Taylor series of $logleft(1+2x^2right)$
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I want to find the Taylor-MacLaurin series of the following function until we find a term that is different from zero. This is the function:
$$f(x)=logleft(1+2x^2right)+4(cosx-1)$$
Finding the derivative of $left(cos x-1right)$ is easy enough. I found:
$$cos(x)-1=-fracx^22+frac14!x^4+o(x^4)$$
Finding the derivatives of $log(1+2x^2)$ up to the $4$th order isn't as straightforward. Can I do something like this:
$$logleft(1+2x^2right)sim2x^2$$
And then find the derivatives from there?
- $0$th order: $f(0)=2(0)^2=0$
- $1$st order: $f'(x)=4ximplies f'(0)=0$
- $2$nd order: $f''(x)=4 implies f''(0)=4$?
Any hints on how to easily find the Tayler series of $logleft(1+2x^2right)$? I have also tried chancing the variables but it leads to unexpected results:
$$logleft(1+2x^2right)=logt$$
- $0$th order: $f(t)=logtimplies f(0)=log0$ isn't defined.
logarithms taylor-expansion
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I want to find the Taylor-MacLaurin series of the following function until we find a term that is different from zero. This is the function:
$$f(x)=logleft(1+2x^2right)+4(cosx-1)$$
Finding the derivative of $left(cos x-1right)$ is easy enough. I found:
$$cos(x)-1=-fracx^22+frac14!x^4+o(x^4)$$
Finding the derivatives of $log(1+2x^2)$ up to the $4$th order isn't as straightforward. Can I do something like this:
$$logleft(1+2x^2right)sim2x^2$$
And then find the derivatives from there?
- $0$th order: $f(0)=2(0)^2=0$
- $1$st order: $f'(x)=4ximplies f'(0)=0$
- $2$nd order: $f''(x)=4 implies f''(0)=4$?
Any hints on how to easily find the Tayler series of $logleft(1+2x^2right)$? I have also tried chancing the variables but it leads to unexpected results:
$$logleft(1+2x^2right)=logt$$
- $0$th order: $f(t)=logtimplies f(0)=log0$ isn't defined.
logarithms taylor-expansion
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to find the Taylor-MacLaurin series of the following function until we find a term that is different from zero. This is the function:
$$f(x)=logleft(1+2x^2right)+4(cosx-1)$$
Finding the derivative of $left(cos x-1right)$ is easy enough. I found:
$$cos(x)-1=-fracx^22+frac14!x^4+o(x^4)$$
Finding the derivatives of $log(1+2x^2)$ up to the $4$th order isn't as straightforward. Can I do something like this:
$$logleft(1+2x^2right)sim2x^2$$
And then find the derivatives from there?
- $0$th order: $f(0)=2(0)^2=0$
- $1$st order: $f'(x)=4ximplies f'(0)=0$
- $2$nd order: $f''(x)=4 implies f''(0)=4$?
Any hints on how to easily find the Tayler series of $logleft(1+2x^2right)$? I have also tried chancing the variables but it leads to unexpected results:
$$logleft(1+2x^2right)=logt$$
- $0$th order: $f(t)=logtimplies f(0)=log0$ isn't defined.
logarithms taylor-expansion
I want to find the Taylor-MacLaurin series of the following function until we find a term that is different from zero. This is the function:
$$f(x)=logleft(1+2x^2right)+4(cosx-1)$$
Finding the derivative of $left(cos x-1right)$ is easy enough. I found:
$$cos(x)-1=-fracx^22+frac14!x^4+o(x^4)$$
Finding the derivatives of $log(1+2x^2)$ up to the $4$th order isn't as straightforward. Can I do something like this:
$$logleft(1+2x^2right)sim2x^2$$
And then find the derivatives from there?
- $0$th order: $f(0)=2(0)^2=0$
- $1$st order: $f'(x)=4ximplies f'(0)=0$
- $2$nd order: $f''(x)=4 implies f''(0)=4$?
