Checking work Taylor Series Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Clever derivation of $arcsin(x)$ Taylor seriesTaylor series of $f(x^2)$Finding taylor expansion of $cos^2x$ and $sin^2x$How would Taylor Series work?Computing a sum with a Taylor seriesTaylor/Maclaurin series type questionFinding taylor series in Real AnalysisCauchy product of two different Taylor series $e^x$ and $cos x$Two Taylor Series for $-sin(x)$ are Equal But One Doesn't Work?How to find the Taylor series of $sin^2(4x)$?
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Checking work Taylor Series
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Clever derivation of $arcsin(x)$ Taylor seriesTaylor series of $f(x^2)$Finding taylor expansion of $cos^2x$ and $sin^2x$How would Taylor Series work?Computing a sum with a Taylor seriesTaylor/Maclaurin series type questionFinding taylor series in Real AnalysisCauchy product of two different Taylor series $e^x$ and $cos x$Two Taylor Series for $-sin(x)$ are Equal But One Doesn't Work?How to find the Taylor series of $sin^2(4x)$?
$begingroup$
I want to check my work, I am constructing a taylor series formula for
$$f(x)=cos(x) + sin(x)$$
So I know
$$cos(x) = sum_n=0^infty (-1)^n fracx^2n2n! $$
then
$$cos(x) + sin(x) = sum_n=0^infty (-1)^n fracx^2n(2n)! + (-1)^n fracx^2n+1(2n+1)!$$
If im wrong could I get some guidance? I haven't done calculus in years.
calculus taylor-expansion
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
$endgroup$
add a comment |
$begingroup$
I want to check my work, I am constructing a taylor series formula for
$$f(x)=cos(x) + sin(x)$$
So I know
$$cos(x) = sum_n=0^infty (-1)^n fracx^2n2n! $$
then
$$cos(x) + sin(x) = sum_n=0^infty (-1)^n fracx^2n(2n)! + (-1)^n fracx^2n+1(2n+1)!$$
If im wrong could I get some guidance? I haven't done calculus in years.
calculus taylor-expansion
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
$endgroup$
1
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57
add a comment |
$begingroup$
I want to check my work, I am constructing a taylor series formula for
$$f(x)=cos(x) + sin(x)$$
So I know
$$cos(x) = sum_n=0^infty (-1)^n fracx^2n2n! $$
then
$$cos(x) + sin(x) = sum_n=0^infty (-1)^n fracx^2n(2n)! + (-1)^n fracx^2n+1(2n+1)!$$
If im wrong could I get some guidance? I haven't done calculus in years.
calculus taylor-expansion
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
$endgroup$
I want to check my work, I am constructing a taylor series formula for
$$f(x)=cos(x) + sin(x)$$
So I know
$$cos(x) = sum_n=0^infty (-1)^n fracx^2n2n! $$
then
$$cos(x) + sin(x) = sum_n=0^infty (-1)^n fracx^2n(2n)! + (-1)^n fracx^2n+1(2n+1)!$$
If im wrong could I get some guidance? I haven't done calculus in years.
calculus taylor-expansion
calculus taylor-expansion
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
edited Mar 31 at 18:59
Temirzhan
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
asked Mar 31 at 18:51
TemirzhanTemirzhan
516314
516314
1
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57
add a comment |
1
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57
1
1
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is better to have a single term.
Use
$$cos(x)=frac12 left(e^i x+e^-i xright)qquad textand qquad sin(x)=-fraci2 left(e^i x-e^-i xright)$$ Now, use the series expansion of the exponentials to get
$$cos(x)+sin(x)=sum_n=0^infty fracleft(frac12+fraci2right) left((-i)^n-i i^nright)n!x^n=sum_n=0^infty fracsin left(fracpi n2right)+cos left(fracpi n2right)n! x^n$$
You could prefer
$$cos(x)+sin(x)=sum_n=0^infty fraci^(n-1) nn!x^n$$
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It is better to have a single term.
Use
$$cos(x)=frac12 left(e^i x+e^-i xright)qquad textand qquad sin(x)=-fraci2 left(e^i x-e^-i xright)$$ Now, use the series expansion of the exponentials to get
$$cos(x)+sin(x)=sum_n=0^infty fracleft(frac12+fraci2right) left((-i)^n-i i^nright)n!x^n=sum_n=0^infty fracsin left(fracpi n2right)+cos left(fracpi n2right)n! x^n$$
You could prefer
$$cos(x)+sin(x)=sum_n=0^infty fraci^(n-1) nn!x^n$$
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
It is better to have a single term.
Use
$$cos(x)=frac12 left(e^i x+e^-i xright)qquad textand qquad sin(x)=-fraci2 left(e^i x-e^-i xright)$$ Now, use the series expansion of the exponentials to get
$$cos(x)+sin(x)=sum_n=0^infty fracleft(frac12+fraci2right) left((-i)^n-i i^nright)n!x^n=sum_n=0^infty fracsin left(fracpi n2right)+cos left(fracpi n2right)n! x^n$$
You could prefer
$$cos(x)+sin(x)=sum_n=0^infty fraci^(n-1) nn!x^n$$
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
It is better to have a single term.
Use
$$cos(x)=frac12 left(e^i x+e^-i xright)qquad textand qquad sin(x)=-fraci2 left(e^i x-e^-i xright)$$ Now, use the series expansion of the exponentials to get
$$cos(x)+sin(x)=sum_n=0^infty fracleft(frac12+fraci2right) left((-i)^n-i i^nright)n!x^n=sum_n=0^infty fracsin left(fracpi n2right)+cos left(fracpi n2right)n! x^n$$
You could prefer
$$cos(x)+sin(x)=sum_n=0^infty fraci^(n-1) nn!x^n$$
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
It is better to have a single term.
Use
$$cos(x)=frac12 left(e^i x+e^-i xright)qquad textand qquad sin(x)=-fraci2 left(e^i x-e^-i xright)$$ Now, use the series expansion of the exponentials to get
$$cos(x)+sin(x)=sum_n=0^infty fracleft(frac12+fraci2right) left((-i)^n-i i^nright)n!x^n=sum_n=0^infty fracsin left(fracpi n2right)+cos left(fracpi n2right)n! x^n$$
You could prefer
$$cos(x)+sin(x)=sum_n=0^infty fraci^(n-1) nn!x^n$$
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
edited Apr 1 at 13:26
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
answered Apr 1 at 6:18
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
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1
$begingroup$
$(2n+1)!$ in the denominator, for $sin$.
$endgroup$
– Clement C.
Mar 31 at 18:53
$begingroup$
Also (2n)!..I think it's typo
$endgroup$
– Believer
Mar 31 at 18:57