Chapter 5 Solutions Question 1 ๐(๐ฅ) as ๐(๐ฅ) = 4log(๐ฅ)sin(๐ฅ3 ) . Then, ๐ ′ (๐ฅ) = 4๐ฅ sin(๐ฅ3 ) + 12๐ฅ2 log(๐ฅ)cos(๐ฅ3 ) . Firstly, let's rewrite Question 2 If we rewrite our function as exp(−๐ฅ) . ๐(๐ฅ) = (1 + exp(−๐ฅ))−1 , then we have ๐ ′ (๐ฅ) = (1+exp(−๐ฅ) )2 Question 3 We have ๐ ′ (๐ฅ) = ๐−๐ฅ๐2 ๐(๐ฅ). Question 4 We compute the ๏ฌrst ๏ฌve derivatives of our function at 0. We have , and . ๐(0) = ๐ ′ (0) = 1, ๐ (2) (0) = ๐ (3) (0) = −1 ๐ (4) (0) = ๐ (5) (0) = 1 The Taylor polynomial ๐5 (๐ฅ) = 1 + ๐ฅ − 1 ๐ฅ2 − 1 ๐ฅ3 + 1 ๐ฅ4 + 1 ๐ฅ5 . The lower-order Taylor polynomials 2 6 24 120 can be found by truncating this expression appropriately. Question 5 Part a We can see below that Part b We have Part c ∂๐1 ∂๐ฅ ∂๐1 ∂๐ฅ has dimension 1 × 2 ; ∂∂๐ฅ๐2 has dimension 1 × ๐ ; and ∂∂๐ฅ๐3 has dimension ๐2 × ๐. = [ cos(๐ฅ1 )cos(๐ฅ2 ) −sin(๐ฅ1 )sin(๐ฅ2 ) ]; ∂∂๐ฅ๐2 = ๐ฆ๐ณ . (Remember, ๐ฆ is a column vector!) ๎๎ ๐ฅ21 ๐ฅ1 ๐ฅ2 โฏ ๐ฅ1 ๐ฅ๐ ๎๎ ๎ ๐ฅ2 ๐ฅ1 ๐ฅ22 โฏ ๐ฅ2 ๐ฅ๐ ๎๎. Thus its derivative will be a higher-order tensor. Note that ๐ฅ๐ฅ๐ณ is the matrix ๎ ๎๎ โฎ โฎ โฑ โฎ ๎๎ ๎ ๐ฅ๐ ๐ฅ1 ๐ฅ๐ ๐ฅ2 2โฏ ๐ฅ2๐ ๎ However, if we consider the matrix to be an ๐ -dimensional object in its own right, we can compute the Jacobian. Its ๏ฌrst row consists of [ 2๐ฅ1 ๐ฅ2 โฏ ๐ฅ๐ | ๐ฅ2 0 โฏ 0 | โฏ |๐ฅ๐ 0 โฏ 0 ], where I have inserted a vertical bar every ๐ columns, to aid readability. Question 6 = cos(log(๐ก๐ณ ๐ก)) ⋅ ๐ก1๐ณ๐ก ⋅ [ 2๐ก1 2๐ก2 โฏ 2๐ก๐ท ] = cos(log(๐ก๐ณ ๐ก)) ⋅ 2๐ก๐ณ๐ก๐ณ๐ก . ๐น ๐ธ For ๐, if we explicitly compute ๐ด๐๐ต and ๏ฌnd its trace, we have that ๐(๐) = ∑๐ท ๐=1 ∑๐=1 ∑๐=1 ๐๐๐ ๐ฅ๐๐ ๐๐๐ . ∂๐ ๐ท Thus we have, ∂๐ฅ๐๐ = ∑๐=1 ๐๐๐ ๐๐๐ , and this is the (๐,๐)-th entry of the required derivative. Hence ๐๐ ๐ณ ๐ณ ๐๐ = ๐ต ๐ด . We have ๐๐ ๐๐ก Question 7 Part a = ๐๐๐๐ง ๐๐ง๐๐ฅ , where ๐๐๐๐ง has dimension 1 × 1 , and ๐๐ง๐๐ฅ has dimension 1 × ๐ท . We ๐๐ = 1 . know ๐๐ง = 2๐ฅ๐ณ from ๐ in Question 6. Also, ๐๐ฅ ๐๐ง 1+๐ง ๐๐ = 2๐ฅ๐ณ . Therefore, ๐๐ฅ 1+๐ฅ๐ณ ๐ฅ The chain rule tells us that Part b ๐๐ ๐๐ฅ ๎๎ cos๐ง1 0 ๐๐ is an ๐ธ × ๐ธ matrix, namely ๎ 0 cos๐ง2 Here we have ๐๐ง ๎๎ โฎ โฎ ๎ 0 0 Also, ๐๐ง is an ๐ธ × ๐ท-dimensional matrix, namely ๐ด itself. ๐๐ฅ โฏ 0 ๎๎ โฏ 0 ๎. โฑ โฎ ๎๎ โฏ cos๐ง๐ธ ๎ The overall derivative is obtained by multiplying these two matrices together, which will again give us an -dimensional matrix. ๐ธ×๐ท Question 8 Part a 1 × 1 , and is simply − 12 exp(− 12 ๐ง). ๐๐ง has dimension 1 × ๐ท , and is given by ๐ฆ๐ณ (๐ −1 + (๐ −1 )๐ณ ) . Now, ๐๐ฆ ๐๐ฆ has dimension ๐ท × ๐ท, and is just the identity matrix. Finally, ๐๐ฅ We have ๐๐ ๐๐ง has dimension Again, we multiply these all together to get our ๏ฌnal derivative. Part b If we explicitly write out ๐๐ ๐๐ฅ Hence, = 2๐ฅ๐ณ . Part c ๎๎ cosh12 ๐ง1 ๎๎ 0 ๐๐ Here, ๐๐ง = ๎ ๎๎ โฎ ๎ 0 Finally, ๐ฅ๐ฅ๐ณ + ๐ 2 ๐ผ , and compute its trace, we ๏ฌnd that ๐(๐ฅ) = ๐ฅ21 + โฏ + ๐ฅ2๐ + ๐๐ 2 . 0 1 cosh2 ๐ง2 โฎ 0 โฏ โฏ โฑ โฏ 0 0 โฎ 1 cosh2 ๐ง๐ ๎๎ ๎๎ ๎๎ , while ๐๐ง๐๐ฅ = ๐ด, as in Question 7b. ๎๎ ๐๐ is given by the product of these two matrices. ๐๐ฅ Question 9 ๐ง with ๐ก(๐,๐) throughout, we have that ๐(๐) = log(๐(๐ฅ,๐ก(๐,๐))) − log(๐(๐ก(๐,๐),๐)) . ๐๐ ๐′ (๐ฅ,๐ก(๐,๐))⋅๐ก′(๐,๐) − ๐′ (๐ก(๐,๐),๐)⋅๐ก′(๐,๐) . Therefore, ๐๐ = ๐(๐ฅ,๐ก(๐,๐)) ๐(๐ก(๐,๐),๐) Piecing this together, replacing
