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

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Plugging in x=pi/4 we get\sin(pi/4)=\sin(\pi/4+y) \sim \sum^\infty_{k=0} \frac{y^k}{k!} \sin^{(k)}(\pi/4)=\sum^\infty_{k=0} \frac{y^k}{y!} \sin(\pi/4+k \, \pi/2)=\sum^\infty_{k=0} \frac{y^k}{k!} (\sin(\pi/4)\cos(k \, \pi/2)+\cos(\pi/4)\sin(k \, \pi/2)) which
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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



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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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  • #2
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.
 
  • #3
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)?
 
  • #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]
 
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  • #5
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].
 
  • #6
^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}
\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(a+b) \sim \sum^\infty_{n=0}
\frac{b^n}{n!} \sin^{(n)}(a)=\sum^\infty_{n=0}
\frac{b^n}{n!} \sin(a+n \, \pi/2)=\sum^\infty_{n=0}
\frac{b^n}{n!} (\sin(a)\cos(n \, \pi/2)+\cos(a)\sin(n \, \pi/2)) \\
\sin(\mathrm{rock}+\mathrm{paper}) \sim \sum^\infty_{\mathrm{scissors}=0}
\frac{\mathrm{paper}^\mathrm{scissors}}{\mathrm{scissors}!} \sin^{(\mathrm{scissors})}(\mathrm{rock})=
\sum^\infty_{\mathrm{scissors}=0}
\frac{\mathrm{paper}^\mathrm{scissors}}{\mathrm{scissors}!} \sin(\mathrm{rock}+\mathrm{scissors} \, \pi/2)=\sum^\infty_{\mathrm{scissors}=0}
\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}
\frac{y^k}{k!} \sin^{(k)}(\pi/4)=\sum^\infty_{k=0}
\frac{y^k}{y!} \sin(\pi/4+k \, \pi/2)=\sum^\infty_{k=0}
\frac{y^k}{k!} (\sin(\pi/4)\cos(k \, \pi/2)+\cos(\pi/4)\sin(k \, \pi/2))[/tex]
 

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

1. What is the Simple Sinx Expansion?

The Simple Sinx Expansion is a mathematical concept used to represent the sine function as a power series. It is a way to express the sine function as an infinite sum of terms, where each term is a polynomial of increasing degree.

2. How is the Simple Sinx Expansion calculated?

The Simple Sinx Expansion is calculated using the Maclaurin series, which is a special case of the Taylor series. The Maclaurin series is a representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at 0.

3. Why is the Simple Sinx Expansion useful?

The Simple Sinx Expansion can be useful in various mathematical and scientific applications, such as solving differential equations, performing integration, and approximating the value of the sine function at different points. It also helps in understanding the behavior of the sine function and its properties.

4. Can the Simple Sinx Expansion be used for other trigonometric functions?

Yes, the concept of power series and Taylor/Maclaurin series can be applied to other trigonometric functions, such as cosine, tangent, and their inverse functions. The Simple Sinx Expansion is just a specific example of this general concept.

5. What are the limitations of the Simple Sinx Expansion?

The Simple Sinx Expansion is an approximation of the sine function and is only valid for certain values of x. As the number of terms in the expansion increases, the accuracy of the approximation also increases. However, for large values of x, the Simple Sinx Expansion may not provide an accurate representation of the sine function.

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