Finding the Maclaurin series of $e^{sin x}$ by comparing coefficients
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I believe I have found a nice way to find the Maclaurin series of $e^{sin x}$. Please check if there are any mistakes with my working. Is this method well known?
calculus proof-verification power-series taylor-expansion
edited Nov 27 at 12:23
Bernard
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asked Nov 27 at 12:12
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up vote
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I believe I have found a nice way to find the Maclaurin series of $e^{sin x}$. Please check if there are any mistakes with my working. Is this method well known?
calculus proof-verification power-series taylor-expansion
edited Nov 27 at 12:23
Bernard
117k637109
asked Nov 27 at 12:12
3684
1277
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I believe I have found a nice way to find the Maclaurin series of $e^{sin x}$. Please check if there are any mistakes with my working. Is this method well known?
calculus proof-verification power-series taylor-expansion
edited Nov 27 at 12:23
Bernard
117k637109
asked Nov 27 at 12:12
3684
1277
I believe I have found a nice way to find the Maclaurin series of $e^{sin x}$. Please check if there are any mistakes with my working. Is this method well known?
calculus proof-verification power-series taylor-expansion
calculus proof-verification power-series taylor-expansion
edited Nov 27 at 12:23
Bernard
117k637109
asked Nov 27 at 12:12
3684
1277
edited Nov 27 at 12:23
Bernard
117k637109
asked Nov 27 at 12:12
3684
1277
edited Nov 27 at 12:23
Bernard
117k637109
edited Nov 27 at 12:23
Bernard
117k637109
edited Nov 27 at 12:23
Bernard
117k637109
117k637109
asked Nov 27 at 12:12
3684
1277
asked Nov 27 at 12:12
3684
1277
asked Nov 27 at 12:12
3684
1277
1277
add a comment |
add a comment |
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(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $cos xapprox1-frac{x^2}{2}+frac{x^4}{24}$.)
Let $f(x)= e^{sin x}$, therefore $f'(x)=cos x*e^{sin x}=cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-frac{x^2}{2}+frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-frac{1}{2})x^2+(d-frac{b}{2})x^3+(e-frac{c}{2}+frac{a}{24})x^4+(f-frac{d}{2}+frac{b}{24})x^5$.
Camparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-frac{1}{2}$
$4e=d-frac{b}{2}$
$5f=e-frac{c}{2}+frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=frac{1}{2}$
$d=0$
$e=-frac{1}{8}$
$f=-frac{1}{15}$.
This means the Maclaurin series of $e^{sin x}=1+x+frac{1}{2}x^2-frac{1}{8}x^4-frac{1}{15}x^5$ which is indeed to correct series.
edited Nov 27 at 12:25
Bernard
117k637109
answered Nov 27 at 12:12
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add a comment |
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1 Answer
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up vote
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(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $cos xapprox1-frac{x^2}{2}+frac{x^4}{24}$.)
Let $f(x)= e^{sin x}$, therefore $f'(x)=cos x*e^{sin x}=cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-frac{x^2}{2}+frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-frac{1}{2})x^2+(d-frac{b}{2})x^3+(e-frac{c}{2}+frac{a}{24})x^4+(f-frac{d}{2}+frac{b}{24})x^5$.
Camparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-frac{1}{2}$
$4e=d-frac{b}{2}$
$5f=e-frac{c}{2}+frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=frac{1}{2}$
$d=0$
$e=-frac{1}{8}$
$f=-frac{1}{15}$.
This means the Maclaurin series of $e^{sin x}=1+x+frac{1}{2}x^2-frac{1}{8}x^4-frac{1}{15}x^5$ which is indeed to correct series.
edited Nov 27 at 12:25
Bernard
117k637109
answered Nov 27 at 12:12
3684
1277
add a comment |
up vote
3
down vote
(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $cos xapprox1-frac{x^2}{2}+frac{x^4}{24}$.)
Let $f(x)= e^{sin x}$, therefore $f'(x)=cos x*e^{sin x}=cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-frac{x^2}{2}+frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-frac{1}{2})x^2+(d-frac{b}{2})x^3+(e-frac{c}{2}+frac{a}{24})x^4+(f-frac{d}{2}+frac{b}{24})x^5$.
Camparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-frac{1}{2}$
$4e=d-frac{b}{2}$
$5f=e-frac{c}{2}+frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=frac{1}{2}$
$d=0$
$e=-frac{1}{8}$
$f=-frac{1}{15}$.
This means the Maclaurin series of $e^{sin x}=1+x+frac{1}{2}x^2-frac{1}{8}x^4-frac{1}{15}x^5$ which is indeed to correct series.
edited Nov 27 at 12:25
Bernard
117k637109
answered Nov 27 at 12:12
3684
1277
add a comment |
up vote
3
down vote
up vote
3
down vote
(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $cos xapprox1-frac{x^2}{2}+frac{x^4}{24}$.)
Let $f(x)= e^{sin x}$, therefore $f'(x)=cos x*e^{sin x}=cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-frac{x^2}{2}+frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-frac{1}{2})x^2+(d-frac{b}{2})x^3+(e-frac{c}{2}+frac{a}{24})x^4+(f-frac{d}{2}+frac{b}{24})x^5$.
Camparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-frac{1}{2}$
$4e=d-frac{b}{2}$
$5f=e-frac{c}{2}+frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=frac{1}{2}$
$d=0$
$e=-frac{1}{8}$
$f=-frac{1}{15}$.
This means the Maclaurin series of $e^{sin x}=1+x+frac{1}{2}x^2-frac{1}{8}x^4-frac{1}{15}x^5$ which is indeed to correct series.
edited Nov 27 at 12:25
Bernard
117k637109
answered Nov 27 at 12:12
3684
1277
(Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $cos xapprox1-frac{x^2}{2}+frac{x^4}{24}$.)
Let $f(x)= e^{sin x}$, therefore $f'(x)=cos x*e^{sin x}=cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-frac{x^2}{2}+frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-frac{1}{2})x^2+(d-frac{b}{2})x^3+(e-frac{c}{2}+frac{a}{24})x^4+(f-frac{d}{2}+frac{b}{24})x^5$.
Camparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-frac{1}{2}$
$4e=d-frac{b}{2}$
$5f=e-frac{c}{2}+frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=frac{1}{2}$
$d=0$
$e=-frac{1}{8}$
$f=-frac{1}{15}$.
This means the Maclaurin series of $e^{sin x}=1+x+frac{1}{2}x^2-frac{1}{8}x^4-frac{1}{15}x^5$ which is indeed to correct series.
edited Nov 27 at 12:25
Bernard
117k637109
answered Nov 27 at 12:12
3684
1277
edited Nov 27 at 12:25
Bernard
117k637109
edited Nov 27 at 12:25
Bernard
117k637109
edited Nov 27 at 12:25
Bernard
117k637109
117k637109
answered Nov 27 at 12:12
3684
1277
answered Nov 27 at 12:12
3684
1277
answered Nov 27 at 12:12
3684
1277
1277
add a comment |
add a comment |
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