Taylor Polynomial (n=4) for g(x): Showing 0<E4(x)<80x^5

AI Thread Summary
The discussion centers on finding the Taylor polynomial of g(x) = (1+5x)^(1/5) and demonstrating that the error term E4(x) satisfies 0 < E4(x) < 80x^5 for x > 0. The user has derived the fourth-degree Taylor polynomial P4(x) and calculated the error term as E4(x) = (399/[5(1+5x)^(24/5)]) * x^5. To show the error is less than 80x^5, the user considers the maximum error occurring when x approaches 0, leading to the conclusion that the fraction approaches a value close to 400. Ultimately, this reasoning confirms the desired inequality for the error term.
sony
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Hi

I have found the following TP (n=4) for g(x) = (1+5x)^1/5
P4(x) = 1+x-2x^2+6x^3-21x^4

Then they ask me to show that 0<E4(x)<80x^5 when x>0.

I don't know how to start, or exactly what I am supposed to show...?

I have found E4(x) to be( 399/[5(1+5X)^24/5] ) *x^5...

And 0<X<x ...?
 
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To show that the error is LESS than something, you want to think about the X that gives the LARGEST possible error. Since X is in the denominator, the largest value of the fraction will be when X= 0. If you take X= 0 what is that value in the parentheses? (Gosh, 399 is awful close to 400!)
 
Oh now I see how I get 0<E4(x)<80x^5

Thanks!
 
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