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The value of 0/0


Raptor115
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  • 2 weeks later...

You can solve these problems your-self

==>what will be velocity when object trevells 0 m distance in 0 seconds

==>what will be acceleration when delta velocity=0 and t=0

==>You did zero work in no time what will be your Power

As a matter of fact some mathematical problems are not applicable in physics

==>Answer to 0/0 is mostly applicable in calculus like this

(x^2-9)/(x-3) becomes 0/0 when x=0 but if we take x->0 then this becomes 6 that's mean that 0/0 is 6 in this case

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You can solve these problems your-self

==>what will be velocity when object trevells 0 m distance in 0 seconds

==>what will be acceleration when delta velocity=0 and t=0

==>You did zero work in no time what will be your Power

No, your velocity is undefined. 0 m in 0 seconds is not defined. See all the reasons listed in the thread above.

 

With an undefined velocity, it makes no sense to talk about concepts like acceleration or power.

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Consider the mirrored functions:

[math]f_1(x) = (x - dx)^2 + dy[/math]

[math]f_2(x) = -(x - dx)^2 + dy[/math]

 

And:

[math]f_3(x) = \frac{f_2(x)}{f_1(x)} \; \; \; \; \; \; f_1(x) \neq 0[/math]

 

Where [math]dx[/math] and [math]dy[/math] are x-axis and y-axis differential shifts.

 

Note that if [math]dy = 0[/math], an indeterminate expression of [math]\frac{0}{0}[/math] occurs in function [math]f_3(x)[/math], as indicated by a flat line in graph02.

 

However, if [math]dy \neq 0[/math] and exists as an infinitesimal, the indeterminate expression is resolved for all values of [math]x[/math] and [math]dx[/math], as indicated in exaggeration in graph03 and graph04.

 

Graph plots and differential shifts:

graph01 - [math]f_1(x),f_2(x), dx = 0, dy = 0[/math]

graph02 - [math]f_3(x), dx = 0, dy = 0[/math]

graph03 - [math]f_1(x),f_2(x), dx = 0, dy = 1[/math]

graph04 - [math]f_3(x), dx = 0, dy = 1[/math]

graph01.bmp

graph02.bmp

graph03.bmp

graph04.bmp

Edited by Orion1
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  • 3 weeks later...

If there's a 0/0 in a function, I don't think we can simply equate 0/0 with an existing limit at that point.

 

Consider a function [math]f(x)=\frac{g(x)}{h(x)}[/math], where [math]g(x)[/math] and [math]h(x)[/math] are non-constant differentiable functions. If there exists a constant [math]a[/math] such that [math]f(a)[/math] results in [math]\frac{0}{0}[/math], is [math]f(a)[/math] not always undefined? Regardless of any limit that may exist?

Edited by Amaton
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If you look at the algebraic equivalent of [math]\frac{0}{0} = x [/math] it's [math]x\times0 = 0 [/math]. In this case, x can literally be any number since anything times 0 is 0.

This is why I think the answer is appropriately called "undefined". You can literally put "0" into something any amount of times you want, the derivative of the vertex of absolute value(x) could literally be anything within the parameters of 360 inus the interior angle , literally any number times 0 could equal 0. It seems to make sense that it is "undefined" because there is nothing in mathematics forcing the result to be any particular number.

Edited by SamBridge
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