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Taylor series: Changing point of differentiation

sweatingbear

Member
May 3, 2013
91
Continuing from http://www.mathhelpboards.com/f10/taylor-series-x-=-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!
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
So basically what you are saying that the coefficients will change after the substitution you made ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
The expansion of \(\displaystyle \sqrt{x+4}\) around \(\displaystyle x =0 \) is according to W|A here while for the function \(\displaystyle \sqrt{x+2}\) around \(\displaystyle x=2\) is here . You can see that the coefficients of the terms are the same so the substitution will not change the composition of the function .
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
A Taylor expansion of the function $f$ at $x=0$ is:
$$f(x)=\sum \frac{f^{(n)}(0)}{n!}x^n \qquad\qquad (1)$$


If you substitute $x=a+h$ in (1), you get:
$$f(a+h)=\sum \frac{f^{(n)}(0)}{n!}(a+h)^n \qquad (2)$$

And if you substitute $f(x)=g(a+x)$ in (1), you get:
$$g(a+x)=\sum \frac{g^{(n)}(a+0)}{n!}x^n \qquad (3)$$

In case (2) your derivatives of f are still at $0$.
In case (3) your derivatives of g are at $a$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
By the way you cannot differentiate with respect to \(\displaystyle x\) then with respect to \(\displaystyle h\) because the function \(\displaystyle f\) is just of one variable if you are referring to the new function after the composition then use another name .
 

sweatingbear

Member
May 3, 2013
91
Ok, so if the derivatives are not taken at \(\displaystyle h=0\) post-substitution, how are we then able to take advantage of known Maclaurin series? I thought the whole point of Maclaurin series (i.e. expansions about an argument equal to zero) were to express functions as a polynomial where the function's derivatives (at the argument equal to zero) are constituents in the coefficients?
 

sweatingbear

Member
May 3, 2013
91
Let my try again. I have taken into consideration the fact that when you make the substiution you effectively end up with a completely new function which is treated differently. I will try to apply that in my arguments; let us see what you guys think:

We have \(\displaystyle f(x) = \sqrt{x+2}\) and wish to expand it about \(\displaystyle x=2\). We can therefore write

\(\displaystyle f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac {f^{(n)}(2)}{n!} (x-2)^n \, .\)

Let us make the substitution \(\displaystyle h = x -2\). We then end up with this new function \(\displaystyle \sqrt{h+4}\) that depends on \(\displaystyle h\) instead of \(\displaystyle x\); let us call this new function \(\displaystyle g(h)\).

The equivalent task is now to find an expansion of \(\displaystyle g(h)\) about \(\displaystyle h=0\) from which we can find the expansion of \(\displaystyle f(x)\) about \(\displaystyle x=2\) when substituting back all instances of \(\displaystyle h\) with instances of \(\displaystyle x\) according to \(\displaystyle h = x-2\).

Since taking derivatives of \(\displaystyle g(h)\) at \(\displaystyle h=0\) is equivalent to taking derivatives of \(\displaystyle f(x)\) at \(\displaystyle x=2\), we can conclude that the coefficients will be the same in either expansion (this is what I was missing earlier, right?).

So, the expansion of \(\displaystyle g(h) = \sqrt{h + 4}\) can be written as

\(\displaystyle g(h) = \sqrt{h + 4} = \sum_{n=0}^\infty \frac {g^{(n)}(0)}{n!} h^n \, ,\)

And now we finally see that this is a Maclaurin series (expansion about an argument at zero; derivatives are taken at an argument at zero) and, due to the nature of the function, we can appropriately use the binomial series!

What do you people think? Did I just have an Heureka-moment?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Yes , excellent :D.
 

sweatingbear

Member
May 3, 2013
91
Great! Thanks to both of you for your patience and help, very much appreciated.