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

  • Thread starter Thread starter bmb2009
  • Start date Start date
  • Tags Tags
    Expansion
bmb2009
Messages
89
Reaction score
0

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
 
Physics news on Phys.org
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))
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top