Solve Uncertainty of Pi: Volume of Cylinder

[rejz55]

[SOLVED] uncertainty of pi?

does pi have any uncertainty? I am trying to solve the volume of a right circular cylinder with h=2.3±0.1 and radius 0.12±0.05m..i cannot continue cause i do not know if pi has an uncertainty..thanks

 

[Gokul43201]

No, it does not.

 

[olgranpappy]

does pi have any uncertainty? I am trying to solve the volume of a right circular cylinder with h=2.3±0.1 and radius 0.12±0.05m..i cannot continue cause i do not know if pi has an uncertainty..thanks

pi has no uncertainty, but it is an infinitely long decimal number, so your calculator doesn't store the actual exact value of pi. but your caculator's value of pi does have enough digits that the error in pi can be ignored.

I.e., just calculate the uncertainty from the uncertainty in 'h' and the uncertainty in 'r'.

 

[nicksauce]

Numerical constants never have uncertainty. Pi, e, 2, 2.75, sqrt(5), etc. are all exact numbers.

 

[rbj]

does pi have any uncertainty? I am trying to solve the volume of a right circular cylinder with h=2.3±0.1 and radius 0.12±0.05m..i cannot continue cause i do not know if pi has an uncertainty..thanks

Numerical constants never have uncertainty. Pi, e, 2, 2.75, sqrt(5), etc. are all exact numbers.

rejz55, if there is conflation of exactness or uncertainty of the value of [itex]\pi[/itex] vs. its not being a rational number, it is true that there are no pair of exact integers, N and D so that

[tex] \pi = \frac{N}{D} [/tex]

but whatever the tolerance of "error" you give me (we'll call this tolerance [itex]\epsilon[/itex]), it is true that one can always find a pair of integers for N and D so that the above is true within that level of tolerance. for whatever [itex]\epsilon>0[/itex] that you (or the devil) tosses at us, we can always find a rational number (a ratio of integers, ^N/_D) that is "within [itex]\epsilon[/itex] of" [itex]\pi[/itex]:

[tex] \left| \pi - \frac{N}{D} \right| < \epsilon [/tex]

or

[tex] \frac{N}{D} - \epsilon < \pi < \frac{N}{D} + \epsilon [/tex]

where

[tex]\epsilon > 0[/tex]

no matter how tiny [itex]\epsilon[/itex] gets (as long as it remains bigger than zero).

 

[rejz55]

thanks guys!