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# It is a useful program for doing symbolic algebra

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Mathematica isn’t covered in this class, it may be useful to see what it can do easily. It is a useful program for doing symbolic algebra, manipulating lists, and for employing a variety of common mathematical functions without having to write too much code. Of course for this problem you could also have written some MATLAB code to accomplish more or less the same thing, or written out more steps by hand. In[389]:= Binomial[0.25, 2] Out[389]= -0.09375 In[390]:= Binomial[0.25, 3] Out[390]= 0.0546875 In[391]:= Binomial[0.25, 4] Out[391]= -0.0375977 In[379]:= f = {1.0, 1.22140, 1.49182, 1.82212, 2.22554} Out[379]= {1., 1.2214, 1.49182, 1.82212, 2.22554} In[383]:= Df = Table[f[[i + 1]] - f[[i]], {i, 4}] Out[383]= {0.2214, 0.27042, 0.3303, 0.40342} In[385]:= DDf = Table[Df[[i + 1]] - Df[[i]], {i, 3}] Out[385]= {0.04902, 0.05988, 0.07312} In[387]:= DDDf = Table[DDf[[i + 1]] - DDf[[i]], {i, 2}] Out[387]= {0.01086, 0.01324} In[388]:= DDDDf = Table[DDDf[[i + 1]] - DDDf[[i]], {i, 1}] Out[388]= {0.00238}

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MATH 128A, SUMMER 2010, HOMEWORK 5 SOLUTION 3 16. For a function f , the Newton divided-difference formula gives the interpolating polynomial P 3 ( x ) = 1 + 4 x + 4 x ( x - 0 . 25) + 16 3 x ( x - 0 . 25)( x - 0 . 5) , on the nodes x 0 = 0, x 1 = 0 . 25, and x 3 = 0 . 75. Find f (0 . 75). solution: Since the interpolating polynomial agrees with f at the x i and we are asked to find f ( x 3 ), it suffices to compute P 3 ( x 3 ). We get P 3 (0 . 75) = 1 + 4(0 . 75) + 4 · 0 . 75 · (0 . 75 - 0 . 25) + 16 3 · 0 . 75(0 . 75 - 0 . 25)(0 . 75 - 0 . 5) = 6. section 3.3 2. Use Theorem 3.9 or Algorithm 3.3 to construct an approximating polynomial for the follow- ing data. a. x f ( x ) f 0 ( x ) 0 1.00000 2.00000 0.5 2.71828 5.43656 solution: I will use Theorem 3.9. It says the Hermite polynomial is given by the formula H 2 n +1 ( x ) = n j =0 f ( x j )[1 - 2( x - x j ) L 0 n,j ( x )] L 2 n,j ( x )+ n j =0 f 0 ( x j )( x - x j ) L 2 n,j ( x ). In our case n = 1 and there are only two points x 0 = 0 and x 1 = 0 . 5. The formula becomes H 3 ( x ) = f ( x 0 )[1 - 2( x - x 0 ) L 0 1 , 0 ( x )] L 2 1 , 0 ( x ) + f ( x 1 )[1 - 2( x - x 1 ) L 0 1 , 1 ( x )] L 2 1 , 1 ( x ) + f 0 ( x 0 )( x - x 0 ) L 2 1 , 0 ( x ) + f 0 ( x 1 )( x - x 1 ) L 2 1 , 1 ( x ) L 1 , 0 ( x ) = x - x 1 x 0 - x 1 = x - 0 . 5 - 0 . 5 = - 2 . 00000( x - 0 . 5) L 2 1 , 0 ( x ) = 4 . 00000( x - 0 . 5) 2 L 0 1 , 0 ( x ) = - 2 . 00000 L 1 , 1 ( x ) = x - x 0 x 1 - x 0 = x - 0 0 . 5 = 2 x L 2 1 , 1 ( x ) = 4 x 2 L 0 1 , 1 ( x ) = 2 So H 3 ( x ) = 1 . 00000[1 - 2 x ( - 2 . 00000)]4 . 00000( x - 0 . 5) 2 + 2 . 71828[1 - 2( x - 0 .
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