Calculating upper and lower bound
Suppose there is a function
$ f(x)= 19n^2/5n +1-n $
I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2
or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.
So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?
functions asymptotics upper-lower-bounds
asked Nov 28 '18 at 11:07
Tom
31
add a comment |
Suppose there is a function
$ f(x)= 19n^2/5n +1-n $
I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2
or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.
So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?
functions asymptotics upper-lower-bounds
asked Nov 28 '18 at 11:07
Tom
31
$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
Suppose there is a function
$ f(x)= 19n^2/5n +1-n $
I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2
or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.
So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?
functions asymptotics upper-lower-bounds
asked Nov 28 '18 at 11:07
Tom
31
Suppose there is a function
$ f(x)= 19n^2/5n +1-n $
I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2
or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.
So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?
functions asymptotics upper-lower-bounds
functions asymptotics upper-lower-bounds
asked Nov 28 '18 at 11:07
Tom
31
asked Nov 28 '18 at 11:07
Tom
31
asked Nov 28 '18 at 11:07
Tom
31
asked Nov 28 '18 at 11:07
Tom
31
asked Nov 28 '18 at 11:07
Tom
31
31
$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31
$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31
add a comment |
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$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 '18 at 11:23
i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 '18 at 11:28
@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 '18 at 11:30
Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 '18 at 11:31