[Bored chemist (OP)]

it's much better to model it as an exponential

An exponential is a polynomial.

Not really, no.
x^e is not the same as e^x

Moderator's Note: This was split off from a thread which looked at current through a diode as an exponential function, see: https://www.thenakedscientists.com/forum/index.php?topic=81689.0

[alancalverd]

10^x is an exponential but not a polynomial. e^-kt is a common exponential decay curve which can be approximated as a polynomial, to any required degree of accuracy. 

[evan_au]

10^x is an exponential but not a polynomial. e^-kt is a common exponential decay curve which can be approximated as a polynomial, to any required degree of accuracy.

Using a Taylor Series, we can calculate the exponential function e^x as a polynomial of infinite degree, to any desired accuracy:
e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...
- This polynomial series converges for all real values of x (and also all imaginary values, but let's leave that for another time...)
See: https://en.wikipedia.org/wiki/Taylor_series#Exponential_function

We can calculate an exponential decay e^-kt using the same polynomial by substituting x=-kt.

We can calculate the exponential 10^z using the same polynomial by substituting x=z/LN(10) = z/2.3026.... (or x=z*2.3026..., depending on which way you are going.)

So they are all variations on the same exponential function and polynomial calculation method.

[Bored chemist (OP)]

So, by doing an indefinitely large amount of work calculating higher and higher order terms, we can show that vhfpmr is still wrong, but only a little bit.

I wonder; how many polynomial terms do you need to use to get a reasonable degree of accuracy for calculating a simple obvious exponential function like 10^x where x=3 using a Taylor series?

As a polynomial, I can calculate 10^3 as being 1000 "by inspection", but I wonder how much work it takes to calculate as an expansion.

[alancalverd]

I was being pedantic, but as we discussed in another thread, the difference between a continuous analytic function and a polynomial approximation can be significant.

[Bored chemist (OP)]

10^x is an exponential but not a polynomial. e^-kt is a common exponential decay curve which can be approximated as a polynomial, to any required degree of accuracy. 

I think your polynomial might struggle if k is complex or imaginary

[alancalverd]

We have no problem calculating  e-^iπ, or even proving that e-^iπ =  e^iπ analytically, but as you say it would be a heck of a task to demonstrate the mutual convergence of series!

[vhfpmr]

So, by doing an indefinitely large amount of work calculating higher and higher order terms, we can show that vhfpmr is still wrong, but only a little bit.

So how do you calculate  e  exactly?

[alancalverd]

Same way as you calculate π - you can't!

The difference is that
π is absolutely defined as the ratio of circumference to diameter of a circle, is provably irrational, and can be approximated by various infinite series, but
e is absolutely defined as the sum of one infinite series (and is provably irrational).

[Bored chemist (OP)]

So, by doing an indefinitely large amount of work calculating higher and higher order terms, we can show that vhfpmr is still wrong, but only a little bit.

So how do you calculate  e  exactly?

Do you mean "exactly how do you go about calculating it?" or do you mean "how do you calculate the exact value of it?"
The answers are
I don't bother and you can't.

[evan_au]

So how do you calculate  e  exactly?

If you need to specify the exact value, use the pronumeral e.

If you want to approximate it, calculate e¹
e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...
e¹ = 1 + 1 + 1²/2! + 1³/3! + 1⁴/4! + 1⁵/5! + ...
= 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...
= 2.718281828...

This series converges quite quickly, because the factorial function n! gets very big, very quickly (ie 1/n! gets small very quickly).
I remember calculating this to over 100 digits accuracy on my first home computer (and it consumed most of the memory...)
(Oops - overlap with Bored Chemist...)

calculating a simple obvious exponential function like 10^x where x=3 using a Taylor series

10 = e^2.3026
10³ = (e^2.3026)³ = e^3x2.3026 = e^6.9

The first 6 terms grow, and it starts to converge after the 7th term = 6.9⁷/7!

Of course, if you had to calculate 10^7.654, it wouldn't be so obvious, and I would recommend using a calculator (log tables being no longer found in every scientist's desk).

I think your polynomial might struggle if k is complex or imaginary

The exponential function converges for all imaginary values, too.
- You can express the answer in terms of COS(θ) + iSIN(θ).

