Computation of Maclaurin Series
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I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
add a comment |
$begingroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
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For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
$begingroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
$endgroup$
I have been working on Maclaurin Series recently and was wondering if there's a more simple and elegant way to obtain series for more complicated functions,say $f(x)=ln(1+2x+2x^2)$ or $g(x)=tan(2x^4-x)$.Using the definition leads to messy derivatives almost immediately.If it was some simple rational function,for example,i would try to use Maclaurin Series of ${1over1+x}$ or ${1over1-x}$ and then manipulate it to get my result,but I can't really think of any shortcut for listed above functions(and many more).
calculus logarithms taylor-expansion
calculus logarithms taylor-expansion
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
asked 3 hours ago
Turan NəsibliTuran Nəsibli
564
564
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
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keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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active
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votes
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
add a comment |
$begingroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
$endgroup$
keep substitution in mind. It may not give the whole infinite series, but you will usually get the first several terms. So,
$$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots $$
$$ log (1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} cdots $$
Taking $t = 2x+2x^2$ correctly gives the first few terms of $log(1+2x+2x^2),$ up to $x^4$
$$ log(1+2x+2x^2) = 2 x - frac{4x^3}{3} + 2 x^4 cdots $$
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
answered 3 hours ago
Will JagyWill Jagy
102k5101199
102k5101199
add a comment |
add a comment |
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$begingroup$
In general, using calculation from "geometric" series method is usually the way to go. However, calculation is at the end of the day just a matter of calculation. So, it really depends on the formula of your function itself. Not to mention, "shortcut" method usually implies "convergence condition" as well.
$endgroup$
– Evan William Chandra
3 hours ago
$begingroup$
For a composition of two functions with known power series you can use the Cauchy product.
$endgroup$
– Ian
3 hours ago
$begingroup$
(Note that the Cauchy product is basically just a way of bookkeeping what happens when you substitute the inner function into the series of the outer function and then combine like terms).
$endgroup$
– Ian
3 hours ago