What does $(dy)(x, Delta x)$ mean? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why saying that “$x$ is an indeterminate real number” is misleading?What is the difference between an implicit ordinary differential equation and a differential algebraic equation?What are the ordinary and singular points of the first order diff. equation?On why we have $dy = f'(x)dx$?What does it mean for a solution to a differential equation to exist on some interval?What is “Phase Space” in differential equations and classical mechanics?What does “modulo equivalence relationship” mean?What is the “differential form” of an ODE?Differentials: What does half-planes defined by… mean?Is “ordinary differential equations in more than two variables” correct? Used in two books
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What does $(dy)(x, Delta x)$ mean?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why saying that “$x$ is an indeterminate real number” is misleading?What is the difference between an implicit ordinary differential equation and a differential algebraic equation?What are the ordinary and singular points of the first order diff. equation?On why we have $dy = f'(x)dx$?What does it mean for a solution to a differential equation to exist on some interval?What is “Phase Space” in differential equations and classical mechanics?What does “modulo equivalence relationship” mean?What is the “differential form” of an ODE?Differentials: What does half-planes defined by… mean?Is “ordinary differential equations in more than two variables” correct? Used in two books
$begingroup$
I am reading ordinary differential equations from a book which says
Hence, $dy = f'(x) Delta x$, we call $dy$ the differential of $y$. As the differential $dy$ is a function of two independent variables $x$ and $Delta x$, we indicate this dependency by $(dy)(x, Delta x)$.
Excuse me but what the hell is this? It goes even further by saying
The differential of $y$, written as $dy$ (or $df$) is defined by $$(dy)(x, Delta x) = f'(x) Delta x$$
And now I am completely confused. Do we define $dy$ twice? Or what is $(dy)(x, Delta x)$? Is it a sort of relation as it says "we write a dependency in this way", so it is a relation in terms of set theory? What is it exactly?
Why do $dy$ and $(dy)(x, Delta x)$ have the same definitions? Can we say that $dy = (dy)(x, Delta x)$ then? If so, what is the point of this mess? Thank you if you read up to this point and I will be more thankful if you help me to understand what I am missing.
ordinary-differential-equations notation terminology
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
$endgroup$
add a comment |
$begingroup$
I am reading ordinary differential equations from a book which says
Hence, $dy = f'(x) Delta x$, we call $dy$ the differential of $y$. As the differential $dy$ is a function of two independent variables $x$ and $Delta x$, we indicate this dependency by $(dy)(x, Delta x)$.
Excuse me but what the hell is this? It goes even further by saying
The differential of $y$, written as $dy$ (or $df$) is defined by $$(dy)(x, Delta x) = f'(x) Delta x$$
And now I am completely confused. Do we define $dy$ twice? Or what is $(dy)(x, Delta x)$? Is it a sort of relation as it says "we write a dependency in this way", so it is a relation in terms of set theory? What is it exactly?
Why do $dy$ and $(dy)(x, Delta x)$ have the same definitions? Can we say that $dy = (dy)(x, Delta x)$ then? If so, what is the point of this mess? Thank you if you read up to this point and I will be more thankful if you help me to understand what I am missing.
ordinary-differential-equations notation terminology
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
$endgroup$
add a comment |
$begingroup$
I am reading ordinary differential equations from a book which says
Hence, $dy = f'(x) Delta x$, we call $dy$ the differential of $y$. As the differential $dy$ is a function of two independent variables $x$ and $Delta x$, we indicate this dependency by $(dy)(x, Delta x)$.
Excuse me but what the hell is this? It goes even further by saying
The differential of $y$, written as $dy$ (or $df$) is defined by $$(dy)(x, Delta x) = f'(x) Delta x$$
And now I am completely confused. Do we define $dy$ twice? Or what is $(dy)(x, Delta x)$? Is it a sort of relation as it says "we write a dependency in this way", so it is a relation in terms of set theory? What is it exactly?
Why do $dy$ and $(dy)(x, Delta x)$ have the same definitions? Can we say that $dy = (dy)(x, Delta x)$ then? If so, what is the point of this mess? Thank you if you read up to this point and I will be more thankful if you help me to understand what I am missing.
ordinary-differential-equations notation terminology
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
$endgroup$
I am reading ordinary differential equations from a book which says
Hence, $dy = f'(x) Delta x$, we call $dy$ the differential of $y$. As the differential $dy$ is a function of two independent variables $x$ and $Delta x$, we indicate this dependency by $(dy)(x, Delta x)$.
Excuse me but what the hell is this? It goes even further by saying
The differential of $y$, written as $dy$ (or $df$) is defined by $$(dy)(x, Delta x) = f'(x) Delta x$$
And now I am completely confused. Do we define $dy$ twice? Or what is $(dy)(x, Delta x)$? Is it a sort of relation as it says "we write a dependency in this way", so it is a relation in terms of set theory? What is it exactly?
