Roger's Equations

This is an amazing formula.

As we all know, cos and sin come from simple properties of right-angled triangles, and i from (among other things) a need to ensure a solution to all quadratic equations.

e was introduced in my maths lessons from differentiation of a common log, leading to (1+x)^1/x as x → 0, which = e. No doubt there are other ways.

These are all from separate branches of maths yet it turns out they share this connection via Euler's formula. It gave me a queer feeling when I first encountered this 40+ years ago, and I still find it fascinating. Those not interested in maths don't know what they're missing!

Here there is also a wandering result:

Take a line, let say L=10 meters, we can divide it in two parts (A and B) been A+B = 10, for example 3 and 7, now lets multiply this numbers and we get M=3*7=21, now take 2 and 8 and multiply and we get M=2*8=16, we can prove easily that the maximum product results when A equals B in this case 5 and 5 and we get M=5*5=25.

This is valid also when we divide the line in more than two pieces for example 5 pieces of 2 if we multiply all this, we get M=32, we can also prove easily that we get the maximum when all the pieces are equal, if we divide in 4 pieces of 2.5 we get M=39.0625, if we take 3 pieces M=37.0369259, the question arise, what is best division to get the maximum of the maximum, we can easily prove that we get the maximum M when we divide the line in "e" pieces ¡wonderfull! isn`t it, in this case M= 39,5986256

I don´t know if there is an application for this, but it is a nice thing to play with.

Regards

Ciro

Hello Ciro

That's interesting as you say, but I think you've made an error. I make max M not when line is divided into "e" pieces, but when the pieces are of length "e", so there are 10/e pieces (for length 10). Continuing your examples would give M = (10/e)^e = 34.49.... but as you point out maximum is 39,5986256.

For general case with line length N, max M = e^N/e.

Difference is greater for larger N, eg for N = 20, (N/e)^e = 227, but e^N/e = 1568

Cheers

Hello Cheers:

Yes, as you point out i made a mistake the length is "e", thanks for your comment

Ciro

I can see where this formula is useful to EE's and Physicists and other trying to analyze waveforms, power factor corrections, etc.

Are there any applications where a Mechanical Design Engineer or a Manufacturing Engineer would find this equation useful? From my own experience, I cannot think of any.

Is this Blog to be focused on those areas (EE/Physics) or will we find equations and formulas which have applications in other engineering areas as well, mechanical, civil, manufacturing, metals and materials, industrial, etc.?

Thanks!

"Are there any applications where a Mechanical Design Engineer or a Manufacturing Engineer would find this equation useful? "

Great Question. In general, the solution of the differential equation in the form:

(d²x/dt²) + ω²x = 0

has a solution in the form

x(t)=Ae^iωt + Be^iωt

A mechanical example would be a spring. The spring equation is:

F=-kx

To solve for the position x, you get,

F=ma=-kx where a=(d²x/dt²) so

(d²x/dt²) + (k/m)x = 0

which has solutions as we stated above

x(t) = Ae^iωt + Be^iωt

Where A and B are determined by Boundary Conditions such as initial position, initial velocity, etc.

So that would be an example for a Mechanical Design Engineer. The Euler equation is used when solving for the boudary conditions. Notice it doesn't have to be a spring, but any situation where the force is negatively proportional to the acceleration (These types of forces are generally called Hookes Law).

I hope that answers your question. Just so you know, this blog is not intended to just be for EE/Physics. The motivation is to try to answer the question "I wonder where the heck that formula/equation/constant, comes from". If you have anything like that, feel free to post a request or email it to me and I'll put it on my list of future derivations.

This may be elementary, but why does ln(e)^x = x? I need to explain this in a way that can be easily understood.

Thanks.

Reply to Brandenb, #6

It doesn't. Log of any number with base same number = 1. eg common log (base 10), log₁₀(10) = 1 and log_e(e) = ln(e) = 1. In general, log_a(a) = 1.

This is from definition of log to any base a, a^log_a(x) = x, so if x = a, log_a(a) = 1.

So ln(e)^x = 1^x = 1. Perhaps you meant x^ln(e) = x^1 = x ?

Sorry if this is confusing, it's the best I've come up with.

There is some abiguity in the equation lne^x since it is unclear which operation should be done first. Convention says lne^x means ln(e^x), not (lne)^x so

Using this convention, ln(e^x)=x is correct.

I'll answer the question why in another post, I have to run.

In reply to #8 - fair enough Roger, I should have spotted that what Brandenb probably meant was ln(e^x), not (ln(e))^x

You have asked why lne^x=x;

To understand the answer, its best we understand what an inverse function is. ln(x), e^x, Sin(x), ArcSin(x), etc. are all functions (f(x)). A function, basically, takes one value, x, and maps it to another value, f(x). Just as numbers have inverses, like;

3 x 1/3 = 1
5 x 1/5 = 1

*Inverse here means the number you muliply by to get one.

Functions have inverses too;

ln(e)=1
Arcsin(Sin)=1

where 1 in the equations above is not the number 1 but the identity function 1, which has the property that anything it acts on doesn't change (1(x)=x). It maps numbers to themselves.

*Inverse here means the function you apply to the original function to get the identity function.

So when you write

lne^x=x
substituting ln(e)=1(x)
1(x)=x
and since 1(x)=x,
x=x

so it's correct.

I've got to run again (sorry). When I come back, I'll try using a series expansion to show you that ln is the inverse of e.

Thanks for this question, it's great.

Ok, I'm back. Please see comment #9 for a preface of this comment.

In this comment I will try to show that;

e^ln(y+1) = y+1

as a way of showing that e is the inverse function of ln. I use y+1 because the series expansion in terms of ln is easier, but it's just as valid as using e^x since we can simply say x=y+1

The series expansion e^x = 1 + x + ¹/₂x² + ¹/₆x³ + ¹/₂₄x⁴ + ......

So e^ln(y+1) = 1 + ln(y+1) + ¹/₂ (ln(y+1))² + ¹/₃ (ln(y+1))³ +......

Using the series expansion ln(y+1)= y - ¹/₂ y² + ¹/₃ y³ - ¹/₄ y⁴ +....
we get;

e^ln(y+1) = 1+(y-¹/₂y²+¹/₃y³-...) + ¹/₂(y-¹/₂y²+¹/₃y³-...)² + ¹/₆(y-¹/₂y²+...)³+...

Next we group like powers of y in the expression above. Since each expansion of ln(y+1) is raised to higher and higher powers, we can be confident of gathering all the terms of a particular power (y, or y², or y³, etc.) and grouping them together.

e^ln(y+1) = 1 + y - ¹/₂y² + ¹/₂y² + ¹/₃y³ - ¹/₂y³ + ¹/₆y³ -.....

When we group all powers of y together, we see that y² terms add up to zero, y³ terms add up to zero, and if we carried it out further we would see all power terms higher than y¹ sum to zero. Canceling out the terms that sum to zero, we are left with;

e^ln(y+1) = 1+ y = y+1

so e^ln behaves like the identity fuction 1(x) that gives x as a result (see post #9).

Thanks for the great question. I hope this explanation helps, please let me know if you have any questions.

too elementary for me to address