$n$th derivative of $\frac 1{f(x)}$

Is there a closed-form solution for $\frac {d^n}{dx^n}\frac 1{f(x)}$? I’ve looked at the first five derivatives in search of some pattern, but I can’t identify anything strong enough to give a closed formula. Answer It’s not exactly a closed form, but Faà di Bruno’s formula $$\frac{d^n}{dx^n} f(g(x))= \sum_{k_1 + 2k_2 + \ldots + n k_n … Read more

n2(n2−1)(n2−4)n^2(n^2-1)(n^2-4) is always divisible by 360 (n>2,n∈N)(n>2,n\in \mathbb{N})

How does one prove that n2(n2−1)(n2−4) is always divisible by 360? (n>2,n∈N) I explain my own way: You can factorize it and get n2(n−1)(n+1)(n−2)(n+2). Then change the condition (n>2,n∈N) into (n>0,n∈N) that is actually equal to (n∈N). Now the statement changes into : n(n+1)(n+2)2(n+3)(n+4) Then I factorized 360 and got 32⋅23⋅5. I don’t know how … Read more

Erroneously Finding the Lagrange Error Bound

Consider $f(x) = \sin(5x + \pi/4)$ and let $P(x)$ be the third-degree Taylor polynomial for $f$ about $0$. I am asked to find the Lagrange error bound to show that $|(f(1/10) – P(1/10))| < 1/100$. Because $P(x)$ is a third-degree polynomial, I know the difference is in the fourth degree term. So I found the … Read more

General solution of the equation sin2015(x)+cos2015(x)=1∀x∈R\sin^{2015}(x)+\cos^{2015}(x) = 1\; \forall \; x\in \mathbb{R}

Calculation of General solution of the equation sin2015(x)+cos2015(x)=1∀x∈R. MyTry: We can Write the equation as sin2015(x)+cos2015(x)≤sin2(x)+cos2(x)=1. And equality hold when sin(x)=1 and cos(x)=0. So we get x=nπ+(−1)n⋅π2 and x=2nπ±π2. Now How can I calculate common solution of sin(x)=0 and cos(x)=0. Please help me. Thanks. Answer Your approach seems correct although sin(x)=cos(x)=0 is just nonsense. You … Read more

calculating a limit of sequence

Can some one help to show that for a>0 lim My tries : I was not able to use riemann sum for calculating. For the special case a=1 the RHS has limit 1 and for the LHS for a=1 we can write \frac{n}{n+1} \le \frac{1}{n+1}+\frac{1}{n+\frac{1}{2}}+…+\frac{1}{n+\frac{1}{n}} \le \frac{n}{n+\frac{1}{n}} so the limit of LHS is 1 . … Read more

Prove cc satisfies the integral

If f:[0,1]→R is continuous, show that there exists c∈[0,1] such that f(c)=∫102tf(t)dt. So it’s pretty clear to me that I have to use Intermediate Value Theorem and Cauchy-Schwarz inequality but I can’t quite get the trick done. Any help appreciated. Answer Since f is continuous there exist a,b∈[0,1] such that f(a)≤f(x)≤f(b),∀x∈[0,1]. Now, t∈[0,1]⟹tf(a)≤tf(x)≤tf(b). So 2f(a)∫10tdt≤2∫10tf(t)dt≤2f(b)∫10tdt. … Read more