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What does $(dy)(x, Delta x)$ mean?

4

1

$begingroup$

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

share|cite|improve this question

asked Apr 8 at 12:08

Turkhan BadalovTurkhan Badalov

489313

$endgroup$

add a comment | 

4

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked Apr 8 at 12:08

Turkhan BadalovTurkhan Badalov

489313

$endgroup$

share|cite|improve this question

asked Apr 8 at 12:08

Turkhan BadalovTurkhan Badalov

489313

share|cite|improve this question

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

2 Answers
2

active

oldest

votes

2

$begingroup$

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...

share|cite|improve this answer

answered Apr 8 at 12:51

GEdgarGEdgar

63.6k269175

$endgroup$

add a comment | 

2

$begingroup$

$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$

$dy(x)$ 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 $dy(x,Delta x)$.

share|cite|improve this answer

edited Apr 8 at 14:16

answered Apr 8 at 12:19

user657324user657324

59510

$endgroup$

add a comment | 

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2

$begingroup$

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...

share|cite|improve this answer

answered Apr 8 at 12:51

GEdgarGEdgar

63.6k269175

$endgroup$

add a comment | 

2

$begingroup$

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...

share|cite|improve this answer

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...

share|cite|improve this answer

answered Apr 8 at 12:51

GEdgarGEdgar

63.6k269175

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$

$dy(x)$ 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 $dy(x,Delta x)$.

share|cite|improve this answer

edited Apr 8 at 14:16

answered Apr 8 at 12:19

user657324user657324

59510

$endgroup$

add a comment | 

2

$begingroup$

$$dy(x,Delta x):= dy(x)(Delta x)=y'(x)Delta x.$$

$dy(x)$ 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 $dy(x,Delta x)$.

share|cite|improve this answer

edited Apr 8 at 14:16

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 $dy(x,Delta x)$.

share|cite|improve this answer

edited Apr 8 at 14:16

answered Apr 8 at 12:19

user657324user657324

59510

$endgroup$

share|cite|improve this answer

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