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."










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  • For one, it would move the center off 0.
    – Randall
    Nov 16 at 12:40















up vote
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down vote

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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."










share|cite|improve this question






















  • For one, it would move the center off 0.
    – Randall
    Nov 16 at 12:40













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."










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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






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asked Nov 16 at 12:35









Henias

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  • 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




For one, it would move the center off 0.
– Randall
Nov 16 at 12:40










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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.






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    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.






    share|cite|improve this answer

























      up vote
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      down vote



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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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 16 at 12:47









        Mees de Vries

        16.3k12654




        16.3k12654






























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