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

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The discussion revolves around calculating the Maclaurin polynomial of grade 3 for the function cos(ln(1+2x-3x²)). The initial approach suggested expanding the logarithm first, but participants debated whether it would be more effective to expand the cosine function instead. The correct method involves substituting the expansion of ln(1+2x-3x²) into the cosine expansion, focusing on the coefficients for x^0, x^1, x^2, and x^3. Ultimately, the participants confirmed the correct coefficients for x² and x³, leading to a successful resolution of the problem. The conversation highlights the importance of carefully considering the order of expansions in Taylor series calculations.
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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,
$$|\pi\rangle$$
 
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Re: Maclaurin polynom

Petrus said:
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...
 
Re: Maclaurin polynom

Prove It said:
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
34e6sld.png

I did translate it in my question
Regards,
$$|\pi\rangle$$
 
Petrus said:
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 ...
 
ZaidAlyafey said:
It is preferable to first expand cosine ...
Hello Zaid,
I don't understand why i can do that..?

Regards,
$$|\pi\rangle$$
 
Suppose that you have the following :

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

How can you find the Maclaurin expansion of the function ?
 
ZaidAlyafey said:
Suppose that you have the following :

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

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

Excellent . Now substitute the value $$f(x)= \ln(1+2x-3x^2)$$
 
ZaidAlyafey said:
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,
$$|\pi\rangle$$
 
  • #10
Petrus said:
$$1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^{4}}{4!}...$$
what shall i do next?

Regards,
$$|\pi\rangle$$

You should place the powers properly ... Next , expand $$\ln(1+2x-3x^2) $$ .
 
  • #11
ZaidAlyafey said:
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,
$$|\pi\rangle$$
 
  • #12
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

$\chi$ $\sigma$
 
  • #13
Petrus said:
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 $$\cos( \ln(1+2x-3x^2)) $$
 
  • #14
ZaidAlyafey said:
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,
$$|\pi\rangle$$
 
  • #15
Petrus said:
$$1-\frac{\ln(1+2x-3x^2)^2}{2!}+\frac{\ln(1+2x-3x^2)^4}{4!}...$$
what shall i do next?

Regards,
$$|\pi\rangle$$

Petrus said:
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,
$$|\pi\rangle$$

$$\cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...$$
 
  • #16
ZaidAlyafey said:
$$\cos\left( \ln(1+2x-3x^2) \right) =1-\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^2}{2!}+\frac{(2x-3x^2-\frac{(2x-3x^2)^2}{2}+\frac{(2x-3x^2)^3}{3}...)^4}{4!}...$$
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,
$$|\pi\rangle$$
 
  • #17
I think the question is find the coefficients of $$x^0,x^1,x^2,x^3 $$

Do you know how to do that ?
 
  • #18
ZaidAlyafey said:
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,
$$|\pi\rangle$$
 
Last edited:
  • #19
Petrus said:
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 ?
 
  • #20
ZaidAlyafey said:
why , wrong ?
the programe say so :P
 
  • #21
ZaidAlyafey said:
[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 ...
 
  • #22
ZaidAlyafey said:
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,
$$|\pi\rangle$$
 
  • #23
Petrus said:
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 :)
 

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