Does f(x,y) = dy/dx or f'(x) in Mathematical Function Notation?

[darkdave3000]

I have a question about math function notation.

f(x) = a function of x

f(x,y) = dy/dx <---- Is this true? What is the logic behind this? I don't understand how a multiple variable function implies that its a derivative

f(x,y) = f ' (x) ? Is this true also? I've been confused by this logic when I watch youtube videos about computational math such as Euler method and Runge Kutta method

 

[malemdk]

Both the above equations are not correct since f(x,y) is function of some other variable, like z.
K

 

[darkdave3000]

Both the above equations are not correct since f(x,y) is function of some other variable, like z.
K

Can you explain how it works then?

 

[malemdk]

If y =f(x)
f'(y)=dy/dx
or
Z=f(x,y)
dz=df/dx x dx +df/dy x dy
Its a partial differention

 

[Mark44]

Both the above equations are not correct since f(x,y) is function of some other variable, like z.

No.
The notation f(x, y) means that f is a function of x and y.

 

[Stephen Tashi]

f(x,y) = dy/dx <---- Is this true?

I don't think you found that statement in a calculus book. Explain the context where you find it. Which particular video? What time in that video?

 

[Mark44]

If y =f(x)
f'(y)=dy/dx

No. dy/dx and f'(x) are alternate notations for the derivative of y or f, with respect to the variable x.

or
Z=f(x,y)
dz=df/dx x dx +df/dy x dy

It's very confusing to use 'x' for multiplication in an expression that also has x as a variable.
If z = f(x, y), then the differential of z is ##dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy##

Its a partial differention

 

[malemdk]

I don't know to type partial derivatives symbol, sorry

 

[Stephen Tashi]

I don't know to type partial derivatives symbol, sorry

Learning LaTex is a project that you might find interesting - https://www.physicsforums.com/help/latexhelp/

It works well on many different scientific forums - including Wikipedia. The method used to set-off LaTex expressions from ordinary text does vary from website to website.

 

[malemdk]

Thank you Stephen, from now I will try to type in Latex

 

[darkdave3000]

Sorry its a pain in the a** to reply here on my phone and I actually thought I already did a few messages back!

I got the notations from literature on wikipedia and youtube about Runge Kutta and Improved Euler methods. Computational Maths.

Look at what k1 equals to in
https://en.wikipedia.org/wiki/Runge–Kutta_methods

 

[Mark44]

Sorry its a pain in the a** to reply here on my phone and I actually thought I already did a few messages back!

I got the notations from literature on wikipedia and youtube about Runge Kutta and Improved Euler methods. Computational Maths.

Look at what k1 equals to in
https://en.wikipedia.org/wiki/Runge–Kutta_methods

From the wiki page in the link above:

Let an initial value problem be specified as follows:
##\dot y = f(t, y), y(t_0) = y_0##
Here y is an unknown function (scalar or vector) of time t, which we would like to approximate; we are told that ##\dot y##, the rate at which y changes, is a function of t and of y itself. At the initial time ##t_0## the corresponding y value is ##y_0## . The function f and the data ##t_0##, ##y_0## are given.

Now pick a step-size h > 0 and define

##y_{n + 1} = y_n + \frac h 6 (k_1 + 2k_2 + 2k_3 + k_4),##
##t_{n + 1} = t_n + h##
for n = 0, 1, 2, 3, ..., using
##k_1 = f(t_n, y_n)##
##k_2 = f(t_n + \frac h 2, y_n + \frac h 2 k_1)##
##k_3 = f(t_n + \frac h 2, y_n + \frac h 2 k_2)##
##k_4= f(t_n +h, y_n + h k_3)##

The problem to solve here is the differential equation ##\dot y = f(t, y)##, where ##\dot y## is the same as ##\frac {dy} {dt}##, and on the right side of the equation is some function of y and t.

You are given ##t_0## and ##y_0##, and you pick a step-size h, typically a small positive number.
To calculate ##y_1## you need to calculate ##k_1, k_2, k_3,## and ##k_4##, using the formulas shown. In the first round of calculations ##k_1 = f(t_0, y_0)##, and similarly for ## k_2, k_3, k_4##.

