# Taylor series: Changing point of differentiation

#### sweatingbear

##### Member
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
So basically what you are saying that the coefficients will change after the substitution you made ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
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
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
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
Yes , excellent .

#### sweatingbear

##### Member
Great! Thanks to both of you for your patience and help, very much appreciated.