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

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The discussion focuses on expanding the function sin(x) around the point x = π/4 using the McLaurin series. Participants clarify that the hint provided in the homework suggests substituting y for (x - π/4) to simplify the expansion. The correct approach involves applying the sine addition formula, leading to the expression sin(x) = sin(y + π/4) = sin(y)cos(π/4) + cos(y)sin(π/4), which can then be expanded using known Taylor series for sin(y) and cos(y).

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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)?
 
\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))

\sin^{(k)}(x)
means the kth derivative of sine evaluated at x which we know to be
\sin(x+k \, \pi/2)
 
Last edited:
lurflurf said:
\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))
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:
\sin(x) = \sin(y + \pi/4) = \sin(y) \cos(\pi/4) + \cos(y) \sin(\pi/4)
And we presumably know the Taylor series for \sin(y) and \cos(y).
 
^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.
<br /> \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))

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

to give
<br /> \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))
 

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