Correcting Maclaurin Series Coefficient of x^4 | Homework Help

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Homework Help Overview

The discussion revolves around the calculation of the coefficient of x^4 in a Maclaurin series expansion. The original poster believes their coefficient is incorrect, having calculated it as -5/96 instead of the expected -1/96. Participants are examining the steps taken to derive these coefficients.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the accuracy of the derivatives calculated at zero, particularly y''(0), and are asking for clarification on how the derivatives y'''(0) and y''''(0) were obtained. There is also a mention of potential algebra mistakes in the calculations.

Discussion Status

The discussion is ongoing, with participants actively seeking to identify errors in the original poster's calculations. Some guidance has been offered regarding the derivatives, but no consensus has been reached on the correct approach or solution.

Contextual Notes

There is mention of a reference to a textbook problem, and some participants note that the original statement is difficult to read, suggesting that clarity in the problem setup may be an issue.

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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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?
 
Last edited:
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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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.
 
can you check my working again? i checked it many times but i just can't don't know where's my mistake
 
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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