Need Help with Differentiation
$begingroup$
I came across below expression where I needed to find the value of K where the maximum value will occur for the below function
$P=binom{k}{2} times 2^{n-k}$
I re-wrote it as
$P=2^{n-1} times {frac{k^2-k}{2^k}}$
differentiating it with respect to k
$frac{dP}{dk}=2^{n-1} times Biggl( frac{(2k-1)-(k^2-k)(log2^k)}{2^k} Biggr)$
My query is, can I keep base of log in the above expression to 2?
I know I need to equate the above value to 0, to find the critical point.
Please help me further I am stuck here.
derivatives
asked Jan 11 at 7:05
user3767495user3767495
3938
$endgroup$
add a comment |
$begingroup$
I came across below expression where I needed to find the value of K where the maximum value will occur for the below function
$P=binom{k}{2} times 2^{n-k}$
I re-wrote it as
$P=2^{n-1} times {frac{k^2-k}{2^k}}$
differentiating it with respect to k
$frac{dP}{dk}=2^{n-1} times Biggl( frac{(2k-1)-(k^2-k)(log2^k)}{2^k} Biggr)$
My query is, can I keep base of log in the above expression to 2?
I know I need to equate the above value to 0, to find the critical point.
Please help me further I am stuck here.
derivatives
asked Jan 11 at 7:05
user3767495user3767495
3938
$endgroup$
1
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54
add a comment |
$begingroup$
I came across below expression where I needed to find the value of K where the maximum value will occur for the below function
$P=binom{k}{2} times 2^{n-k}$
I re-wrote it as
$P=2^{n-1} times {frac{k^2-k}{2^k}}$
differentiating it with respect to k
$frac{dP}{dk}=2^{n-1} times Biggl( frac{(2k-1)-(k^2-k)(log2^k)}{2^k} Biggr)$
My query is, can I keep base of log in the above expression to 2?
I know I need to equate the above value to 0, to find the critical point.
Please help me further I am stuck here.
derivatives
asked Jan 11 at 7:05
user3767495user3767495
3938
$endgroup$
I came across below expression where I needed to find the value of K where the maximum value will occur for the below function
$P=binom{k}{2} times 2^{n-k}$
I re-wrote it as
$P=2^{n-1} times {frac{k^2-k}{2^k}}$
differentiating it with respect to k
$frac{dP}{dk}=2^{n-1} times Biggl( frac{(2k-1)-(k^2-k)(log2^k)}{2^k} Biggr)$
My query is, can I keep base of log in the above expression to 2?
I know I need to equate the above value to 0, to find the critical point.
Please help me further I am stuck here.
derivatives
derivatives
asked Jan 11 at 7:05
user3767495user3767495
3938
asked Jan 11 at 7:05
user3767495user3767495
3938
asked Jan 11 at 7:05
user3767495user3767495
3938
asked Jan 11 at 7:05
user3767495user3767495
3938
asked Jan 11 at 7:05
user3767495user3767495
3938
3938
1
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54
add a comment |
1
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54
1
1
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54
add a comment |
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1
$begingroup$
The derivative of $2^k$ is $2^klog 2$.
$endgroup$
– Gaffney
Jan 11 at 7:09
$begingroup$
Yeah in the above shown step I have removed $2^k$ from both numerator and denominator.
$endgroup$
– user3767495
Jan 11 at 7:10
$begingroup$
Why not take logs and use $$log(m!) = sum_{j=1}^{m} log(j)$$ It should make life easier.
$endgroup$
– Mattos
Jan 11 at 7:12
$begingroup$
You will get $log2$, not $log2^k$, in the numerator. On equating to $0$, you will get a quadratic in $k$, which can be solved easily, though integral solutions are not guaranteed
$endgroup$
– Shubham Johri
Jan 11 at 7:54