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Theorem 1: a n x n + ... + a 1 x + a 0 is O ( x n ) for any real numbers a n , ..., a 0 and any nonnegative number n .
an xn + ... + a1 x + a0 is O( xn ) for any real numbers an , ..., a0 and any nonnegative number n .
Theorem 1: calling makeLayer with valid inputs (a list of weights and two non-zero natural numbers) returns a valid layer recognized by isLayer. (“layers.lisp”)
Theorem 2: calling makeNetwork with valid inputs (a list of non-zero natural numbers and a list of weights) returns a valid network recognized by isNetwork. (“networks.lisp”)
calling makeLayer with valid inputs (a list of weights and two non-zero natural numbers) returns a valid layer recognized by isLayer. (“layers.lisp”)
calling makeNetwork with valid inputs (a list of non-zero natural numbers and a list of weights) returns a valid network recognized by isNetwork. (“networks.lisp”)
Theorem 4: a n x n + ... + a 1 x + a 0 is θ ( x n ) for any real numbers a n , ..., a 0 and any nonnegative number n .
Let f ( x ) and g ( x ) be functions from a set of real numbers to a set of real numbers.
Then
1. If $f(x)/g(x)=0$ , then f ( x ) is o ( g ( x ) ) . Note that if f ( x ) is o ( g ( x ) ), then f ( x ) is O ( g ( x ) ).
2. If $f(x)/g(x)=\infty $ , then g ( x ) is o ( f ( x ) ) .
3. If $f(x)/g(x)<\infty $ , then f ( x ) is θ ( g ( x ) ) .
4. If $f(x)/g(x)<\infty $ , then f ( x ) is O ( g ( x ) ) .
For example,
$$ (4x ^{3} + 3x ^{2} + 5)/(x ^{4} – 3x ^{3} – 5x -4)
= $$ ( 4/x + 3/x 2 + 5/x 4 )/( 1 - 3/x - 5/x 3 - 4/x 4 ) = 0 .
Hence
( 4x 3 + 3x 2 + 5 ) is o ( x 4 - 3x 3 - 5x - 4 ),
or equivalently, ( x 4 - 3x 3 - 5x - 4 ) is ω ( 4x 3 + 3x 2 + 5 ) .
Let us see why these rules hold. Here we give a proof for 4. Others can be proven similarly.
Proof : Suppose $f(x)/g(x){\text{=C}}_{1}<\infty $ .
By the definition of limit this means that
∀ε>0, ∃ n _{0} such that | f(x)/g(x) – C 1 |<ε whenever x> n _{0}
Hence –ε _{} < f(x)/g(x) – C 1 _{} <ε
Hence –ε _{} + C 1 _{} < f(x)/g(x) _{} <ε + C 1
In particular f(x)/g(x) _{} <ε + C 1
Hence f(x) _{} <(ε + C 1 ) g(x)
Let C = ε + C 1 , then f(x) _{} < Cg(x) whenever x> n _{0 .}
Since we are interested in non-negative functions f and g , this means that | f ( x ) | ≤ C | g ( x ) |
Hence f ( x ) = O ( g ( x ) ) .
an xn + ... + a1 x + a0 is θ( xn ) for any real numbers an , ..., a0 and any nonnegative number n .
Let f(x) and g(x) be functions from a set of real numbers to a set of real numbers.
Then
1. If $f(x)/g(x)=0$ , then f(x) is o( g(x) ) . Note that if f(x) is o( g(x) ), then f(x) is O( g(x) ).
2. If $f(x)/g(x)=\infty $ , then g(x) is o( f(x) ) .
3. If $f(x)/g(x)<\infty $ , then f(x) is θ( g(x) ) .
4. If $f(x)/g(x)<\infty $ , then f(x) is O( g(x) ) .
For example,
$$ (4x3 + 3x2 + 5)/(x4 – 3x3 – 5x -4)
= $$ ( 4/x + 3/x2 + 5/x4 )/(1 - 3/x - 5/x3 - 4/x4 ) = 0 .
Hence
( 4x3 + 3x2 + 5 ) is o(x4 - 3x3 - 5x - 4 ),
or equivalently, (x4 - 3x3 - 5x - 4 ) is ω(4x3 + 3x2 + 5 ) .
Let us see why these rules hold. Here we give a proof for 4. Others can be proven similarly.
Proof: Suppose $f(x)/g(x){\text{=C}}_{1}<\infty $ .
By the definition of limit this means that
∀ε>0, ∃n0 such that |f(x)/g(x) – C1|<ε whenever x>n0
Hence –ε<f(x)/g(x) – C1<ε
Hence –ε +C1<f(x)/g(x)<ε +C1
In particular f(x)/g(x)<ε +C1
Hence f(x)<(ε +C1)g(x)
Let C = ε +C1 , then f(x)<Cg(x) whenever x>n0 .
Since we are interested in non-negative functions f and g, this means that |f(x) | ≤C | g(x) |
Hence f(x) = O( g(x) ) .
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