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

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    Maclaurin Polynomial
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Discussion Overview

The discussion revolves around calculating the Maclaurin polynomial of grade 3 for the function $$\cos(\ln(1+2x-3x^2))$$. Participants explore various approaches to derive the polynomial, including the expansion of the logarithmic function and the cosine function, while addressing potential errors in the initial problem statement.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest starting with the Taylor expansion of $$\ln(1+2x-3x^2)$$, while others question the correctness of the initial problem statement.
  • There is a proposal to first expand the cosine function instead of the logarithm, with some participants expressing confusion about this approach.
  • Participants discuss the general form of the Maclaurin expansion and the need to compute derivatives of the function $$\cos(\ln(1+2x-3x^2))$$.
  • There is a focus on finding coefficients for specific powers of $$x$$, with participants attempting to identify contributions from various terms in the expansion.
  • Some participants express uncertainty about the correct method to derive the coefficients for $$x^2$$ and $$x^3$$, leading to further exploration of the terms involved.
  • Corrections and refinements to earlier claims are made, with participants acknowledging mistakes and adjusting their calculations accordingly.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to calculate the Maclaurin polynomial, with multiple competing views on whether to expand the logarithm or the cosine function first. The discussion remains unresolved regarding the most effective method to derive the polynomial.

Contextual Notes

Some participants note the importance of correctly identifying contributions from all terms in the expansion, indicating that earlier calculations may have overlooked certain factors. There is also mention of the potential for infinite series in the expansions, which adds complexity to the problem.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in series expansions, particularly in the context of calculus and mathematical analysis, as well as those exploring the properties of logarithmic and trigonometric functions.

Petrus
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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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