You can see this in the similarity between the Taylor series for the exponential function EXP(x), SIN(x) & COS(x).
- SIN and COS are what you get if you use the EXP function with complex inputs
- Remembering that i² = -1

EXP(x) = 1 + x + x²/2! + x³/3! + x⁴/4! + x⁵/5! + ...
SIN(x) =     x         - x³/3!        + x⁵/5! - ...
COS(x) = 1     - x²/2!         + x⁴/4! - ...

[Bored chemist (OP)]

The first 6 terms grow, and it starts to converge after the 7th term = 6.97/7!

But how long does it take before it converges to within 1% of 1000 ?
Also, I think we may have possibly drifted away from the core of the topic slightly.

[alancalverd]

10 = e^2.3026

Er, no, it's closer to e^{2.3025850929940456840179914546844} or, as we say in my world, e^2.3025+1/-0

[evan_au]

My observation is that most people get surprised by the speed of exponential growth.
- Once it is big enough to be noticeable, it just explodes on the scene
- Until it runs into some fundamental limit (like a virus that infects the whole population)

A comment from some epidemiologists: For every day you delay the start of a lockdown, you add a week to the time to lift it
Rationale:
- At the start of an outbreak, the number of infections is growing exponentially; with a 5-day lag between infection and showing symptoms, the number of cases is far higher than you see
- As you try to control an outbreak, the number of cases falls off as a negative exponential. The rate of fall is far lower than the rate of increase (if you plot case numbers on a logarithmic graph)


In the following thread, there is some spreadsheet modeling of the near-exponential growth of the early days of the COVID-19 pandemic (warning: New Theories, with sometimes dodgy assumptions!):
https://www.thenakedscientists.com/forum/index.php?topic=81624.msg629376#msg629376

[vhfpmr]

Same way as you calculate π - you can't!

I know. That's my point.

He was carping that a polynomial is a lot of work for an approximate answer, but it's a lot of work summing a series all the way to infinity, so any attempt at numerical evaluation of a transcendental function is an approximation.

[evan_au]

But how long does it take before it converges to within 1% of 1000 ?

As shown below, the Taylor series for e^x takes 17 terms to converge within 1% of 10³
- But 26 terms to converge within 1% of 10^-3: The terms are the same size, but alternating + & -
- And it's not a good idea to subtract large numbers that are close to each other: You lose a lot of accuracy
- In theory, the Taylor series is infinitely accurate (within a "radius of convergence"), but it takes forever to get there
- In practice, different coefficients are used that give you a finite accuracy, but in a finite time

Taylor_Convergence.png (31.15 kB . 419x618 - viewed 2721 times)

[Bored chemist (OP)]

The exponential function converges for all imaginary values, too.

Yes it does- sort of.
Euler's formula tells you that exp (i x)  = i sin (x) + cos (x)

And the thing about that is that (even just the real part of it)  waves  up and down forever as x increases.
And the only way to get a polynomial to do that is to have a polynomial of infinite order.
Even a bad approximation to the function needs an infinite number of terms.

Over the range 0 to  tenth of a radian you can approximate sin (x) as x and not get a bad answer.
You can get a bit more sophisticated and subtract x^3/3! and get a pretty good approximation over the range 0 to 1

But if you want to cover the range 0 to 3 you need a 7th order polynomial.

And it gets rapidly worse, which is why I said

your polynomial might struggle if k is complex or imaginary

An ordinary exponential function (of a real number) is monotonic but if you include i in the equation it can bounce up and down forever.
An nth order polynomial can  only bounce about n times

[evan_au]

Playing around with it, I graphed the convergence of the Taylor series for EXP(x) when x=10^3 and x=10^-3.
- The terms have the same values, but oscillate + & - when x < 0.

Taylor_convergence_graph.png (22.99 kB . 586x353 - viewed 2721 times)

The magnitude of positive and negative terms are both around 500; but their difference introduces significant errors.
- The following table only looks at the first 26 terms in the Taylor series for EXP(x)
- It's better to calculate EXP(10^3) in both cases, and then calculate EXP(10^-3) = 1/EXP(10^3)

Sum_26_Terms.png (10.27 kB . 312x299 - viewed 2643 times)

Most of us use computers which store floating point numbers in binary.
- In some ways, it's more straightforward to calculate functions like 2^x and LOG₂(x) and convert the base
- than calculate e^x and LOG_e(x) or 10^x and LOG₁₀(x)