meaning of a phrase in mathematic paragraph
$begingroup$
Suppose $x(t) inmathbb{R}^n$ and denote $dot{x} $ its derivative.
Let $F(x, dot{x}, t)$ be a function. Suppose we can find an $x(t)$ so that the integral
$$ intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
is a minimum. Now, consider the vector $x(t) + delta x$ where $delta x$ is a small change. So that the following better be positive
$$ intlimits_{t_1}^{t_2} F(x + delta x, dot{x} + delta dot{x}, t) d t - intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
Expanding this in powers of variations $delta x $ and its trivially, one trivially see that the leading terms are of the first order.
Question:
In the last sentence, when it says trivially, I dont know if it is really trivial this statement because I cant see it. When they say expand in terms of $delta x$ and $delta dot{x}$ , does he mean that we have to expand with taylor series expansion? Also, leadings terms are of the first order, does he mean that the leading terms are linear? But, how can we expand such a function? I mean $F$ is a function of $2n +1$ variables.
calculus
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
$endgroup$
add a comment |
$begingroup$
Suppose $x(t) inmathbb{R}^n$ and denote $dot{x} $ its derivative.
Let $F(x, dot{x}, t)$ be a function. Suppose we can find an $x(t)$ so that the integral
$$ intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
is a minimum. Now, consider the vector $x(t) + delta x$ where $delta x$ is a small change. So that the following better be positive
$$ intlimits_{t_1}^{t_2} F(x + delta x, dot{x} + delta dot{x}, t) d t - intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
Expanding this in powers of variations $delta x $ and its trivially, one trivially see that the leading terms are of the first order.
Question:
In the last sentence, when it says trivially, I dont know if it is really trivial this statement because I cant see it. When they say expand in terms of $delta x$ and $delta dot{x}$ , does he mean that we have to expand with taylor series expansion? Also, leadings terms are of the first order, does he mean that the leading terms are linear? But, how can we expand such a function? I mean $F$ is a function of $2n +1$ variables.
calculus
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
$endgroup$
$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11
add a comment |
$begingroup$
Suppose $x(t) inmathbb{R}^n$ and denote $dot{x} $ its derivative.
Let $F(x, dot{x}, t)$ be a function. Suppose we can find an $x(t)$ so that the integral
$$ intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
is a minimum. Now, consider the vector $x(t) + delta x$ where $delta x$ is a small change. So that the following better be positive
$$ intlimits_{t_1}^{t_2} F(x + delta x, dot{x} + delta dot{x}, t) d t - intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
Expanding this in powers of variations $delta x $ and its trivially, one trivially see that the leading terms are of the first order.
Question:
In the last sentence, when it says trivially, I dont know if it is really trivial this statement because I cant see it. When they say expand in terms of $delta x$ and $delta dot{x}$ , does he mean that we have to expand with taylor series expansion? Also, leadings terms are of the first order, does he mean that the leading terms are linear? But, how can we expand such a function? I mean $F$ is a function of $2n +1$ variables.
calculus
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
$endgroup$
Suppose $x(t) inmathbb{R}^n$ and denote $dot{x} $ its derivative.
Let $F(x, dot{x}, t)$ be a function. Suppose we can find an $x(t)$ so that the integral
$$ intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
is a minimum. Now, consider the vector $x(t) + delta x$ where $delta x$ is a small change. So that the following better be positive
$$ intlimits_{t_1}^{t_2} F(x + delta x, dot{x} + delta dot{x}, t) d t - intlimits_{t_1}^{t_2} F(x, dot{x}, t) dt $$
Expanding this in powers of variations $delta x $ and its trivially, one trivially see that the leading terms are of the first order.
Question:
In the last sentence, when it says trivially, I dont know if it is really trivial this statement because I cant see it. When they say expand in terms of $delta x$ and $delta dot{x}$ , does he mean that we have to expand with taylor series expansion? Also, leadings terms are of the first order, does he mean that the leading terms are linear? But, how can we expand such a function? I mean $F$ is a function of $2n +1$ variables.
calculus
calculus
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
asked Dec 19 '18 at 22:26
NeymarNeymar
423214
423214
$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11
add a comment |
$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11
$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11
add a comment |
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$begingroup$
Yes... expand $F(x + delta x, ...)$ in a Taylor series. After subtraction, the leading term is "trivially" linear (first-order).
$endgroup$
– David G. Stork
Dec 20 '18 at 0:38
$begingroup$
but it is the regular taylor series? we have more than one variables that it what is causin confusion
$endgroup$
– Neymar
Dec 20 '18 at 1:02
$begingroup$
Taylor series apply perfectly well to $n$-variable functions: en.wikipedia.org/wiki/…
$endgroup$
– David G. Stork
Dec 20 '18 at 1:04
$begingroup$
yes I see now, just working out the details is cumbersome for this problem
$endgroup$
– Neymar
Dec 20 '18 at 1:11