Proving Big theta for polynomials using quantificational definition
How can I prove Big Theta using quantificational definition? I know that u have to find 2 constants such that c1*g(n)<= f(n)<= c2*g(n)- but how do you find these constants?
Could anyone help me prove the following to show an example 5x3 − 7x2 + 5x + 1 = Θ(x3)?
big-o polynomial-math
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
add a comment |
How can I prove Big Theta using quantificational definition? I know that u have to find 2 constants such that c1*g(n)<= f(n)<= c2*g(n)- but how do you find these constants?
Could anyone help me prove the following to show an example 5x3 − 7x2 + 5x + 1 = Θ(x3)?
big-o polynomial-math
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
add a comment |
How can I prove Big Theta using quantificational definition? I know that u have to find 2 constants such that c1*g(n)<= f(n)<= c2*g(n)- but how do you find these constants?
Could anyone help me prove the following to show an example 5x3 − 7x2 + 5x + 1 = Θ(x3)?
big-o polynomial-math
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
How can I prove Big Theta using quantificational definition? I know that u have to find 2 constants such that c1*g(n)<= f(n)<= c2*g(n)- but how do you find these constants?
Could anyone help me prove the following to show an example 5x3 − 7x2 + 5x + 1 = Θ(x3)?
big-o polynomial-math
big-o polynomial-math
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
asked Nov 20 '18 at 12:56
MunchiesOatsMunchiesOats
62
62
add a comment |
add a comment |
1 Answer
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Let's assume x > 0, which is usually what we have.
5x^3 − 7x^2 + 5x + 1 <= 5x^3 + 5x + 1
<= 5x^3 + 5x^3 + x^3 ; x >= 1
= 11x^3
On the other hand
5x^3 − 7x^2 + 5x + 1 >= 5x^3 - 7x^2
>= 5x^3 - 4x^3 ; if 7x^2 <= 4x3, i.e. x >= 7/4
= x^3
In conclusion, for x >= 7/4 we have:
x^3 <= (5x^3 − 7x^2 + 5x + 1) <= 11x^3
and we are done.
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let's assume x > 0, which is usually what we have.
5x^3 − 7x^2 + 5x + 1 <= 5x^3 + 5x + 1
<= 5x^3 + 5x^3 + x^3 ; x >= 1
= 11x^3
On the other hand
5x^3 − 7x^2 + 5x + 1 >= 5x^3 - 7x^2
>= 5x^3 - 4x^3 ; if 7x^2 <= 4x3, i.e. x >= 7/4
= x^3
In conclusion, for x >= 7/4 we have:
x^3 <= (5x^3 − 7x^2 + 5x + 1) <= 11x^3
and we are done.
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
add a comment |
Let's assume x > 0, which is usually what we have.
5x^3 − 7x^2 + 5x + 1 <= 5x^3 + 5x + 1
<= 5x^3 + 5x^3 + x^3 ; x >= 1
= 11x^3
On the other hand
5x^3 − 7x^2 + 5x + 1 >= 5x^3 - 7x^2
>= 5x^3 - 4x^3 ; if 7x^2 <= 4x3, i.e. x >= 7/4
= x^3
In conclusion, for x >= 7/4 we have:
x^3 <= (5x^3 − 7x^2 + 5x + 1) <= 11x^3
and we are done.
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
add a comment |
Let's assume x > 0, which is usually what we have.
5x^3 − 7x^2 + 5x + 1 <= 5x^3 + 5x + 1
<= 5x^3 + 5x^3 + x^3 ; x >= 1
= 11x^3
On the other hand
5x^3 − 7x^2 + 5x + 1 >= 5x^3 - 7x^2
>= 5x^3 - 4x^3 ; if 7x^2 <= 4x3, i.e. x >= 7/4
= x^3
In conclusion, for x >= 7/4 we have:
x^3 <= (5x^3 − 7x^2 + 5x + 1) <= 11x^3
and we are done.
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
Let's assume x > 0, which is usually what we have.
5x^3 − 7x^2 + 5x + 1 <= 5x^3 + 5x + 1
<= 5x^3 + 5x^3 + x^3 ; x >= 1
= 11x^3
On the other hand
5x^3 − 7x^2 + 5x + 1 >= 5x^3 - 7x^2
>= 5x^3 - 4x^3 ; if 7x^2 <= 4x3, i.e. x >= 7/4
= x^3
In conclusion, for x >= 7/4 we have:
x^3 <= (5x^3 − 7x^2 + 5x + 1) <= 11x^3
and we are done.
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
edited Nov 20 '18 at 20:47
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
answered Nov 20 '18 at 16:11
Leandro CanigliaLeandro Caniglia
8,85122035
8,85122035
add a comment |
add a comment |
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