A Mechanized Proof of the Curve Length of a Rectifiable Curve - PowerPoint PPT Presentation

A Mechanized Proof of the Curve Length of a Rectifiable Curve Jagadish Bapanapally and Ruben Gamboa ACL2 Workshop 2017 May 23, 2017

Theory L ≈ � n i =1 | P i − P i − 1 |

Deriving length of a continuously differentiable curve � n L = lim n →∞ i =1 | P i − P i − 1 |

Deriving length of a continuously differentiable curve � n L = lim n →∞ i =1 | P i − P i − 1 | � n = lim n →∞ i =1 | f ( t i ) − f ( t i − 1 ) | { f ( t ) = x ( t )+ i ∗ y ( t ) , t 0 ≤ t ≤ t n }

Deriving length of a continuously differentiable curve � n L = lim n →∞ i =1 | P i − P i − 1 | � n = lim n →∞ i =1 | f ( t i ) − f ( t i − 1 ) | { f ( t ) = x ( t )+ i ∗ y ( t ) , t 0 ≤ t ≤ t n } i =1 | f ( t i ) − f ( t i − 1 ) � n = lim n →∞ | ∆ t ∆ t

Deriving length of a continuously differentiable curve � n L = lim n →∞ i =1 | P i − P i − 1 | � n = lim n →∞ i =1 | f ( t i ) − f ( t i − 1 ) | { f ( t ) = x ( t )+ i ∗ y ( t ) , t 0 ≤ t ≤ t n } i =1 | f ( t i ) − f ( t i − 1 ) � n = lim n →∞ | ∆ t ∆ t � t n t 0 | f ′ ( t ) | dt =

Deriving length of a continuously differentiable curve � n L = lim n →∞ i =1 | P i − P i − 1 | � n = lim n →∞ i =1 | f ( t i ) − f ( t i − 1 ) | { f ( t ) = x ( t )+ i ∗ y ( t ) , t 0 ≤ t ≤ t n } i =1 | f ( t i ) − f ( t i − 1 ) � n = lim n →∞ | ∆ t ∆ t � t n t 0 | f ′ ( t ) | dt = �� dx � 2 � 2 � dy � t n = + dt t 0 dt dt

Continuously Differentiable curve ( encapsulate (( c ( x ) t ) ( c − d e r i v a t i v e ( x ) t )) ; ; Our witness continuous f u n c t i o n i s the i d e n t i t y ( l o c a l ( defun c ( x ) x )) ( l o c a l ( defun c − d e r i v a t i v e ( x ) ( d e c l a r e ( i g n o r e x )) 1 )) ; ( i − close (/ ( − ( c x ) ( c y )) ( − x y )) ; ( c − d e r i v a t i v e x )) ; ( i − close ( c − d e r i v a t i v e x ) ( c − d e r i v a t i v e y )) )

Norm of the derivative of a continuous function is continuous ◮ Continuously differentiable function means it’s derivative is continuous.

Norm of the derivative of a continuous function is continuous ◮ Continuously differentiable function means it’s derivative is continuous. ◮ real and imaginary parts of a continuous function are continuous. ( dx dt , dy dt are continuous)

Norm of the derivative of a continuous function is continuous ◮ Continuously differentiable function means it’s derivative is continuous. ◮ real and imaginary parts of a continuous function are continuous. ( dx dt , dy dt are continuous) ◮ Square of a continuous function is continuous. � 2 � 2 � dx � dy , are continuous dt dt

Norm of the derivative of a continuous function is continuous ◮ Continuously differentiable function means it’s derivative is continuous. ◮ real and imaginary parts of a continuous function are continuous. ( dx dt , dy dt are continuous) ◮ Square of a continuous function is continuous. � 2 � 2 � dx � dy , are continuous dt dt ◮ Sum of 2 continuous functions is continuous. � 2 � 2 � dx � dy + is continuous dt dt

Square root of a continuous function is Continuous ( i m p l i e s ( and ( realp y1 ) ( realp y2 ) ( i − l i m i t e d y1 ) ( i − l i m i t e d y2 ) ( > = y1 0) ( > = y2 0) ( not ( i − close y1 y2 ) ) ) ( not (= ( standard − part ( square y1 )) ( standard − part ( square y2 ) ) ) ) ) ( i m p l i e s ( and ( realp y1 ) ( realp y2 ) ( i − l i m i t e d y1 ) ( i − l i m i t e d y2 ) ( > = y1 0) ( > = y2 0) ( not ( i − close y1 y2 ) ) ) ( not ( i − close ( square y1 ) ( square y2 ) ) ) )

