# 2 corrected april 12 the radius of conver gence is 3

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2. (corrected April 12) The radius of conver- gence is 3, the interval of convergence is [ - 3 , 3 ) . 3. The radius of convergence is 1, the interval of convergence is [ 1 , 3 ] . 4. the radius of convergence is 4, the interval of convergence is ( - 5 , 3 ) . Taylor and MacLaurin series Give (quoting from textbook allowed!) the MacLaurin series and their radii of convergence for: 10
1. 1 1 - x 2. e x 3. sin x 4. cos x 5. arctan x 6. ln(1 + x ) 7. (1 + x ) k Compute the MacLaurin series (by using the definition OR by subsituting into a known series) for the following functions: 1. e - 2 x 2 2. x · cos( x ) 3. cos( x 2 ) 4. x · cos( 1 2 x 2 ) 5. (1 + 2 x ) 1 / 4 Solution : 1. n =0 ( - 2 x 2 ) n n ! = n =0 ( - 1) n 2 n x 2 n n ! 2. = n =0 ( - 1) n x 2 n +1 (2 n )! 3. = n =0 ( - 1) n x 4 n (2 n )! 4. = n =0 ( - 1) n x 4 n +1 2 2 n · (2 n )! 5. = n =0 ( 1 / 4 n ) (2 x ) n Use the MacLaurin series expansions to compute the following integrals: 1. sin( x 2 ) dx 2. ln(1 + x 2 ) dx 3. e 3 x dx Solution : 1. = n =0 ( - 1) n x 4 n +3 (2 n +1)!(4 n +3) 2. This requires two steps. The first one to find the MacLaurin series of ln(1 + x 2 ) and then the second one to compute the integral. In total we get ln(1 + x 2 ) dx = n =1 ( - 1) n - 1 x 2 n +1 n · (2 n +1) . 3. = n =0 3 n · x n 2 +1 n ! · ( n 2 +1) = n =0 2 · 3 n · x n 2 +1 n ! · ( n +2) What is the general formula of the MacLaurin series? What is the general formula for the Taylor Series at a point a ? 11
Functions of several variables Level Curves, Traces Sketch the graph of the following functions (it is recommended to begin with determining the domain, the range, then sketching some well chosen level curves and traces). 1. f ( x, y ) = xy + 2. 2. f ( x, y ) = 1 x 2 +4 y 2 . 3. f ( x, y ) = ln( y - 3 x 2 ) Solution : 1. The domain is R 2 , and the range is R . The curve of level k has equation xy = k - 2: if k ̸ = 2, this is an hyperbola: y = k - 2 x , if k = 2 this is the reunion of the two lines x = 0 and y = 0. The trace of the graph in the yz plane has equation z = 0 + 2, this is the horizontal line at height 2. The trace of the graph in the plane x = 1 has equation z = y + 2. 2. The domain is R 2 - { (0 , 0) } , the range is (0 , ). The curve of level k > 0 has equation x 2 + 4 y 2 = 1 k 2 , this is an ellipse passing through the points ( ± 1 k , 0) and (0 , ± 1 2 k ). The trace of the graph in the yz plane has equation z = 1 | 2 y | . 3. (corrected April 12) f ( x, y ) = ln( y - 3 x 2 ) The domain is { ( x, y ) R 2 , y > 3 x 2 } (above the parabola y = 3 x 2 ). The range is R . The curve of level k has equation y = 3 x 2 + e k this is a parabola. The trace of the graph in the yz plane has equation z = ln( y ). The trace of the graph in the plane y = 1 has equation z = ln(1 - 3 x 2 ) (only defined for | x | < 1 3 , with maximum 0 at x = 0). Partial Derivatives Find all second partial derivatives of f at the given points. 1. f ( x, y ) = xy sin( y 2 ) at (3 , 2). 2. f ( x, y ) = y 5 - 3 x 2 y at (4 , 1). 3. f ( x, y ) = x ( x 2 + y ) 2 at (1 , 1). 4. f ( x, y ) = x ln( y x ) at (1 , 4).
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