MHB Extremely Difficult Extremum Problem.

[mathbalarka]

$g(u(x))$ is certainly a function of $x$.

 

[Amad27]

$g(u(x))$ is certainly a function of $x$.

I reread the post, and it helped a lot, I'm almost clear with this concept.

$g(u(x))$ is a function of $x$.

$g(u(x))$ is not a function of $u$

Then how can you differentiate a function with respect to $u$, if it is not a function of $u$ itself??

Thanks

 

[MarkFL]

$g(u)$ may be differentiated with respect to $u$. If $u$ is a function of $x$, then we may differentiate $g(u)$ with respect to $x$ via the chain rule.

 

[mathbalarka]

I reread the post, and it helped a lot, I'm almost clear with this concept.

I'm glad I could have been of some help. You are encouraged to post any doubts you have on anything in MHB, we would be happy to help you! (Yes)

$g(u(x))$ is a function of $x$.

$g(u(x))$ is not a function of $u$

Then how can you differentiate a function with respect to $u$, if it is not a function of $u$ itself??

Repeating what MarkFL, said, we are differentiating $g(u(x))$ with respect to $u(x)$, i.e.,

$$\frac{d\left(g(u(x)) \right)}{d\left(u(x)\right)}$$

The chain rule says that you can naively "cancel out" the differentials, i.e.,

$$\frac{d \left(g(u(x)) \right)}{dx} = \frac{d\left(g(u(x)) \right)}{\cancel{d\left(u(x)\right)}} \cdot \frac{\cancel{d \left (u(x) \right)}}{dx} = g'(u(x)) \cdot u'(x)$$

 

[Amad27]

I'm glad I could have been of some help. You are encouraged to post any doubts you have on anything in MHB, we would be happy to help you! (Yes)
Repeating what MarkFL, said, we are differentiating $g(u(x))$ with respect to $u(x)$, i.e.,

$$\frac{d\left(g(u(x)) \right)}{d\left(u(x)\right)}$$

The chain rule says that you can naively "cancel out" the differentials, i.e.,

$$\frac{d \left(g(u(x)) \right)}{dx} = \frac{d\left(g(u(x)) \right)}{\cancel{d\left(u(x)\right)}} \cdot \frac{\cancel{d \left (u(x) \right)}}{dx} = g'(u(x)) \cdot u'(x)$$

Thank you mathbalarka and to MarkFL. both were extremely helpful, and I appreciate every bit of your assistance ;)

So all this is just an application of the chain rule?

We get

$f(x)= g(u(x))$
$f(x) = u + \frac{2}{u-1}$
$f'(x) = (1 - \frac{2}{(u-1)^2} )(\d{u}{du})$
$\d{u}{du} = cos(x) - sin(x) = v$
$f'(x) = (1 - \frac{2}{(u-1)^2})(v)$

Then you set $f'(x) = 0$

But the solution on the anemone linked page doesn't consider $v$ as part of the derivative...?

Thank you very very much =D

 

[Amad27]

$g(u)$ may be differentiated with respect to $u$. If $u$ is a function of $x$, then we may differentiate $g(u)$ with respect to $x$ via the chain rule.

Hey MarkFL,

Didnt notice about this.

Then if we define

$v = u' = cos(x) - sin(x)$

When isn't $v$ considered part of the derivate as the chain rule? Like
$du/dx$ in the chain rule part ?

Thanks

 

[MarkFL]

Hey MarkFL,

Didnt notice about this.

Then if we define

$v = u' = cos(x) - sin(x)$

When isn't $v$ considered part of the derivate as the chain rule? Like
$du/dx$ in the chain rule part ?

Thanks

I'm not sure I understand your question. (Wondering)

 

[Amad27]

I'm not sure I understand your question. (Wondering)

Okay, sorry, I should've been more specific :)

We had the following,

$g(u(x)) = u + \frac{2}{(u-1)}$

What others suggested was,

$g'(u(x)) = 1 - \frac{2}{(u-1)^2}$

But since you had suggested the chain rule, the above $g'(u(x))$ is incorrect.

