MHB Calc MacLaurin Polynom Grade 3 for \cos(\ln(1+2x-3x^2))

[Petrus]

Calculate MacLaurin-polynom of grade 3 to function $$\cos(\ln(1+2x-3x^2))$$if i make Taylor expansion in that ln first is this correct
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$
Is that correct?

Regards,
$$|\pi\rangle$$

 

[Prove It]

Re: Maclaurin polynom

Calculate MacLaurin-polynom of grade 3 to function $$\cos(\ln(1+2x-3x^2))$$if i make Taylor expansion in that ln first is this correct
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$
Is that correct?

Regards,
$$|\pi\rangle$$

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

 

[Petrus]

Re: Maclaurin polynom

Are you sure you copied the question down right? I have a feeling it might actually be [math]\displaystyle \begin{align*} \ln{ \left[ \cos{ \left( 1 + 2x - 3x^2 \right) } \right] } \end{align*}[/math]? 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,
$$|\pi\rangle$$

 

[alyafey22]

Calculate MacLaurin-polynom of grade 3 to function $$\cos(\ln(1+2x-3x^2))$$if i make Taylor expansion in that ln first is this correct
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$
Is that correct?

Regards,
$$|\pi\rangle$$

It is preferable to first expand cosine ...

 

[Petrus]

It is preferable to first expand cosine ...

Hello Zaid,
I don't understand why i can do that..?

Regards,
$$|\pi\rangle$$

 

[alyafey22]

Suppose that you have the following :

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

How can you find the Maclaurin expansion of the function ?

 

[Petrus]

Suppose that you have the following :

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

How can you find the Maclaurin expansion of the function ?

$$1-\frac{f(x)^2}{2!}+\frac{f(x)^4}{4!}...$$

 

[alyafey22]

$$1-\frac{f(x)^2}{2!}+\frac{f(x)^4}{4!}...$$

Excellent . Now substitute the value $$f(x)= \ln(1+2x-3x^2)$$

 

[Petrus]

Excellent . Now substitute the value $$f(x)= \ln(1+2x-3x^2)$$

$$1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...$$
what shall i do next?

Regards,
$$|\pi\rangle$$

 

[alyafey22]

$$1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^{4}}{4!}...$$
what shall i do next?

Regards,
$$|\pi\rangle$$

You should place the powers properly ... Next , expand $$\ln(1+2x-3x^2) $$ .

 

[Petrus]

You should place the powers properly ... Next , expand $$\ln(1+2x-3x^2) $$ .

Is that correct what I did in first post where I expand ln
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$

Regards,
$$|\pi\rangle$$

 

[chisigma]

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

$\chi$ $\sigma$

 

[alyafey22]

Is that correct what I did in first post where I expand ln
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$

Regards,
$$|\pi\rangle$$

Now substitute the expansion in the expansion of $$\cos( \ln(1+2x-3x^2)) $$

 

[Petrus]

Now substitute the expansion in the expansion of $$\cos( \ln(1+2x-3x^2)) $$

hmm... It will go on to infinity and I don't see how I shall answer the question?

Regards,
$$|\pi\rangle$$

 

[alyafey22]

$$1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...$$
what shall i do next?

Regards,
$$|\pi\rangle$$

Is that correct what I did in first post where I expand ln
$$\ln(1+2x-3x^2)=2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...$$

Regards,
$$|\pi\rangle$$

$$\cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...$$

 

[Petrus]

$$\cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...$$

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,
$$|\pi\rangle$$

 

[alyafey22]

I think the question is find the coefficients of $$x^0,x^1,x^2,x^3 $$

Do you know how to do that ?

 

[Petrus]

I think the question is find the coefficients of $$x^0,x^1,x^2,x^3 $$

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 $$1x^0,0x$$
If I think correct we should only get $$x^2,x^3$$ from
$$-\frac{(2x-3x^2)^2}{2!}$$ and I get $$-\frac{4}{2!}x^2,\frac{12}{2!}x^3$$ but that is not correct so I think wrong

Regards,
$$|\pi\rangle$$

 

[alyafey22]

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 $$1x^0,0x$$
If I think correct we should only get $$x^2,x^3$$ from
$$-\frac{(2x-3x^2)^2}{2!}$$ and I get $$-\frac{4}{2!}x^2,\frac{12}{2!}x^3$$ but that is not correct so I think wrong

Regards,
$$|\pi\rangle$$

why , wrong ?

 

[Petrus]

why , wrong ?

the programe say so :P

 

[alyafey22]

[math] \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]

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
$$-\frac{4}{2}x^2,\frac{20}{2}x^3$$
Thanks once again!:)

Regards,
$$|\pi\rangle$$

 

[alyafey22]

ofcourse...! I did just not think correct... now I get correct! Thanks for the help
$$-\frac{4}{2}x^2,\frac{20}{2}x^3$$
Thanks once again!:)

Regards,
$$|\pi\rangle$$

Finally , that is good . Happy to help :)