We usually abuse notation and use the numbers themselves as higher order functions, and we also...
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We usually abuse notation and use the numbers themselves as higher order functions, and we also replace “application of a higher order func- tion" by "multiplication from left". As examples, we write ( a +1) f(x), where we actually mean d (1+1). f(x), dx d (x-2) ƒ(x) dx d (1-27). f(x) dx (1.9) (1.10) In the rest of this homework (and in the lectures), I will keep abusing the language as long as it is clear what we mean. Another place we abuse the notation is where we chain applications of derivatives. For instance, the expression d (²x + ³3) ( ₁/² - 2) F(x) dx (1.11) only makes sense when we interpret it as whereas does not make sense: (+³)-((-2)/(x)) dx d dx +3 ((²x + ³). (²x − ²)) · ƒ (x) dx (1.12) d (² + 3) ² +3 as we defined takes a type R → R as the dx d (1.13) input; it cannot act on dx - 2 which has the type (R → R) → (R → R). There are actually objects that can act on higher order functions (these are themselves called higher order too); for instance, consider the following operator: = [(x → ƒ(x)) → (x → ƒ'(x))] → [(x → ƒ(x)) → (x → ƒ”(x))] (1.14a) d d² c. = dx dx² which turns a first derivative higher order function to a second derivative higher order function, i.e. C ::??? (1.15) What is the type of this operator C when acting on real variables and real functions, i.e (1.16) We usually abuse notation and use the numbers themselves as higher order functions, and we also replace “application of a higher order func- tion" by "multiplication from left". As examples, we write ( a +1) f(x), where we actually mean d (1+1). f(x), dx d (x-2) ƒ(x) dx d (1-27). f(x) dx (1.9) (1.10) In the rest of this homework (and in the lectures), I will keep abusing the language as long as it is clear what we mean. Another place we abuse the notation is where we chain applications of derivatives. For instance, the expression d (²x + ³3) ( ₁/² - 2) F(x) dx (1.11) only makes sense when we interpret it as whereas does not make sense: (+³)-((-2)/(x)) dx d dx +3 ((²x + ³). (²x − ²)) · ƒ (x) dx (1.12) d (² + 3) ² +3 as we defined takes a type R → R as the dx d (1.13) input; it cannot act on dx - 2 which has the type (R → R) → (R → R). There are actually objects that can act on higher order functions (these are themselves called higher order too); for instance, consider the following operator: = [(x → ƒ(x)) → (x → ƒ'(x))] → [(x → ƒ(x)) → (x → ƒ”(x))] (1.14a) d d² c. = dx dx² which turns a first derivative higher order function to a second derivative higher order function, i.e. C ::??? (1.15) What is the type of this operator C when acting on real variables and real functions, i.e (1.16)
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The operator C takes a function that takes a real number as input and returns a real number and retu... View the full answer
Related Book For
Smith and Roberson Business Law
ISBN: 978-0538473637
15th Edition
Authors: Richard A. Mann, Barry S. Roberts
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