17. Show that if g is a function in C transformation from C a linear (a, b), then Tg(x) is (a, b) to C (a, b).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Problem 17:**

Show that if \( g \) is a function in \( C^\infty(a, b) \), then \( T_{g(x)} \) is a linear transformation from \( C^\infty(a, b) \) to \( C^\infty(a, b) \).

**Explanation:**

This problem involves demonstrating the linearity of a transformation \( T_{g(x)} \) in the context of smooth functions. A function is said to be in \( C^\infty(a, b) \) if it is infinitely differentiable on the interval \( (a, b) \). The goal is to show that the transformation \( T_{g(x)} \) maps smooth functions to other smooth functions and satisfies the properties of linearity.
Transcribed Image Text:**Problem 17:** Show that if \( g \) is a function in \( C^\infty(a, b) \), then \( T_{g(x)} \) is a linear transformation from \( C^\infty(a, b) \) to \( C^\infty(a, b) \). **Explanation:** This problem involves demonstrating the linearity of a transformation \( T_{g(x)} \) in the context of smooth functions. A function is said to be in \( C^\infty(a, b) \) if it is infinitely differentiable on the interval \( (a, b) \). The goal is to show that the transformation \( T_{g(x)} \) maps smooth functions to other smooth functions and satisfies the properties of linearity.
**Proof:**

We will prove the associativity property in part (6) and leave the proofs of the remaining parts as exercises (Exercise 16). In fact, this associativity property holds for all functions, not just linear transformations. To prove it (which is the approach used to prove all of these properties), we take a vector \( v \) in the domain and verify that the functions on each side of the equation applied to \( v \) give us the same result: Applying the definition of the composite of functions, we have

\[ R(ST)(v) = R(ST(v)) = R(S(T(v))) \]

and

\[ (RS)T(v) = RS(T(v)) = R(S(T(v))). \]

Therefore, the functions \( R(ST) \) and \( (RS)T \) are the same.

One other similarity between matrices and linear transformations we mention at this time involves exponents. Recall that if \( A \) is a square matrix and \( n \) is a positive integer, we have defined the \( n \)th power of \( A \), \( A^n \), as the product of \( n \) factors of \( A \). Notice that we can do likewise for a linear transformation that has the same domain and codomain. That is, if \( T : V \to V \) is a linear transformation and \( n \) is a positive integer, then \( T^n \) is the product (composite) of \( n \) factors of \( T \).

We are now going to see how we can use the ideas we have developed involving linear transformations to obtain a particularly elegant approach to the study of linear differential equations. To simplify our presentation, let us assume that our functions have derivatives of all orders on an open interval \( (a, b) \); that is, we will use \( C^\infty (a, b) \) as our vector space of functions. We will make use of two basic types of linear transformations. One type involves the differential operator \( D : C^\infty (a, b) \to C^\infty (a, b) \) and its powers \( D^n \), which amount to taking the \( n \)th derivative. The other type involves multiplying functions \( f \) in \( C^\infty (a, b) \) by a fixed function \( g \) in \( C
Transcribed Image Text:**Proof:** We will prove the associativity property in part (6) and leave the proofs of the remaining parts as exercises (Exercise 16). In fact, this associativity property holds for all functions, not just linear transformations. To prove it (which is the approach used to prove all of these properties), we take a vector \( v \) in the domain and verify that the functions on each side of the equation applied to \( v \) give us the same result: Applying the definition of the composite of functions, we have \[ R(ST)(v) = R(ST(v)) = R(S(T(v))) \] and \[ (RS)T(v) = RS(T(v)) = R(S(T(v))). \] Therefore, the functions \( R(ST) \) and \( (RS)T \) are the same. One other similarity between matrices and linear transformations we mention at this time involves exponents. Recall that if \( A \) is a square matrix and \( n \) is a positive integer, we have defined the \( n \)th power of \( A \), \( A^n \), as the product of \( n \) factors of \( A \). Notice that we can do likewise for a linear transformation that has the same domain and codomain. That is, if \( T : V \to V \) is a linear transformation and \( n \) is a positive integer, then \( T^n \) is the product (composite) of \( n \) factors of \( T \). We are now going to see how we can use the ideas we have developed involving linear transformations to obtain a particularly elegant approach to the study of linear differential equations. To simplify our presentation, let us assume that our functions have derivatives of all orders on an open interval \( (a, b) \); that is, we will use \( C^\infty (a, b) \) as our vector space of functions. We will make use of two basic types of linear transformations. One type involves the differential operator \( D : C^\infty (a, b) \to C^\infty (a, b) \) and its powers \( D^n \), which amount to taking the \( n \)th derivative. The other type involves multiplying functions \( f \) in \( C^\infty (a, b) \) by a fixed function \( g \) in \( C
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