After calculating ##k_1, k_2, k_3, k_4##, you use them to calculate ##y_1## from ##y_0## and the k values and h, and then calculate ##t_1##.

For the second round, you recalculate the four k values, and use these values to calculate ##y_2## and ##t_2##.

By the time you get up to ##y_n##, you will have n + 1 function values, corresponding to the n + 1 time values. A plot of these coordinate pairs represents an approximation to the solution of the differential equation.

What part of this don't you understand?

 

[darkdave3000]

From the wiki page in the link above:

The problem to solve here is the differential equation ##\dot y = f(t, y)##, where ##\dot y## is the same as ##\frac {dy} {dt}##, and on the right side of the equation is some function of y and t.

You are given ##t_0## and ##y_0##, and you pick a step-size h, typically a small positive number.
To calculate ##y_1## you need to calculate ##k_1, k_2, k_3,## and ##k_4##, using the formulas shown. In the first round of calculations ##k_1 = f(t_0, y_0)##, and similarly for ## k_2, k_3, k_4##.

After calculating ##k_1, k_2, k_3, k_4##, you use them to calculate ##y_1## from ##y_0## and the k values and h, and then calculate ##t_1##.

For the second round, you recalculate the four k values, and use these values to calculate ##y_2## and ##t_2##.

By the time you get up to ##y_n##, you will have n + 1 function values, corresponding to the n + 1 time values. A plot of these coordinate pairs represents an approximation to the solution of the differential equation.

What part of this don't you understand?

What I don't understand is how k1 = f(to,yo) = dy/dt Because that's basicly what their saying isn't it?

 

[DrClaude]

What I don't understand is how k1 = f(to,yo) = dy/dt Because that's basicly what their saying isn't it?

Runge-Kutta methods are used to solve first-order ordinary differential equations, so
$$
\frac{dy}{dt} = f(t,y)
$$
is the ODE that needs to be solved. ##k_1## is simply the value of the rhs of the differential equation for ##t = t_0## and ##y_0 = y(t=t_0)##.

 

[darkdave3000]

Runge-Kutta methods are used to solve first-order ordinary differential equations, so
$$
\frac{dy}{dt} = f(t,y)
$$
is the ODE that needs to be solved. ##k_1## is simply the value of the rhs of the differential equation for ##t = t_0## and ##y_0 = y(t=t_0)##.

Ok so youre agreeing with me. But what I don't understand is the logic behind how a multi variable function is the same as a derivative of one of those variable over the other.

 

[DrClaude]

Ok so youre agreeing with me. But what I don't understand is the logic behind how a multi variable function is the same as a derivative of one of those variable over the other.

It's a generic expression. The result of dy/dt is a function of y and t, generally speaking. I have a textbook here that even uses as a definition of a first-order differential equation:
$$
\frac{d y}{dx} = f(x,y)
$$

Take for example
$$
y = \frac{C}{t} + 2
$$
and calculate dy/dt.

 

[Mark44]

What I don't understand is how k1 = f(to,yo) = dy/dt Because that's basicly what their saying isn't it?

No, that's not what they are saying.
(Reply to DrClaude's post #14)

Ok so youre agreeing with me. But what I don't understand is the logic behind how a multi variable function is the same as a derivative of one of those variable over the other.

Here's an example that is probably more enlightening than DrClaude's example of ##y = \frac C t + 2##.

##y' = 2xy^2, y(0) = y_0 = .5##
Here y' should be taken to mean ##\frac {dy}{dx}##. In this example ##f(x, y) = 2xy^2##, which is a function of both x and y, and the goal is to find y as a function of x alone. By using Runge-Kutta one can obtain a set of ##(x_i, y_i)## pairs that can be graphed. This method does not produce a formula for y as a function of x. Instead, it produces a set of points that give a good idea of the relationship between x and y on the interval ##[a = x_0, b = x_0 + nh]##.

 

[Mark44]

I should add that the first equality in your thread title, "f(x,y) = f'(x) = dy/dx ?" does not accurately reflect what you're asking. We didn't find out what you were really asking until post #11, in which you provided a link to the wiki article.

