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q-derivative of binomial
The 2019 Stack Overflow Developer Survey Results Are InA contradiction involving derivativeInequality for the Binomial TheoremLimit of Binomial distributionexplaining the derivative of $x^x$Proof derivative equals zero?Derivative of conditional expectationFlaw in proof involving binomial theoremWhat is the derivative of $frac1(1-x)^2$?Proof by derivative that 2=1Error in the derivative proof
$begingroup$
Problem
Find the q-derivative of $(a-x)_q^n$ for $n ge 1$.
Answer
This is actually equation (3.11) of Kac and Cheung's Quantum Calculus
$$D_q(a-x)_q^n=-[n](a-qx)_q^n-1.tag3.11$$
My question
I've got a different answer $-[n](a-x)_q^n-1$ by applying the fact that
$$D_q(x-a)_q^n=-[n](x-a)_q^n-1.tag3.5$$
with $a$ replaced by $-a$ (since $a$ can be any number) and the chain rule for $u(x)=alpha x^beta$
$$D_q f(u(x)) = (D_q^beta f) (u(x)) ; D_q u(x) tag1.15$$
with $u(x)=-x$ and $f(x)=(x+a)_q^n$.
Where's my mistake?
I understand the book's derivation (3.11) line by line, so please don't repeat the book's proof. Instead, please point out my mistakes.
fake-proofs quantum-calculus
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
add a comment |
$begingroup$
Problem
Find the q-derivative of $(a-x)_q^n$ for $n ge 1$.
Answer
This is actually equation (3.11) of Kac and Cheung's Quantum Calculus
$$D_q(a-x)_q^n=-[n](a-qx)_q^n-1.tag3.11$$
My question
I've got a different answer $-[n](a-x)_q^n-1$ by applying the fact that
$$D_q(x-a)_q^n=-[n](x-a)_q^n-1.tag3.5$$
with $a$ replaced by $-a$ (since $a$ can be any number) and the chain rule for $u(x)=alpha x^beta$
$$D_q f(u(x)) = (D_q^beta f) (u(x)) ; D_q u(x) tag1.15$$
with $u(x)=-x$ and $f(x)=(x+a)_q^n$.
Where's my mistake?
I understand the book's derivation (3.11) line by line, so please don't repeat the book's proof. Instead, please point out my mistakes.
fake-proofs quantum-calculus
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
add a comment |
$begingroup$
Problem
Find the q-derivative of $(a-x)_q^n$ for $n ge 1$.
Answer
This is actually equation (3.11) of Kac and Cheung's Quantum Calculus
$$D_q(a-x)_q^n=-[n](a-qx)_q^n-1.tag3.11$$
My question
I've got a different answer $-[n](a-x)_q^n-1$ by applying the fact that
$$D_q(x-a)_q^n=-[n](x-a)_q^n-1.tag3.5$$
with $a$ replaced by $-a$ (since $a$ can be any number) and the chain rule for $u(x)=alpha x^beta$
$$D_q f(u(x)) = (D_q^beta f) (u(x)) ; D_q u(x) tag1.15$$
with $u(x)=-x$ and $f(x)=(x+a)_q^n$.
Where's my mistake?
I understand the book's derivation (3.11) line by line, so please don't repeat the book's proof. Instead, please point out my mistakes.
fake-proofs quantum-calculus
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
Problem
Find the q-derivative of $(a-x)_q^n$ for $n ge 1$.
Answer
This is actually equation (3.11) of Kac and Cheung's Quantum Calculus
$$D_q(a-x)_q^n=-[n](a-qx)_q^n-1.tag3.11$$
My question
I've got a different answer $-[n](a-x)_q^n-1$ by applying the fact that
$$D_q(x-a)_q^n=-[n](x-a)_q^n-1.tag3.5$$
with $a$ replaced by $-a$ (since $a$ can be any number) and the chain rule for $u(x)=alpha x^beta$
$$D_q f(u(x)) = (D_q^beta f) (u(x)) ; D_q u(x) tag1.15$$
with $u(x)=-x$ and $f(x)=(x+a)_q^n$.
Where's my mistake?
I understand the book's derivation (3.11) line by line, so please don't repeat the book's proof. Instead, please point out my mistakes.
fake-proofs quantum-calculus
fake-proofs quantum-calculus
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
asked Mar 21 at 20:28
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
14k82651
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = begincases
1 & text if n = 0, tag3.4 \
(x-a)(x-qa) cdots (x-q^n-1 a) & text if n ge 1.
endcases$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^n-1$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^n-1$" instead.
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = begincases
1 & text if n = 0, tag3.4 \
(x-a)(x-qa) cdots (x-q^n-1 a) & text if n ge 1.
endcases$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^n-1$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^n-1$" instead.
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
add a comment |
$begingroup$
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = begincases
1 & text if n = 0, tag3.4 \
(x-a)(x-qa) cdots (x-q^n-1 a) & text if n ge 1.
endcases$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^n-1$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^n-1$" instead.
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
add a comment |
$begingroup$
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = begincases
1 & text if n = 0, tag3.4 \
(x-a)(x-qa) cdots (x-q^n-1 a) & text if n ge 1.
endcases$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^n-1$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^n-1$" instead.
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
$endgroup$
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = begincases
1 & text if n = 0, tag3.4 \
(x-a)(x-qa) cdots (x-q^n-1 a) & text if n ge 1.
endcases$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^n-1$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^n-1$" instead.
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
answered Mar 30 at 10:15
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
14k82651
14k82651
add a comment |
add a comment |
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