Differential of iterative function
$begingroup$
I previously got great help with understanding that
$$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$
always had a positive differential in H (differentiation of difficult function with unknown constants)
Now I wonder if we can generalize - the general next entry in the numerator would be
$$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be
$$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?
I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.
calculus
asked Jan 14 at 8:39
user37217user37217
12
$endgroup$
add a comment |
$begingroup$
I previously got great help with understanding that
$$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$
always had a positive differential in H (differentiation of difficult function with unknown constants)
Now I wonder if we can generalize - the general next entry in the numerator would be
$$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be
$$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?
I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.
calculus
asked Jan 14 at 8:39
user37217user37217
12
$endgroup$
add a comment |
$begingroup$
I previously got great help with understanding that
$$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$
always had a positive differential in H (differentiation of difficult function with unknown constants)
Now I wonder if we can generalize - the general next entry in the numerator would be
$$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be
$$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?
I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.
calculus
asked Jan 14 at 8:39
user37217user37217
12
$endgroup$
I previously got great help with understanding that
$$C = -Afrac{frac{K1}{f1^2H}+2frac{K1K2}{f1^2f2H^2}+3frac{K1K2K3}{f1^3f3H^3}}{1+frac{K1}{f1^2H}+frac{K1K2}{f1^2f2H^2}+frac{K1K2K3}{f1^3f3H^3}}$$
always had a positive differential in H (differentiation of difficult function with unknown constants)
Now I wonder if we can generalize - the general next entry in the numerator would be
$$ifrac{K1K2...Ki}{f1^ifiH^i}$$ and the next entry in the denominator would be
$$frac{K1K2...Ki}{f1^ifiH^i}$$ - it seems also from actual numeric experimenttion that it is indeed true that the general differential is still positive - but can we prove that?
I should add that all K and f values are positive and H (proton concentration) is also positive. K values are true thermodynamic constants, and f values, activity coefficients are assumed fixed for each value of total buffer concentration (A) as a first approximation.
calculus
calculus
asked Jan 14 at 8:39
user37217user37217
12
asked Jan 14 at 8:39
user37217user37217
12
edited Jan 18 at 16:02
user37217
asked Jan 14 at 8:39
user37217user37217
12
asked Jan 14 at 8:39
user37217user37217
12
asked Jan 14 at 8:39
user37217user37217
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not a proof.
Let $H=frac 1x$ and consider the function
$$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
$$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
x^iright)$$ The rhs is then a polynomial of degree $2(n-1)$ that is to say
$$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
$$left(
begin{array}{cc}
i & b_i \
0 & a_1 \
1 & 4 a_2 \
2 & a_1 a_2+9 a_3 \
3 & 4 a_1 a_3+16 a_4 \
4 & a_2 a_3+9 a_1 a_4+25 a_5 \
5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
15 & 4 a_7 a_9+16 a_6 a_{10} \
16 & a_8 a_9+9 a_7 a_{10} \
17 & 4 a_8 a_{10} \
18 & a_9 a_{10}
end{array}
right)$$
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is not a proof.
Let $H=frac 1x$ and consider the function
$$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
$$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
x^iright)$$ The rhs is then a polynomial of degree $2(n-1)$ that is to say
$$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
$$left(
begin{array}{cc}
i & b_i \
0 & a_1 \
1 & 4 a_2 \
2 & a_1 a_2+9 a_3 \
3 & 4 a_1 a_3+16 a_4 \
4 & a_2 a_3+9 a_1 a_4+25 a_5 \
5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
15 & 4 a_7 a_9+16 a_6 a_{10} \
16 & a_8 a_9+9 a_7 a_{10} \
17 & 4 a_8 a_{10} \
18 & a_9 a_{10}
end{array}
right)$$
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
add a comment |
$begingroup$
This is not a proof.
