interpretation "Rplus" 'plus x y =
(cic:/matita/demo/power_derivative/Rplus.con x y).
+
+notation "hvbox(a break \middot b)"
+ left associative with precedence 55
+for @{ 'times $a $b }.
+
interpretation "Rmult" 'times x y =
(cic:/matita/demo/power_derivative/Rmult.con x y).
λf,g:R→R.λx:R.f x + g x.
definition Fmult ≝
- λf,g:R→R.λx:R.f x * g x.
+ λf,g:R→R.λx:R.f x · g x.
interpretation "Fplus" 'plus x y =
(cic:/matita/demo/power_derivative/Fplus.con x y).
let rec Rpower (x:R) (n:nat) on n ≝
match n with
[ O ⇒ 1
- | S n ⇒ x * (Rpower x n)
+ | S n ⇒ x · (Rpower x n)
].
interpretation "Rpower" 'exp x n =
axiom f_eq_extensional:
∀f,g:R→R.(∀x:R.f x = g x) → f=g.
-lemma Fmult_one_f: ∀f:R→R.1*f=f.
+lemma Fmult_one_f: ∀f:R→R.1·f=f.
intro;
unfold Fmult;
simplify;
auto.
qed.
-lemma Fmult_zero_f: ∀f:R→R.0*f=0.
+lemma Fmult_zero_f: ∀f:R→R.0·f=0.
intro;
unfold Fmult;
simplify;
qed.
lemma monomio_product:
- ∀n,m.monomio (n+m) = monomio n * monomio m.
+ ∀n,m.monomio (n+m) = monomio n · monomio m.
intros;
unfold monomio;
unfold Fmult;
| simplify;
apply f_eq_extensional;
intro;
- cut (x\sup (n1+m) = x \sup n1 * x \sup m);
+ cut (x\sup (n1+m) = x \sup n1 · x \sup m);
[ rewrite > Hcut;
auto
- | change in ⊢ (? ? % ?) with ((λx:R.(x)\sup(n1+m)) x);
+ | change in ⊢ (? ? % ?) with ((λx:R.x\sup(n1+m)) x);
rewrite > H;
reflexivity
]
axiom derivative_x0: D[x \sup 0] = 0.
axiom derivative_x1: D[x] = 1.
-axiom derivative_mult: ∀f,g:R→R. D[f*g] = D[f]*g + f*D[g].
+axiom derivative_mult: ∀f,g:R→R. D[f·g] = D[f]·g + f·D[g].
alias symbol "times" = "Fmult".
-theorem derivative_power: ∀n:nat. D[x \sup n] = n*x \sup (pred n).
+theorem derivative_power: ∀n:nat. D[x \sup n] = n·x \sup (pred n).
assume n:nat.
we proceed by induction on n to prove
- (D[x \sup n] = n*x \sup (pred n)).
+ (D[x \sup n] = n · x \sup (pred n)).
case O.
- the thesis becomes (D[x \sup 0] = 0*x \sup (pred 0)).
+ the thesis becomes (D[x \sup 0] = 0·x \sup (pred 0)).
by _
done.
case S (m:nat).
by induction hypothesis we know
- (D[x \sup m] = m*x \sup (pred m)) (H).
+ (D[x \sup m] = m·x \sup (pred m)) (H).
the thesis becomes
- (D[x \sup (1+m)] = (1+m)*x \sup m).
+ (D[x \sup (1+m)] = (1+m) · x \sup m).
we need to prove
- (m * (x \sup (1+ pred m)) = m * x \sup m) (Ppred).
+ (m · (x \sup (1+ pred m)) = m · x \sup m) (Ppred).
by _ we proved (0 < m ∨ 0=m) (cases).
we proceed by induction on cases
- to prove (m * (x \sup (1+ pred m)) = m * x \sup m).
+ to prove (m · (x \sup (1+ pred m)) = m · x \sup m).
case left.
suppose (0 < m) (m_pos).
by (S_pred m m_pos) we proved (m = 1 + pred m) (H1).
suppose (0=m) (m_zero). by _ done.
conclude
(D[x \sup (1+m)])
- = (D[x * x \sup m]) by _.
- = (D[x] * x \sup m + x * D[x \sup m]) by _.
- = (x \sup m + x * (m * x \sup (pred m))) by _.
+ = (D[x · x \sup m]) by _.
+ = (D[x] · x \sup m + x · D[x \sup m]) by _.
+ = (x \sup m + x · (m · x \sup (pred m))) by _.
clear H.
- = (x \sup m + m * (x \sup (1 + pred m))) by _.
- = (x \sup m + m * x \sup m) by _.
- = ((1+m)*x \sup m) by _ (timeout=30)
+ = (x \sup m + m · (x \sup (1 + pred m))) by _.
+ = (x \sup m + m · x \sup m) by _.
+ = ((1+m) · x \sup m) by _ (timeout=30)
done.
qed.
interpretation "Rderivative" 'derivative f =
(cic:/matita/demo/power_derivative/derivative.con f).
-theorem derivative_power': ∀n:nat. D[x \sup (1+n)] = (1+n)*x \sup n.
+theorem derivative_power': ∀n:nat. D[x \sup (1+n)] = (1+n) · x \sup n.
assume n:nat.
we proceed by induction on n to prove
- (D[x \sup (1+n)] = (1+n)*x \sup n).
+ (D[x \sup (1+n)] = (1+n) · x \sup n).
case O.
- the thesis becomes (D[x \sup 1] = 1*x \sup 0).
+ the thesis becomes (D[x \sup 1] = 1 · x \sup 0).
by _
done.
case S (m:nat).
by induction hypothesis we know
- (D[x \sup (1+m)] = (1+m)*x \sup m) (H).
+ (D[x \sup (1+m)] = (1+m) · x \sup m) (H).
the thesis becomes
- (D[x \sup (2+m)] = (2+m)*x \sup (1+m)).
+ (D[x \sup (2+m)] = (2+m) · x \sup (1+m)).
conclude
(D[x \sup (2+m)])
- = (D[x \sup 1 * x \sup (1+m)]) by _.
- = (D[x \sup 1] * x \sup (1+m) + x * D[x \sup (1+m)]) by _.
- = (x \sup (1+m) + x * (costante (1+m) * x \sup m)) by _.
+ = (D[x \sup 1 · x \sup (1+m)]) by _.
+ = (D[x \sup 1] · x \sup (1+m) + x · D[x \sup (1+m)]) by _.
+ = (x \sup (1+m) + x · (costante (1+m) · x \sup m)) by _.
clear H.
- = (x \sup (1+m) + costante (1+m) * x \sup (1+m)) by _.
- = (x \sup (1+m) * (costante (2 + m))) by _
+ = (x \sup (1+m) + costante (1+m) · x \sup (1+m)) by _.
+ = (x \sup (1+m) · (costante (2 + m))) by _
done.
-qed.
+qed.
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