Expand sinx about pi/4 with McLaurin series | Simple Homework Solution

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Homework Help Overview

The problem involves expanding the sine function, specifically sin(x), about the point x = π/4 using the McLaurin series. The original poster introduces a hint suggesting to represent sin(x) as sin(y + π/4) with y being a small variable, which raises questions about the implications of this substitution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's attempt to apply the McLaurin expansion directly and express confusion regarding the hint. Some suggest substituting y for (x - π/4) as a simplification, while others question the validity of the initial expansion due to its inconsistency with the value of sin(π/4).

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants have provided guidance on how to approach the hint, while others are questioning the assumptions and definitions involved in the problem setup.

Contextual Notes

There is a noted concern regarding the correctness of the expansion when evaluated at the point of interest, as well as the implications of using different variables in the series expansion. The participants are navigating through these constraints without reaching a definitive conclusion.

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Homework Statement



Expand sinx about the point x= pi/4. Hint: Represent the function as sinx= sin(y+pi/4) and assume y to be small



Homework Equations





The Attempt at a Solution



I thought the problem was simply asking to expand sinx with the McLauran expansions about the point pi/4 and get something like... (x-pi/4) - ((x-pi/4)^3)/3! + ((x-pi/4)^5)/5!...so on and so forth.

But the hint throws me off? What does that mean? Any help?

Thanks
 
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The hint is telling you to do precisely what you thought. It's just trying to make life simpler by substituting y for (x-pi/4) everywhere.
 
Your expansion clearly cannot be correct: if you plug in x = pi/4, you get 0. But sin(pi/4) is not 0.

What happens if you apply a trig identity to sin(y + pi/4)?
 
[tex]\sin(x+h) \sim \sum^\infty_{k=0} \frac{h^k}{k!} \sin^{(k)}(x)=\sum^\infty_{k=0} \frac{h^k}{k!} \sin(x+k \, \pi/2)=\sum^\infty_{k=0} \frac{h^k}{k!} (\sin(x)\cos(k \, \pi/2)+\cos(x)\sin(k \, \pi/2))[/tex]

[tex]\sin^{(k)}(x)[/tex]
means the kth derivative of sine evaluated at x which we know to be
[tex]\sin(x+k \, \pi/2)[/tex]
 
Last edited:
lurflurf said:
[tex]\sin(x+h) \sim \sum^\infty_{k=0} \frac{h^k}{k!} \sin^{(k)}(x)=\sum^\infty_{k=0} \frac{h^k}{k!} \sin(x+k \, \pi/2)=\sum^\infty_{k=0} \frac{h^k}{k!} (\sin(x)\cos(k \, \pi/2)+\cos(x)\sin(k \, \pi/2))[/tex]
I'm not sure what good that does. This isn't a Taylor series in x. I guess it's a Taylor series in h, but centered at h=0, which isn't what is asked for.

I think the hint is intended to lead to the following:
[tex]\sin(x) = \sin(y + \pi/4) = \sin(y) \cos(\pi/4) + \cos(y) \sin(\pi/4)[/tex]
And we presumably know the Taylor series for [itex]\sin(y)[/itex] and [itex]\cos(y)[/itex].
 
^It does not matter what variables are used.
In general h is not zero that is an uninteresting case.

These all mean exactly the same thing.
[tex] \sin(x+h) \sim \sum^\infty_{k=0} <br /> \frac{h^k}{k!} \sin^{(k)}(x)=\sum^\infty_{k=0} <br /> \frac{h^k}{k!} \sin(x+k \, \pi/2)=\sum^\infty_{k=0} <br /> \frac{h^k}{k!} (\sin(x)\cos(k \, \pi/2)+\cos(x)\sin(k \, \pi/2)) \\<br /> \sin(a+b) \sim \sum^\infty_{n=0} <br /> \frac{b^n}{n!} \sin^{(n)}(a)=\sum^\infty_{n=0} <br /> \frac{b^n}{n!} \sin(a+n \, \pi/2)=\sum^\infty_{n=0} <br /> \frac{b^n}{n!} (\sin(a)\cos(n \, \pi/2)+\cos(a)\sin(n \, \pi/2)) \\<br /> \sin(\mathrm{rock}+\mathrm{paper}) \sim \sum^\infty_{\mathrm{scissors}=0} <br /> \frac{\mathrm{paper}^\mathrm{scissors}}{\mathrm{scissors}!} \sin^{(\mathrm{scissors})}(\mathrm{rock})=<br /> \sum^\infty_{\mathrm{scissors}=0} <br /> \frac{\mathrm{paper}^\mathrm{scissors}}{\mathrm{scissors}!} \sin(\mathrm{rock}+\mathrm{scissors} \, \pi/2)=\sum^\infty_{\mathrm{scissors}=0} <br /> \frac{\mathrm{paper}^\mathrm{scissors}}{\mathrm{scissors}!} (\sin(\mathrm{rock})\cos(\mathrm{scissors} \, \pi/2)+\cos(\mathrm{rock})\sin(\mathrm{scissors} \, \pi/2))[/tex]

In the given exercise we can take
x=rock+paper
y=paper
pi/4=rock
k=scissors

to give
[tex] \sin(x)=\sin(\pi/4+y) \sim \sum^\infty_{k=0} <br /> \frac{y^k}{k!} \sin^{(k)}(\pi/4)=\sum^\infty_{k=0} <br /> \frac{y^k}{y!} \sin(\pi/4+k \, \pi/2)=\sum^\infty_{k=0} <br /> \frac{y^k}{k!} (\sin(\pi/4)\cos(k \, \pi/2)+\cos(\pi/4)\sin(k \, \pi/2))[/tex]
 

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