"How does exp(x) keep showing up in mathematics When I first learnt about e, I just treated it as another number (as far as I know, it doesn't even have a natural definition), but how is it that exp(x) is so important and keeps showing up at various places in mathematics?"

Slovenujozk 2022-09-12 Answered
How does exp(x) keep showing up in mathematics
When I first learnt about e, I just treated it as another number (as far as I know, it doesn't even have a natural definition), but how is it that exp(x) is so important and keeps showing up at various places in mathematics?
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Answers (2)

Dalton Erickson
Answered 2022-09-13 Author has 10 answers
e is very important because it keeps showing up in various places in mathematics. Specifically, calculus though.
Calculus is the study of the rate of change, and the first time you have possibly learned about the number e, you spot how a whole lot it performs a key role within the charge of exchange of bank interest, or exponential boom/decay. given that calculus is the examine of rate of variety, e is certain to appear in many factors across calculus, whether or not that be derivatives, integration, polynomials, many superior topics too like differential equations.

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Julian Werner
Answered 2022-09-14 Author has 1 answers
One beautiful occurrence I just found out today:
Consider the equation:
x y = y x
x , y R +
For every positive value of x, there are two values of y that satisfy the above equation ONLY after x= e and vice versa.
To admire this amazing result, look at the graph of the equation on Desmos.

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For which α does the integral 2 + e α x ( x 1 ) α ln x d x converge?
For which values of α does the integral
2 + e α x ( x 1 ) α ln x d x
converge?
I'm lost here.
For simplicity, let us denote the above integral as I 1
I was able to prove (I noticed that it's kind of standard exercise though) that:
I 2 = 2 + 1 ( x 1 ) α ln β x
I2 converges for α > 1 and   β.
converges for α = 1 and β > 1. For β 1 it diverges.
diverges for α < 1 and   β.
So my strategy was to use I 2 to prove the converges/divergence of I 1 , by means of inequalities, but I get nothing from this.
For instance I did the following:
For α > 1 we have that
0 < 2 + d x x α ln x 2 + e α x ( x 1 ) α ln x d x
That is, for β = 1
0 < I 2 I 1
From 1. we know that I 2 converges, but this doesn't tell us anything about the convergence of I 1 .
The answer given by my textbook is that I 1 converges for α < 0, but I don't know how to arrive at that conclusion.

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