Why do $dy$ and $(dy)(x, Delta x)$ have the same definitions? Can we say that $dy = (dy)(x, Delta x)$ then? If so, what is the point of this mess? Thank you if you read up to this point and I will be more thankful if you help me to understand what I am missing.
ordinary-differential-equations notation terminology
ordinary-differential-equations notation terminology
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
asked Apr 8 at 12:08
Turkhan BadalovTurkhan Badalov
489313
489313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Say $y=e^2x$. Then we write $dy = 2 e^2x;dx$. This exhibits $dy$ as a function of two variables, namely $x$ and $dx$. Perhaps this author wrote $Delta x$ in place of $dx$, hoping that this would reduce confusion...
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
$begingroup$
$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$
$dy(x)$ is a linear application : it's the linear approximation of $y$ in a neighborhood of $x$, i.e. $dy(x):mathbb Rto mathbb R$ if defined by $$dy(x)(h):= y'(x)h.$$
But $h=x+h-x=:Delta x$, so you have to see $Delta x$ as a "variable" (namely as the distance from $x$), and we denote $$dy(x,Delta x):=dy(x)(Delta x)=y'(x)Delta x.$$
Normally, we write $dy$ instead of $dy(x,Delta x)$.
answered Apr 8 at 12:19
user657324user657324
59510
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Say $y=e^2x$. Then we write $dy = 2 e^2x;dx$. This exhibits $dy$ as a function of two variables, namely $x$ and $dx$. Perhaps this author wrote $Delta x$ in place of $dx$, hoping that this would reduce confusion...
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
$begingroup$
Say $y=e^2x$. Then we write $dy = 2 e^2x;dx$. This exhibits $dy$ as a function of two variables, namely $x$ and $dx$. Perhaps this author wrote $Delta x$ in place of $dx$, hoping that this would reduce confusion...
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
$endgroup$
add a comment |
$begingroup$
Say $y=e^2x$. Then we write $dy = 2 e^2x;dx$. This exhibits $dy$ as a function of two variables, namely $x$ and $dx$. Perhaps this author wrote $Delta x$ in place of $dx$, hoping that this would reduce confusion...
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
$endgroup$
Say $y=e^2x$. Then we write $dy = 2 e^2x;dx$. This exhibits $dy$ as a function of two variables, namely $x$ and $dx$. Perhaps this author wrote $Delta x$ in place of $dx$, hoping that this would reduce confusion...
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
answered Apr 8 at 12:51
GEdgarGEdgar
63.6k269175
63.6k269175
add a comment |
add a comment |
$begingroup$
$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$
$dy(x)$ is a linear application : it's the linear approximation of $y$ in a neighborhood of $x$, i.e. $dy(x):mathbb Rto mathbb R$ if defined by $$dy(x)(h):= y'(x)h.$$
But $h=x+h-x=:Delta x$, so you have to see $Delta x$ as a "variable" (namely as the distance from $x$), and we denote $$dy(x,Delta x):=dy(x)(Delta x)=y'(x)Delta x.$$
Normally, we write $dy$ instead of $dy(x,Delta x)$.
answered Apr 8 at 12:19
user657324user657324
59510
$endgroup$
add a comment |
$begingroup$
$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$
$dy(x)$ is a linear application : it's the linear approximation of $y$ in a neighborhood of $x$, i.e. $dy(x):mathbb Rto mathbb R$ if defined by $$dy(x)(h):= y'(x)h.$$
But $h=x+h-x=:Delta x$, so you have to see $Delta x$ as a "variable" (namely as the distance from $x$), and we denote $$dy(x,Delta x):=dy(x)(Delta x)=y'(x)Delta x.$$
Normally, we write $dy$ instead of $dy(x,Delta x)$.
answered Apr 8 at 12:19
user657324user657324
59510
$endgroup$
add a comment |
$begingroup$
$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$
$dy(x)$ is a linear application : it's the linear approximation of $y$ in a neighborhood of $x$, i.e. $dy(x):mathbb Rto mathbb R$ if defined by $$dy(x)(h):= y'(x)h.$$
But $h=x+h-x=:Delta x$, so you have to see $Delta x$ as a "variable" (namely as the distance from $x$), and we denote $$dy(x,Delta x):=dy(x)(Delta x)=y'(x)Delta x.$$
Normally, we write $dy$ instead of $dy(x,Delta x)$.
answered Apr 8 at 12:19
user657324user657324
59510
$endgroup$
$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$
$dy(x)$ is a linear application : it's the linear approximation of $y$ in a neighborhood of $x$, i.e. $dy(x):mathbb Rto mathbb R$ if defined by $$dy(x)(h):= y'(x)h.$$
But $h=x+h-x=:Delta x$, so you have to see $Delta x$ as a "variable" (namely as the distance from $x$), and we denote $$dy(x,Delta x):=dy(x)(Delta x)=y'(x)Delta x.$$
Normally, we write $dy$ instead of $dy(x,Delta x)$.
answered Apr 8 at 12:19
user657324user657324
59510
edited Apr 8 at 14:16
answered Apr 8 at 12:19
user657324user657324
59510
answered Apr 8 at 12:19
user657324user657324
59510
answered Apr 8 at 12:19
user657324user657324
59510
59510
add a comment |
add a comment |
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