Then I defined $v = cos(x) - sin(x) = u' = \d{u}{x}$

So that by the chain rule the derivative,

$=g'(u(x))u'(x)$

To get

$(1 - \frac{2}{(u-1)^2})(v)$

Which should be the derivative no? According to the chain rule..

Thanks

 

[I like Serena]

We had the following,

$g(u(x)) = u + \frac{2}{(u-1)}$

What others suggested was,

$g'(u(x)) = 1 - \frac{2}{(u-1)^2}$

But since you had suggested the chain rule, the above $g'(u(x))$ is incorrect.

This is correct.

$g'(u(x))$ means that you differentiate $g$ with respect to its argument (in this case $u$), and afterwards you fill in $u(x)$ for its argument.
As such it is not an application of the chain rule.

Now suppose we define the function $G(x)=g(u(x))$.
Note that this is a different function, although it is often written (sloppily) as $g(x)$.
Then we have by the chain rule that:
$$G'(x)=g'(u(x))\cdot u'(x)$$

 

[Amad27]

This is correct.

$g'(u(x))$ means that you differentiate $g$ with respect to its argument (in this case $u$), and afterwards you fill in $u(x)$ for its argument.
As such it is not an application of the chain rule.

Now suppose we define the function $G(x)=g(u(x))$.
Note that this is a different function, although it is often written (sloppily) as $g(u)$.
Then we have by the chain rule that:
$$G'(x)=g'(u(x))\cdot u'(x)$$

Ah - I am starting to see the difference here (more clearly).

So you are differentiating with respect to $u$
You are NOT differentiating with respect to $x$ ?

But keep in mind that $u = u(x)$ is a function.

How can you differentiate with respect to another function?
Thank you!

 

[I like Serena]

Ah - I am starting to see the difference here (more clearly).

So you are differentiating with respect to $u$
You are NOT differentiating with respect to $x$ ?

But keep in mind that $u = u(x)$ is a function.

How can you differentiate with respect to another function?
Thank you!

We do not differentiate with respect to a function.
We differentiate with respect to the implicit (unnamed) argument.
And afterwards, we replace the implicit argument in the function definition with what is specified to be substituted.

Btw, strictly speaking, $u(x)$ is not a function.
It's the application of a function $u$ to an unknown value $x$.
The result is a value, and it is this value that gets substituted.Let's make it more explicit, applying it to the functions at hand.

The function $g$ is given by:
$$g(v)=v+\frac {2} {v-1} \tag{1}$$
Note that I've used a different letter for the argument to avoid confusion with the other symbols.
The function $u$ is given by:
$$u(x)=\sin(x) + \cos(x) \tag{2}$$If follows that for instance:
$$g(u(x)) = \big(\sin(x) + \cos(x)\big)+\frac {2} {\big(\sin(x) + \cos(x)\big)-1}$$

Furthermore, the derivative $g'$ is given by:
$$g'(v)=1-\frac {2} {(v-1)^2}$$
Substituting $v=u(x)$, we get:
$$g'(u(x))=1-\frac {2} {(u(x)-1)^2}$$
To highlight the difference we can also write this as:
$$g'(\sin(x) + \cos(x))=1-\frac {2} {\left(\big(\sin(x) + \cos(x)\big)-1\right)^2}$$

 

[Amad27]

We do not differentiate with respect to a function.
We differentiate with respect to the implicit (unnamed) argument.
And afterwards, we replace the implicit argument in the function definition with what is specified to be substituted.

Btw, strictly speaking, $u(x)$ is not a function.
It's the application of a function $u$ to an unknown value $x$.
The result is a value, and it is this value that gets substituted.Let's make it more explicit, applying it to the functions at hand.

