Correcting Maclaurin Series Coefficient of x^4 | Homework Help

In summary, the student is asking for help with a problem where they have found that their coefficient of x^4 is incorrect. They applied the MacLaurin series formula and got a coefficient of -5/96, but the correct answer is -1/96. They are asking for someone to check their work and identify where they made a mistake. They mention having trouble with y''(0) and y'''(0) values and attach a photo, but it seems to be from a different problem. The helper suggests checking the algebra and redoing the problem carefully to find the mistake. The student acknowledges this, but asks for their working to be checked again. The helper reminds them that y''(0) should be -
  • #1
delsoo
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Homework Statement



for this question, i found that my coefficient of x^4 is wrong... after applying the maclaurin series formula, i would get the coefficient of X^4 is -5/96... but the exact ans is -1/96... can anyone check which part is wrong?

Homework Equations





The Attempt at a Solution

 

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  • #2
delsoo said:

Homework Statement



for this question, i found that my coefficient of x^4 is wrong... after applying the maclaurin series formula, i would get the coefficient of X^4 is -5/96... but the exact ans is -1/96... can anyone check which part is wrong?

Homework Equations


The Attempt at a Solution


For one thing y''(0)=(-1/2) not (+1/2). Can you check how you got that? I think you made an algebra mistake. Then can you spell out how you got y'''(0) and y''''(0) a little more clearly?
 
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  • #3
Dick said:
For one thing y''(0)=(-1/2) not (+1/2). Can you check how you got that? I think you made an algebra mistake. Then can you spell out how you got y'''(0) and y''''(0) a little more clearly?

this is actually the question from my book(gt it from my senior), do u mean the proving part, the third term is -1/2 y^2 ?? for y'''(0) and y''''(0) , i differentiate the equation y(d2y/dx2) + (dy/dx)^2 + 1/2y^2 a few times to get y''' and y'''' ... refer to the photo attached...
 

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  • #4
delsoo said:
this is actually the question from my book(gt it from my senior), do u mean the proving part, the third term is -1/2 y^2 ?? for y'''(0) and y''''(0) , i differentiate the equation y(d2y/dx2) + (dy/dx)^2 + 1/2y^2 a few times to get y''' and y'''' ... refer to the photo attached...

The photo you attached seems to be from a different problem and what you have in the original statement is pretty hard to read. I think if you do it again more carefully you should get the correct answer.
 
  • #5
can you check my working again? i checked it many times but i just can't don't know where's my mistake
 
  • #6
delsoo said:
can you check my working again? i checked it many times but i just can't don't know where's my mistake

I told you a long time ago. y''(0) should be -1/2, not +1/2. Didn't you check that?
 
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1. What is a Maclaurin series?

A Maclaurin series is a way to represent a function as an infinite sum of terms, where each term is a polynomial of increasing degree. It is a special case of a Taylor series, where the series is centered at x=0.

2. Why is it important to correct the Maclaurin series coefficient of x^4?

The coefficient of x^4 in the Maclaurin series determines the behavior of the function near the origin. If this coefficient is incorrect, the approximation of the function will be inaccurate and may not match the actual function.

3. How is the Maclaurin series coefficient of x^4 corrected?

The Maclaurin series coefficient of x^4 can be corrected by using the formula for the nth derivative of a function evaluated at x=0, divided by n!. This will give the correct coefficient for the x^4 term in the Maclaurin series.

4. Can the Maclaurin series coefficient of x^4 be negative?

Yes, the Maclaurin series coefficient of x^4 can be negative. This means that the function has a decreasing behavior near the origin, and the curve will be concave downwards.

5. What are some applications of correcting Maclaurin series coefficients?

Correcting Maclaurin series coefficients is important in various fields of science, such as physics, engineering, and economics. It allows for more accurate approximations of functions, which can be used to model and predict real-world phenomena. It is also used in numerical methods to solve differential equations and in optimization problems.

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