2. Let S(x)==0and that for x € 1,(-1) k2k+1(2k+1)!∞S'(x) =>k=0for x € I. Prove that S is differentiable on I(-1) k2k(2k)!1x²2!+x44!

Answered: Image /qna-images/answer/b5e03507-a9a3-4ba7-a86c-615be1d6ac73.jpg

Answered: 2. Let S(x) = Σº (-1)+¹ for x € I.…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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if it was F(X+Y-Z₁ X² + y²) =o where Z = z(x,y) Prove that x2² - y2/²/² = x-y 1
1. Assume that f, g: RR are differentiable, f(0) = 3, f'(0) = 2, g(3) = 7 and g'(3) 4. Compute (go f)(0) and (gof)'(0). =
= [¹ f(4Tan= ¹(u)) - du given that * f(x) dx = 12 1+u² Evaluate
Let f (z) = (z²+1)(z²-4z+3) and I = P.V. f)G-te43) dx. (x²+1)(x²-4x+3) Given Res f(z)|z=i = 2-i and Res f(z)|z=1 =-0.25. Then 20 None of these Res f(z)|z=3 = 0 and I = 5 the above holds Res f(z)|z=3 = and I = 10 the above holds 1 Res f(z)|z=3 20 and I = 10 the above holds I=0 ||
prove that:
Q2) show that the function f is continuous at the given point 23 – 1 | |z| #1 1+iv3 z²+z +1° -1+iv3 f(2) = ; z0 = 2 |2| = 1 2 1) Re(z) 222; zo = e" f(2) = in/4 z + iz 2)
Let z = 3e-ir/2. Which statement is not correct? (a) |z*| = 3 (c) Re(2*) = 3 (b) Zz* = T/2 (d) Im(2*) = 3
min f(x, y) = (x– 4)² + (y – 4)2 subject to x+ y<4 and x+ 3y <9 a. Find the max through First order condition (FoC). b. Check second order condition (SoC).
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Let y"+p(x)y'+q(x)y=g(x) (2)and y'+p(x)y'+q(x)y= 0(3) where p,q and g are continuous on R. Indicate whether the following statements are true or false. Write out the word"true" or“false" in the blank. (a) If p and q are constants, we know how to find the general solution of (2) for any g(x) (b) If and p are constants, we know how to find the general solution of (2) for any q(x) (c) The general solution of (2) is the sum of the general solution of (3) and a particular solution of (2) (d) If y1 and y2 are two solutions of (2), their sum is also a solution of (2) (e) If yland y2 are two solutions of (3), their sum is also a solution of (3) (f) If y1 and y2 are two solutions of (2), their difference is a solution of (3)
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2. Let S(x) = Σº (-1)+¹ for x € I. Prove that S is differentiable on I k=0 (2k+1)! and that for x = I, ∞ S'(x) = Σ k=0 (-1) k2k x² (2k)! 1 x² 2! x4 4!