- #1
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 $$h = x -a$$? 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 $$f(x) := \sqrt{x+2}$$ about $$x = 2$$. 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
$$\Large f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2} }{n!} (x-2)^n \, .$$
Now, we let $$h = x - 2$$. This tells us that when $$x$$ is $$2$$, $$h$$ is $$0$$. Thus derivatives will be taken at $$h = 0$$ post-substitution (which is after all what allows us to take advantage of Maclaurin series!). Moreover, we can infer $$h + 4 = x + 2$$. So, let us replace all instances of $$x$$ with instances of $$h$$.
$$\Large f(2 + h) = \sqrt{h+4} = \underbrace{2\sqrt{1 + \frac{h}{4}} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0} }{n!} h^n}_{\text{We can use Maclaurin series!}} \, .$$
Now we see from the sum in the right-hand side that $$2\sqrt{1 + \frac {h}{4} }$$ has a power series about $$h = 0$$ i.e. a Maclaurin series! Great, we can now use the binomial series and re-substitute for $$x$$, thusly arriving at the Taylor series for $$\sqrt{x+2}$$ about $$x=2$$. Awesome!
Is there something erroneous in my line of reasoning? I think "changing the point of differenation" from $$x=2$$ from $$h=0$$ 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 $$\left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2}$$ to $$\left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0}$$ (i.e. taking the liberty to shift the functions dependence from $$x$$ to $$h$$), but I am not able to pinpoint the error. Perhaps because $$f$$ depends on $$x$$ and not on $$h$$? 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 $$h = x -a$$? 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 $$f(x) := \sqrt{x+2}$$ about $$x = 2$$. 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
$$\Large f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2} }{n!} (x-2)^n \, .$$
Now, we let $$h = x - 2$$. This tells us that when $$x$$ is $$2$$, $$h$$ is $$0$$. Thus derivatives will be taken at $$h = 0$$ post-substitution (which is after all what allows us to take advantage of Maclaurin series!). Moreover, we can infer $$h + 4 = x + 2$$. So, let us replace all instances of $$x$$ with instances of $$h$$.
$$\Large f(2 + h) = \sqrt{h+4} = \underbrace{2\sqrt{1 + \frac{h}{4}} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0} }{n!} h^n}_{\text{We can use Maclaurin series!}} \, .$$
Now we see from the sum in the right-hand side that $$2\sqrt{1 + \frac {h}{4} }$$ has a power series about $$h = 0$$ i.e. a Maclaurin series! Great, we can now use the binomial series and re-substitute for $$x$$, thusly arriving at the Taylor series for $$\sqrt{x+2}$$ about $$x=2$$. Awesome!
Is there something erroneous in my line of reasoning? I think "changing the point of differenation" from $$x=2$$ from $$h=0$$ 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 $$\left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2}$$ to $$\left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0}$$ (i.e. taking the liberty to shift the functions dependence from $$x$$ to $$h$$), but I am not able to pinpoint the error. Perhaps because $$f$$ depends on $$x$$ and not on $$h$$? Or maybe it depends on both? Hm, much appreciated if somebody helps me see things clearer!
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