<p>Defining the following terms where <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are vectors:</p>
<p><span class="math-container">$∆x = x - x^*$</span></p>
<p><span class="math-container">$\therefore x = ∆x + x^*$</span></p>
<p><span class="math-container">$∆y = ∆f(x) = f(x) - f(x^*) = f(x^* + ∆x) - f(x^*)$</span></p>
<p>We can draw a fresh axis in terms of <span class="math-container">$∆x$</span> and <span class="math-container">$∆y$</span> essentially treating <span class="math-container">$x^*$</span> and <span class="math-container">$f(x^*)$</span> as the origin. This makes sense as <span class="math-container">$∆x$</span> and <span class="math-container">$∆f(x)$</span> evaluated at <span class="math-container">$x^* = 0$</span>. We then call this function <span class="math-container">$F(∆x)$</span> (See attached diagram)</p>
<p>This gives the required: <span class="math-container">$∆y = F(∆x) = f(x) - f(x^*) = f(x^* + ∆x) - f(x^*)$</span></p>
<p>The  <span class="math-container">$∆x$</span> and <span class="math-container">$∆y$</span>  axis imposed on the original graph can be seen at this point in the <a href="https://youtu.be/KesZslq5uVg?t=882" rel="nofollow noreferrer">video</a>, the use of <span class="math-container">$F$</span> instead of <span class="math-container">$f$</span> can be seen <a href="https://youtu.be/iPuMrI8wAl4?t=713" rel="nofollow noreferrer">here</a>.</p>
<p><strong>Answering Questions</strong></p>
<ol>
<li>I believe that <span class="math-container">$∆F(∆x)$</span> seen <a href="https://youtu.be/KesZslq5uVg?t=1165" rel="nofollow noreferrer">here</a> was a typo and this should have either been <span class="math-container">$∆f(x)$</span> or <span class="math-container">$F(∆x)$</span></li>
<li><strong>What's the benefit of <span class="math-container">$F(∆x)$</span>?</strong> I'm not entirely sure, but I believe the author does this to allow him to talk about Limits as <span class="math-container">$(∆x) \to 0$</span></li>
<li>What's going on with  <span class="math-container">$∆y = F(λ{∆x}) = f(x^* + λ{∆x}) - f(x^*)$</span>? Here he notes we are using a &quot;fixed&quot; <span class="math-container">$∆x$</span> and then taking λ scalar multiples of it. I suppose this is still just some arbitrary change representing <span class="math-container">$∆y$</span>.</li>
<li>He uses λ so that he can then divide through by <span class="math-container">$λ^2$</span> which he says is the approximate length squared of the vector <span class="math-container">$(∆x)$</span> so that we get: <span class="math-container">$\frac{1}{λ^2}[\nabla f(x^*)•λ(∆x) + \frac{1}{2}λ(∆x)^TH(x^*)λ(∆x)  + R_2(λ{∆x})] = \frac{1}{2}(∆x)^TH(x^*)(∆x)$</span></li>
</ol>
<ul>
<li>I have some questions about how we remove the remainder term <span class="math-container">$\lim \limits_{λ \to 0} \frac{R_2(λ∆x)}{(λ)^2} = 0$</span> , but I have added these in a <a href="https://economics.stackexchange.com/questions/54457/remainder-term-in-linear-approximations-going-to-0">new post</a>!</li>
<li>Any other thoughts about the notation in general and the use of <span class="math-container">$F(λ{∆x})$</span> are welcome.</li>
</ul>
<p><strong><span class="math-container">$∆x$</span> and <span class="math-container">$∆y$</span> axis</strong>
<a href="https://i.sstatic.net/d13Cb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/d13Cb.jpg" alt="enter image description here" /></a></p>
<p><strong>Clearer explanation of the links:</strong></p>
<ol>
<li>The <a href="https://youtu.be/KesZslq5uVg?t=882" rel="nofollow noreferrer">First Link</a> is showing an example of the axis using <span class="math-container">$∆x$</span> and <span class="math-container">$∆y$</span>.</li>
<li>The <a href="https://youtu.be/iPuMrI8wAl4?t=713" rel="nofollow noreferrer">Second Link</a> if you look at the top line (LHS) of his notation we see the use of <span class="math-container">$F(∆x)$</span> which is different to <span class="math-container">$∆f(∆x)$</span> which is shown in the next link. These two notation are implied to mean the same thing and is a source of confusion discussed here.</li>
<li>The <a href="https://youtu.be/KesZslq5uVg?t=1165" rel="nofollow noreferrer">Third Link</a> shows the use of <span class="math-container">$∆f(∆x)$</span> notation.</li>
</ol>
<ul>
<li>Here is an <a href="https://youtu.be/E_wkLs1uv2E?t=1177" rel="nofollow noreferrer">additional example</a> of <span class="math-container">$(∆x), (∆y)$</span> axis, when he first uses it. However it's more clear what's happening in the first link (because he gives arrows to the new axis there).</li>
</ul>

