A clarification of (Ferrario 2011, page 43). This is a trivial example, but according to a member of the National Academy of the Sciences, it's difficult to write a page of mathematics without errors.

We say that $(d+ 1)$ points $(x_0, x_1, \ldots,x_ d)$ belonging to the Euclidean space $(\mathbb{R}^n)$ are linearly independent (from the affine point of view) if the vectors $(x_1 − x_0, x_2 − x_0,\ldots,x_d − x_0)$ are linearly independent. A vector $(x − x_0)$ of the vector space generated by these vectors can be written as a sum $(x-x_0=\sum_{i=1}^d r_i(x_i-x_0))$ with real coefficients $(r_i)$; notice that if we write $(x)$ as $(x=\sum_{i=0}^d \alpha_i x_i)$ , then $(\sum_{i=0}^d \alpha_i =1)$."
—Ferrario 2011, page 43.

The assertion after the semicolon is false as stated, unless it is assumed that $(r_i = \alpha_i)$ for $(1\le i\le d)$. Are the authors wrong? Who cares—here's a restatement.

Suppose that $(x- x_0 = \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right).)$ Then $(x = \sum_{i=0}^d \alpha_i x_i)$ if and only if $(\sum_{i=0}^d \alpha_i =1)$.

Proof. Substituting, we have $(\sum_{i=0}^d \alpha_i x_i - x_0 = \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right).)$ Therefore $(\alpha_0 x_0 - x_0 = -\left(\sum_{i=1}^d \alpha_i\right)x_0)$, which implies $(\alpha_0 x_0 = \left( 1- \sum_{i=1}^d \alpha_i\right)x_0)$. Since the vector $(x_0)$ is nonzero, it is independent, and $(\alpha_0 = 1- \sum_{i=1}^d \alpha_i)$. It follows that $(\sum_{i=0}^d \alpha_i =1)$.

Ferrario, D.L., and R.A. Piccinini. 2010. “The Category of Simplicial Complexes.” In Simplicial Structures in Topology., 43–97. CMS Books in Mathematics (Ouvrages de Mathématiques de La SMC). Springer, New York, NY.

Post edited by ZettelDistraction on

Erdős #2. ZK software components. “If you’re thinking without writing, you only think you’re thinking.” -- Leslie Lamport. Replies sometimes delayed since life is short.

If it may help, Morganeua made a YT video on this topic. The tips and trichs she gives may not fit to anyone, but... could for some people!
The video is here.

I have a couple of notes on this, but I'll pull from a few of them:

To write in your own words is to express in your own way

To say something in your own words is just to express it in the way you can communicate.

It might just mean that the way you heard it is the exact way you can express it.

It's not about being different; it's about being you.

And from another note:

In your own words is not just your take on something, it is your take on that something tied to a conversation you feel is worth having.

That might just mean you'll have a beat for beat interpretation of something because you're using that same something as a conversation starter.

It may just be an interpretation with relation to a inquiry that you have in mind.

Very often you will have not have the exact same use for a piece of information as did the source you learned from. It may be dramatically or slightly different, but that difference would make all the difference for you to communicate something with your own voice.

## Comments

## 20211121160510 A technical example

#mathematics #inyourownwords

A clarification of (Ferrario 2011, page 43). This is a trivial example, but according to a member of the National Academy of the Sciences, it's difficult to write a page of mathematics without errors.

The assertion after the semicolon is false as stated, unless it is assumed that $(r_i = \alpha_i)$ for $(1\le i\le d)$. Are the authors wrong? Who cares—here's a restatement.

Suppose that $(x- x_0 = \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right).)$ Then $(x = \sum_{i=0}^d \alpha_i x_i)$ if and only if $(\sum_{i=0}^d \alpha_i =1)$.

Proof.Substituting, we have $(\sum_{i=0}^d \alpha_i x_i - x_0 = \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right).)$ Therefore $(\alpha_0 x_0 - x_0 = -\left(\sum_{i=1}^d \alpha_i\right)x_0)$, which implies $(\alpha_0 x_0 = \left( 1- \sum_{i=1}^d \alpha_i\right)x_0)$. Since the vector $(x_0)$ is nonzero, it is independent, and $(\alpha_0 = 1- \sum_{i=1}^d \alpha_i)$. It follows that $(\sum_{i=0}^d \alpha_i =1)$.Conversely, if $(\sum_{i=0}^d \alpha_i =1)$, then

$$(\begin{align}{}

x &= \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right) + x_0\\

&= \sum_{i=1}^d \alpha_i \left( x_i - x_0 \right) + \left(\sum_{i=0}^d \alpha_i\right)x_0\\

&= \sum_{i=0}^d \alpha_i x_i .

\end{align})$$

QED## References

Ferrario, D.L., and R.A. Piccinini. 2010. “The Category of Simplicial Complexes.” In

Simplicial Structures in Topology., 43–97. CMS Books in Mathematics (Ouvrages de Mathématiques de La SMC). Springer, New York, NY.Erdős #2. ZK software components. “If you’re thinking without writing, you only think you’re thinking.” -- Leslie Lamport. Replies sometimes delayed since life is short.

If it may help, Morganeua made a YT video on this topic. The tips and trichs she gives may not fit to anyone, but... could for some people!

The video is here.

I have a couple of notes on this, but I'll pull from a few of them:

And from another note:

Very often you will have not have the exact same use for a piece of information as did the source you learned from. It may be dramatically or slightly different, but that difference would make all the difference for you to communicate something with your own voice.