Let $H=frac 1x$ and consider the function
$$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
$$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
x^iright)$$ The rhs is then a polynomial of degree $2(n-1)$ that is to say
$$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
$$left(
begin{array}{cc}
i & b_i \
0 & a_1 \
1 & 4 a_2 \
2 & a_1 a_2+9 a_3 \
3 & 4 a_1 a_3+16 a_4 \
4 & a_2 a_3+9 a_1 a_4+25 a_5 \
5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
15 & 4 a_7 a_9+16 a_6 a_{10} \
16 & a_8 a_9+9 a_7 a_{10} \
17 & 4 a_8 a_{10} \
18 & a_9 a_{10}
end{array}
right)$$
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
add a comment |
$begingroup$
This is not a proof.
Let $H=frac 1x$ and consider the function
$$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
$$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
x^iright)$$ The rhs is then a polynomial of degree $2(n-1)$ that is to say
$$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
$$left(
begin{array}{cc}
i & b_i \
0 & a_1 \
1 & 4 a_2 \
2 & a_1 a_2+9 a_3 \
3 & 4 a_1 a_3+16 a_4 \
4 & a_2 a_3+9 a_1 a_4+25 a_5 \
5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
15 & 4 a_7 a_9+16 a_6 a_{10} \
16 & a_8 a_9+9 a_7 a_{10} \
17 & 4 a_8 a_{10} \
18 & a_9 a_{10}
end{array}
right)$$
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
This is not a proof.
Let $H=frac 1x$ and consider the function
$$f=frac{sum _{i=1}^n i, a_i, x^i}{1+sum _{i=1}^n a_i, x^i}$$ Compute the derivative to get
$$left(1+sum _{i=1}^n a_i, x^i right)^2, f'=left(sum _{i=1}^n i^2 a_i x^{i-1}right) left(1+sum _{i=1}^n a_i
x^iright)-left(sum _{i=1}^n i a_i x^{i-1}right)left( sum _{i=1}^n i a_i
x^iright)$$ The rhs is then a polynomial of degree $2(n-1)$ that is to say
$$text{rhs}=sum_{i=0}^{2(n-1)} b_i,x^i$$ in which the coefficients are all positive since all $a_i$'s are non negative. For example, using $n=10$, we should get
$$left(
begin{array}{cc}
i & b_i \
0 & a_1 \
1 & 4 a_2 \
2 & a_1 a_2+9 a_3 \
3 & 4 a_1 a_3+16 a_4 \
4 & a_2 a_3+9 a_1 a_4+25 a_5 \
5 & 4 a_2 a_4+16 a_1 a_5+36 a_6 \
6 & a_3 a_4+9 a_2 a_5+25 a_1 a_6+49 a_7 \
7 & 4 a_3 a_5+16 a_2 a_6+36 a_1 a_7+64 a_8 \
8 & a_4 a_5+9 a_3 a_6+25 a_2 a_7+49 a_1 a_8+81 a_9 \
9 & 4 a_4 a_6+16 a_3 a_7+36 a_2 a_8+64 a_1 a_9+100 a_{10} \
10 & a_5 a_6+9 a_4 a_7+25 a_3 a_8+49 a_2 a_9+81 a_1 a_{10} \
11 & 4 a_5 a_7+16 a_4 a_8+36 a_3 a_9+64 a_2 a_{10} \
12 & a_6 a_7+9 a_5 a_8+25 a_4 a_9+49 a_3 a_{10} \
13 & 4 a_6 a_8+16 a_5 a_9+36 a_4 a_{10} \
14 & a_7 a_8+9 a_6 a_9+25 a_5 a_{10} \
15 & 4 a_7 a_9+16 a_6 a_{10} \
16 & a_8 a_9+9 a_7 a_{10} \
17 & 4 a_8 a_{10} \
18 & a_9 a_{10}
end{array}
right)$$
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 22 at 13:00
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
add a comment |
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
Thanks a lot - but what is then the interpretation in regard tothe question of whether we caan know if thederivativeis positive?
$endgroup$
– user37217
Jan 26 at 8:06
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
@user37217. You have sums of only positive terms.
$endgroup$
– Claude Leibovici
Jan 26 at 8:43
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
$begingroup$
Thanks a lot - so this is indeed a proof? - nice!!!
$endgroup$
– user37217
Jan 26 at 13:35
add a comment |
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