Question on taylor series
$begingroup$
When we try to to find the Taylor series of let's say $e^x$ in $x=-1$ and of order up to $3$, I would think that we just find the Maclaurin series of $e^x$ and then instead of simply $x$ we put $x+1$.But my book as a first step does the following: $t=x+1$ so $x=t-1$
so $f (t − 1) = e^{(t−1)}$. I feel that this is pretty simple so not understanding it is pretty frustrating.What is going on here?
power-series taylor-expansion
edited Dec 21 '18 at 19:32
caverac
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
$endgroup$
add a comment |
$begingroup$
When we try to to find the Taylor series of let's say $e^x$ in $x=-1$ and of order up to $3$, I would think that we just find the Maclaurin series of $e^x$ and then instead of simply $x$ we put $x+1$.But my book as a first step does the following: $t=x+1$ so $x=t-1$
so $f (t − 1) = e^{(t−1)}$. I feel that this is pretty simple so not understanding it is pretty frustrating.What is going on here?
power-series taylor-expansion
edited Dec 21 '18 at 19:32
caverac
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
$endgroup$
add a comment |
$begingroup$
When we try to to find the Taylor series of let's say $e^x$ in $x=-1$ and of order up to $3$, I would think that we just find the Maclaurin series of $e^x$ and then instead of simply $x$ we put $x+1$.But my book as a first step does the following: $t=x+1$ so $x=t-1$
so $f (t − 1) = e^{(t−1)}$. I feel that this is pretty simple so not understanding it is pretty frustrating.What is going on here?
power-series taylor-expansion
edited Dec 21 '18 at 19:32
caverac
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
$endgroup$
When we try to to find the Taylor series of let's say $e^x$ in $x=-1$ and of order up to $3$, I would think that we just find the Maclaurin series of $e^x$ and then instead of simply $x$ we put $x+1$.But my book as a first step does the following: $t=x+1$ so $x=t-1$
so $f (t − 1) = e^{(t−1)}$. I feel that this is pretty simple so not understanding it is pretty frustrating.What is going on here?
power-series taylor-expansion
power-series taylor-expansion
edited Dec 21 '18 at 19:32
caverac
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
edited Dec 21 '18 at 19:32
caverac
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
edited Dec 21 '18 at 19:32
caverac
14.5k31130
edited Dec 21 '18 at 19:32
caverac
14.5k31130
edited Dec 21 '18 at 19:32
caverac
14.5k31130
14.5k31130
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
asked Dec 21 '18 at 19:27
Michael P.Michael P.
222
222
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write
$$
begin{align}
e^x
&=frac1ecdot e^{x+1}\
&=frac1eleft(sum_{n=0}^inftyfrac{(x+1)^n}{n!}right)\
&=sum_{n=0}^inftyfrac{(x+1)^n}{e,n!}
end{align}
$$
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.
let $g(x)=f(x+a)$.
$g$ is then defined arround $0$.
we expand $g(x)$ arround zero at order $n$ as
$$g(x)=P_n(x)+x^nepsilon(x)$$
then we get the expansion of $f(x)$ arround $a$ using that
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^nepsilon(x)$$ with
$$lim_{xto a}epsilon(x)=0$$
In your case
$$f(x)=e^x,;; a=-1, ;; g(x)=e^{x-1}$$
$$g(x)=e^{-1}Bigl(1+x+frac{x^2}{2}+frac{x^3}{6}Bigr)+x^3epsilon(x)$$
therefore
$$f(x)=e^{-1}Bigl(1+(x+1)+frac{(x+1)^2}{2}+frac{(x+1)^3}{6}Bigr)+(x+1)^3epsilon(x)$$
with $$lim_{xto -1}epsilon(x)=0$$
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
You could start with the Maclaurin series.
$e^x = 1 + x + frac 12x^2 + cdots$
Then plug $x+1$
$e^{x+1} = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e(e^{x}) = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e^{x} = e^{-1} + e^{-1}(x+1) + frac {e^{-1}}2(x+1)^2 + cdots\
$
But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.
$a_n = frac {f^{(n)}(-1)}{n!}$
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write
$$
begin{align}
e^x
&=frac1ecdot e^{x+1}\
&=frac1eleft(sum_{n=0}^inftyfrac{(x+1)^n}{n!}right)\
&=sum_{n=0}^inftyfrac{(x+1)^n}{e,n!}
end{align}
$$
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write
$$
begin{align}
e^x
&=frac1ecdot e^{x+1}\
&=frac1eleft(sum_{n=0}^inftyfrac{(x+1)^n}{n!}right)\
&=sum_{n=0}^inftyfrac{(x+1)^n}{e,n!}
end{align}
$$
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write
$$
begin{align}
e^x
&=frac1ecdot e^{x+1}\
&=frac1eleft(sum_{n=0}^inftyfrac{(x+1)^n}{n!}right)\
&=sum_{n=0}^inftyfrac{(x+1)^n}{e,n!}
end{align}
$$
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
$endgroup$
We want to write $e^x$ in a series with $(x+1)^n$. To do this we can write
$$
begin{align}
e^x
&=frac1ecdot e^{x+1}\
&=frac1eleft(sum_{n=0}^inftyfrac{(x+1)^n}{n!}right)\
&=sum_{n=0}^inftyfrac{(x+1)^n}{e,n!}
end{align}
$$
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
answered Dec 21 '18 at 19:36
robjohn♦robjohn
267k27308631
267k27308631
add a comment |
add a comment |
$begingroup$
Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.
let $g(x)=f(x+a)$.