Square root of a continuous function is Continuous ( defthmd root − close − f ( i m p l i e s ( and ( standardp x1 ) ( realp x1 ) ( realp x2 ) ( > = x1 0) ( > = x2 0) ( i − close x1 x2 )) ( i − close ( acl2 − sqrt x1 ) ( acl2 − sqrt x2 ) ) ) ; h i n t s omitted ) �� dx � 2 � 2 � dy + is continuous ∴ dt dt

Riemann sum of the lengths of the chords �� dx � 2 � 2 � dy ◮ Let h ( t ) = + dt dt

Riemann sum of the lengths of the chords �� dx � 2 � 2 � dy ◮ Let h ( t ) = + dt dt ◮ Riemann sum for the partition ( t 0 , t 1 , t 2 , ..., t n ) is h ( t 1 ) . ∆ t + h ( t 2 ) . ∆ t + ..... h ( t n ) . ∆ t

Riemann sum of the lengths of the chords �� dx � 2 � 2 � dy ◮ Let h ( t ) = + dt dt ◮ Riemann sum for the partition ( t 0 , t 1 , t 2 , ..., t n ) is h ( t 1 ) . ∆ t + h ( t 2 ) . ∆ t + ..... h ( t n ) . ∆ t ◮ We can prove this is limited using limited − riemann − rcfn − small − partition in continuous − function book

Riemann sum of the lengths of the chords �� dx � 2 � 2 � dy ◮ Let h ( t ) = + dt dt ◮ Riemann sum for the partition ( t 0 , t 1 , t 2 , ..., t n ) is h ( t 1 ) . ∆ t + h ( t 2 ) . ∆ t + ..... h ( t n ) . ∆ t ◮ We can prove this is limited using limited − riemann − rcfn − small − partition in continuous − function book ◮ Thus as n → ∞ , ∆ t is infinitely small and riemann sum is � t n equal to t 0 h ( t ) dt

Circumference of a circle with radius r Circle with radius r (standard and real number) can defined as f ( t ) = r ∗ e it = r ∗ (cos t + i ∗ sin t ) , 0 ≤ t ≤ 2 π

Circumference of a circle with radius r Circle with radius r (standard and real number) can defined as f ( t ) = r ∗ e it = r ∗ (cos t + i ∗ sin t ) , 0 ≤ t ≤ 2 π let, g ( t ) = r ∗ ( − sin t + i ∗ cos t )

Circumference of a circle with radius r Circle with radius r (standard and real number) can defined as f ( t ) = r ∗ e it = r ∗ (cos t + i ∗ sin t ) , 0 ≤ t ≤ 2 π let, g ( t ) = r ∗ ( − sin t + i ∗ cos t ) Since, d d dt cos t = − sin t and dt sin t = cos t , f ′ ( t ) ≈ g ( t )

Circumference of a circle with radius r Circle with radius r (standard and real number) can defined as f ( t ) = r ∗ e it = r ∗ (cos t + i ∗ sin t ) , 0 ≤ t ≤ 2 π let, g ( t ) = r ∗ ( − sin t + i ∗ cos t ) Since, d d dt cos t = − sin t and dt sin t = cos t , f ′ ( t ) ≈ g ( t ) Since sin t and cos t are continuous, g ( t ) is continuous.

Circumference of a circle with radius r Circle with radius r (standard and real number) can defined as f ( t ) = r ∗ e it = r ∗ (cos t + i ∗ sin t ) , 0 ≤ t ≤ 2 π let, g ( t ) = r ∗ ( − sin t + i ∗ cos t ) Since, d d dt cos t = − sin t and dt sin t = cos t , f ′ ( t ) ≈ g ( t ) Since sin t and cos t are continuous, g ( t ) is continuous. Thus by using above proof length of f ( t ) is equal to � 2 π | g ( t ) | dt 0

Applying second Fundamental Theorem of Calculus g ( t ) = r ∗ ( − sin t + i ∗ cos t ) � 2 π � 2 π | g ( t ) | = r ; | g ( t ) | dt = r dt ∴ 0 0 h ′ ( t ) = | g ( t ) | Let, h ( t ) = r ∗ t , ∴ Using second fundamental theorem of calculus � 2 π | g ( t ) | dt = h (2 π ) − h (0) = r ∗ 2 ∗ π 0