Any hints on how to easily find the Tayler series of $logleft(1+2x^2right)$? I have also tried chancing the variables but it leads to unexpected results:
$$logleft(1+2x^2right)=logt$$
- $0$th order: $f(t)=logtimplies f(0)=log0$ isn't defined.
logarithms taylor-expansion
logarithms taylor-expansion
edited Sep 3 at 19:00
Jendrik Stelzner
7,69121137
7,69121137
asked Sep 3 at 14:33
Cesare
734410
734410
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2 Answers
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Since$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots,$$you havebeginalignlog(1+2x^2)&=2x^2-frac(2x^2)^22+frac(2x^2)^33-frac(2x^2)^44+cdots\&=2x^2-frac4x^42+frac8x^63-frac16x^44+cdotsendalign
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
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Let $g(x)=ln 1+x$ therefore $$g'(x)=dfrac11+x\g''(x)=-dfrac1(1+x)^2\g'''(x)=dfrac2(1+x)^3\.\.\.\g^(n)(x)=(-1)^n+1dfrac(n-1)!(1+x)^n$$which means that$$g^(n)(0)=(-1)^n+1(n-1)!$$and we have $$ln(1+x)=sum_n=1^inftydfrac(-1)^n+1nx^n$$therefore $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Since$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots,$$you havebeginalignlog(1+2x^2)&=2x^2-frac(2x^2)^22+frac(2x^2)^33-frac(2x^2)^44+cdots\&=2x^2-frac4x^42+frac8x^63-frac16x^44+cdotsendalign
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
add a comment |Â
up vote
5
down vote
accepted
Since$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots,$$you havebeginalignlog(1+2x^2)&=2x^2-frac(2x^2)^22+frac(2x^2)^33-frac(2x^2)^44+cdots\&=2x^2-frac4x^42+frac8x^63-frac16x^44+cdotsendalign
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Since$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots,$$you havebeginalignlog(1+2x^2)&=2x^2-frac(2x^2)^22+frac(2x^2)^33-frac(2x^2)^44+cdots\&=2x^2-frac4x^42+frac8x^63-frac16x^44+cdotsendalign
Since$$log(1+x)=x-fracx^22+fracx^33-fracx^44+cdots,$$you havebeginalignlog(1+2x^2)&=2x^2-frac(2x^2)^22+frac(2x^2)^33-frac(2x^2)^44+cdots\&=2x^2-frac4x^42+frac8x^63-frac16x^44+cdotsendalign
edited Sep 3 at 14:39
answered Sep 3 at 14:36
José Carlos Santos
122k16101186
122k16101186
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
add a comment |Â
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
@DavideMorgante I've edited my answer. Thank you.
â José Carlos Santos
Sep 3 at 14:40
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
Thanks, I didn't realize that I could do it this way.
â Cesare
Sep 3 at 14:42
add a comment |Â
up vote
0
down vote
Let $g(x)=ln 1+x$ therefore $$g'(x)=dfrac11+x\g''(x)=-dfrac1(1+x)^2\g'''(x)=dfrac2(1+x)^3\.\.\.\g^(n)(x)=(-1)^n+1dfrac(n-1)!(1+x)^n$$which means that$$g^(n)(0)=(-1)^n+1(n-1)!$$and we have $$ln(1+x)=sum_n=1^inftydfrac(-1)^n+1nx^n$$therefore $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$
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up vote
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Let $g(x)=ln 1+x$ therefore $$g'(x)=dfrac11+x\g''(x)=-dfrac1(1+x)^2\g'''(x)=dfrac2(1+x)^3\.\.\.\g^(n)(x)=(-1)^n+1dfrac(n-1)!(1+x)^n$$which means that$$g^(n)(0)=(-1)^n+1(n-1)!$$and we have $$ln(1+x)=sum_n=1^inftydfrac(-1)^n+1nx^n$$therefore $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$
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up vote
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up vote
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down vote
Let $g(x)=ln 1+x$ therefore $$g'(x)=dfrac11+x\g''(x)=-dfrac1(1+x)^2\g'''(x)=dfrac2(1+x)^3\.\.\.\g^(n)(x)=(-1)^n+1dfrac(n-1)!(1+x)^n$$which means that$$g^(n)(0)=(-1)^n+1(n-1)!$$and we have $$ln(1+x)=sum_n=1^inftydfrac(-1)^n+1nx^n$$therefore $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$
Let $g(x)=ln 1+x$ therefore $$g'(x)=dfrac11+x\g''(x)=-dfrac1(1+x)^2\g'''(x)=dfrac2(1+x)^3\.\.\.\g^(n)(x)=(-1)^n+1dfrac(n-1)!(1+x)^n$$which means that$$g^(n)(0)=(-1)^n+1(n-1)!$$and we have $$ln(1+x)=sum_n=1^inftydfrac(-1)^n+1nx^n$$therefore $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$
answered Sep 3 at 14:44
Mostafa Ayaz
10.4k3730
10.4k3730
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