# Maclaurin polynomial

#### Petrus

##### Well-known member
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
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
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
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
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
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
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
$$\displaystyle 1-\frac{f(x)^2}{2!}+\frac{f(x)^4}{4!}...$$
Excellent . Now substitute the value $$\displaystyle f(x)= \ln(1+2x-3x^2)$$

#### Petrus

##### Well-known member
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
$$\displaystyle 1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^{4}}{4!}...$$
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
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
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
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
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
$$\displaystyle 1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...$$
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
$$\displaystyle \cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...$$
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
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
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
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
why , wrong ?
the programe say so #### ZaidAlyafey

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

#### Petrus

##### Well-known member
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
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 