Introduction
This is a relatively quick post as I just wanted to share with you an interesting problem that I heard about and attempted to solve. I hope that you find it equally interesting as I did, and learn a thing or two in the process.
Y = x^x
Before attempting to solve the full problem, it is useful to consider how to differentiate the function, f(x) = x^x. This is not a typical question that you may have encountered as the variable, x, is in the exponent. Sketching the graph here would be very useful to begin with. At x=0, the graph is undefined as 0^0 is undefined. However, if x is infinitesimally small, the value of f(x) will essentially be 1 (it will tend to 1). As x increases in the positive direction (still considering small values), f(x) begins to decrease, until it reaches what seems like a minimum value and then begins to increase again, at an incredibly quick rate. We can get the rough shape of the curve by choosing to plot some points. Keep in mind, we are only focusing on when x≥0 as the graph is very strange when x becomes negative - more on this later.
x = 0.001; f(x) = 0.993
x = 0.010; f(x) = 0.955
x = 0.050; f(x) = 0.861
x = 0.100; f(x) = 0.794
x = 0.250; f(x) = 0.707
x = 0.400; f(x) = 0.693
x = 0.500; f(x) = 0.707
x = 0.750; f(x) = 0.806
x = 1; f(x) = 1.000
x = 2; f(x) = 4.000
x = 3; f(x) = 27.000
x = 5; f(x) = 3125.000
x = 10; f(x) = 10000000000
By plugging in a few points, we can get a general 'feeling' what the graph looks like.
I have included a rough, hand-drawn sketch of the function below (where x≥0):
It is also worth putting somewhere that there is a vertical asymptote at x = 0; however, I am not too worried about this for the moment as the sketch is here only to help us visualise the problem at hand.
You may be asking 'what about when x is negative?'. The issue with when x < 0 is that the graph is not continuous so it is practically impossible to sketch. In fact, any graphing software out there would simply not graph any points for when x < 0, even if the graph is theoretically defined at certain points.
Let me give you an example. Let's suppose that x = -1/2. If we substitute in this value for x into our equation, it will look like the following:
We know that we can simplify this expression as a negative in the exponent simply means that we have to take the reciprocal of the expression. This leaves us with this:
Finally, we have a 1/2 to deal with in the exponent. This simply means that we must take a square root of the entire expression. However, this will result us in taking the square root of a negative number. There is no real solution for this and we aren't sketching a complex graph with an imaginary and real axis; therefore, we simply restrict the domain to x > 0. Some negative values for x do come out with a real answer, such as x = -3; however, as aforementioned, the graph is not continuous when x < 0 and thus we will simply not sketch it - it is of no use to us.
We are now aware of the general shape for y = x^x and can thus begin to differentiate it.
The key to finding dy/dx for y = x^x is the use of the natural logarithm - so a logarithm with base e. We can take the natural log of both sides, allowing us to rewrite the expression like so.
We can further simplify the right-hand side by using one of our logarithm rules; we can bring the exponent x, within the natural logarithm, to the front like so:
From here, we can differentiate both sides with respect to x. Since the left-hand side is a function of y rather than simply y itself, we will need to perform implicit differentiation. Effectively, you just differentiate both sides like per normal; however, you multiply by dy/dx, whenever you differentiate a function of y with respect to x - this is a consequence of the chain rule of differentiation.
So, if we focus our attention to differentiating the left-hand side of the equation, we can do so quite easily. We know that the standard derivative of ln(y) is simply 1/y - this can be proved; however, it is absolutely essential for one to just remember it as a standard derivative to save time in an exam. As aforementioned, we must also multiply the derivative (1/y) by dy/dx as we are differentiating implicitly - and not explicitly, like we 'normally' do.
Turning our attention to the right-hand side of the equation, we can differentiate the right-hand side using the product rule. We know the derivative of x is simply 1 and that the derivative of ln(x) is simply 1/x. Combining all this together, we can write our equation like so:
We aren't yet finished - almost there - as we have to rearrange the equation to get dy/dx as the subject. Firstly, we can eliminate the 1/y by multiplying both sides by y. Moreover, we can substitute in x^x for y in order to eliminate the y variable that has now 'appeared' on the right-hand side of the equation. Et voila! We have figured out what dy/dx is of x^x!
Now we can apply this same approach to the problem that the title of this post focuses on.
Y = x^x^x^x^x^x... to infinity
This question is more challenging as it involves an infinite series; however, we implement a similar principle. Our equation firstly looks like this:
However, since x is raised to the power of x to the power of x again, an infinite number of times, we can simply substitute in y as the exponent. This is due to the fact that y is already equal to x^x^x^x^x to infinity. You may be objecting to this as you believe that the original value of y contains the base x; however, remember, we are dealing with infinities. Infinity minus one is still infinity as infinity is a concept rather than a number. This therefore allows us to write the equation like so:
Like before, we take the natural logarithm of both sides. A lot of people make the mistake of taking log base 10 of both sides; however, log(x) does not have a nice, standard derivative! Moreover, as before, we can bring the exponent y to the front of the logarithm like so:
Here, we apply implicit differentiation like we did when tackled dy/dx of x^x. Since, we already know how to differentiate y, x and ln(x), we can easily figure out the respective derivatives. I have included an image below doing so.
From here, we just have to rearrange for dy/dx. So, we group any terms of dy/dx together, using standard algebraic manipulation, and make dy/dx the subject.
That wasn't so difficult now, was it? Unlike before, I have left y in the final answer as it is much neater than substituting in an infinite power series. By focusing on the 'elementary' case of x^x, it made differentiating x^x^x^x...∞ a lot easier as we literally did the exact same thing. The only difference was noticing that we can replace the infinite exponent series of x with just y.
I hope that you enjoyed this post - I certainly enjoyed answering the question at hand. It is a great question to wrap your head around as it utilising a lot of maths principles that are commonly taught in A-Level Maths (especially in the A2 side of things). Moreover, it could easily come up in an admissions interview for any maths-related subject - so pretty much any STEM subject out there.
Please let me know in the comments if you have any queries regarding the question. Also, don't forget to sign up to OnlyPhysics - it comes with loads of benefits, like free tutoring. Thank you for reading!
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