# Simple Substitution Taylor Polynomials

• friendbobbiny
In summary, the basic form of a function's Maclaurin series allows for finding complex forms of the same series. For example, the first three terms for arctan(x) are x-x^3/3 + x^5/5, which can be applied to arctan(x^2+1) by substituting x with (x^2+1). However, this method does not always work, as seen in the example of arctan(1+x^2). The general case for this method involves expanding f(g(x)) about x=0 and evaluating the nth derivative of f(g(x)). It is a labor intensive process and
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

Welcome to PF;
friendbobbiny said:
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.

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

friendbobbiny said:
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
$$\arctan(x^2 + 1) = \sum_{n=0}^\infty \frac{(-1)^{n} (x^2 + 1)^{2n+1}}{(2n + 1)!} = \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n + 1)!} \sum_{k=0}^{2n+1} \frac{(2n+1)!}{(2n + 1 - k)!k!}x^{2k} \\ = \sum_{n=0}^\infty \sum_{k=0}^{2n+1} \frac{(-1)^n}{(2n + 1 - k)!k!}x^{2k}$$
so that the first three terms in the series for $\arctan(1 + x^2)$ are
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n + 1)!} + x^2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} + x^4 \sum_{n=1}^\infty \frac{(-1)^n}{(2n - 1)!2!}.$$

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.

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.

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.

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.

## 1. What is a Simple Substitution Taylor Polynomial?

A Simple Substitution Taylor Polynomial is a mathematical expression that approximates a function using a series of polynomials. It involves substituting the value of a variable into the polynomial and calculating the resulting polynomial value. This method is commonly used in calculus and numerical analysis to simplify complex functions and make them easier to work with.

## 2. How is a Simple Substitution Taylor Polynomial different from a regular Taylor Polynomial?

The main difference between a Simple Substitution Taylor Polynomial and a regular Taylor Polynomial is that in the former, the value of the variable is substituted into the polynomial before any calculations are done, while in the latter, the calculations are done first and then the value of the variable is substituted. This makes the Simple Substitution method faster and more efficient for certain types of functions.

## 3. What are the benefits of using Simple Substitution Taylor Polynomials?

Using Simple Substitution Taylor Polynomials can make complex functions easier to work with and understand. It can also help in solving problems that involve infinite series, as it allows for a simpler representation of the function. Additionally, this method can improve the accuracy of approximations and make calculations more efficient.

## 4. What are some common applications of Simple Substitution Taylor Polynomials?

Simple Substitution Taylor Polynomials are commonly used in physics, engineering, and other fields of science to approximate and simplify functions. They are also used in numerical analysis to solve problems involving differential equations, optimization, and interpolation. In addition, they play a crucial role in calculus and are used to find derivatives and integrals of functions.

## 5. What are some limitations of Simple Substitution Taylor Polynomials?

While Simple Substitution Taylor Polynomials can be very useful in certain situations, they also have some limitations. For example, this method may not work well for functions with a lot of discontinuities or sharp changes in slope. In addition, the accuracy of the approximation may decrease as the number of terms in the polynomial increases. It is important to use caution and consider the limitations when using this method for calculations.

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