How can the Taylor expansion of x^x at x=1 be simplified to make solving easier?

In summary, the conversation discusses finding the Taylor expansion of x^x around x=1, with the individual attempting to solve it through brute force before finding a more efficient method using the fact that e^x is its own derivative. They still inquire about any potentially easier methods for solving the problem.
  • #1
nezahualcoyot
5
1

Homework Statement


Find the Taylor expansion up to four order of x^x around x=1.

Homework Equations

The Attempt at a Solution


I first tried doing this by brute force (evaluating f(1), f'(1), f''(1), etc.), but this become too cumbersome after the first derivative. I then tried writing: $$x^x = e^{x \ln(x)}$$

And found the Taylor expansion of x*ln(x) (which I can do), and the "plug" that into the Taylor expansion of e^x, and carefully only keep the terms up to four order. I checked the final result with Wolfram Alpha and I got it correct, but this procedure took me way too long (specially the last step) and feels way harder than the rest of my course.

My question is, is there an alternative / easier way of solving this problem?
 
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  • #2
You seemed to have found an easy way to do it without brute force. In my opinion, you've used quite an efficient way using the fact that ##e^x## is its own derivative.

But await on some more answers from blokes with experience in Calculus. I've only just started learning...
 
  • #3
The way you solved it seems pretty straightforward. Of course, a lot of it depends on how you did the algebra. You can end up doing a lot of unnecessary work.
 

Related to How can the Taylor expansion of x^x at x=1 be simplified to make solving easier?

1.

What is a Taylor series?

A Taylor series is a representation of a mathematical function as an infinite sum of terms. It is named after the mathematician Brook Taylor, who first described this method of approximation in the early 18th century.

2.

How is the Taylor series of x^x at x=1 calculated?

The Taylor series of x^x at x=1 can be calculated using the general formula for a Taylor series, which is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ..., where f(a) is the value of the function at the point a and f'(a), f''(a), etc. are the derivatives of the function evaluated at the point a. In this case, a=1 and the derivatives of x^x can be found using the power rule.

3.

What is the significance of the Taylor series of x^x at x=1?

The Taylor series of x^x at x=1 can be used to approximate the value of the function x^x at values close to x=1. This can be useful in situations where it is difficult to calculate the exact value of the function, such as for irrational or complex numbers.

4.

What is the convergence of the Taylor series of x^x at x=1?

The Taylor series of x^x at x=1 has a radius of convergence of 1, which means that it will only converge for values of x that are within 1 unit of the center point, x=1. This means that the series will converge for values of x between 0 and 2.

5.

Can the Taylor series of x^x at x=1 be used to find the exact value of x^x?

No, the Taylor series of x^x at x=1 is only an approximation of the function and will never give an exact value. As more terms are added to the series, the approximation will become more accurate, but it will never be exact. The exact value of x^x can only be found using methods such as calculator, computer, or numerical methods.

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