manipulate well known Taylor series

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How can you manipulate a known power series when $x Ã¢Â‰Â 0$ ?

For example $f(x)=3x e^x$
What is the Taylor series to the first 5 terms generated by $f$ at $x=4$ ?

I know
$$3xe^x = 3x + 3x^2 + frac3x^32 + frac3x^43! + frac3x^54! +... frac 3x^n+1n!$$
because of the well known Taylor series for

$$ e^x = sum_n=0^infty fracx^nn!$$

but not sure how to rework this at $x=4$

edited Jul 23 at 11:25

BCLC

6,89721973

asked Jul 22 at 21:01

Jake

add a commentÂ |Â

up vote
-2
down vote

favorite

How can you manipulate a known power series when $x Ã¢Â‰Â 0$ ?

For example $f(x)=3x e^x$
What is the Taylor series to the first 5 terms generated by $f$ at $x=4$ ?

I know
$$3xe^x = 3x + 3x^2 + frac3x^32 + frac3x^43! + frac3x^54! +... frac 3x^n+1n!$$
because of the well known Taylor series for

$$ e^x = sum_n=0^infty fracx^nn!$$

but not sure how to rework this at $x=4$

edited Jul 23 at 11:25

BCLC

6,89721973

asked Jul 22 at 21:01

Jake

add a commentÂ |Â

up vote
-2
down vote

favorite

How can you manipulate a known power series when $x Ã¢Â‰Â 0$ ?

For example $f(x)=3x e^x$
What is the Taylor series to the first 5 terms generated by $f$ at $x=4$ ?

I know
$$3xe^x = 3x + 3x^2 + frac3x^32 + frac3x^43! + frac3x^54! +... frac 3x^n+1n!$$
because of the well known Taylor series for

$$ e^x = sum_n=0^infty fracx^nn!$$

but not sure how to rework this at $x=4$

edited Jul 23 at 11:25

BCLC

6,89721973

asked Jul 22 at 21:01

Jake

How can you manipulate a known power series when $x Ã¢Â‰Â 0$ ?

For example $f(x)=3x e^x$
What is the Taylor series to the first 5 terms generated by $f$ at $x=4$ ?

I know
$$3xe^x = 3x + 3x^2 + frac3x^32 + frac3x^43! + frac3x^54! +... frac 3x^n+1n!$$
because of the well known Taylor series for

$$ e^x = sum_n=0^infty fracx^nn!$$

but not sure how to rework this at $x=4$

edited Jul 23 at 11:25

BCLC

6,89721973

asked Jul 22 at 21:01

Jake

edited Jul 23 at 11:25

BCLC

6,89721973

edited Jul 23 at 11:25

BCLC

6,89721973

edited Jul 23 at 11:25

BCLC

6,89721973

asked Jul 22 at 21:01

Jake

asked Jul 22 at 21:01

Jake

asked Jul 22 at 21:01

Jake

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
3
down vote

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

answered Jul 22 at 21:07

Robert Israel

304k22201441

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up vote
2
down vote

You can do it like this.

$3xe^x=3(x-4+4)e^x-4+4=3e^4(x-4)e^x-4+12e^4e^x-4$

Then use the series that you know $e^x=sum_n=0^inftyfracx^nn!$ to instead write $e^x-4=sum_n=0^inftyfrac(x-4)^nn!$

Putting this into the expression of your function above yields

$$beginalign3e^4(x-4)e^x-4+12e^4e^x-4&=3e^4(x-4)sum_n=0^inftyfracx^nn!+12e^4sum_n=0^inftyfracx^nn!\&=12e^4+sum_n=1^infty3e^4left(frac1(n-1)!+frac4n!right)(x-4)^nendalign$$

answered Jul 22 at 21:13

user578878

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up vote
0
down vote

Taylor theorem states that $dots$

$$f(x) = sum_n=0^infty fracf^(n)(a)(x-a)^nn!$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

answered Jul 22 at 21:11

bkarthik

320212

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

add a commentÂ |Â

up vote
0
down vote

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

$$f(4)=3sum_n=0^inftyfrac4^n+1n!,
\f'(4)=3sum_n=0^infty(n+1)frac4^nn!=3sum_n=1^inftyfrac4^n(n-1)!+3sum_n=0^inftyfrac4^nn!,
\f''(4)=3sum_n=2^inftyfrac4^n-1(n-2)!+3sum_n=1^inftyfrac4^n-1(n-1)!,
\cdots$$

answered Jul 22 at 21:20

Yves Daoust

111k665203

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
3
down vote

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

answered Jul 22 at 21:07

Robert Israel

304k22201441

add a commentÂ |Â

up vote
3
down vote

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

answered Jul 22 at 21:07

Robert Israel

304k22201441

add a commentÂ |Â

up vote
3
down vote

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

answered Jul 22 at 21:07

Robert Israel

304k22201441

Let $t = x - 4$. Then $f(x) = f(t+4) = 3e^4 (t+4) e^t$. Can you take it from there?

answered Jul 22 at 21:07

Robert Israel

304k22201441

answered Jul 22 at 21:07

Robert Israel

304k22201441

answered Jul 22 at 21:07

Robert Israel

304k22201441

answered Jul 22 at 21:07

Robert Israel

304k22201441

add a commentÂ |Â

up vote
2
down vote

You can do it like this.

