<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>ralferoo</author><text>It&amp;#x27;s a pretty good way of explaining it.&lt;p&gt;I&amp;#x27;ve seen (1+1&amp;#x2F;n)^n before, but never seen an explanation of why I might ever want to use something of that form. I&amp;#x27;ve used the e^ix notation extensively, but again I&amp;#x27;ve never really cared, because to me it was just a compact representation of sin and cos together. Likewise, all the proofs in that article are still a bit &amp;quot;so at this point on this carefully chosen graph, the gradient is e&amp;quot; and I think &amp;quot;Who cares? You carefully chose the graph to prove a point, I&amp;#x27;ll never see it in the real world.&amp;quot;&lt;p&gt;And then the example with compounding interest - immediately I can see the application. It&amp;#x27;s definitely a good way of explaining it, although maybe it&amp;#x27;d be even more grounded if it had n=12 and n=365 as examples. When you notice that the actual values seem to be converging, then you can try plugging in ever bigger and bigger numbers. This way you can discover the value of e for yourself and that process of discovery leads to a better understanding than rote learning of an abstract thing you haven&amp;#x27;t mentally visualised yet. All the other explanations are useful later, and they allow you to see it in different situations, but having at least one &amp;quot;this is why it&amp;#x27;s tangibly useful&amp;quot; hook at the start is definitely a massive help in understanding something.</text><parent_chain><item><author>lisper</author><text>That may be how it was arrived at historically, but it is not the best way to explain it.&lt;p&gt;e arises when you ask the question: is there a function that is its own derivative?  And it turns out the answer is yes.  It is this infinite series:&lt;p&gt;1 + x + x^2&amp;#x2F;2 + x^3&amp;#x2F;6 + ... x^n&amp;#x2F;n! ...&lt;p&gt;which you can easily verify is its own derivative simply by differentiating it term-by-term.  When you evaluate this function at x=1, the result is e.  In general, when you evaluate this function at any real value of x, the result is e^x.&lt;p&gt;But the Euler equation e^iπ = -1 has nothing to do with exponentiating e, it&amp;#x27;s just a &lt;i&gt;notational convention&lt;/i&gt; that is &lt;i&gt;defined&lt;/i&gt; to be the series above.  When you evaluate that series at x=iπ, the result is -1.  In general, the value of the series for any x=iy is cos(y) + i*sin(y).&lt;p&gt;It&amp;#x27;s that simple.</text></item><item><author>mihaic</author><text>Exactly the same for me and I&amp;#x27;m pretty sure this is how Jakob Bernoulli came to define the number as well, trying to see what the upper bound for infinitesimal compounding was.</text></item><item><author>munchler</author><text>This is a frustrating article because it never explains why e is the natural logarithm base. To me, the easiest way to understand it is via continuous compound interest:&lt;p&gt;* If you invest $1 at 100% interest for 1 year, you get $2 at the end&lt;p&gt;* Compounded 2 times in a year, you get 100&amp;#x2F;2 = 50% interest every 1&amp;#x2F;2 year, which amounts to $2.25&lt;p&gt;* Compounded 4 times in a year, you get 100&amp;#x2F;4 = 25% interest every 1&amp;#x2F;4 year, which amounts to $2.44&lt;p&gt;* Compounded n times in a year, you get 100&amp;#x2F;n percent interest every 1&amp;#x2F;n year, which amounts to (1+1&amp;#x2F;n)^n dollars&lt;p&gt;* So continuous compound interest is the limit as n approaches infinity, which amounts to $2.71828 at the end of the year&lt;p&gt;(This is a great problem to give to pre-calc students to see if they can figure out the calculation for themselves.)</text></item></parent_chain></comment><story><title>What makes e natural? (2004)</title><url>https://www.komal.hu/cikkek/2004-ang/e.e.shtml</url></story>
