Integration of Taylor Series with Higher Derivatives

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SUMMARY

The discussion focuses on the challenges of finding the nth derivative of complex functions, specifically x^x, and the integration of Taylor series. Users express frustration with the cumbersome nature of deriving higher-order derivatives, particularly for x^x, which involves increasingly complex expressions. The conversation also touches on the conceptual hierarchy of mathematical operations, questioning whether x^x represents a higher form of exponentiation. Additionally, the integration of a Taylor polynomial of a Taylor polynomial is deemed difficult, often leading to non-elementary integrals.

PREREQUISITES
  • Understanding of Taylor series and their properties
  • Familiarity with differentiation techniques, including higher derivatives
  • Knowledge of implicit differentiation
  • Basic concepts of exponential functions and their properties
NEXT STEPS
  • Research the formula for the nth derivative of x^x using advanced calculus techniques
  • Explore methods for integrating Taylor series, particularly nested series
  • Study the concept of repeated exponentiation and its implications in mathematics
  • Learn about non-elementary integrals and conditions under which they arise
USEFUL FOR

Mathematicians, calculus students, and educators seeking to deepen their understanding of higher derivatives and Taylor series integration techniques.

okkvlt
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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?
 
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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.
 
There isn't anything special about x^x, when you see that x^x = e^(x ln x).
 

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