Dokazi bez rije ci - matematika.hr
Transcript of Dokazi bez rije ci - matematika.hr
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Dokazi bez rijeci
doc.dr.sc. Julije [email protected]
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Sume
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Suma prvih n neparnih brojeva
1 + 3 + · · · + (2n− 1) = n2
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Suma prvih n neparnih brojeva
1 + 3 = 22
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Suma prvih n neparnih brojeva
1 + 3 + 5 = 32
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Identitet za sumu prvih n brojeva
Sn = 1 + 2 + · · · + n
8Sn + 1 = (2n + 1)2
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Identitet za sumu prvih n brojeva
8 · 1 + 1 = 32
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Identitet za sumu prvih n brojeva
8 · (1 + 2) + 1 = 52
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Identitet za sumu prvih n brojeva
8 · (1 + 2 + 3) + 1 = 72
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Suma geometrijskog reda I
1/2
1/2
1/2
1/2
1/4
1/4
1/8
1/8
1
4+
1
16+
1
64+ · · · = 1
3
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Suma geometrijskog reda II
P
S T
Q
R1
1
r
r
r2
r2
r3
r3. . .
4PRQ ∼ 4STP
1 + r + r2 + · · · = 1
1− r
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Trisekcija kuta u beskonacnokoraka
1
2
34
1
3=
1
2− 1
4+
1
8− 1
16+ · · ·
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Ekvipotentnost
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#(a, b) = #R
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Identiteti za trokut
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Pitagorin poucak
c
a
b
c c− a
b
c + a=c− ab⇒ c2 = a2 + b2
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Kosinusov poucak
aa
a
2a cos(γ)− b
bc
a−c
γ
(a + c)(a− c) = (2a cos γ − b)b⇒ c2 = a2 + b2 − 2ab cos γ
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Suma arkus tangensa I
arctg1
2+ arctg
1
3=π
4
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Suma arkus tangensa II
arctg 1 + arctg 2 + arctg 3 = π
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Nejednakosti
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A-G nejednakost I
a b
√ab
a+b2
√ab ≤ a + b
2
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Nejednakost sredina
a b
HS
AS KS
GS
21a + 1
b
≤√ab ≤ a + b
2≤√a2 + b2
2
Povrsina trokuta = ab
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A-G nejednakost II-Lema
abbc
ac
⊆ a2
b2c2
ab + bc + ac ≤ a2 + b2 + c2
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A-G nejednakost II
abc
abc
abc
a b c
bc
ac
ab
⊆a3
b3
c3
a b c
a2
b2
c2
3abc ≤ a3 + b3 + c3
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Napierova nejednakost-prvi dokaz
0 < a < b⇒ 1
b<
ln b− ln a
b− a <1
ay
xa b
y = lnx
p1
p2
p3
n(p3) < n(p2) < n(p1)
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Napierova nejednakost-drugidokaz
y
xa b
y = 1x
1
b(b− a) <
b∫
a
1
xdx <
1
a(b− a)
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Eksponencijalna nejednakostAB > BA za e ≤ A < B
y = nAx
y = nBx
1 e A B
y = lnx
nA > nB ⇒lnA
A>
lnB
B⇒ AB > BA
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Limesi
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limn→∞
(1 +
1
n
)n= e
y
x1 1 + 1n
y = 1x
n
n + 1
1
n< ln
(1 +
1
n
)<
1
nn
n + 1< n ln
(1 +
1
n
)< 1
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√
2 +
√2 +
√2 +√· · · = 2
y
x2
√2
2 +√2
√2 +
√2
2 +√2 +
√2
√2 +
√2 +
√2
4
2
y=x− 2
y =√x
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Parcijalna integracija
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Parcijalna integracijav
up = f(a) q = f(b)
r = g(a)
s = g(b)u = f(x)
v = g(x)
s∫
r
udv +
q∫
p
vdu = uv∣∣∣(q,s)
(p,r)
b∫
a
f (x)g′(x)dx = f (x)g(x)∣∣∣b
a−
b∫
a
g(x)f ′(x)dx
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Dandelinove kugle
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Dandelinove kugle
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Hvala na paznji!