change when one makes the substitution \(\displaystyle h = x -a\)? I like Serena affirmed this "conjecture but ZaidAlyafey seemingly disagreed.
Let me illustrate my argument by an example: Suppose we wish to find the Taylor series of \(\displaystyle f(x) := \sqrt{x+2}\) about \(\displaystyle x = 2\). Instead of tediously computing a plethora of derivatives, let us somehow take advantage of any Maclaurin series we know of. Judiciously so, we can attempt to use the binomial series. As a general statement, we can write
\(\displaystyle \Large f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2} }{n!} (x-2)^n \, .\)
Now, we let \(\displaystyle h = x - 2\). This tells us that when \(\displaystyle x\) is \(\displaystyle 2\), \(\displaystyle h\) is \(\displaystyle 0\). Thus derivatives will be taken at \(\displaystyle h = 0\) post-substitution (which is after all what allows us to take advantage of Maclaurin series!). Moreover, we can infer \(\displaystyle h + 4 = x + 2\). So, let us replace all instances of \(\displaystyle x\) with instances of \(\displaystyle h\).
\(\displaystyle \Large f(2 + h) = \sqrt{h+4} = \underbrace{2\sqrt{1 + \frac{h}{4}} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0} }{n!} h^n}_{\text{We can use Maclaurin series!}} \, .\)
Now we see from the sum in the right-hand side that \(\displaystyle 2\sqrt{1 + \frac {h}{4} }\) has a power series about \(\displaystyle h = 0\) i.e. a Maclaurin series! Great, we can now use the binomial series and re-substitute for \(\displaystyle x\), thusly arriving at the Taylor series for \(\displaystyle \sqrt{x+2}\) about \(\displaystyle x=2\). Awesome!
Is there something erroneous in my line of reasoning? I think "changing the point of differenation" from \(\displaystyle x=2\) from \(\displaystyle h=0\) makes perfect sense but maybe there is something I am not seeing.
Addition: I suspect there is some kind of mathematical error when I go from \(\displaystyle \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2}\) to \(\displaystyle \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0}\) (i.e. taking the liberty to shift the functions dependence from \(\displaystyle x\) to \(\displaystyle h\)), but I am not able to pinpoint the error. Perhaps because \(\displaystyle f\) depends on \(\displaystyle x\) and not on \(\displaystyle h\)? Or maybe it depends on both? Hm, much appreciated if somebody helps me see things clearer!
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