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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 - - 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 =
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Deriving length of a continuously differentiable curve
L = limn→∞ n
i=1 | Pi − Pi−1 |
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Deriving length of a continuously differentiable curve
L = limn→∞ n
i=1 | Pi − Pi−1 |
= limn→∞ n
i=1 | f (ti) −f (ti−1) | {f (t) = x(t)+i ∗y(t), t0 ≤ t ≤ tn}
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Deriving length of a continuously differentiable curve
L = limn→∞ n
i=1 | Pi − Pi−1 |
= limn→∞ n
i=1 | f (ti) −f (ti−1) | {f (t) = x(t)+i ∗y(t), t0 ≤ t ≤ tn}
= limn→∞ n
i=1 | f (ti) − f (ti−1)
∆t | ∆t
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Deriving length of a continuously differentiable curve
L = limn→∞ n
i=1 | Pi − Pi−1 |
= limn→∞ n
i=1 | f (ti) −f (ti−1) | {f (t) = x(t)+i ∗y(t), t0 ≤ t ≤ tn}
= limn→∞ n
i=1 | f (ti) − f (ti−1)
∆t | ∆t = tn
t0 |f ′(t)| dt
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Deriving length of a continuously differentiable curve
L = limn→∞ n
i=1 | Pi − Pi−1 |
= limn→∞ n
i=1 | f (ti) −f (ti−1) | {f (t) = x(t)+i ∗y(t), t0 ≤ t ≤ tn}
= limn→∞ n
i=1 | f (ti) − f (ti−1)
∆t | ∆t = tn
t0 |f ′(t)| dt
= tn
t0
dx dt 2 + dy dt 2 dt
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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 )) )
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Norm of the derivative of a continuous function is continuous
◮ Continuously differentiable function means it’s derivative is
continuous.
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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)
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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.
dx dt 2 , dy dt 2 are continuous
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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.
dx dt 2 , dy dt 2 are continuous
◮ Sum of 2 continuous functions is continuous.
dx dt 2 + dy dt 2 is continuous
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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 ) ) ) )
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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
- mitted
) ∴ dx dt 2 + dy dt 2 is continuous
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Riemann sum of the lengths of the chords
◮ Let h(t) =
dx dt 2 + dy dt 2
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Riemann sum of the lengths of the chords
◮ Let h(t) =
dx dt 2 + dy dt 2
◮ Riemann sum for the partition (t0, t1, t2, ..., tn) is
h(t1).∆t + h(t2).∆t + .....h(tn).∆t
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Riemann sum of the lengths of the chords
◮ Let h(t) =
dx dt 2 + dy dt 2
◮ Riemann sum for the partition (t0, t1, t2, ..., tn) is
h(t1).∆t + h(t2).∆t + .....h(tn).∆t
◮ We can prove this is limited using
limited − riemann − rcfn − small − partition in continuous − function book
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Riemann sum of the lengths of the chords
◮ Let h(t) =
dx dt 2 + dy dt 2
◮ Riemann sum for the partition (t0, t1, t2, ..., tn) is
h(t1).∆t + h(t2).∆t + .....h(tn).∆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
equal to tn
t0 h(t)dt
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Circumference of a circle with radius r
Circle with radius r (standard and real number) can defined as f (t) = r ∗ eit = r ∗ (cos t + i ∗ sin t), 0 ≤ t ≤ 2π
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Circumference of a circle with radius r
Circle with radius r (standard and real number) can defined as f (t) = r ∗ eit = r ∗ (cos t + i ∗ sin t), 0 ≤ t ≤ 2π let, g(t) = r ∗ (− sin t + i ∗ cos t)
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Circumference of a circle with radius r
Circle with radius r (standard and real number) can defined as f (t) = r ∗ eit = r ∗ (cos t + i ∗ sin t), 0 ≤ t ≤ 2π let, g(t) = r ∗ (− sin t + i ∗ cos t) Since, d
dt cos t = − sin t
and
d dt sin t = cos t,
f ′(t) ≈ g(t)
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Circumference of a circle with radius r
Circle with radius r (standard and real number) can defined as f (t) = r ∗ eit = r ∗ (cos t + i ∗ sin t), 0 ≤ t ≤ 2π let, g(t) = r ∗ (− sin t + i ∗ cos t) Since, d
dt cos t = − sin t
and
d dt sin t = cos t,
f ′(t) ≈ g(t) Since sin t and cos t are continuous, g(t) is continuous.
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Circumference of a circle with radius r
Circle with radius r (standard and real number) can defined as f (t) = r ∗ eit = r ∗ (cos t + i ∗ sin t), 0 ≤ t ≤ 2π let, g(t) = r ∗ (− sin t + i ∗ cos t) Since, d
dt cos t = − sin t
and
d 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 2π |g(t)| dt
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