CC BY-SA 4.0
2023-02-21T13:01:11.290Z
I find it , $ F(\Delta x)$, a quite strange notation, but it isn't possible to say anything more precise without seeing the video entirely. We have usaually the Taylor's formula with $x*=0$, that is called the formula of MacLaurin,  but there isn't  $ F(\Delta x)$, nor a translation af axes, even if could be the same. Maybe the professor here wants you to have a geometric insight, but it is impossible to say it without seeing the video. By the way, the video you linked in the other question maybe is wrong, it isn't the video to which the question refers to.
BakerStreet
CC BY-SA 4.0
2023-02-27T13:51:58.897Z
Hello there! Yes it is unusual I agree, the $F(∆x)$ makes some sense to me now, and I suppose he wanted to write things as a function of $∆x$. That being said I am confused still on $F(∆x)$ vs $∆f(∆x)$...I have made the video links explicit at the end, so that you understand the purpose of them and what to look for in the video. He's a very very good professor, like your explanations usually his are excellent...so perhaps if you do find the time you will get something from it! I'm not too worried now, but the explicit guide to the links at the bottom should help if you're curious! Many thanks!
CormJack
CC BY-SA 4.0
2023-02-27T14:27:45.873Z
I will try to listen to the video, I hope it colud be enough , because when there is a particular notation it is explained at the beginning of the lessons.  I've never seen in my books such a notation when explaining maximum conditions.
BakerStreet
CC BY-SA 4.0
2023-02-16T04:09:12.943Z
2023-02-27T13:49:28.390Z
2023-02-27T13:49:28.390Z
CormJack
<p><strong>Edit:</strong> I have updated the link so that it works!</p>
<p>I was watching a lecture for a  proof that if <span class="math-container">$x^*$</span> is a local maximiser of <span class="math-container">$f$</span> then necessarily the hessian is negative semi definite.</p>
<p>However i've got stuck on this notation: <a href="https://youtu.be/ArqDkIQLqEw?t=329" rel="nofollow noreferrer">Time stamped Source</a></p>
<p><span class="math-container">$∆y = f(λ∆x) = f(x^* + λ∆x) - f(x^*)$</span></p>
<ul>
<li>The <span class="math-container">$x's$</span> are vectors:</li>
<li>He mentions later that &quot;Dividing by <span class="math-container">$λ^2$</span>  is basically dividing through by the square of the length of the vector <span class="math-container">$∆x$</span>,&quot; which i guess implies that <span class="math-container">$∆x$</span> is very small and hence deriving most of it's length from λ?</li>
</ul>
<ol>
<li>I'm more used to defining a <span class="math-container">$∆x = x^* - x$</span> and <span class="math-container">$∆y = f(x^*) - f(x)$</span></li>
<li>I suppose given <span class="math-container">$y = f(x)$</span> we could take <span class="math-container">$y$</span> as <span class="math-container">$∆y$</span> and <span class="math-container">$x$</span> as <span class="math-container">$∆x$</span> hence we'd have <span class="math-container">$f(∆x)$</span> and then really <span class="math-container">$λ∆x = ∆x$</span> for some arbitrary change?</li>
<li>Then setting setting this = <span class="math-container">$f(x^* + λ∆x) - f(x^*)$</span> is equivalent to something like <span class="math-container">$f(x^* + hx) - f(x^*)$</span></li>
</ol>
<p>But for example if you actually sub in <span class="math-container">$∆x = x^* - x$</span> you get something that looks like the convex combination of <span class="math-container">$x^*$</span> and <span class="math-container">$x$</span>, so I'm really not convinced i know whats going on.</p>
<p>If anyone could walk me through this notation that would be great...</p>
<p><strong>Additional Context</strong></p>
<p>For context he then goes onto use the second degree Taylor polynomial:
<span class="math-container">$\nabla fλ∆x^* + \frac{1}{2}λ∆xH(x^*)λ∆x  + R_2(λ∆x)$</span>
Where the final term is just his notation for the remainder term for the second degree Taylor series with respect to our augment, and then divides by <span class="math-container">$λ^2$</span> as mentioned, to reduce the remainder to 0.</p>

