Is there a simple formula when for getting the right sig-digs?

  • Thread starter Thread starter student34
  • Start date Start date
  • Tags Tags
    Formula
AI Thread Summary
Determining the correct significant digits in complex equations involving addition, subtraction, multiplication, and division can be challenging. The discussion highlights a specific example where the user struggles with applying the correct rules for significant digits, particularly when combining operations. An error in the initial calculation was identified, emphasizing the importance of following the correct order of operations (PEMDAS). The user expresses frustration over their misunderstanding and seeks clarification on the proper method for handling significant digits in such calculations. Understanding the intermediate steps and rules for significant digits is crucial for accurate results in mathematical computations.
student34
Messages
639
Reaction score
21

Homework Statement



I have no problem knowing how to find the right sig-digs when just doing one of: adding, subtracting, multiplying or dividing. But, I know doing them all at once in more complicated equations gets trickier.

For example, something like 110.0 + (500434.0 - 34.0)/0.37

Homework Equations


The Attempt at a Solution



This is my thought process for finding sig-digs when addition/subtraction is combined with multiplication/division. The answer may be correct by luck, but I know that this is not the right way to come up with the correct number of sig-digs.

110.0 + (500434.0 - 34.0)/0.37
= 110.0 + 500434.0/0.37
= 110.0 + 1.4*10^6 (I make 500434/0.37 equal 2 sig digs because the denominator only has 2 sig-digs.)
= 1.4001100*10^6 (Then I increase the sig-digs by 5 because I am adding 110.0.)

I know this process is wrong because my professor tried to explain to me that there is an intermediate step that I am missing. I can't grasp what it is though.

If it is correct, it is because of pure luck.
 
Last edited:
Physics news on Phys.org
You made a sign error in the very first step. You should fix that first.
 
vela said:
You made a sign error in the very first step. You should fix that first.

Thanks, I meant to put 5400 - 34 instead of 34 - 5400, so I can use both + and - in the example.
 
How did you go from ##110 + 1.3437\times10^5## to ##111.3437\times10^5##? Remember PEMDAS.
 
vela said:
How did you go from ##110 + 1.3437\times10^5## to ##111.3437\times10^5##? Remember PEMDAS.

I don't know why I am so careless today (it's very frustrating being in university and not knowing this yet.).

I changed the question too.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top