The wiki article, about the Runge-Kutta algorithm, does NOT say that f(x, y) = f'(x). What the article says (with a slight change), is that y' = f(x, y). Your thread title makes no sense because the notation f(x, y) means that f is a function of two variables. The notation f'(x) means that f' is a function of a single variable, and from that we can infer that f itself is a function of one variable. My point is that f can be a function of one variable OR f can be a function of two variables, but both of these can't be true.

 

[Stephen Tashi]

I have a question about math function notation.

f(x) = a function of x

f(x,y) = dy/dx <---- Is this true?

If you mean to say that "f(x) is a function of x" is given information in some situation then you are correct to object to the use of "f(x,y)" to denote a function of two real variables in the same situation. (However, such ambiguous notation is sometimes used.)

If you mean to say that , by tradition, "f" is always used as the name of a function of one real variable, that, of course, is not correct. It also not correct to think that if "y" is used to denote an "unknown function of x" that we must use the notation "y = f(x)" and name that unknown function "f".You will find that people writing about applications of mathematics , especially applications of mathematics to physics, use notation and terminology that is technically wrong because it is ambiguous. Ambiguous notation is convenient, but it can also be confusing.

For example, let ##f(x,y)## be a function of two variables defined by ##f(x,y) = x + xy ##. The function has a domain consisting of pairs of real numbers. However, a book that uses that notation will often introduce other information without changing notation. For example, suppose we add the information ##y = 3x##. It is very tempting to continue to use the name "##f##" to denote the function of one real variable given by ## x + x(3x)##. Technically, this is incorrect notation because it is an ambiguous use of the name "##f##". Nevertheless, in such situation a book might use notation like ##\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}##, where "##f##" on the left hand side of the equation denotes a function of a single variable and "##f##" on the right hand side denotes a function of two real variables.

f(x,y) = f ' (x) ? Is this true also? I've been confused by this logic when I watch youtube videos about computational math such as Euler method and Runge Kutta method

In the material you are watching, I think the idea might be ##f(x,y) = \frac{dy}{dx} = y'(x) ## The "unknown function" is named "##y##", not "##f##".

A typical scenario for differential equations is that "##y##" denotes a ( perhaps unknown) function of a single real variable. We could denote this function as "##y(x)##". However, that notation would get confusing because when we wish to denote the function "##y##" times an algebraic expression, the "##( )##" could be mistaken for multiplication. For example, to denote the concept "The function ##y## evaluated at the value ##t## times the expression ##(t+3)## evaluated at that same value", such notation would give us ##y(t)(t+3)## which might be mistakenly manipulated into ##y(t^2 + 3t)##.

To define a differential equation, we can begin with a real valued function ##f## of several real variables. For example let ##f(a,b,c) = c + ab + 3##, which is a function of 3 real variables. Using the definition of ##f##, we write can an equation ##f(a,b,c) = 0## (Don't confuse the concept of an equation with the concept of a function.) The differential equation is defined using ""##y##" to denote an unknown function and "##y'##" to denote the derivative of that function. A solution to the differential equation "##f(x,y,y') = 0##" is defined to be a function ##y## of a single real variable such that when ##y## is evaluated at the value ##x##, we have ##f(x,y(x),y'(x)) = 0## In the particular case of ##f(a,b,c) = c + ab + 3##, the differential equation would be ##y'(x) + xy(x) + 3 = 0##. However, as mentioned above, the argument "##(x)##" is traditionally omitted when we write the function ##y##, so the equation would be written ## y' + xy + 3 = 0## or ##\frac{dy}{dx} + xy + 3 = 0 ##

Notation in applied mathematics shows no mercy to people who do not understand the concept of the "scope" (i.e. the context) of variables. In the above example, I used the notation ##f(a,b,c) = c + ab + 3##" in a effort to be clear. However, it would be permissible to define ##f## by the notation "##f(x,y,z) = z + xy + 3##" or even (to be confusing) as "##f(z,y,x) = x + zy + 3##", because variable names are arbitrary within a given "scope" of the variables. The names "##x##" and "##y##" used in the scope for defining ##f## can be unrelated to the use of those names when they appear farther down the page in the discussion of "##y##" being a function of "##x##".