Higher order derivatives with help of Taylor expansion?

[nikolafmf]

Homework Statement

Function f(x) = x^2/(x-1) should be expanded by Taylor method around point x=2 and 17th order derivative at that point should be calculated.

Homework Equations

Taylor formula: f(x)=f(x₀)+f'(x₀)*(x-x₀)+f''(x₀)*(x-x₀)^2+...

The Attempt at a Solution

I expand this function by the given formula. However, the problem is that it is tedious work to get 17th order derivative by simply differentiating the function 17 times! I can't imagine any other method. Any suggestion, please?

 

[maajdl]

Develop 1/(1-x) to sufficient number of terms.
This is very easy, as this is a simple geometric series.
Finding the 17th derivative will then be a piece of cake.

 

[nikolafmf]

Develop 1/(1-x) to sufficient number of terms.
This is very easy, as this is a simple geometric series.
Finding the 17th derivative will then be a piece of cake.

But my function is x²/(x-1) and I need the derivative of that function.

 

[maajdl]

Develop 1/(1-x) to sufficient number of terms and then multiply by x².

 

[maajdl]

Sorry, I did not pay attention that you were asked to develop f = x²*1/(x-1) around x=2.
The principle remains the same, however.
You just need make a change of variable: y=2-x.
Then, you need to develop f = (2-y)²*1/(1-y) around y=0.
By developing 1/(1-y) in Taylor series around y=0 you will find the result easily.

You could also start as follows:

x²/(x-1) = (2-y)²/(1-y) = (1+y + 1)²/(1-y) = (1+y) + 2 + 1/(1-y)

and develop 1/(1-y) to get the final result even more easily.

 

[nikolafmf]

Now I know how to calculate particular member in the Taylor expansion. But, the question here is, how to find the form of a general member. The answer of this problem says that the n-th member in Taylor expansion is (-1)^n*(x-2)^n, for n=2 or larger. How can I get this? I have calculated up to the third derivative and my calculation agrees with the given answer, but unfortunately I have no time to calculate up to infinity, nor up to 17th derivative :(.

 

[Ray Vickson]

Sorry, I did not pay attention that you were asked to develop f = x²*1/(x-1) around x=2.
The principle remains the same, however.
You just need make a change of variable: y=2-x.
Then, you need to develop f = (2-y)²*1/(1-y) around y=0.
By developing 1/(1-y) in Taylor series around y=0 you will find the result easily.

You could also start as follows:

x²/(x-1) = (2-y)²/(1-y) = (1+y + 1)²/(1-y) = (1+y) + 2 + 1/(1-y)

and develop 1/(1-y) to get the final result even more easily.

I guess that in principle we need to know/prove that if an infinite series ##\sum_n c_n y^n## converges in an interval I containing y = 0 in its interior, then that series is, indeed, the Maclauren series expansion of the sum. It might be fun to try to prove this, or to find a counterexample.

 

[nikolafmf]

Then, you need to develop f = (2-y)²*1/(1-y) around y=0.
By developing 1/(1-y) in Taylor series around y=0 you will find the result easily.

Now the question is to develop f = (2-y)²*1/(1-y). How could developing just 1/(1-y) lead mi to the answer for (2-y)²*1/(1-y)? I can't see...

 

[Ray Vickson]

Now the question is to develop f = (2-y)²*1/(1-y). How could developing just 1/(1-y) lead mi to the answer for (2-y)²*1/(1-y)? I can't see...

Expand the numerator, then do the multiplication and see what you get.

 

[Office_Shredder]

Now the question is to develop f = (2-y)²*1/(1-y). How could developing just 1/(1-y) lead mi to the answer for (2-y)²*1/(1-y)? I can't see...

Can you do 1/(1-y) and write down the general term? Once you have that we can work with finishing the question, but it's very hard to describe how to continue the problem without that part done.

As a very basic example, can you write down the Taylor series for e^x? Once you have that, how do you write down the Taylor series for xe^x? It shouldn't involve taking any derivatives at all.

 

[pasmith]

I guess that in principle we need to know/prove that if an infinite series ##\sum_n c_n y^n## converges in an interval I containing y = 0 in its interior, then that series is, indeed, the Maclauren series expansion of the sum. It might be fun to try to prove this, or to find a counterexample.

The key propositions are

• a convergent power series converges uniformly on its interval of convergence, and
• if [itex](f_n)[/itex] is a sequence of differentiable functions converging pointwise to [itex]f[/itex], and if the derivatives [itex]f'_n[/itex] converge uniformly to [itex]g[/itex], then [itex]f[/itex] is differentiable and its derivative is [itex]g[/itex].

From these one can prove that if [itex]f_N(y) = \displaystyle\sum_{n=0}^N c_ny^n \to f[/itex] is such that there exists an open interval [itex]I[/itex] containing zero on which [itex]f_N^{(k)}[/itex] converges for all [itex]k \in \mathbb{N}[/itex], then for all [itex]k \in \mathbb{N}[/itex],
[tex]
f^{(k)}(0) = c_k k!.
[/tex]

It may be possible to show that [itex](f'_N)[/itex] converges wherever [itex](f_N)[/itex] does, which would show that a convergent power series is always its Maclaurin series.