One of the basic notions of functions that’s taught in Precalculus and in Discrete Mathematics is the notion of an inverse function: if is a one-to-one and onto function, then there is an inverse function so that
for all and
for all .
If , this is commonly taught in high school as a function that satisfies the horizontal line test.
In other words, if the function is applied to , the result is . When the inverse function is applied to that, the answer is the original number . Therefore, I’ll tell my class, “By applying the function , we uh-uh-uh-uh-uh-uh-uh-undo it.”
If I have a few country music fans in the class, this always generates a bit of a laugh.
See also the amazing duet with Carrie Underwood and Steven Tyler at the 2011 ACM awards:
The second definition of the log function is something which I found rather unsatisfactory at the time, and it has been bugging the back of my mind ever since. The reason is “How on earth did they come across the idea that the integral of 1/x had anything to do with logs.?”
But last year, while getting upset at school definitions of the exponential function as a power series (really gross), I thought “Use a bit of calculus”. So here you are:
Let y = e^x
Since it is a “nice” function it has an inverse g
So g(y) = x …… goes both ways
Differentiate wrt x
g’ (y) * y’ = 1
But y’ = y
So g'(y) = 1/y
and since y is “just another variable – the psychological sticking point!)
we have g'(x) = 1/x
And so g(x) = integral of 1/x wrt x
Not a proof, it’s not meant to be, but it does show how 1/x got into the story.
The integration of 1/x now gives the product into sum form as is needed.
I sleep much better now !
Thanks!
I had another approach, looking at the derivative formula for log(x) and making the “next” point not x + h but x + xh. The the ratio reduces to 1/x * log(1+h)/h, but this requires the acceptance that log(1+h)/h has a limit as h —> 0. Messy.
howardat58
/ February 6, 2016The second definition of the log function is something which I found rather unsatisfactory at the time, and it has been bugging the back of my mind ever since. The reason is “How on earth did they come across the idea that the integral of 1/x had anything to do with logs.?”
But last year, while getting upset at school definitions of the exponential function as a power series (really gross), I thought “Use a bit of calculus”. So here you are:
Let y = e^x
Since it is a “nice” function it has an inverse g
So g(y) = x …… goes both ways
Differentiate wrt x
g’ (y) * y’ = 1
But y’ = y
So g'(y) = 1/y
and since y is “just another variable – the psychological sticking point!)
we have g'(x) = 1/x
And so g(x) = integral of 1/x wrt x
Not a proof, it’s not meant to be, but it does show how 1/x got into the story.
The integration of 1/x now gives the product into sum form as is needed.
I sleep much better now !
John Quintanilla
/ February 6, 2016For what it’s worth, this is something that I wrote about some time ago: https://meangreenmath.com/2015/01/28/different-definitions-of-logarithm-index/
John Quintanilla
/ February 6, 2016But I agree that your intuitive explanation gets to the heart of the matter pretty quickly.
howardat58
/ February 6, 2016Thanks!
I had another approach, looking at the derivative formula for log(x) and making the “next” point not x + h but x + xh. The the ratio reduces to 1/x * log(1+h)/h, but this requires the acceptance that log(1+h)/h has a limit as h —> 0. Messy.