real analysis, math homework help

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errz2015

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hello,

Could you please explain question 2 page 290 by more details especially what I selected by the red lines?

Because it is not clear for me how you get it.

https://alansinyal.files.wordpress.com/2012/08/method-of-real-analysis.pdf

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Question # 1 page287: For f ( x ) = x 2 , where 0  x  1 , the Bernstein polynomials are calculated and graphed as follow 1 1 1− k k B1 ( x ) =    x k (1 − x ) f   1 k =0  k  1 1− 0 1−1  0   1 =   x 0 (1 − x ) f   +   x1 (1 − x )  1   1 0 1 f  1 = 0+ x =x 2  2 2−k k B2 ( x ) =    x k (1 − x ) f   2 k =0  k   2 2−0 2 −1  0   2 =   x 0 (1 − x ) f   +   x1 (1 − x )  2  1 0 2 2− 2  1   2 2 f   +   x 2 (1 − x ) f    2   2 2 1 2 = 1  (1 − x )  0 + 2  x (1 − x )    + 1  x 2    2 2 1 = x ( x + 1) 2 2 2 2 3  3 3− k k B3 ( x ) =    x k (1 − x ) f   3 k =0  k   3 3− 0 3−1  0   3 =   x 0 (1 − x ) f   +   x1 (1 − x )  3  1 0 2 3− 2 3− 3  1   3  2   3  3 f   +   x 2 (1 − x ) f   +   x 3 (1 − x ) f    3   2  3   3  3 2 1 2 = 1  (1 − x )  0 + 3  x (1 − x )    + 3  x 2 (1 − x )    + 1 x3  x 2 3 3 2 1 = x5 − x3 + x 2 + x 3 3 3 2 2 Question # 11 page 298: It is suffice to prove that f 2 over 0,1 is 0 . This will implies that f  0 , because if f  0 , then f 2  0 on some interval and thus f 2  0 on 0,1 . We can approximate f uniformly by polynomials Pn so that Pn − f  1 in the sup norm, using Stonen 1 Weierstrass theorem. The given condition clearly follows that  f ( x ) P ( x ) = 0 is zero for any 0 polynomial P , by linearity, we have 1  f ( x ) f ( x ) dx = 0 1  0 1 1 0 0 f ( x ) f ( x ) dx −  f ( x ) Pn ( x ) dx   f ( x ) f ( x ) − Pn ( x ) dx 1  1 f ( x ) dx n 0 Since the last integral is constant, so by taking n →  we can conclude that f ( x ) = 0 for all x   a, b Question # 12 page 298: Let f  C 0,1 and let   0 . We first approximate f by a polygonal function. Since f is continuous, we can find a sufficiently large n so that f ( x ) − f ( y )   whenever x − y  k k  = f   for k = 0,1, 2, n n the polygonal function defined by g   k k +1   ,  satisfies f − g n n   1 , which means that n , n , and g is linear on each interval  . Next, we modify our approximating function: Let h be another polygonal function that also has nodes at k for k = 0,1, 2, n k k k , n , but with h   rational and satisfying h   − g     for each k . Then n n n g − h    and consequently, f − h   2 . n k  for some n , is  n k = 0 The set of all polygonal functions taking only rational values at the node   countable. Hence, C  0,1 has a countable dense subset, as desired. Question # 2 page 290: x Define T as follows: For any  let T  =  where  ( x ) =  ( t ) dt .  0 Let 0 = 1 , if 1 = T0 , then 1 ( x ) = 1 . If 2 = T 1 , then  2 ( x ) = 1 + x −1 1! x − 1 ( x − 1) ( x − 1) + + + In general, if n = Tn−1 , then n ( x ) = 1 + 1! 2! 3! 2 3 Therefore, we have lim n ( x ) = lim T n0 ( x ) = e x−1 n→ n→ Thus,  ( x ) = e x −1 is the solution to the original problem y = y, y (1) = 1 . ( x − 1) + n! n
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Explanation & Answer

here's my explanation, sorry for the wrong solution earlier
here's my explanation, sorry for the wrong solution earlier

Question # 2 page 290:
For x0 = 1, y0 = 1 , define T as follows: For any  let T  =  where
x

x

x0

1

 ( x ) = y0 +   ( t ) dt = 1 +   ( t ) dt .
Let 0 = 1 , if 1 = T0 , then
x

x

( )

1 ( x ) = 1 +  0 ( t )dt = 1 +  1dt = 1 + t 1 = 1 + ( x − 1) = 1 +
x

1

1

x −1
.
1!

If 2 = T 1 , then
x

x

 1 ( t − 1)2 
 t −...