Simple Substitution Taylor Polynomials

[friendbobbiny]

I've been taught that with the basic form of a function's maclaurin series, complex forms of the same series can be found. For example, the first three terms for arctan(x) are x-x^3/3 + x^5/5, meaning the first three terms for arctan(x^2+1) at a=0 should be (x^2+1) - ((x^2+1)^3)/3 + ((x^2+1)^5/5). If this is the case, why is it?

I can understand why the taylor series for e^2x and other, repetitive functions follow this simple substitution.

But for those functions whose derivatives require repeated application of the chain rule, why does " simple substitution" work.

Thanks

 

[Simon Bridge]

Welcome to PF;

I've been taught that with the basic form of a function's maclaurin series, complex forms of the same series can be found.

For example, the first three terms for arctan(x) are x-x^3/3 + x^5/5, meaning the first three terms for arctan(x^2+1) at a=0 should be (x^2+1) - ((x^2+1)^3)/3 + ((x^2+1)^5/5).

Just making sure I understand you:

By "complex" you mean "more complicated"?
(the word "complex" has a special meaning in maths.)

In particular, you mean that the Maclaurin series expansion of f(g(x)) can be just the expansion for f(x) with g(x) substituted for every x?

If this is the case, why is it?

... for the specific example where f(x)=arctan(x), you sound like you don't know that this is actually the case - have you tried calculating it from scratch?

I can understand why the taylor series for e^2x and other, repetitive functions follow this simple substitution.

But for those functions whose derivatives require repeated application of the chain rule, why does " simple substitution" work.

... it sounds like you have yet to show that it does work.
Start there - have a go working out the general case of f(g(x)) and see where you get to.

Notice:
http://www.sosmath.com/calculus/tayser/tayser02/tayser02.html
... the substitution approach is pretty general, but does it work for all substitutions?
i.e. if you want the series for f(x)=1/(1+x^2) can you just use the series for f(q)=1/q?

Working these out should help you understand what is going on.

 

[friendbobbiny]

Hi,
Thanks for the help. Pardon the formatting below. I tried the following:

To see whether substitution of u=g(x) for the taylor polynomial f(g(x) worked, I compared the constant term of the taylor polynomial by expansion with the constat term of the taylor polynomial by substitution. I could not see how else to compare the two polynomials, given what I know.

the constant term by expansion is +/- (pi/4). arctan(1+(0)^2) = ...

Is this equal to the constat term for ∑((1+x^2)^(2n+1)*(-1)^(n+1))/(2n+1)! from n=0 to infiniti ?

the constant term for the above polynomial is given by:

∑ (-1)^(n+1)/(n+2) from n=1 to infiniti. This is because expansion of the above polynomial gives (for constant terms): 1-1/3+1/5...

But this sum is not equal to pi/4. So I'm not sure what I did wrong here.

how should I set up the general case?

many thanks

 

[pasmith]

I've been taught that with the basic form of a function's maclaurin series, complex forms of the same series can be found. For example, the first three terms for arctan(x) are x-x^3/3 + x^5/5, meaning the first three terms for arctan(x^2+1) at a=0 should be (x^2+1) - ((x^2+1)^3)/3 + ((x^2+1)^5/5). If this is the case, why is it?

That's not correct; the presence of the +1 means this method doesn't work.
$$\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^{n} x^{2n+1}}{(2n + 1)!}$$
so
$$\arctan(x^2 + 1) = \sum_{n=0}^\infty \frac{(-1)^{n} (x^2 + 1)^{2n+1}}{(2n + 1)!}<br /> = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n + 1)!} <br /> \sum_{k=0}^{2n+1} \frac{(2n+1)!}{(2n + 1 - k)!k!}x^{2k} \\<br /> = \sum_{n=0}^\infty \sum_{k=0}^{2n+1} \frac{(-1)^n}{(2n + 1 - k)!k!}x^{2k}$$
so that the first three terms in the series for $\arctan(1 + x^2)$ are
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n + 1)!} <br /> + x^2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}<br /> + x^4 \sum_{n=1}^\infty \frac{(-1)^n}{(2n - 1)!2!}.$$

Working out the Taylor series for $f(x) = \arctan(1 + x) = \sum_{n=0}^\infty a_nx^n$ and then calculating $f(x^2) = \sum_{n=0}^\infty a_nx^{2n}$ is the better method here.

 

[Simon Bridge]

What pasmith has done is part of what I was suggesting you do ...

Your "1st 3 terms" expansion:
(x^2+1) - ((x^2+1)^3)/3 + ((x^2+1)^5/5)

going off the actual calculation (post #4):
n=0 gives 1+x^2 ... same as your zero term
n=1 gives -1/6 - (1/2)x^2 - (1/2)x^4 ... which is way off yours, but have you calculated all the coefficients of x^2 and x^4 yet?

n=3 requires more terms in both expansions, so we cannot compare them.

What this shows you though is that the substitution approach is not a term-by-term equivalence
... the first three "terms" in the substitution method are not equal to the first three Taylor series terms.

... it is the overall summation that is supposed to be equal - so to do the term-by-term comparison I was suggesting needs more substitution terms to be calculated - so they can be compared with the taylor series terms.

You asked about the general case:

In general: f(g(x)) expanded about x=0:$$f\circ g = \sum_{n=0} \frac{x^n}{n!}\frac{d^n}{dx^n}[f\circ g](x=0)$$
So you need to be able to evaluate the nth derivative:
$$(f\circ g)'=g'[f'\circ g]\\ (f\circ g)'' = g'' [f'\circ g]+(g')^2[f''\circ g]\\ \cdots$$... etc. Look for a pattern.

Then try it for $$f(g)=\sum_{n=0} \frac{g^n}{n!}\frac{d^n}{dg^n}f(g=0)$$... and compare.

Note: your examples have g(x) as a polynomial ... are the above expansions special if g is a polynomial? What about other functions?

(Aside: you can always make a polynomial expansion of any arbitrary g...)

All this is a lot of work to get through - but it is labor intensive rather than difficult.
Just take it slowly and take care, you may find some sort of restrictions for how the substituton approach works.

 

[HomogenousCow]

For finite polynomial g it is not much work since the n-th derivatives of g all vanish after some power n.
Not quite sure what the general pattern is for infinite polynomials though.