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.









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    up vote
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    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.









    share|cite|improve this question

























      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.









      share|cite|improve this question















      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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      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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          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






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          • @DavideMorgante I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 3 at 14:40










          • Thanks, I didn't realize that I could do it this way.
            – Cesare
            Sep 3 at 14:42

















          up vote
          0
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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
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            2 Answers
            2






            active

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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






            share|cite|improve this answer






















            • @DavideMorgante I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 3 at 14:40










            • Thanks, I didn't realize that I could do it this way.
              – Cesare
              Sep 3 at 14:42














            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






            share|cite|improve this answer






















            • @DavideMorgante I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 3 at 14:40










            • Thanks, I didn't realize that I could do it this way.
              – Cesare
              Sep 3 at 14:42












            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






            share|cite|improve this answer














            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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            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.
              – José Carlos Santos
              Sep 3 at 14:40










            • Thanks, I didn't realize that I could do it this way.
              – Cesare
              Sep 3 at 14:42
















            • @DavideMorgante I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 3 at 14:40










            • Thanks, I didn't realize that I could do it this way.
              – Cesare
              Sep 3 at 14:42















            @DavideMorgante I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 3 at 14:40




            @DavideMorgante I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 3 at 14:40












            Thanks, I didn't realize that I could do it this way.
            – Cesare
            Sep 3 at 14:42




            Thanks, I didn't realize that I could do it this way.
            – Cesare
            Sep 3 at 14:42










            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 $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$






            share|cite|improve this answer
























              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 $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$






              share|cite|improve this answer






















                up vote
                0
                down vote










                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 $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$






                share|cite|improve this answer












                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 $$ln(1+2x^2)=sum_n=1^inftydfrac(-1)^n+1n2^nx^2n=sum_n=1^infty-dfrac(-2)^n+1nx^2n=$$







                share|cite|improve this answer












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                share|cite|improve this answer










                answered Sep 3 at 14:44









                Mostafa Ayaz

                10.4k3730




                10.4k3730



























                     

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