Question

How to find the taylor polynomial of degree 5 of f(x)=cos(x)ln(x+1) at x=0?

Answer 1

Idk how to do calc but you can go here if that helps : http://www.wolframalpha.com/input/?i=How%20to%20find%20the%20taylor%20polynomial%20of%20degree%205%20of%20f(x)%3Dcos(x)ln(x%2B1)%20at%20x%3D0%3F&t=crmtb01&f=rc

Answer 2

Thanks but I really need to figure out how to calculate it :(

Answer 3

$$\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k}\\\log (1+x)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^k\\\cos(x)\log(1+x)=\sum_{j=1}^\infty\sum_{k=1}^j\left(\frac{(-1)^k}{(2k)!}x^{2k}\times\frac{(-1)^{j-k}}{j-k}x^{j-k}\right)=\sum_{j=1}^\infty\sum_{k=1}^j\frac{(-1)^j}{(j-k)(2k)!}x^{j+k}$$

For the fifth-degree approximation, we consider:$$\cos(x)\log(1+x)\approx\sum_{j=1}^4\sum_{k=1}^j\frac{(-1)^j}{(j-k)(2k)!}x^{j+k}$$

Answer 4

Smplify that and only keep the terms that are up to fifth-degree

Answer 5

@oldrin.bataku I'm about to start Calc in a while^^ is this complicated once I get to it?

Answer 6

Alternatively, we start off with:$$\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k}\approx1-\frac12x^2+\frac1{24}x^4\\\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}kx^k\approx x-\frac12x^2+\frac13x^3-\frac14x^4+\frac15x^5$$Now multiply out:$$\cos(x)\log(1+x)\approx\left(1-\frac12x^2+\frac1{24}x^4\right)\left(x-\frac12 x^2+\frac13x^3-\frac14x^4+\frac15x^5\right)$$... and keep only terms up to fifht-degree

Answer 7

@doulikepiecauseidont nope this is second-semester calculus anyways :p