Taylor Polynomial for f(x)=sec(x)

  • Thread starter Thread starter ehilge
  • Start date Start date
  • Tags Tags
    Polynomial Taylor
ehilge
Messages
160
Reaction score
0
Hey all, so I need to find 4th degree taylor polynomial of f(x)=sec(x) centered at c=0

Can I just use substitution to find the answer since sec(x) = 1/cos(x) and I know the taylor series for cos(x). I guess, essentially, can I take the reciprocal of the taylor series of cosx to get sec(x). (in the same way for sin ((x^3)) you can just plug in x^3 wherever there is an x in the taylor series)

I know if this doesn't work I can keep taking the derivatives of secx and plug everything in and expand it out and all of that jazz. I did start to take the derivatives of sec(x) before I realized that it got excessively complicated and cumbersome.

So if my original idea doesn't work, what would be a better way to solve the problem without having to differentiate a whole bunch?

Thanks!
 
Physics news on Phys.org
Do not spam your questions -- ask them once.
 
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