CC BY-SA 4.0
2023-02-15T02:43:38.013Z
I've also noticed in this video he writes $∆y = ∆f(∆x) = f(x) -  f(x^*)$. In general I've found this lecturer to be very good, and usually he even responds to comments, so I'm nervous to call the discrepancy a typo...but I don't really understand what's going on here: https://youtu.be/KesZslq5uVg?t=1158
CormJack
CC BY-SA 4.0
2023-02-15T11:02:01.467Z
AS for $f(\Delta y)$, I saw a very short part of the video of this lesson at minute 19: , it is strange, you are right, one had to see the video from the beginning, but there is something unusual. Coluld be a typo.
BakerStreet
CC BY-SA 4.0
2023-02-15T13:49:51.697Z
Hi BakerStreet, your comment reads as if thee was another part to it before the As?...Perhaps you were referencing the Brocken link in the video which I have now fixed!
CormJack
CC BY-SA 4.0
2023-02-15T13:52:48.030Z
Yes, the minute 19 of the video, where the professor writes $\Delta f( \Delta x)$
BakerStreet
CC BY-SA 4.0
2023-02-15T13:54:24.493Z
Ah okay that link in the comments works, you'll notice there is a different link which was Brocken (but now works) in the video which links to the notation I was trying to understand in the question! Subtly different to the notation in the comment...Look forward to your response whenever you get the time!
CormJack
CC BY-SA 4.0
2023-02-15T13:57:04.303Z
The link didn't work, actually. In the video , in my opinion, that is a typo, $f(\Delta x)$ makes no sense, as far as I saw.
BakerStreet
CC BY-SA 4.0
2023-02-15T14:02:06.070Z
Does the link for the first video work in the original post? It should take you to 5:30 in this video: https://www.youtube.com/watch?v=ArqDkIQLqEw&t=329s. He used $f(∆x)$ consistently here, but never $∆f(∆x)$ which definitely feels like a typo with the additional ∆ in front of the $f$
CormJack
CC BY-SA 4.0
2023-02-15T14:03:35.450Z
Yes, it works now.
BakerStreet
CC BY-SA 4.0
2023-02-15T14:08:40.310Z
Awesome! Also if you don't mind me asking what time zone is it for you now? You seem to be solving questions at all hours!
CormJack
CC BY-SA 4.0
2023-02-15T14:36:08.613Z
It is 3:35 p.m.
BakerStreet
CC BY-SA 4.0
2023-02-15T17:45:06.947Z
Ah cool you’re in Europe! I’m normally based in London, Peru atm. I’ve also made great progress with this, will post an answer soon. But there are still some grey areas
CormJack
CC BY-SA 4.0
2023-02-15T18:01:34.780Z
Ah you're in London, beautiful! I'll read what you will write as soon as possible!
BakerStreet
CC BY-SA 4.0
2023-02-16T04:09:30.347Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/142907/discussion-between-cormjack-and-bakerstreet).
CormJack
<mathematical-economics><optimization>
Notation for ∆y used in Necessary Second Order Condition Proof
CormJack
# Notation for ∆y used in Necessary Second Order Condition Proof

(Asked by: username_0 on 2023-02-15)

**Edit:** I have updated the link so that it works!

I was watching a lecture for a proof that if $x^\*$ is a local maximiser of $f$ then necessarily the hessian is negative semi definite.

