taeto 93 Posted April 10, 2020 Share Posted April 10, 2020 1 hour ago, DimaMazin said: What will be when ( Pi/infinity)*infinity ? From usual definitions it would be \(\pi/\infty = 0\) and \(0 \cdot \infty = 0,\) so that means \((\pi/\infty)\cdot \infty = 0.\) But it seems you are trying to treat \(\infty\) as if it were a natural number, and that has only small chance or working out. I do not think that you need to introduce \(\infty\) in the first place anyway. Link to post Share on other sites

DimaMazin 93 Posted July 25, 2020 Author Share Posted July 25, 2020 On 4/10/2020 at 10:17 PM, taeto said: From usual definitions it would be π/∞=0 and 0⋅∞=0, so that means (π/∞)⋅∞=0. But it seems you are trying to treat ∞ as if it were a natural number, and that has only small chance or working out. I do not think that you need to introduce ∞ in the first place anyway. Let's consider that stupid rule instead of true science because it will appear again and again . (Pi/million)*million=Pi (Pi/billion)*billion=Pi There is no trend to reduce result therefore so called scientists should prove that billion is not nearer to infinity than million. -1 Link to post Share on other sites

DimaMazin 93 Posted July 26, 2020 Author Share Posted July 26, 2020 On 4/10/2020 at 3:16 PM, DimaMazin said: Yes. Let's consider next thing: Arc Pi -2 a is divided for infinite quantity equal parts and its chord is divided for infinite quantity of equal parts.The straight line intersects the nearest points of the divisions to its middle . The point on arc has coordinates([Pi-2a]/infinity ; 1), the point on chord has coordinates (cos(a)/infinity ; sin(a)). Their straight line intersects y axis in point(0; y_{1}) . Increased side (Pi - 2a)/infinity of the triangle in infinite quantity of times on tangent in point(0 ; 1) creates new point on its edge with coordinates ([Pi - 2a] ; 1). Increased side cos(a)/infinity of the smaller triangle in infinite quantity times has edge point(cos(a) ; sin(a)). The new straight line of the new points intersects y axis in the same point (0 ; y_{1}). y_{1} = (2a - Pi*sin^{2}(a)) / (2a*sin(a) - Pi*sin(a)+2a*cos(a)) Some similar thing we can make with edge points of the division. Then their derivative points lie on straight line which intersects y axis in point(0 ; y_{2}) I did not make equation for y_{2}. But feature exists there : y_{1} is lower than y_{2} when arc is bigger than Pi - 2a and y_{1} is upper than y_{2} when arc is smaller than Pi - 2a It can prove that the arc of definition of trigonometric functions exists. I mistaken there y_{1}=[sin(a)*(Pi - 2a) - 2cos(a)] / [Pi-2cos(a)-2a] y_{2}= [2sin(a)*cos(a)+2a - Pi ] / [2cos(a) - sin(a)*(Pi - 2a)] Now everyone can define is any arc less or more than the arc of definition. Link to post Share on other sites

DimaMazin 93 Posted December 18, 2020 Author Share Posted December 18, 2020 (edited) I have got such complex equation a= ((3^{1/2} - 2^{1/2})cos(a)+(2*2^{1/2} - 3)sin(a)+(3*2^{1/2} -2*6^{1/2})/2))*Pi / ((4*2^{1/2} - 6)sin(a) - 2*6^{1/2}+3*2^{1/2})= ={4Pi - 4cos(a) + 4sin(a)cos(a) - 4Pi*sin^{2}(a) - [(4cos(a) - 4Pi - 4sin(a)cos(a)+4Pi*sin^{2}(a))^{2} - (16 - 16sin^{2}(a))*(Pi^{2} - 2Pi*cos(a)+4sin(a)cos^{2}(a) - Pi^{2}sin^{2}(a)+2Pi*sin(a)cos(a))]^{1/2}} / {8-8sin^{2}(a)} Theoretically it can be solved, then sin(a) , cos(a) and a will be known. Then we can not only exactly define trigonometric functions but and write them down as numbers. Ofcourse we can describe trigonometric functions which relate to Pi by nambers with exponents and mathematical actions only relative to Pi. But it maybe usefule for solve of Travelling salesman problem. 😛 Edited December 18, 2020 by DimaMazin Link to post Share on other sites

DimaMazin 93 Posted January 18 Author Share Posted January 18 The idea is wrong because next formula must work but it is not working t/(1-y)=sin t/(cos t -y) y=(t cost-sint)/(t-sint) Link to post Share on other sites

DimaMazin 93 Posted February 27 Author Share Posted February 27 On 1/18/2021 at 11:24 AM, DimaMazin said: The idea is wrong because next formula must work but it is not working t/(1-y)=sin t/(cos t -y) y=(t cost-sint)/(t-sint) Rather that is wrong because t is not obligated to be proportional to ( Pi-2a)/2 at level 1. Link to post Share on other sites

Country Boy 69 Posted March 5 Share Posted March 5 On 4/10/2020 at 2:17 PM, taeto said: From usual definitions it would be π/∞=0 and 0⋅∞=0, so that means (π/∞)⋅∞=0. But it seems you are trying to treat ∞ as if it were a natural number, and that has only small chance or working out. I do not think that you need to introduce ∞ in the first place anyway. No. [tex]0*\infty[/tex] is NOT 0, it is "undefined". [tex]\frac{\pi}{\infty}*\infty[/tex] is "undefined". Link to post Share on other sites

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