Evaluating ln(x + sqrt[1+x2]) - sin(x) at x=.001

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SUMMARY

The discussion focuses on evaluating the expression ln(x + sqrt[1+x²]) - sin(x) at x = 0.001. Participants emphasize the use of power series expansions for both the natural logarithm and the sine function. The relevant equations include the Taylor series for sin(x) and ln(1 + x), which are utilized to simplify the evaluation. The consensus is that for small values of x, sin(x) can be approximated as x, facilitating the calculation.

PREREQUISITES
  • Understanding of Taylor series expansions
  • Familiarity with natural logarithm properties
  • Basic knowledge of calculus, specifically limits and approximations
  • Ability to manipulate algebraic expressions involving square roots
NEXT STEPS
  • Study Taylor series for ln(1 + x) and sin(x) in detail
  • Practice evaluating limits using power series approximations
  • Explore numerical methods for evaluating functions at small values
  • Learn about convergence of power series and their applications
USEFUL FOR

Students in calculus or mathematical analysis, educators teaching series expansions, and anyone interested in evaluating complex functions at specific points.

psyklon
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Homework Statement


The problem I'm attempting to solve asks you to evaluate the function at a given point. In this case:

ln(x + sqrt[1+x2] ) - sin(x)
@ x = .001


Homework Equations


sin(x) = x - (x3/3!) + (x5/5!) - (x7/7!) + ...
AND
ln(1 + x) = x - (x2/2) + (x3/3) - (x4/4) + ...


The Attempt at a Solution


I know I'm supposed to substitute in the power series expansions, but I'm not sure how to begin modifying the second expansion to suit my needs in solving the problem. Can anyone point me in the right direction?
 
Physics news on Phys.org
Since x is small, you can approximate sin(x) as just x, you can do the same with the other functions.
 

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