What Are the Taylor Polynomials for Specific Functions?

Click For Summary
SUMMARY

The discussion focuses on deriving Taylor polynomials for specific functions, specifically \( T_n(a^x) \) and \( T_n\left(\frac{1}{1+x}\right) \). The Taylor polynomial for \( a^x \) is expressed as \( T_n = \sum_{k=0}^{n} \frac{(\log a)^k}{k!} x^k \), while the polynomial for \( \frac{1}{1+x} \) is given by \( T_n = \sum_{k=0}^{n} (-1)^k x^k \). The discussion emphasizes the importance of evaluating derivatives at zero and recognizing the relationship between power series and Taylor series.

PREREQUISITES
  • Understanding of Taylor series and Maclaurin series
  • Knowledge of derivatives and their evaluation at specific points
  • Familiarity with exponential functions and logarithms
  • Basic concepts of geometric series
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Learn how to compute derivatives of exponential functions
  • Explore the convergence criteria for Taylor series
  • Investigate the applications of Taylor polynomials in approximation theory
USEFUL FOR

Students studying calculus, mathematicians interested in series expansions, and educators teaching Taylor series concepts.

Jimmy84
Messages
190
Reaction score
0

Homework Statement


Obtain the Taylor polynomials Tnf(x) as indicated. In each case, it
is understood that f(x) is defined for a11 x for which f(x) is meaningful.


Problem one
Tn = (a^x) = sigma from k = 0 to n of ((log a)^k)/k! x^k

Problem two
Tn = (1/(1+x)) = sigma from k = o to n of (-1)^k x^k


Homework Equations





The Attempt at a Solution



Im totally lost and I don't know where to start can anyone help me please?
 
Physics news on Phys.org
Did you really have to use "a11" when you have a perfectly good "l" key?

They Taylor series for function f(x) is [tex]\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n[/tex] where [itex]f^{(n)}(0)[/itex] indicates the nth derivative of f evaluated at 0.

(Actually, that is the "Taylor series at 0" or "MacLaurin" series.)

So for [itex]a^x[/itex] you only need to find the derivatives of [itex]a^x[/itex] and evaluate at x= 0. That can be done most efficiently by writing [itex]a^x= e^{ln a^x}= e^{x ln a}[/itex].

It is also true that any power series equal to a given function is a Taylor series. If you remember that the sum of the geometric series [itex]\sum_{n=0}^\infty r^n[/itex] is equal to 1/(1- r), then the Taylor series for 1/(1+x) should be obvious.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K