For Maclaurin Series, why can $kx$ be substituted to obtain $f(kx)$ but not applicable for $f(x+k)$?
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So this is taken from my textbook, but I cant seem to understand why is it so.
"When the Maclaurin series for a function f(x) is known, we can substitute $kx$ (where $k$ is a constant) for $x$ to obtain the series for $f(kx)$. This may not be applicable to the other forms such as $f(k+x)$ for practical reasons."
sequences-and-series taylor-expansion
asked Nov 16 at 12:35
Henias
615
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up vote
0
down vote
favorite
So this is taken from my textbook, but I cant seem to understand why is it so.
"When the Maclaurin series for a function f(x) is known, we can substitute $kx$ (where $k$ is a constant) for $x$ to obtain the series for $f(kx)$. This may not be applicable to the other forms such as $f(k+x)$ for practical reasons."
sequences-and-series taylor-expansion
asked Nov 16 at 12:35
Henias
615
For one, it would move the center off 0.
– Randall
Nov 16 at 12:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So this is taken from my textbook, but I cant seem to understand why is it so.
"When the Maclaurin series for a function f(x) is known, we can substitute $kx$ (where $k$ is a constant) for $x$ to obtain the series for $f(kx)$. This may not be applicable to the other forms such as $f(k+x)$ for practical reasons."
sequences-and-series taylor-expansion
asked Nov 16 at 12:35
Henias
615
So this is taken from my textbook, but I cant seem to understand why is it so.
"When the Maclaurin series for a function f(x) is known, we can substitute $kx$ (where $k$ is a constant) for $x$ to obtain the series for $f(kx)$. This may not be applicable to the other forms such as $f(k+x)$ for practical reasons."
sequences-and-series taylor-expansion
sequences-and-series taylor-expansion
asked Nov 16 at 12:35
Henias
615
asked Nov 16 at 12:35
Henias
615
asked Nov 16 at 12:35
Henias
615
asked Nov 16 at 12:35
Henias
615
asked Nov 16 at 12:35
Henias
615
615
For one, it would move the center off 0.
– Randall
Nov 16 at 12:40
add a comment |
For one, it would move the center off 0.
– Randall
Nov 16 at 12:40
For one, it would move the center off 0.
– Randall
Nov 16 at 12:40
For one, it would move the center off 0.
– Randall
Nov 16 at 12:40
add a comment |
1 Answer
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If you know that the Maclaurin series for $f(x)$ is given by the sequence $(a_i)$, that is,
$$
f(x) = sum_{i=0}^n a_ix^i,
$$
then we have that
$$
f(kx) = sum_{i=0}^n a_i k^i x^i.
$$
So by setting $b_i = k^i a_i$, we have that the Maclaurin series for $f(x)$ is given by $(b_i)$. This is simple. However, looking at $f(x + k)$, we get
$$
f(x+k) = sum_{i=0}^n a_i(x + k)^i = sum_{i=0}^infty a_isum_{j=0}^ibinom{i}{j} x^jk^{i-j} = sum_{i = 0}^infty left(sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^lright)x^i.
$$
So now the coefficients are given by
$$
c_i = sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^l,
$$
and it's not even obvious if and when these series converge, never mind what value they have.
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If you know that the Maclaurin series for $f(x)$ is given by the sequence $(a_i)$, that is,
$$
f(x) = sum_{i=0}^n a_ix^i,
$$
then we have that
$$
f(kx) = sum_{i=0}^n a_i k^i x^i.
$$
So by setting $b_i = k^i a_i$, we have that the Maclaurin series for $f(x)$ is given by $(b_i)$. This is simple. However, looking at $f(x + k)$, we get
$$
f(x+k) = sum_{i=0}^n a_i(x + k)^i = sum_{i=0}^infty a_isum_{j=0}^ibinom{i}{j} x^jk^{i-j} = sum_{i = 0}^infty left(sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^lright)x^i.
$$
So now the coefficients are given by
$$
c_i = sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^l,
$$
and it's not even obvious if and when these series converge, never mind what value they have.
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
add a comment |
up vote
0
down vote
accepted
If you know that the Maclaurin series for $f(x)$ is given by the sequence $(a_i)$, that is,
$$
f(x) = sum_{i=0}^n a_ix^i,
$$
then we have that
$$
f(kx) = sum_{i=0}^n a_i k^i x^i.
$$
So by setting $b_i = k^i a_i$, we have that the Maclaurin series for $f(x)$ is given by $(b_i)$. This is simple. However, looking at $f(x + k)$, we get
$$
f(x+k) = sum_{i=0}^n a_i(x + k)^i = sum_{i=0}^infty a_isum_{j=0}^ibinom{i}{j} x^jk^{i-j} = sum_{i = 0}^infty left(sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^lright)x^i.
$$
So now the coefficients are given by
$$
c_i = sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^l,
$$
and it's not even obvious if and when these series converge, never mind what value they have.
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If you know that the Maclaurin series for $f(x)$ is given by the sequence $(a_i)$, that is,
$$
f(x) = sum_{i=0}^n a_ix^i,
$$
then we have that
$$
f(kx) = sum_{i=0}^n a_i k^i x^i.
$$
So by setting $b_i = k^i a_i$, we have that the Maclaurin series for $f(x)$ is given by $(b_i)$. This is simple. However, looking at $f(x + k)$, we get
$$
f(x+k) = sum_{i=0}^n a_i(x + k)^i = sum_{i=0}^infty a_isum_{j=0}^ibinom{i}{j} x^jk^{i-j} = sum_{i = 0}^infty left(sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^lright)x^i.
$$
So now the coefficients are given by
$$
c_i = sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^l,
$$
and it's not even obvious if and when these series converge, never mind what value they have.
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
If you know that the Maclaurin series for $f(x)$ is given by the sequence $(a_i)$, that is,
$$
f(x) = sum_{i=0}^n a_ix^i,
$$
then we have that
$$
f(kx) = sum_{i=0}^n a_i k^i x^i.
$$
So by setting $b_i = k^i a_i$, we have that the Maclaurin series for $f(x)$ is given by $(b_i)$. This is simple. However, looking at $f(x + k)$, we get
$$
f(x+k) = sum_{i=0}^n a_i(x + k)^i = sum_{i=0}^infty a_isum_{j=0}^ibinom{i}{j} x^jk^{i-j} = sum_{i = 0}^infty left(sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^lright)x^i.
$$
So now the coefficients are given by
$$
c_i = sum_{l=0}^inftybinom{i+l}{i}a_{i+l}k^l,
$$
and it's not even obvious if and when these series converge, never mind what value they have.
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
answered Nov 16 at 12:47
Mees de Vries
16.3k12654
16.3k12654
add a comment |
add a comment |
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For one, it would move the center off 0.
– Randall
Nov 16 at 12:40