Integration of Taylor Series with Higher Derivatives

In summary: So, by the chain rule, x^x ln x. So, by the product rule, x^x(1 ln x) = x^x + x^x ln x. Which is what we wanted. But, then the question is what is the derivative of ln x.
  • #1
okkvlt
53
0
Questions:

Is there a quicker way to find the formula for the nth derivative of a function, instead of finding the first several derivatives and trying to find a pattern, and using that pattern to form the equation for the nth derivative?

Also, is there a formula for the nth derivative of x^x? I can't find a formula for its derivatives so i can't find the polynomial. By the third derivative it starts to get complicated, and by the 5th i need to use 4 lines on a sheet of notebook paper just to write the derivative. Then i gave up because it got too cumbersome. when x^x is diffferentiated, it gives a x^x(1+lnx), and when that's differentiated, it gives a x^(x-1), then that gives a x^(x-1)*(1+lnx-1/x) Those terms keep producing more and more things, all the while the x^x(1+lnx)^n and x^(x-1)*(1+lnx-1/x)^n keep adding more and more things to the equation until it becomes a monster.

maybe if i tried differentiating logx(y)=x, but i can't figure out how to do that in the first place. I have a feeling that it will be even more complicated, because implicit differentation would be required.

Also I am curious, is x^x a function one step higher than exponentation? Just as exponentation is a step higher than multiplication and multiplication is a step higher than addition.



And, how do i integrate the series of a series? like if the x of a taylor polynomial is another taylor polynomial. If that integration is impossible, then how do i find the taylor polynomial that represents a taylor polynomial of a taylor polynomial?
 
Physics news on Phys.org
  • #2
okkvlt said:
Questions:

Is there a quicker way to find the formula for the nth derivative of a function, instead of finding the first several derivatives and trying to find a pattern, and using that pattern to form the equation for the nth derivative?

Not in any "useful" form. Of course, [tex] \frac{d^n}{dx^n}[/tex] does it quite well.

Also, is there a formula for the nth derivative of x^x?

Same answer.
Also I am curious, is x^x a function one step higher than exponentation? Just as exponentation is a step higher than multiplication and multiplication is a step higher than addition.

Multiplication is *repeated* addition, Exponentiation is repeated *multiplication.* Is this repeated exponentiation .. ? =]

And, how do i integrate the series of a series? like if the x of a taylor polynomial is another taylor polynomial. If that integration is impossible, then how do i find the taylor polynomial that represents a taylor polynomial of a taylor polynomial?

For a general function that would seem quite difficult and would most likely result in integrals with non -elementary derivatives. For some special function that may reduce to something "nice" though.
 
  • #3
There isn't anything special about x^x, when you see that x^x = e^(x ln x).
 

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is a polynomial function of the variable. It is named after the mathematician Brook Taylor who introduced the concept in the 18th century.

2. How is a Taylor series used for integration?

A Taylor series can be used to approximate the value of a function at a specific point. By integrating the Taylor series term by term, we can obtain an approximation of the original function's integral. This is known as Taylor series integration.

3. What are the benefits of using Taylor series integration?

Taylor series integration allows us to approximate the value of a function's integral without needing to know the exact form of the function. This makes it a helpful tool in solving integrals that cannot be solved using traditional methods. Additionally, it provides a more accurate result compared to other numerical integration methods.

4. What are the limitations of Taylor series integration?

Taylor series integration relies on the assumption that the function can be represented as a power series. This means that it may not be applicable to all types of functions. Additionally, the accuracy of the approximation depends on the number of terms used in the series, so a large number of terms may be required for highly complex functions.

5. How is Taylor series integration different from other numerical integration methods?

Traditional numerical integration methods, such as the trapezoidal rule or Simpson's rule, use a finite number of points to approximate the integral. In contrast, Taylor series integration uses an infinite number of points, resulting in a more accurate approximation. However, it may also be more computationally intensive. Additionally, Taylor series integration can only be used for functions that can be represented as a power series, while other methods can be applied to a wider range of functions.

Similar threads

I How to Derive the Taylor Series for log(x)?
    • Calculus
Replies
5
Views
12K
B Why do I not understand what seems to be basic Calculus?
    • Calculus
Replies
11
Views
2K
I Looking for references on this form of a Taylor series
    • Calculus
Replies
3
Views
980
I Taylor series and variable substitutions
    • Calculus
Replies
3
Views
1K
MHB Using Standard Taylor Series to build other Taylor Series
    • Calculus
Replies
2
Views
1K
I What's the point of Taylor/Maclaurin series?
    • Calculus
Replies
17
Views
3K
I Different series than Taylor series for a function
    • Calculus
Replies
9
Views
1K
I How to understand Taylor/Mclaurin series?
    • Calculus
Replies
8
Views
2K
I Spivak, Ch. 20: Understanding a step in the proof of lemma
    • Calculus
Replies
1
Views
1K
I Regarding Error Bound of Taylor Series
    • Calculus
Replies
8
Views
1K

Hot Threads

Recent Insights

Back
Top