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Theorem 11.3.5. The Weierstrass-M Test Let (fn)-1 be a sequence of functions defined on S C R and suppose that (Mn)n=1 is a sequence of nonnegative real numbers such that n=1 8. |fn(x)| < Mn, Væ € S, n = 1, 2, 3, .... If En-1 Mn converges then En-1 fn(x) converges uniformly on S to some function (which we will denote by f(x)). n=1 Sketch of Proof. Since the crucial feature of the theorem is the function f(x) that our series converges to, our plan of attack is to first define f(x) and then show that our series, E1 fn(æ), converges to it uniformly. n=1 First observe that for any a E S, En-1 fn(x) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by f(x). This actually defines the function f(x) for all æ E S. It follows that Ei fn(x) converges pointwise to f(x). =1 Next, let e > 0 be given. Notice that since , M, converges, say to M, then there is a real number, N, such that if n > N, then in=1 n Σ Μ Σ Μ. M - ) Mk < e. k=n+1 k=n+1 k=1 You should be able to use this to show that if n > N, then f(x) – fr(x) < e, Væ e S. k=1 Problem 11.3.6. Use the ideas above to provide a formal proof of Theorem 11.3.5.

Theorem 11.3.5. The Weierstrass-M Test Let (fn)-1 be a sequence of functions defined on S C R and suppose that (Mn)n=1 is a sequence of nonnegative real numbers such that n=1 8. |fn(x)| < Mn, Væ € S, n = 1, 2, 3, .... If En-1 Mn converges then En-1 fn(x) converges uniformly on S to some function (which we will denote by f(x)). n=1 Sketch of Proof. Since the crucial feature of the theorem is the function f(x) that our series converges to, our plan of attack is to first define f(x) and then show that our series, E1 fn(æ), converges to it uniformly. n=1 First observe that for any a E S, En-1 fn(x) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by f(x). This actually defines the function f(x) for all æ E S. It follows that Ei fn(x) converges pointwise to f(x). =1 Next, let e > 0 be given. Notice that since , M, converges, say to M, then there is a real number, N, such that if n > N, then in=1 n Σ Μ Σ Μ. M - ) Mk < e. k=n+1 k=n+1 k=1 You should be able to use this to show that if n > N, then f(x) – fr(x) < e, Væ e S. k=1 Problem 11.3.6. Use the ideas above to provide a formal proof of Theorem 11.3.5.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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Real math analysis, I need help with problem 11.3.6

thank you so much.

Theorem 11.3.5. The Weierstrass-M Test Let (fn) be a sequence of functions defined on SCR and suppose that (Mn)n-1 is a sequence of nonnegative real numbers such that n=1 |fn(x)| < Mn, Væ E S, n = 1, 2, 3, .... If E Mn converges then fn(x) converges uniformly on S to some function (which we will denote by f(x)). 'n=1 n=1 Sketch of Proof. Since the crucial feature of the theorem is the function f(x) that our series converges to, our plan of attack is to first define f(x) and then show that our series, E fn(x), converges to it uniformly. in= First observe that for any a E S, E fn(x) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by f(x). This actually defines the function f(x) for all æ E S. It follows that E1 fn(x) converges pointwise to f(x). 'n=1 n=1 Next, let e > 0 be given. Notice that since E-1 Mn converges, say to M, then there is a real number, N, such that if n > N, then n=1 E Mk = Mk Mk = M < E. k=n+1 k=n+1 k=1 You should be able to use this to show that if n > N, then n f(x) – > f«(x) < ɛ, Væ E S. - k=1 Problem 11.3.6. Use the ideas above to provide a formal proof of Theorem 11.3.5.
Transcribed Image Text:Theorem 11.3.5. The Weierstrass-M Test Let (fn) be a sequence of functions defined on SCR and suppose that (Mn)n-1 is a sequence of nonnegative real numbers such that n=1 |fn(x)| < Mn, Væ E S, n = 1, 2, 3, .... If E Mn converges then fn(x) converges uniformly on S to some function (which we will denote by f(x)). 'n=1 n=1 Sketch of Proof. Since the crucial feature of the theorem is the function f(x) that our series converges to, our plan of attack is to first define f(x) and then show that our series, E fn(x), converges to it uniformly. in= First observe that for any a E S, E fn(x) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by f(x). This actually defines the function f(x) for all æ E S. It follows that E1 fn(x) converges pointwise to f(x). 'n=1 n=1 Next, let e > 0 be given. Notice that since E-1 Mn converges, say to M, then there is a real number, N, such that if n > N, then n=1 E Mk = Mk Mk = M < E. k=n+1 k=n+1 k=1 You should be able to use this to show that if n > N, then n f(x) – > f«(x) < ɛ, Væ E S. - k=1 Problem 11.3.6. Use the ideas above to provide a formal proof of Theorem 11.3.5.
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