The function $g$ is given by:
$$g(v)=v+\frac {2} {v-1} \tag{1}$$
Note that I've used a different letter for the argument to avoid confusion with the other symbols.
The function $u$ is given by:
$$u(x)=\sin(x) + \cos(x) \tag{2}$$If follows that for instance:
$$g(u(x)) = \big(\sin(x) + \cos(x)\big)+\frac {2} {\big(\sin(x) + \cos(x)\big)-1}$$

Furthermore, the derivative $g'$ is given by:
$$g'(v)=1-\frac {2} {(v-1)^2}$$
Substituting $v=u(x)$, we get:
$$g'(u(x))=1-\frac {2} {(u(x)-1)^2}$$
To highlight the difference we can also write this as:
$$g'(\sin(x) + \cos(x))=1-\frac {2} {\left(\big(\sin(x) + \cos(x)\big)-1\right)^2}$$

Ah - I see this system.

So you are suppose we had a function of $t$ as an independent variable.

$g(t) = t + \frac{2}{(t-1)}$

So $t$ is not related to anything else as a function; it is independent.

$g'(t) = 1 - \frac{2}{(t-1)^2}$

Then you are saying we have

$u(x) = sin(x) + cos(x)$ and that like any other function we can substitute.

$g'(u(x)) = 1 - \frac{2}{(u(x) - 1)^2}$

Ok this part seems to be in the clear now, let's move on.

When you set $g'(u(x)) = 0$ and then you find

$u(x) = 0$ at $u = {a}$

We had

$g(t) = t + \frac{2}{t-1}$

How/ whywill

$g(a) = a + \frac{2}{a-1}$ give you the minimum of the original,

$y$ trig function??

Thanks

 

[I like Serena]

AWhen you set $g'(u(x)) = 0$ and then you find

$u(x) = 0$ at $u = {a}$

The symbol $a$ is introduced here without definition.
I'm guessing it is defined as a solution of $g'(a)=0$.

However, that means that:
$$u(x) = a$$

Writing it as $u=a$ is a short hand notation.

How/ whywill

$g(a) = a + \frac{2}{a-1}$ give you the minimum of the original,

$y$ trig function??

Let's define the original trig function as:
$$G(x) = g(u(x)) \tag 1$$

Then $G(x)$ takes its minimum value when $G'(x)=0$.
That means that:
$$G'(x) = g'(u(x))\cdot u'(x) = 0$$
This happens if $g'(u(x))=0$ or $u'(x) = 0$.

With $a$ as a solution of $g'(a)=0$, we get that $G$ takes a minimum value at $x=u^{-1}(a)$.

The corresponding minimum is:
$$G\Big(u^{-1}(a)\Big) = g\Big(u\big(u^{-1}(a)\big)\Big) = g(a)$$

 

[Amad27]

The symbol $a$ is introduced here without definition.
I'm guessing it is defined as a solution of $g'(a)=0$.

However, that means that:
$$u(x) = a$$

Writing it as $u=a$ is a short hand notation.

Let's define the original trig function as:
$$G(x) = g(u(x)) \tag 1$$

Then $G(x)$ takes it minimum value when $G'(x)=0$.
That means that:
$$G'(x) = g'(u(x))\cdot u'(x) = 0$$
This happens if $g'(u(x))=0$ or $u'(x) = 0$.

With $a$ as the solution of $g'(a)=0$, we get that $G$ takes a minimum value at $x=u^{-1}(a)$.

The corresponding minimum is:
$$G\Big(u^{-1}(a)\Big) = g\Big(u\big(u^{-1}(a)\big)\Big) = g(a)$$

Okay, so a few things here.

This was a very, very, very, very, very good explanation. No doubts on that. Kudos to I Like Serena!

So we set $g'(u(x)) = 0$ and supposed it happened at $a$

So,

$g'(u(x)) = 0$ at $u(x) = a$ And like you said

$G(x) = g(u(x))$ which is the original trig equation.

$u(x) = a$ would mean

$x = u^{-1}(a)$

$G(u^{-1}(a)) = g(u(u^{-1}(a)))$

By definition,

$u(u^{-1}(a)) = a$ therefore we get,

$G(u^{-1}(a)) = g(u(u^{-1}(a))) = g(a)$

Wow, that is amazing how this works. Because essentially

We (you) have proved that the minimum is indeed at $u(x) = a$

In problems though, it gets very confusing.

