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Maclaurin polynomial

Petrus

Well-known member
Feb 21, 2013
739
Calculate MacLaurin-polynom of grade 3 to function \(\displaystyle \cos(\ln(1+2x-3x^2))\)


if i make Taylor expansion in that ln first is this correct
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)
Is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: Maclaurin polynom

Calculate MacLaurin-polynom of grade 3 to function \(\displaystyle \cos(\ln(1+2x-3x^2))\)


if i make Taylor expansion in that ln first is this correct
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)
Is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
Are you sure you copied the question down right? I have a feeling it might actually be \(\displaystyle \displaystyle \begin{align*} \ln{ \left[ \cos{ \left( 1 + 2x - 3x^2 \right) } \right] } \end{align*}\)? That would be much easier because the derivative is much more simple to find a series for...
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Maclaurin polynom

Are you sure you copied the question down right? I have a feeling it might actually be \(\displaystyle \displaystyle \begin{align*} \ln{ \left[ \cos{ \left( 1 + 2x - 3x^2 \right) } \right] } \end{align*}\)? That would be much easier because the derivative is much more simple to find a series for...
I am sure I did copy it

I did translate it in my question
Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Calculate MacLaurin-polynom of grade 3 to function \(\displaystyle \cos(\ln(1+2x-3x^2))\)


if i make Taylor expansion in that ln first is this correct
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)
Is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
It is preferable to first expand cosine ...
 

Petrus

Well-known member
Feb 21, 2013
739
It is preferable to first expand cosine ...
Hello Zaid,
I don't understand why i can do that..?

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Suppose that you have the following :

\(\displaystyle g(x)= \cos(f(x)) \)

How can you find the Maclaurin expansion of the function ?
 

Petrus

Well-known member
Feb 21, 2013
739
Suppose that you have the following :

\(\displaystyle g(x)= \cos(f(x)) \)

How can you find the Maclaurin expansion of the function ?
\(\displaystyle 1-\frac{f(x)^2}{2!}+\frac{f(x)^4}{4!}...\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle 1-\frac{f(x)^2}{2!}+\frac{f(x)^4}{4!}...\)
Excellent . Now substitute the value \(\displaystyle f(x)= \ln(1+2x-3x^2)\)
 

Petrus

Well-known member
Feb 21, 2013
739
Excellent . Now substitute the value \(\displaystyle f(x)= \ln(1+2x-3x^2)\)
\(\displaystyle 1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...\)
what shall i do next?

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle 1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^{4}}{4!}...\)
what shall i do next?

Regards,
\(\displaystyle |\pi\rangle\)
You should place the powers properly ... Next , expand \(\displaystyle \ln(1+2x-3x^2) \) .
 

Petrus

Well-known member
Feb 21, 2013
739
You should place the powers properly ... Next , expand \(\displaystyle \ln(1+2x-3x^2) \) .
Is that correct what I did in first post where I expand ln
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)

Regards,
\(\displaystyle |\pi\rangle\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
May be it is necessary to remember how is written the McLaurin expansion of a function...

$$ f(x)= f(0) + f^{\ '} (0)\ x + \frac{f^{\ ''} (0)}{2}\ x^{2} + ... + \frac{f^{(n)} (0)}{n!}\ x^{n} + ...\ (1)$$

What is f(x) in this case?... is...

$$ f(x)= \cos \{\ln (1 + 2\ x - 3\ x^{2})\}\ (2)$$

... so that the first term of (1) is $f(0)= 1$. The successive step is the computation of the derivative of (2)...

$$ f^{\ '} (x) = - \sin \{\ln (1 + 2\ x - 3\ x^{2})\}\ \frac{2 - 6\ x}{1 + 2\ x - 3\ x^{2}}\ (3)$$

... so that the second term of (1) is $f^{\ '} (0)\ x = 0$. The sucessive step is the computation of the derivative of (3) and then of the third term of (1)...

All right?...

Kind regards

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Is that correct what I did in first post where I expand ln
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)

Regards,
\(\displaystyle |\pi\rangle\)
Now substitute the expansion in the expansion of \(\displaystyle \cos( \ln(1+2x-3x^2)) \)
 

Petrus

Well-known member
Feb 21, 2013
739
Now substitute the expansion in the expansion of \(\displaystyle \cos( \ln(1+2x-3x^2)) \)
hmm... It will go on to infinity and I don't see how I shall answer the question?

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle 1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...\)
what shall i do next?

Regards,
\(\displaystyle |\pi\rangle\)
Is that correct what I did in first post where I expand ln
\(\displaystyle \ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...\)

Regards,
\(\displaystyle |\pi\rangle\)
\(\displaystyle \cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...\)
 

Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle \cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...\)
Well it was like I did write :S I will try put it to the programe we are supposed to input the answer somehow. Thanks for taking your time and helping me as well!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think the question is find the coefficients of \(\displaystyle x^0,x^1,x^2,x^3 \)

Do you know how to do that ?
 

Petrus

Well-known member
Feb 21, 2013
739
I think the question is find the coefficients of \(\displaystyle x^0,x^1,x^2,x^3 \)

Do you know how to do that ?
I don't know the 'trick' if there is one... I try find all which will give me x^2 etc what I know that we got \(\displaystyle 1x^0,0x\)
If I think correct we should only get \(\displaystyle x^2,x^3\) from
\(\displaystyle -\frac{(2x-3x^2)^2}{2!}\) and I get \(\displaystyle -\frac{4}{2!}x^2,\frac{12}{2!}x^3\) but that is not correct so I think wrong

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I don't know the 'trick' if there is one... I try find all which will give me x^2 etc what I know that we got \(\displaystyle 1x^0,0x\)
If I think correct we should only get \(\displaystyle x^2,x^3\) from
\(\displaystyle -\frac{(2x-3x^2)^2}{2!}\) and I get \(\displaystyle -\frac{4}{2!}x^2,\frac{12}{2!}x^3\) but that is not correct so I think wrong

Regards,
\(\displaystyle |\pi\rangle\)
why , wrong ?
 

Petrus

Well-known member
Feb 21, 2013
739

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2})^2}{2!}\)
I think we missed other things , since the third term also contributes ...
 

Petrus

Well-known member
Feb 21, 2013
739
I think we missed other things , since the third term also contributes ...
ofcourse....! I did just not think correct... now I get correct! Thanks for the help
\(\displaystyle -\frac{4}{2}x^2,\frac{20}{2}x^3\)
Thanks once again!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
ofcourse....! I did just not think correct... now I get correct! Thanks for the help
\(\displaystyle -\frac{4}{2}x^2,\frac{20}{2}x^3\)
Thanks once again!:)

Regards,
\(\displaystyle |\pi\rangle\)
Finally , that is good . Happy to help :)