First Few Terms in a Maclaurin Series
$begingroup$
This is the question
I found the derivatives of ln(1+sinx) and I keep getting $x$ $-$ $x^2$$/2$ $+$
$x^3$$/3$ $-$ $x^4$$/4$ as my terms. None of the answers have those so I'm not sure what I'm doing wrong here.
integration sequences-and-series derivatives taylor-expansion
asked Dec 19 '18 at 2:46
krauser126krauser126
636
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add a comment |
$begingroup$
This is the question
I found the derivatives of ln(1+sinx) and I keep getting $x$ $-$ $x^2$$/2$ $+$
$x^3$$/3$ $-$ $x^4$$/4$ as my terms. None of the answers have those so I'm not sure what I'm doing wrong here.
integration sequences-and-series derivatives taylor-expansion
asked Dec 19 '18 at 2:46
krauser126krauser126
636
$endgroup$
add a comment |
$begingroup$
This is the question
I found the derivatives of ln(1+sinx) and I keep getting $x$ $-$ $x^2$$/2$ $+$
$x^3$$/3$ $-$ $x^4$$/4$ as my terms. None of the answers have those so I'm not sure what I'm doing wrong here.
integration sequences-and-series derivatives taylor-expansion
asked Dec 19 '18 at 2:46
krauser126krauser126
636
$endgroup$
This is the question
I found the derivatives of ln(1+sinx) and I keep getting $x$ $-$ $x^2$$/2$ $+$
$x^3$$/3$ $-$ $x^4$$/4$ as my terms. None of the answers have those so I'm not sure what I'm doing wrong here.
integration sequences-and-series derivatives taylor-expansion
integration sequences-and-series derivatives taylor-expansion
asked Dec 19 '18 at 2:46
krauser126krauser126
636
asked Dec 19 '18 at 2:46
krauser126krauser126
636
asked Dec 19 '18 at 2:46
krauser126krauser126
636
asked Dec 19 '18 at 2:46
krauser126krauser126
636
asked Dec 19 '18 at 2:46
krauser126krauser126
636
636
add a comment |
add a comment |
2 Answers
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$$ f(x) = ln (1+sin x) implies f(0)=0$$
$$f'(x) = frac {cos x}{1+sin x} implies f'(0) = 1$$
$$ f''(x) = frac { -sin x (1+sin x)-cos^2 x}{(1+sin x )^2} implies f''(0) = -1$$
Continue and you get
$$ ln( 1 + sin x ) = x-x^2/2 + x^3/6 -...$$
Which is the choice (e)
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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add a comment |
$begingroup$
To get the series for composed functions up to a small degree, you can compose series. Let's see,
$$ log(1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} + ; mbox{more} $$
in the above, set
$$ t approx sin x = x - frac{x^3}{6} + ; mbox{more} $$
So, just put $t = x - frac{x^3}{6}$ into $t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
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2 Answers
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2 Answers
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$begingroup$
$$ f(x) = ln (1+sin x) implies f(0)=0$$
$$f'(x) = frac {cos x}{1+sin x} implies f'(0) = 1$$
$$ f''(x) = frac { -sin x (1+sin x)-cos^2 x}{(1+sin x )^2} implies f''(0) = -1$$
Continue and you get
$$ ln( 1 + sin x ) = x-x^2/2 + x^3/6 -...$$
Which is the choice (e)
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
$$ f(x) = ln (1+sin x) implies f(0)=0$$
$$f'(x) = frac {cos x}{1+sin x} implies f'(0) = 1$$
$$ f''(x) = frac { -sin x (1+sin x)-cos^2 x}{(1+sin x )^2} implies f''(0) = -1$$
Continue and you get
$$ ln( 1 + sin x ) = x-x^2/2 + x^3/6 -...$$
Which is the choice (e)
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
$$ f(x) = ln (1+sin x) implies f(0)=0$$
$$f'(x) = frac {cos x}{1+sin x} implies f'(0) = 1$$
$$ f''(x) = frac { -sin x (1+sin x)-cos^2 x}{(1+sin x )^2} implies f''(0) = -1$$
Continue and you get
$$ ln( 1 + sin x ) = x-x^2/2 + x^3/6 -...$$
Which is the choice (e)
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$$ f(x) = ln (1+sin x) implies f(0)=0$$
$$f'(x) = frac {cos x}{1+sin x} implies f'(0) = 1$$
$$ f''(x) = frac { -sin x (1+sin x)-cos^2 x}{(1+sin x )^2} implies f''(0) = -1$$
Continue and you get
$$ ln( 1 + sin x ) = x-x^2/2 + x^3/6 -...$$
Which is the choice (e)
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 19 '18 at 3:09
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
To get the series for composed functions up to a small degree, you can compose series. Let's see,
$$ log(1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} + ; mbox{more} $$
in the above, set
$$ t approx sin x = x - frac{x^3}{6} + ; mbox{more} $$
So, just put $t = x - frac{x^3}{6}$ into $t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
To get the series for composed functions up to a small degree, you can compose series. Let's see,
$$ log(1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} + ; mbox{more} $$
in the above, set
$$ t approx sin x = x - frac{x^3}{6} + ; mbox{more} $$
So, just put $t = x - frac{x^3}{6}$ into $t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
To get the series for composed functions up to a small degree, you can compose series. Let's see,
$$ log(1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} + ; mbox{more} $$
in the above, set
$$ t approx sin x = x - frac{x^3}{6} + ; mbox{more} $$
So, just put $t = x - frac{x^3}{6}$ into $t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
$endgroup$
To get the series for composed functions up to a small degree, you can compose series. Let's see,
$$ log(1+t) = t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4} + ; mbox{more} $$
in the above, set
$$ t approx sin x = x - frac{x^3}{6} + ; mbox{more} $$
So, just put $t = x - frac{x^3}{6}$ into $t - frac{t^2}{2} + frac{t^3}{3} - frac{t^4}{4}$ and keep only terms with $x$ exponent up to 4
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
answered Dec 19 '18 at 2:55
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
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