<jupyter_start><jupyter_text>Its Own SlopeSo we're looking for a function whose value is equal to the size of its slope. One such trivial function would bef(x) = 0But are there any more interesting ones<jupyter_code>import math<jupyter_output><empty_output><jupyter_text>So what would it look like?Start out detective work by cutting put a few triangles...So there are plenty of functions that have a value that match their slope. In fact there are infinite ones.Add diagramSo what else do we knowIf there is a function f(x) that yields its own slope then the function Af(x) would also yield its own slope. Setting A to zero gives the trivial case that is zero everywhere. So let's say f(x) is the function that f(0) = 1$$f(0) = 1$$This is the taylor expansion we saw from Wigglometer:$$f(x) = f(0) + f'(0).x/1! + f''(0).x^2/2! + f'''(0).x^3/3!$$Given that the slope is the same value as the function if we were to plot the slope then the slope of that slope would also be the same as the function; and the slope of the slope of the slope. That is to say:$$f(0) \equiv f'(0) \equiv f''(0) \equiv f'''(0) \equiv 1$$Hence f(x) would be:$$f(x) = 1 + x/1! + x^2/2! + x^3/3!$$So f(1) would be:$$f(1) = 1 + 1/1! + 1/2! + 1/3!$$As we saw before in the Tug of War these factorials get big very quickly so if this is a finite number we may expect it to appear after not many iterations.Let's look:<jupyter_code>def self_slope_at_one_generator():
    factorial = 0
    summation = 0
    while True:
        summation += 1 / math.factorial(factorial)
        yield summation
        factorial += 1  
gen = self_slope_at_one_generator()
for x in range(20):
    print('Up to {:2} factorial: {}'.format(x, next(gen)))
math.e<jupyter_output><empty_output>
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