<instructions>Your goal is to analyze the following comment and estimate how highly it will be upvoted by the Hacker News community.</instructions><comment><author>moefh</author><text>&amp;gt; But the Euler equation e^iπ = -1 has nothing to do with exponentiating e, it&amp;#x27;s just a notational convention that is &lt;i&gt;defined&lt;/i&gt; to be the series above.&lt;p&gt;Can&amp;#x27;t the same thing be said about using fractions on the exponent? Exponentiation is &lt;i&gt;actually&lt;/i&gt; just repeated multiplication (a^n=a*a*...*a, repeated n times), but you can&amp;#x27;t do that when n is a fraction or irrational anymore than you can do it when it&amp;#x27;s imaginary.&lt;p&gt;We have to define what it means for an exponent to be non-integer: for fractions we might define a^(b&amp;#x2F;c) as the root of the equation x^c=a^b, and to allow irrationals I think you need some real analysis (it&amp;#x27;s been a while, but I think the usual way is to first define exp and log, and then say that a^b=exp(b*log(a)), which is kind of cheating because we have to define exp first!).&lt;p&gt;There&amp;#x27;s a very intuitive way to &amp;quot;see&amp;quot; that e^ix=cos(x)+i*sin(x): all you have to do is to treat complex numbers like you would any other number, and &amp;quot;believe&amp;quot; the derivative rule for complex numbers (so (e^(ix))&amp;#x27;=ie^(ix)). Then you can just graph f(x)=e^(ix) for real x by starting at x=0 (when clearly f(x)=1) and from there take small steps in the x axis and use the derivative to find the value of the next step with the usual formula f(x+dx)=f(x)+f&amp;#x27;(x)*dx.&lt;p&gt;Doing that you realize the image of e^(ix) just traces a circle in the complex plane because every small step in the x direction makes e^(ix0) walk a small step perpendicular to the line going from 0 to e^(ix0), simply because multiplying by i means rotating 90 degrees.</text><parent_chain><item><author>lisper</author><text>That may be how it was arrived at historically, but it is not the best way to explain it.&lt;p&gt;e arises when you ask the question: is there a function that is its own derivative?  And it turns out the answer is yes.  It is this infinite series:&lt;p&gt;1 + x + x^2&amp;#x2F;2 + x^3&amp;#x2F;6 + ... x^n&amp;#x2F;n! ...&lt;p&gt;which you can easily verify is its own derivative simply by differentiating it term-by-term.  When you evaluate this function at x=1, the result is e.  In general, when you evaluate this function at any real value of x, the result is e^x.&lt;p&gt;But the Euler equation e^iπ = -1 has nothing to do with exponentiating e, it&amp;#x27;s just a &lt;i&gt;notational convention&lt;/i&gt; that is &lt;i&gt;defined&lt;/i&gt; to be the series above.  When you evaluate that series at x=iπ, the result is -1.  In general, the value of the series for any x=iy is cos(y) + i*sin(y).&lt;p&gt;It&amp;#x27;s that simple.</text></item><item><author>mihaic</author><text>Exactly the same for me and I&amp;#x27;m pretty sure this is how Jakob Bernoulli came to define the number as well, trying to see what the upper bound for infinitesimal compounding was.</text></item><item><author>munchler</author><text>This is a frustrating article because it never explains why e is the natural logarithm base. To me, the easiest way to understand it is via continuous compound interest:&lt;p&gt;* If you invest $1 at 100% interest for 1 year, you get $2 at the end&lt;p&gt;* Compounded 2 times in a year, you get 100&amp;#x2F;2 = 50% interest every 1&amp;#x2F;2 year, which amounts to $2.25&lt;p&gt;* Compounded 4 times in a year, you get 100&amp;#x2F;4 = 25% interest every 1&amp;#x2F;4 year, which amounts to $2.44&lt;p&gt;* Compounded n times in a year, you get 100&amp;#x2F;n percent interest every 1&amp;#x2F;n year, which amounts to (1+1&amp;#x2F;n)^n dollars&lt;p&gt;* So continuous compound interest is the limit as n approaches infinity, which amounts to $2.71828 at the end of the year&lt;p&gt;(This is a great problem to give to pre-calc students to see if they can figure out the calculation for themselves.)</text></item></parent_chain></comment><story><title>What makes e natural? (2004)</title><url>https://www.komal.hu/cikkek/2004-ang/e.e.shtml</url></story>