Taylor series of “function of functions”
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I have crossed a problem, and it took me about 3 days. I wonder how I can find the Taylor series of "function of functions".
To be explicit, I am looking for the Taylor series of
$$y(s)=f(u_1(s),u_2(s),s)$$
I think the Taylor series at t in the T neighborhood of t-T (Where T is small enough) is as follows. Am I right?
$$y(t)=y(t-T)+left.frac{partial f}{partial u_1}frac{partial u_1}{partial s}right|_{s=t-T}u_1(t-T)+left.frac{partial f}{partial u_2}frac{partial u_2}{partial s}right|_{s=t-T}u_2(t-T)+left.frac{partial f}{partial s}right|_{s=t-T}(t-T)+cdots $$
I want it to become as follows:
$$y(t)=theta_1(t) u_1(t)+theta_2(t) u_2(t)+theta_3(t) $$
taylor-expansion
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
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add a comment |
$begingroup$
I have crossed a problem, and it took me about 3 days. I wonder how I can find the Taylor series of "function of functions".
To be explicit, I am looking for the Taylor series of
$$y(s)=f(u_1(s),u_2(s),s)$$
I think the Taylor series at t in the T neighborhood of t-T (Where T is small enough) is as follows. Am I right?
$$y(t)=y(t-T)+left.frac{partial f}{partial u_1}frac{partial u_1}{partial s}right|_{s=t-T}u_1(t-T)+left.frac{partial f}{partial u_2}frac{partial u_2}{partial s}right|_{s=t-T}u_2(t-T)+left.frac{partial f}{partial s}right|_{s=t-T}(t-T)+cdots $$
I want it to become as follows:
$$y(t)=theta_1(t) u_1(t)+theta_2(t) u_2(t)+theta_3(t) $$
taylor-expansion
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
$endgroup$
3
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You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
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– Ian
Dec 25 '18 at 15:29
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@Ian would you please post the complete answer?
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– V.Ajall
Dec 25 '18 at 16:13
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This is Faa di Bruno's formula, basically. See the wikipedia article
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– kimchi lover
Dec 25 '18 at 16:31
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Still looking for the solution
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– V.Ajall
Dec 27 '18 at 11:38
add a comment |
$begingroup$
I have crossed a problem, and it took me about 3 days. I wonder how I can find the Taylor series of "function of functions".
To be explicit, I am looking for the Taylor series of
$$y(s)=f(u_1(s),u_2(s),s)$$
I think the Taylor series at t in the T neighborhood of t-T (Where T is small enough) is as follows. Am I right?
$$y(t)=y(t-T)+left.frac{partial f}{partial u_1}frac{partial u_1}{partial s}right|_{s=t-T}u_1(t-T)+left.frac{partial f}{partial u_2}frac{partial u_2}{partial s}right|_{s=t-T}u_2(t-T)+left.frac{partial f}{partial s}right|_{s=t-T}(t-T)+cdots $$
I want it to become as follows:
$$y(t)=theta_1(t) u_1(t)+theta_2(t) u_2(t)+theta_3(t) $$
taylor-expansion
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
$endgroup$
I have crossed a problem, and it took me about 3 days. I wonder how I can find the Taylor series of "function of functions".
To be explicit, I am looking for the Taylor series of
$$y(s)=f(u_1(s),u_2(s),s)$$
I think the Taylor series at t in the T neighborhood of t-T (Where T is small enough) is as follows. Am I right?
$$y(t)=y(t-T)+left.frac{partial f}{partial u_1}frac{partial u_1}{partial s}right|_{s=t-T}u_1(t-T)+left.frac{partial f}{partial u_2}frac{partial u_2}{partial s}right|_{s=t-T}u_2(t-T)+left.frac{partial f}{partial s}right|_{s=t-T}(t-T)+cdots $$
I want it to become as follows:
$$y(t)=theta_1(t) u_1(t)+theta_2(t) u_2(t)+theta_3(t) $$
taylor-expansion
taylor-expansion
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
edited Dec 25 '18 at 16:15
V.Ajall
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
asked Dec 25 '18 at 15:26
V.AjallV.Ajall
124
124
3
$begingroup$
You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
$endgroup$
– Ian
Dec 25 '18 at 15:29
$begingroup$
@Ian would you please post the complete answer?
$endgroup$
– V.Ajall
Dec 25 '18 at 16:13
$begingroup$
This is Faa di Bruno's formula, basically. See the wikipedia article
$endgroup$
– kimchi lover
Dec 25 '18 at 16:31
$begingroup$
Still looking for the solution
$endgroup$
– V.Ajall
Dec 27 '18 at 11:38
add a comment |
3
$begingroup$
You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
$endgroup$
– Ian
Dec 25 '18 at 15:29
$begingroup$
@Ian would you please post the complete answer?
$endgroup$
– V.Ajall
Dec 25 '18 at 16:13
$begingroup$
This is Faa di Bruno's formula, basically. See the wikipedia article
$endgroup$
– kimchi lover
Dec 25 '18 at 16:31
$begingroup$
Still looking for the solution
$endgroup$
– V.Ajall
Dec 27 '18 at 11:38
3
3
$begingroup$
You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
$endgroup$
– Ian
Dec 25 '18 at 15:29
$begingroup$
You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
$endgroup$
– Ian
Dec 25 '18 at 15:29
$begingroup$
@Ian would you please post the complete answer?
$endgroup$
– V.Ajall
Dec 25 '18 at 16:13
$begingroup$
@Ian would you please post the complete answer?
$endgroup$
– V.Ajall
Dec 25 '18 at 16:13
$begingroup$
This is Faa di Bruno's formula, basically. See the wikipedia article
$endgroup$
– kimchi lover
Dec 25 '18 at 16:31
$begingroup$
This is Faa di Bruno's formula, basically. See the wikipedia article
$endgroup$
– kimchi lover
Dec 25 '18 at 16:31
$begingroup$
Still looking for the solution
$endgroup$
– V.Ajall
Dec 27 '18 at 11:38
$begingroup$
Still looking for the solution
$endgroup$
– V.Ajall
Dec 27 '18 at 11:38
add a comment |
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You also need to multiply the derivatives by $T$ of course. And this is only first order, the multivariable chain rule gets messier at higher order.
$endgroup$
– Ian
Dec 25 '18 at 15:29
$begingroup$
@Ian would you please post the complete answer?
$endgroup$
– V.Ajall
Dec 25 '18 at 16:13
$begingroup$
This is Faa di Bruno's formula, basically. See the wikipedia article
$endgroup$
– kimchi lover
Dec 25 '18 at 16:31
$begingroup$
Still looking for the solution
$endgroup$
– V.Ajall
Dec 27 '18 at 11:38