$g$ is then defined arround $0$.
we expand $g(x)$ arround zero at order $n$ as
$$g(x)=P_n(x)+x^nepsilon(x)$$
then we get the expansion of $f(x)$ arround $a$ using that
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^nepsilon(x)$$ with
$$lim_{xto a}epsilon(x)=0$$
In your case
$$f(x)=e^x,;; a=-1, ;; g(x)=e^{x-1}$$
$$g(x)=e^{-1}Bigl(1+x+frac{x^2}{2}+frac{x^3}{6}Bigr)+x^3epsilon(x)$$
therefore
$$f(x)=e^{-1}Bigl(1+(x+1)+frac{(x+1)^2}{2}+frac{(x+1)^3}{6}Bigr)+(x+1)^3epsilon(x)$$
with $$lim_{xto -1}epsilon(x)=0$$
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.
let $g(x)=f(x+a)$.
$g$ is then defined arround $0$.
we expand $g(x)$ arround zero at order $n$ as
$$g(x)=P_n(x)+x^nepsilon(x)$$
then we get the expansion of $f(x)$ arround $a$ using that
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^nepsilon(x)$$ with
$$lim_{xto a}epsilon(x)=0$$
In your case
$$f(x)=e^x,;; a=-1, ;; g(x)=e^{x-1}$$
$$g(x)=e^{-1}Bigl(1+x+frac{x^2}{2}+frac{x^3}{6}Bigr)+x^3epsilon(x)$$
therefore
$$f(x)=e^{-1}Bigl(1+(x+1)+frac{(x+1)^2}{2}+frac{(x+1)^3}{6}Bigr)+(x+1)^3epsilon(x)$$
with $$lim_{xto -1}epsilon(x)=0$$
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.
let $g(x)=f(x+a)$.
$g$ is then defined arround $0$.
we expand $g(x)$ arround zero at order $n$ as
$$g(x)=P_n(x)+x^nepsilon(x)$$
then we get the expansion of $f(x)$ arround $a$ using that
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^nepsilon(x)$$ with
$$lim_{xto a}epsilon(x)=0$$
In your case
$$f(x)=e^x,;; a=-1, ;; g(x)=e^{x-1}$$
$$g(x)=e^{-1}Bigl(1+x+frac{x^2}{2}+frac{x^3}{6}Bigr)+x^3epsilon(x)$$
therefore
$$f(x)=e^{-1}Bigl(1+(x+1)+frac{(x+1)^2}{2}+frac{(x+1)^3}{6}Bigr)+(x+1)^3epsilon(x)$$
with $$lim_{xto -1}epsilon(x)=0$$
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
Assume we want the expansion of $f(x)$ arround $x=a$. $f$ is defined arround $a$.
let $g(x)=f(x+a)$.
$g$ is then defined arround $0$.
we expand $g(x)$ arround zero at order $n$ as
$$g(x)=P_n(x)+x^nepsilon(x)$$
then we get the expansion of $f(x)$ arround $a$ using that
$$f(x)=g(x-a)=$$
$$P_n(x-a)+(x-a)^nepsilon(x)$$ with
$$lim_{xto a}epsilon(x)=0$$
In your case
$$f(x)=e^x,;; a=-1, ;; g(x)=e^{x-1}$$
$$g(x)=e^{-1}Bigl(1+x+frac{x^2}{2}+frac{x^3}{6}Bigr)+x^3epsilon(x)$$
therefore
$$f(x)=e^{-1}Bigl(1+(x+1)+frac{(x+1)^2}{2}+frac{(x+1)^3}{6}Bigr)+(x+1)^3epsilon(x)$$
with $$lim_{xto -1}epsilon(x)=0$$
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
edited Dec 21 '18 at 19:45
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 21 '18 at 19:34
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
$begingroup$
You could start with the Maclaurin series.
$e^x = 1 + x + frac 12x^2 + cdots$
Then plug $x+1$
$e^{x+1} = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e(e^{x}) = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e^{x} = e^{-1} + e^{-1}(x+1) + frac {e^{-1}}2(x+1)^2 + cdots\
$
But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.
$a_n = frac {f^{(n)}(-1)}{n!}$
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
You could start with the Maclaurin series.
$e^x = 1 + x + frac 12x^2 + cdots$
Then plug $x+1$
$e^{x+1} = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e(e^{x}) = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e^{x} = e^{-1} + e^{-1}(x+1) + frac {e^{-1}}2(x+1)^2 + cdots\
$
But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.
$a_n = frac {f^{(n)}(-1)}{n!}$
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
$endgroup$
add a comment |
$begingroup$
You could start with the Maclaurin series.
$e^x = 1 + x + frac 12x^2 + cdots$
Then plug $x+1$
$e^{x+1} = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e(e^{x}) = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e^{x} = e^{-1} + e^{-1}(x+1) + frac {e^{-1}}2(x+1)^2 + cdots\
$
But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.
$a_n = frac {f^{(n)}(-1)}{n!}$
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
$endgroup$
You could start with the Maclaurin series.
$e^x = 1 + x + frac 12x^2 + cdots$
Then plug $x+1$
$e^{x+1} = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e(e^{x}) = 1 + (x+1) + frac 12(x+1)^2 + cdots\
e^{x} = e^{-1} + e^{-1}(x+1) + frac {e^{-1}}2(x+1)^2 + cdots\
$
But for other functions, it is not so easy to factor out that $+1$ and you should start from scratch.
$a_n = frac {f^{(n)}(-1)}{n!}$
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
edited Dec 21 '18 at 21:15
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
answered Dec 21 '18 at 19:34
Doug MDoug M
44.7k31854
44.7k31854
add a comment |
add a comment |
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