COURSE : Real/Mathematical AnalysisTOPIC: Taylor Series We now just need to show that lim ( n-pn-c0 (p+1)= 0 since0s k" < (P) n-p.(p+1)'Thus by the squeeze theorem if lim\x\ yn-p= 0 then(p+1)lim|(x)"= 0.n+00 (n)!< 1, and- is a constant, if we can just show that lim a" = 0 ifSince(p+1)n-00lal < 1, we will be done.We must show given any e> 0 we can find N >0 such that if n > N then|a" – 0| < e.|a|" < €n(In |al) < In(e)In(e)n>since In Ja| < 0.InſalIn(e)If e > 1 then"Injal max(0, mO).In|a|Now let's show this N works.In(e) thenIfn 2 N > max(0,injaIn(c)In(c)|a" – 0| = |a|" < l|a|ina = (elnlalyinja= elnlel = €.So limn-00 (n+1)!(1x)n+1= 0 and lim R, (x,0) = 0.So f(x) = e* = En-o%3Dn! **SOLUTIONS MUST FOLLOW THE EXAMPLE PROVIDED BELOW**OTHERWISE, ANSWERS WILL NOT BE ACCEPTEDQUESTION:Prove that the Taylor series around a = 0 for f(x) = e-2= converges to f(x) forall r €R.EXAMPLE:Ex. Prove that the Taylor Series around a = 0 for f (x) = e* converges tof(x) = e* for all xeR.%3Df(x) = exf(0) = e° = 1f'(x) = e*f'(0) = 1f"(x) = e*f"(0) = 1f( (x) = e*f(n)(0) = 1x2x3Tn(x) = 1+x++n!2!3!|R„(x, 0)| = | (x)n+1| = |(n+1)!; (x)까1|;(n+1)!where c is between O and x.We need to show that lim(x)n+1| = 0 for any xeR.n-0' (n+1)!Thus we just have to show for any fixed number x,|(x)n+1lim= 0, since en00 (n+1)!is just a constant once x is fixed.Fix x and let p = [|x|] =the greatest integer less than or equal to lxl.Notice that:= (부) (필) () ().. .() s ()(y-P, where <1.n!(p+1)(p+1)

Answered: Prove that the Taylor series around a =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

COURSE : Real/Mathematical Analysis

TOPIC: Taylor Series

We now just need to show that lim ( n-p
n-c0 (p+1)
= 0 since
0s k" < (P) n-p.
(p+1)'
Thus by the squeeze theorem if lim
\x\ yn-p
= 0 then
(p+1)
lim
|(x)"
= 0.
n+00 (n)!
< 1, and
- is a constant, if we can just show that lim a" = 0 if
Since
(p+1)
n-00
lal < 1, we will be done.
We must show given any e> 0 we can find N >0 such that if n > N then
|a" – 0| < e.
|a|" < €
n(In |al) < In(e)
In(e)
n>
since In Ja| < 0.
Inſal
In(e)
If e > 1 then
"Injal
<O so let's choose N > max(0, mO).
In|a|
Now let's show this N works.
In(e) then
Ifn 2 N > max(0,inja
In(c)
In(c)
|a" – 0| = |a|" < l|a|ina = (elnlalyinja
= elnlel = €.
So lim
n-00 (n+1)!
(1x)n+1
= 0 and lim R, (x,0) = 0.
So f(x) = e* = En-o
%3D
n!

**SOLUTIONS MUST FOLLOW THE EXAMPLE PROVIDED BELOW**
OTHERWISE, ANSWERS WILL NOT BE ACCEPTED
QUESTION:
Prove that the Taylor series around a = 0 for f(x) = e-2= converges to f(x) for
all r €R.
EXAMPLE:
Ex. Prove that the Taylor Series around a = 0 for f (x) = e* converges to
f(x) = e* for all xeR.
%3D
f(x) = ex
f(0) = e° = 1
f'(x) = e*
f'(0) = 1
f"(x) = e*
f"(0) = 1
f( (x) = e*
f(n)(0) = 1
x2
x3
Tn(x) = 1+x+
+
n!
2!
3!
|R„(x, 0)| = | (x)n+1| = |
(n+1)!
; (x)까1|;
(n+1)!
where c is between O and x.
We need to show that lim
(x)n+1| = 0 for any xeR.
n-0' (n+1)!
Thus we just have to show for any fixed number x,
|(x)n+1
lim
= 0, since e
n00 (n+1)!
is just a constant once x is fixed.
Fix x and let p = [|x|] =the greatest integer less than or equal to lxl.
Notice that:
= (부) (필) () ().. .() s ()(y-P, where <1.
n!
(p+1)
(p+1)

Expert Solution

Step 1

To prove that the Taylor series around $a = 0$ for $f (x) = e^{- 2 x}$ converges to $f (x)$ for all $x \in R$ .

$f^{'} (x) = (- 2) e^{- 2 x}$	$f^{'} (0) = - 2$
$f^{''} (x) = {(- 2)}^{2} e^{- 2 x}$	$f^{''} (0) = {(- 2)}^{2}$
$⋮$	$⋮$
$f^{n} (x) = {(- 2)}^{n} e^{- 2 x}$	$f^{n} (0) = {(- 2)}^{n}$
$f^{n + 1} (x) = {(- 2)}^{n + 1} e^{- 2 x}$	$f^{n + 1} (0) = {(- 2)}^{n + 1}$

$\begin{array}{rcl} T_{n} (x) & = & f (0) + \frac{f^{'} (0)}{1!} (x - 0) + \frac{f'' {(0)}^{}}{2!} {(x - 0)}^{2} + \dots \dots + \frac{f^{n} (0)}{n!} {(x - 0)}^{n} \\ = & 1 + \frac{(- 2 x)}{1!} + \frac{{(- 2 x)}^{2}}{2!} + \frac{{(- 2 x)}^{3}}{3!} + \dots \dots + \frac{{(- 2 x)}^{n}}{n!} \end{array}$

Further,

$\begin{array}{rcl} |R_{n} (x, 0)| & = & |\frac{f^{n + 1} (c)}{(n + 1)!} {(x)}^{n + 1}| \\ = & |\frac{{(- 2)}^{n + 1} e^{- 2 c}}{(n + 1)!} x^{n + 1}| \\ = & \frac{{|- 2 x|}^{n + 1} e^{^{- 2 c}}}{(n + 1)!} \\ = & \frac{{|2 x|}^{n + 1} e^{^{- 2 c}}}{(n + 1)!} \end{array}$

where $c$ is between $0$ and $x$ .

To show $\lim_{n \to \infty} \begin{array}{rcl} \frac{{|2 x|}^{n + 1} e^{^{- 2 c}}}{(n + 1)!} \end{array} = 0$ for any $x \in R$ .

Here, it is enough to show that $\lim_{n \to \infty} \begin{array}{rcl} \frac{{|2 x|}^{n + 1}}{(n + 1)!} \end{array} = 0$ since $e^{- 2 c}$ is a constant once $x$ is fixed.

Fix $x$ and let $p = [|2 x|] =$ the greatest integer less than or equal to $|2 x|$ .

Then,

$\frac{{|2 x|}^{n}}{n!} = (\frac{|2 x|}{1}) (\frac{|2 x|}{n}) \dots (\frac{|2 x|}{p}) (\frac{|2 x|}{p + 1}) \dots (\frac{|2 x|}{n}) \leq (\frac{{|2 x|}^{p}}{p!}) {(\frac{|2 x|}{p + 1})}^{n - p};$ where $\frac{|2 x|}{(p + 1)} < 1$ .

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

Knowledge Booster

$Series$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions