MHB Derivative of a Real-Valued Function of Several Variables: Junghenn Defn 9.1.3

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In the discussion about Definition 9.1.3 from Junghenn's "A Course in Real Analysis," the focus is on understanding the existence of the derivative vector \( f'(a) \) in \( \mathbb{R}^n \). It is clarified that the definition itself does not require a proof of existence for the vector; rather, it states that if such a vector satisfies the conditions of the definition, it is termed the derivative. The participants emphasize that the definition is not asserting that the derivative always exists for every function. The conversation highlights the distinction between definitions and theorems in mathematical analysis. Overall, the key takeaway is the understanding that proving existence is not necessary within the context of definitions.
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I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on $\mathbb{R}^n$"

I need some help with an aspect of Definition 9.1.3 ...

Definition 9.1.3 and the relevant accompanying text read as follows:
https://www.physicsforums.com/attachments/7865
View attachment 7866

At the top of the above text, in Definition 9.1.3 we read the following text:

" ... ... there exists a vector $$f'(a)$$ in $$\mathbb{R}^n$$ ... ... "My question is as follows:

How (arguing from the definition of derivative) do we indicate\demonstrate\prove that $$f'(a) \in \mathbb{R}^n$$ ... ...?Hope someone can help ... ...

Peter
 
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You don't have to prove it. This is a definition (no proofs allowed). The definition is saying that if there's a vector $f'(\mathbf{a})$ satisfying the definition, then we call it the derivative. The definition is not actually claiming that the vector exists (it doesn't always, depending on $f$).
 
Ackbach said:
You don't have to prove it. This is a definition (no proofs allowed). The definition is saying that if there's a vector $f'(\mathbf{a})$ satisfying the definition, then we call it the derivative. The definition is not actually claiming that the vector exists (it doesn't always, depending on $f$).
Thanks Ackbach ...

Appreciate the help...

Peter
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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