Chapter 4: Problem 13
(a) If \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}^{m}\) and \(g: \mathbf{R}^{m} \rightarrow \mathbf{R}^{p}\), show that \((g \circ f)_{*}=g_{*} \circ f_{*}\) and \((g \circ f)^{*}=f^{*} \circ g^{*}\) (b) If \(f, g: \mathbf{R}^{n} \rightarrow \mathbf{R}\), show that \(d(f \cdot g)=f \cdot d g+g \cdot d f\).
Short Answer
Step by step solution
Understand the Definitions
Show Pushforward Identity \((g \circ f)_{*} = g_{*} \circ f_{*}\)
Show Pullback Identity \((g \circ f)^{*} = f^{*} \circ g^{*}\)
Understand the Product Rule
Apply the Product Rule
Combine and Simplify
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pushforward in Calculus
If you have a compound function like \( g \, \circ \, f \), the pushforward can be derived using the chain rule as follows: \( (g \circ f)_{*}(x) = D(g \circ f)(x) = Dg(f(x)) \, \cdot \, Df(x) \). This equals \( g_{*}(f(x)) \, \circ \, f_{*}(x) \). This straightforward linear representation helps us understand how transformations occur under the function \( f \) and their implications on the output of \( g \).
Here's a key takeaway:
- The pushforward linearizes a function, giving insight into how objects move under transformation.
- This is crucial when dealing with composite functions, as it ensures continuity and predictability through the Jacobian matrices involved.
Pullback in Calculus
Consider a function \( h: \mathbb{R}^p \rightarrow \mathbb{R} \). When pulling back using the pullback map of the composite function \( (g \circ f) \), it translates to \( (g \circ f)^{*}(h) = f^{*}(g^{*}(h)) \). This operation ensures that the function \( h \) is evaluated at every pre-image in the domain of \( f \) via the function \( g \). The cascade of transformations ensures that meaningful computations are possible by anchoring everything back to the space where they originated.
- Pullbacks are essential for transforming functions backward across composite transformations.
- It ensures that functions can be reliably evaluated through intermediate steps in a composite transformation.
Product Rule in Calculus
\[ d(f \cdot g) = f \cdot dg + g \cdot df. \] If we have two functions \( f \) and \( g \) on \( \mathbb{R}^n \), their differential product is given by:
\[ d(f \cdot g) = d(f) \cdot g + f \cdot d(g). \]
This formula is incredibly useful, especially in multivariable calculus where functions are complex. The principle can be extended when dealing with exterior derivatives and differential forms.
- The product rule ensures the correct breakdown of derivatives in product functions.
- It applies universally to multi-variable functions, aiding in complex differential computations.
Chain Rule
This is formalized as:
\[ D(g \circ f)(x) = Dg(f(x)) \cdot Df(x). \] The chain rule shows that to differentiate a composite function, you multiply the derivative of the outer function by the derivative of the inner function evaluated at the inner function’s point.
- The chain rule is vital for differentiating nested functions, ensuring continuous derivative accuracy.
- It is foundational for advanced calculus, particularly useful in multi-step transformations.
Differentiable Maps
This makes analysis and computations more tractable, as one can rely on the linear approximation for proofs and problem-solving. Consider how the pushforward and pullback operate; both rely on the existence of a well-defined derivative, underscoring the importance of differentiable maps.
- Differentiable maps ensure that transformations remain smooth without abrupt changes.
- These maps allow the use of linear approximations locally, simplifying complex analyses.