$3xe^x=3(x-4+4)e^x-4+4=3e^4(x-4)e^x-4+12e^4e^x-4$

Then use the series that you know $e^x=sum_n=0^inftyfracx^nn!$ to instead write $e^x-4=sum_n=0^inftyfrac(x-4)^nn!$

Putting this into the expression of your function above yields

$$beginalign3e^4(x-4)e^x-4+12e^4e^x-4&=3e^4(x-4)sum_n=0^inftyfracx^nn!+12e^4sum_n=0^inftyfracx^nn!\&=12e^4+sum_n=1^infty3e^4left(frac1(n-1)!+frac4n!right)(x-4)^nendalign$$

answered Jul 22 at 21:13

user578878

add a commentÂ |Â

up vote
2
down vote

You can do it like this.

$3xe^x=3(x-4+4)e^x-4+4=3e^4(x-4)e^x-4+12e^4e^x-4$

Then use the series that you know $e^x=sum_n=0^inftyfracx^nn!$ to instead write $e^x-4=sum_n=0^inftyfrac(x-4)^nn!$

Putting this into the expression of your function above yields

$$beginalign3e^4(x-4)e^x-4+12e^4e^x-4&=3e^4(x-4)sum_n=0^inftyfracx^nn!+12e^4sum_n=0^inftyfracx^nn!\&=12e^4+sum_n=1^infty3e^4left(frac1(n-1)!+frac4n!right)(x-4)^nendalign$$

answered Jul 22 at 21:13

user578878

add a commentÂ |Â

up vote
2
down vote

You can do it like this.

$3xe^x=3(x-4+4)e^x-4+4=3e^4(x-4)e^x-4+12e^4e^x-4$

Then use the series that you know $e^x=sum_n=0^inftyfracx^nn!$ to instead write $e^x-4=sum_n=0^inftyfrac(x-4)^nn!$

Putting this into the expression of your function above yields

$$beginalign3e^4(x-4)e^x-4+12e^4e^x-4&=3e^4(x-4)sum_n=0^inftyfracx^nn!+12e^4sum_n=0^inftyfracx^nn!\&=12e^4+sum_n=1^infty3e^4left(frac1(n-1)!+frac4n!right)(x-4)^nendalign$$

answered Jul 22 at 21:13

user578878

You can do it like this.

$3xe^x=3(x-4+4)e^x-4+4=3e^4(x-4)e^x-4+12e^4e^x-4$

Then use the series that you know $e^x=sum_n=0^inftyfracx^nn!$ to instead write $e^x-4=sum_n=0^inftyfrac(x-4)^nn!$

Putting this into the expression of your function above yields

$$beginalign3e^4(x-4)e^x-4+12e^4e^x-4&=3e^4(x-4)sum_n=0^inftyfracx^nn!+12e^4sum_n=0^inftyfracx^nn!\&=12e^4+sum_n=1^infty3e^4left(frac1(n-1)!+frac4n!right)(x-4)^nendalign$$

answered Jul 22 at 21:13

user578878

answered Jul 22 at 21:13

user578878

answered Jul 22 at 21:13

user578878

answered Jul 22 at 21:13

user578878

add a commentÂ |Â

up vote
0
down vote

Taylor theorem states that $dots$

$$f(x) = sum_n=0^infty fracf^(n)(a)(x-a)^nn!$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

answered Jul 22 at 21:11

bkarthik

320212

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

add a commentÂ |Â

up vote
0
down vote

Taylor theorem states that $dots$

$$f(x) = sum_n=0^infty fracf^(n)(a)(x-a)^nn!$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

answered Jul 22 at 21:11

bkarthik

320212

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

add a commentÂ |Â

up vote
0
down vote

Taylor theorem states that $dots$

$$f(x) = sum_n=0^infty fracf^(n)(a)(x-a)^nn!$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

answered Jul 22 at 21:11

bkarthik

320212

Taylor theorem states that $dots$

$$f(x) = sum_n=0^infty fracf^(n)(a)(x-a)^nn!$$

So what you want to do is take $f(x)$ to be $3xe^x$ and $a = 4$. From there it is just a plug and chug to find the polynomial expansion.

answered Jul 22 at 21:11

bkarthik

320212

answered Jul 22 at 21:11

bkarthik

320212

answered Jul 22 at 21:11

bkarthik

320212

answered Jul 22 at 21:11

bkarthik

320212

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

add a commentÂ |Â

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

Understood, but I'm working on a question that asks for the coefficient of the 17th term so I'm not sure how to do that without taking 17 derivatives.
â€“Â Jake
Jul 22 at 21:14

Take the first few and you will see a pattern
â€“Â bkarthik
Jul 22 at 21:15

add a commentÂ |Â

up vote
0
down vote

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

answered Jul 22 at 21:20

Yves Daoust

111k665203

add a commentÂ |Â

up vote
0
down vote

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

answered Jul 22 at 21:20

Yves Daoust

111k665203

add a commentÂ |Â

up vote
0
down vote

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

answered Jul 22 at 21:20

Yves Daoust

111k665203

If the radius of convergence of the development of the function and its derivatives around $x=0$ contains the new point ($x=4$), you can evaluate the new Taylor coefficients from the series:

answered Jul 22 at 21:20

Yves Daoust

111k665203

answered Jul 22 at 21:20

Yves Daoust

111k665203

answered Jul 22 at 21:20

Yves Daoust

111k665203

answered Jul 22 at 21:20

Yves Daoust

111k665203

add a commentÂ |Â

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