However i've got stuck on this notation: [Time stamped Source](https://youtu.be/ArqDkIQLqEw?t=329)

$∆y = f(λ∆x) = f(x^\* + λ∆x) - f(x^\*)$

* The $x's$ are vectors:
* He mentions later that "Dividing by $λ^2$ is basically dividing through by the square of the length of the vector $∆x$," which i guess implies that $∆x$ is very small and hence deriving most of it's length from λ?

1. I'm more used to defining a $∆x = x^\* - x$ and $∆y = f(x^\*) - f(x)$
2. I suppose given $y = f(x)$ we could take $y$ as $∆y$ and $x$ as $∆x$ hence we'd have $f(∆x)$ and then really $λ∆x = ∆x$ for some arbitrary change?
3. Then setting setting this = $f(x^\* + λ∆x) - f(x^\*)$ is equivalent to something like $f(x^\* + hx) - f(x^\*)$

But for example if you actually sub in $∆x = x^\* - x$ you get something that looks like the convex combination of $x^\*$ and $x$, so I'm really not convinced i know whats going on.

If anyone could walk me through this notation that would be great...

**Additional Context**

For context he then goes onto use the second degree Taylor polynomial:
$\nabla fλ∆x^\* + \frac{1}{2}λ∆xH(x^\*)λ∆x  + R\_2(λ∆x)$
Where the final term is just his notation for the remainder term for the second degree Taylor series with respect to our augment, and then divides by $λ^2$ as mentioned, to reduce the remainder to 0.

**Tags:** mathematical-economics, optimization
> **Comments:**
> **username_0**: I've also noticed in this video he writes $∆y = ∆f(∆x) = f(x) -  f(x^*)$. In general I've found this lecturer to be very good, and usually he even responds to comments, so I'm nervous to call the discrepancy a typo...but I don't really understand what's going on here: https://youtu.be/KesZslq5uVg?t=1158  
> **username_1**: AS for $f(\Delta y)$, I saw a very short part of the video of this lesson at minute 19: , it is strange, you are right, one had to see the video from the beginning, but there is something unusual. Coluld be a typo.  
> **username_0**: Hi BakerStreet, your comment reads as if thee was another part to it before the As?...Perhaps you were referencing the Brocken link in the video which I have now fixed!  
> **username_1**: Yes, the minute 19 of the video, where the professor writes $\Delta f( \Delta x)$  
> **username_0**: Ah okay that link in the comments works, you'll notice there is a different link which was Brocken (but now works) in the video which links to the notation I was trying to understand in the question! Subtly different to the notation in the comment...Look forward to your response whenever you get the time!  
> **username_1**: The link didn't work, actually. In the video , in my opinion, that is a typo, $f(\Delta x)$ makes no sense, as far as I saw.  
> **username_0**: Does the link for the first video work in the original post? It should take you to 5:30 in this video: https://www.youtube.com/watch?v=ArqDkIQLqEw&t=329s. He used $f(∆x)$ consistently here, but never $∆f(∆x)$ which definitely feels like a typo with the additional ∆ in front of the $f$  
> **username_1**: Yes, it works now.  
> **username_0**: Awesome! Also if you don't mind me asking what time zone is it for you now? You seem to be solving questions at all hours!  
> **username_1**: It is 3:35 p.m.  
> **username_0**: Ah cool you’re in Europe! I’m normally based in London, Peru atm. I’ve also made great progress with this, will post an answer soon. But there are still some grey areas  
> **username_1**: Ah you're in London, beautiful! I'll read what you will write as soon as possible!  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/142907/discussion-between-cormjack-and-bakerstreet).  

---

## 1 answers

### Answer by username_0 on 2023-02-16. Score: 0 | Accepted answer

Defining the following terms where $x$ and $y$ are vectors:

$∆x = x - x^\*$

$\therefore x = ∆x + x^\*$

$∆y = ∆f(x) = f(x) - f(x^\*) = f(x^\* + ∆x) - f(x^\*)$

We can draw a fresh axis in terms of $∆x$ and $∆y$ essentially treating $x^\*$ and $f(x^\*)$ as the origin. This makes sense as $∆x$ and $∆f(x)$ evaluated at $x^\* = 0$. We then call this function $F(∆x)$ (See attached diagram)

This gives the required: $∆y = F(∆x) = f(x) - f(x^\*) = f(x^\* + ∆x) - f(x^\*)$

The $∆x$ and $∆y$ axis imposed on the original graph can be seen at this point in the [video](https://youtu.be/KesZslq5uVg?t=882), the use of $F$ instead of $f$ can be seen [here](https://youtu.be/iPuMrI8wAl4?t=713).

