Working with values below 1

[DevilSolution]

Having a hard time quantifying certain equations because most the values are below 1. One issue is standard deviation, simple enough but when squaring these numbers they get smaller etc...Also working with angles below 1 degree. For some reason doing opp/sin(x) gives a hypotenuse barely 4^-10 over the adjacent.

 

I've tried a few basic things..multiply by say 100000 then divide by a multiplier that counts the resulting exponent (with standard deviation). Is there a particular trick to working with values below 1?

 

As a side note 3162 is my new favourite number.

 

Regards.

Edited August 31, 2015 by DevilSolution

[timo]

The standard way for dealing with numbers smaller than one is not seeing a problem with doing so. I don't see the problem with the standard deviation: You square numbers to get the mean square distance, and then you take the square root of this which sort of cancels the square operation.

[ajb]

Is there a particular trick to working with values below 1?

Nothing that I can think of that is generally useful. You seem to be aware of the immediate and common misconceptions eg. 'how can you multiply a number by itself and get a smaller number?'

[studiot]

Even with the examples you give you can use the Engineer's solution to this.

 

Work with smaller units.

 

This is also effective with large numbers that can also be problematic.

 

A simple example

 

Most batteries have a voltage of a few volts.

Most modern circuits draw small amounts of current when measured in amps say 0.01 amps.

 

So Engineers use this form of Ohm's Law

 

Volts = milliamps x kilo ohms

 

or even

 

Volts = microamps times megohms.

 

The game plan is to choose units so that the numbers in the calculation falls between 1 and 1000.

 

Other ways to cope are to put every quantity in to the form a number between 1 and 10 multiplied by a power of ten, and collect all the powers of ten together at the end.

This is implemented in two ways,

 

directly as in 1.3 x 10^-3 amps for hand calculation

 

and

 

using the E notation for computers and calculators

 

as in 1.3 E-3 amps.

 

This won't solve all your problems, but it has helped many people over many years.

Edited August 31, 2015 by studiot

[Commander]

Having a hard time quantifying certain equations because most the values are below 1. One issue is standard deviation, simple enough but when squaring these numbers they get smaller etc...Also working with angles below 1 degree. For some reason doing opp/sin(x) gives a hypotenuse barely 4^-10 over the adjacent.

 

I've tried a few basic things..multiply by say 100000 then divide by a multiplier that counts the resulting exponent (with standard deviation). Is there a particular trick to working with values below 1?

 

As a side note 3162 is my new favourite number.

 

Regards.

 

Didn't know that ! Deleted !

Edited August 31, 2015 by Commander

[imatfaal]

!

Moderator Note

 

 

 

If you are quantifying a rational value between 0 and 1 it has to be the Sum of the Subset of Elements of the Walker's Equation Matrix of Double infinite Series as I have explained in my Posts !

 

Commander

 

Oi! No advertising your own ideas and threads in another member's thread. This is considered hijacking and is against the rules.

 

Do not do this again. Do not respond to this moderation - report the post if you think it is unfair

 

[swansont]

As noted, working with scientific notation is one standard method. Many scientists want to make terms smaller than 1 so that they can use approximations from expansions which only converge if the term is <1. Then e.g. sin (theta) = theta (using radians), and higher order terms in the expansion can be ignored.

[DevilSolution]

Timo was right about standard deviation i think, just the precision of my phone.

 

Another question as far as angles go. For my purposes i can use pythag to find the hyp instead of opp/sin(x) however for example the adj is always 1 and the opp very small such as 0.004. Which again 0.004 * 0.004 = 0.000016. Now if a = adj, b = opp and c = ? Then a^2 - c^2 = b^2 wouldnt work and b^2 - c^2 = a^2 would make c less than 1 everytime. Doing c^2 =((a*10^5)^2 + (b*10^5)^2) / 10^5 and finally sqrt© gives a number marginally bigger than 1, 1.000079997 to be exact. Does that seem correct?

 

Im starting to understand the relationship between numbers less than 0 in relation to numbers above. Its just quantifying things in the form you want.

Edited September 5, 2015 by DevilSolution

[ajb]

Doing c^2 =((a*10^5)^2 + (b*10^5)^2) / 10^5 and finally sqrt© gives a number marginally bigger than 1, 1.000079997 to be exact. Does that seem correct?

That seems okay.

 

You have a right angled triangle with hyp = c (say) and then the other lengths are 1 and 0.004. Using Pythagoras we have

 

[math]c = \sqrt{1+ (0.004)^{2}} \hspace{20pt} ``=" 1+ \textnormal{small correction}[/math]

[John Cuthber]

Be careful with approximations like that. I just worked out that the moon does not exist.

I calculated it's diameter based on the fact that it subtends a small angle (about half a degree).

 

It has a diameter that's given by the sine of that angle times the distance to the moon.

Well, it's a small angle so I can assume that the "small angle" approximations hold.

I can find the Taylor series for cosine

1- 1/2 x^2 + 1/24 x^4...

Since x is small I can assume that x^2 is too small to bother with and the other terms are even smaller.

So the cosine of a small angle is 1

Then from sine² + Cosine² =1 I can conclude that sine (x) = 0 for small x

 

So the diameter of the moon is zero time the distance to the moon.

So that diameter is zero.

So the moon isn't there.

 

In general, the answer to the problem of small number is not to worry about them.

Even if you are not happy with them, the maths still works just fine.

[studiot]

I posted a reply based on half angle formulae, yesterday and it seems to have disappeared, along with the post from DS that I replied to.

 

What is happening please?

Edited September 5, 2015 by studiot