I think the best way to explain why e matters without direct mention of calculus concepts is (as Wikipedia says in its page on natural logarithm)

“for any numeric value close to the number one, the natural logarithm can be mentally computed by subtracting the number one from the numeric value. For example, the natural logarithm of 1.01 is 0.01 to an accuracy better than 5 parts per thousand. With similar accuracy one can assert that the natural logarithm of 0.99 is minus 0.01. The accuracy of this concept increases as one approaches the number one ever more closely, and reaches completeness of accuracy precisely there.”

In other words, for small values of theta, e ^ theta approximates to 1 + theta. The relevance of this to Euler’s formula is apparent when we consider that sine theta approximates to theta for small values of theta expressed in radians. So cos theta + sine theta approximates to 1 + theta.

If this characteristic of e is generalised to complex numbers, then it follows that in your equation

e ^ i theta = cos (n theta) + i sine (n theta),

the factor “n” must equal 1.

To express these points with full rigour, one would admittedly need to call epsilon and delta back from their vacation.

]]>all of which equal 0 at the end of the day. (or any other time of day)

right? Or am I off?

]]>Re what you said about the relation between cos and cosh functions. You describe the statement “cosh θ = cos i θ / 2 ” as “strange but not illegal”. (I think actually the implied statement is “cosh θ = cos 2 i θ”, but that is by the way.) Whether such a statement is “illegal” or not depends on how we define the cos and cosh functions, and on how we define circular and hyperbolic angles which are the arguments of those functions.

If circular angle is defined as twice the area of the section of the unit circle, and if hyperbolic angle is defined as twice the area of the section of the unit hyperbola, then for both sin and sinh, as the angle becomes small, sin θ approximates to θ, that is, sin θ / θ approaches a limit of 1, and sinh θ / θ does the same, as can be seen intuitively by drawing the unit circle and unit hyperbola on a sheet of paper. Extending that result into the realm of complex numbers, sin (i θ) becomes approximately equal to i θ when θ becomes small. All of which is consistent with the statement “sinh θ = – i sin i θ”. Deriving cos from sin and cosh from sinh by means of the formulas (sin θ)^2 + (cos θ)^2 = 1, and (cosh θ)^2 – (sinh θ)^2 = 1, the formula relating cosh to cos then has to be cosh θ = cosh i θ. Not “cosh θ = cos i θ / 2 ” or “cosh θ = cos 2 i θ” or any other such variation.

]]>Consider the orange region, which is the area under 1/x between x=2 and x=4. Make it half as wide by replacing x with 2x; now we’re looking at the area under 1/2x between x=1 and x=2. Then make it twice as tall by doubling it; now we’re looking at the area under 2/2x between x=1 and x=2. But 2/2x = 1/x. So if we make the green thing half as wide and twice as tall, we get exactly the orange thing, without using any calculus at all.

]]>I know it’s been quite a while, but I’m going to answer Eugenio’s question anyway. The mistake is in the x(1/x)=1 part: you forgot to evaluate this term at the integration limits. If the integral is from a to b, then this term becomes f(b) – f(a) = 1 – 1 = 0.

So you’ve used integration by parts to show that ∫(1/x)dx = ∫(1/x)dx, which is something you hopefully already knew. :)

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