**Answering Questions**

1. I believe that $∆F(∆x)$ seen [here](https://youtu.be/KesZslq5uVg?t=1165) was a typo and this should have either been $∆f(x)$ or $F(∆x)$
2. **What's the benefit of $F(∆x)$?** I'm not entirely sure, but I believe the author does this to allow him to talk about Limits as $(∆x) \to 0$
3. What's going on with $∆y = F(λ{∆x}) = f(x^\* + λ{∆x}) - f(x^\*)$? Here he notes we are using a "fixed" $∆x$ and then taking λ scalar multiples of it. I suppose this is still just some arbitrary change representing $∆y$.
4. He uses λ so that he can then divide through by $λ^2$ which he says is the approximate length squared of the vector $(∆x)$ so that we get: $\frac{1}{λ^2}[\nabla f(x^\*)•λ(∆x) + \frac{1}{2}λ(∆x)^TH(x^\*)λ(∆x)  + R\_2(λ{∆x})] = \frac{1}{2}(∆x)^TH(x^\*)(∆x)$

* I have some questions about how we remove the remainder term $\lim \limits\_{λ \to 0} \frac{R\_2(λ∆x)}{(λ)^2} = 0$ , but I have added these in a [new post](https://economics.stackexchange.com/questions/54457/remainder-term-in-linear-approximations-going-to-0)!
* Any other thoughts about the notation in general and the use of $F(λ{∆x})$ are welcome.

**$∆x$ and $∆y$ axis**
[![enter image description here](https://i.sstatic.net/d13Cb.jpg)](https://i.sstatic.net/d13Cb.jpg)

**Clearer explanation of the links:**

1. The [First Link](https://youtu.be/KesZslq5uVg?t=882) is showing an example of the axis using $∆x$ and $∆y$.
2. The [Second Link](https://youtu.be/iPuMrI8wAl4?t=713) if you look at the top line (LHS) of his notation we see the use of $F(∆x)$ which is different to $∆f(∆x)$ which is shown in the next link. These two notation are implied to mean the same thing and is a source of confusion discussed here.
3. The [Third Link](https://youtu.be/KesZslq5uVg?t=1165) shows the use of $∆f(∆x)$ notation.

* Here is an [additional example](https://youtu.be/E_wkLs1uv2E?t=1177) of $(∆x), (∆y)$ axis, when he first uses it. However it's more clear what's happening in the first link (because he gives arrows to the new axis there).

> **Comments:**
> **username_1**: I find it , $ F(\Delta x)$, a quite strange notation, but it isn't possible to say anything more precise without seeing the video entirely. We have usaually the Taylor's formula with $x*=0$, that is called the formula of MacLaurin,  but there isn't  $ F(\Delta x)$, nor a translation af axes, even if could be the same. Maybe the professor here wants you to have a geometric insight, but it is impossible to say it without seeing the video. By the way, the video you linked in the other question maybe is wrong, it isn't the video to which the question refers to.  
> **username_0**: Hello there! Yes it is unusual I agree, the $F(∆x)$ makes some sense to me now, and I suppose he wanted to write things as a function of $∆x$. That being said I am confused still on $F(∆x)$ vs $∆f(∆x)$...I have made the video links explicit at the end, so that you understand the purpose of them and what to look for in the video. He's a very very good professor, like your explanations usually his are excellent...so perhaps if you do find the time you will get something from it! I'm not too worried now, but the explicit guide to the links at the bottom should help if you're curious! Many thanks!  
> **username_1**: I will try to listen to the video, I hope it colud be enough , because when there is a particular notation it is explained at the beginning of the lessons.  I've never seen in my books such a notation when explaining maximum conditions.  

---