When people say

$g(u)$ rather than $g(u(x))$

$g(u)$ means a function of $u$ as $u$ being an independent variable.
$g(u(x))$ means a function of a function $u(x)$ as a composite function.

Because by definition, a single letter would mean an independent variable.

-- This was the source of my confusion, which seems to be better, now. Wow, @I Like Serena, that was a million dollar explanation, I wish I could give a million thanks through MHB, excellent. Thank you very much!

(The rest of the problem is quite simple now)!

 

[I like Serena]

By definition,

$u(u^{-1}(a)) = a$

Just an observation, we are making the implicit assumption that $u$ is an invertible function.
And furthermore that the range of $u$ covers the domain of $g$, because otherwise we might "miss" a minimum.

When people say

$g(u)$ rather than $g(u(x))$

$g(u)$ means a function of $u$ as $u$ being an independent variable.
$g(u(x))$ means a function of a function $u(x)$ as a composite function.

Because by definition, a single letter would mean an independent variable.

-- This was the source of my confusion, which seems to be better, now. Wow, @I Like Serena, that was a million dollar explanation, I wish I could give a million thanks through MHB, excellent. Thank you very much!

(The rest of the problem is quite simple now)!

In practice the distinction between independent variables, functions, and composite functions is often not made - especially in physics or engineering.

So you'll typically see $y$, $y(x)$, $y(u)$, and $y(u(x))$ be used interchangeably, even though strictly speaking they are all different.
Not to mention that $x$, $y^{-1}$, and $x(u)$ are used interchangeably, which can be a serious source of confusion.

Normally this works out fine, but sometimes one needs to be alert about what it really means.It does allow us to write a chain rule simply as:
$$\d y x = \d y u \cdot \d u x$$
which makes a lot of intuitive sense, and is more readable than:
$$\d {y^*(x)} x = \d {y(u)} u \cdot \d {u(x)} x$$
where $y^*(x)=y(u(x))$.And it also allows us to write the inverse derivative rule as:
$$\d x y = \frac{1}{\d y x}$$
instead of the cumbersome:
$$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$

 

[Amad27]

In practice the distinction between independent variables, functions, and composite functions is often not made - especially in physics or engineering.

That is disappointing.

This means,

$f(x) = x^2$ could mean
$f(x) = sin^{2}(t)$

where $x = sin(t)$

Isnt it quite important to make the distinction though?

There is a critical difference between

$f(x) = x^2$ and $f(x) = sin^{2}(t)$

Thanks

 

[I like Serena]

This means,

$f(x) = x^2$ could mean
$f(x) = sin^{2}(t)$

where $x = sin(t)$

Isnt it quite important to make the distinction though?

There is a critical difference between

$f(x) = x^2$ and $f(x) = sin^{2}(t)$

Writing it like that is considered wrong, since $f(x) = \sin^{2}(t)$ implies that $x$ and $t$ are unrelated.

We could see for instance:
$$y=x^2$$
$$x=\sin(t)$$
$$y=\sin^2(t)$$
This represents some harmonic oscillation on a parabolic trajectory.
Keeping the physical meaning in mind, we can be sure that no mistakes will creep in.

Btw, to distinguish the derivatives, you'll see $y'$ to represent $y'(x)$, and $\dot y$ to represent $\d {} {t}y(x(t))$.

 

[Amad27]

Writing it like that is considered wrong, since $f(x) = sin^{2}(t)$ implies that $x$ and $t$ are unrelated.

We could see for instance:
$$y=x^2$$
$$x=\sin(t)$$
$$y=\sin^2(t)$$
This represents some harmonic oscillation on a parabolic trajectory.
Keeping the physical meaning in mind, we can be sure that no mistakes will creep in.

Btw, to distinguish the derivatives, you'll see $y'$ to represent $y'(x)$, and $\dot y$ to represent $\d {} {t}y(x(t))$.

Hi,

I suppose this is right, but when you see,

$f(x) = x^3$

You would suspect $x$ is independent.

Its not wrong to write $f(u)$

But my point is that it will help avoid confusion. Thats all ! THANKS!