A well known Taylor polynomial is the approximation for the exponential about x=0. The nth ex≈1+x+2x2+3!x3+⋯+n!xn=k=0
Posted: Wed Jul 13, 2022 5:06 am
A well known Taylor polynomial is the approximation for the exponential about x=0. The nth ex≈1+x+2x2+3!x3+⋯+n!xn=k=0∑nk!xk The larger n is, and the closer x is to 0 , the better this approximation becomes. We may obtain various other useful Taylor polynomials which are closely related to this one.
(b) A Taylor polynomial that you may encounter in the wild is related to the normal distribution from probability and statistics; specifically, the Taylor polynomial of the function defined by e−x2 at x=0. Substituting −x2 into the nth Taylor polynomial for ex gives e−x2≈k=0∑nh(k)x2k where h(k)= ( ) (in terms of k ). Notice that this is the (2n) th Taylor polynomial for e−x2 at x=0.
(c) A somewhat trickier example is found in deriving the Taylor polynomials of the hyperbolic trigonometric functions coshx=2ex+e−x,sinhx=2ex−e−x from the Taylor polynomial for ex at x=0. Specifically, if we substitute −x into the nth Taylor polynomial for ex, we obtain the nth Taylor polynomial for e−x. Hence, taking the average of the two sums and adding the entries term-wise gives 2ex+e−x≈k=0∑nw(k)xk, where w(k)= 匀 (in terms of k ). Notice that w(k)=0 whenever k is This is unsurprising, since cosh is and so its nth Taylor polynomial about x=0 only contains powers of x. A similar argument applies for the Taylor polynomial of sinh; give it a go!
(b) A Taylor polynomial that you may encounter in the wild is related to the normal distribution from probability and statistics; specifically, the Taylor polynomial of the function defined by e−x2 at x=0. Substituting −x2 into the nth Taylor polynomial for ex gives e−x2≈k=0∑nh(k)x2k where h(k)= ( ) (in terms of k ). Notice that this is the (2n) th Taylor polynomial for e−x2 at x=0.
(c) A somewhat trickier example is found in deriving the Taylor polynomials of the hyperbolic trigonometric functions coshx=2ex+e−x,sinhx=2ex−e−x from the Taylor polynomial for ex at x=0. Specifically, if we substitute −x into the nth Taylor polynomial for ex, we obtain the nth Taylor polynomial for e−x. Hence, taking the average of the two sums and adding the entries term-wise gives 2ex+e−x≈k=0∑nw(k)xk, where w(k)= 匀 (in terms of k ). Notice that w(k)=0 whenever k is This is unsurprising, since cosh is and so its nth Taylor polynomial about x=0 only contains powers of x. A similar argument applies for the Taylor polynomial of sinh; give it a go!