Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Let \(n\in\mathbb N\). For\[e^xf_n(x)=\sum_{k=1}^\infty\frac{k^nx^k}{\left(k-1\right)!}\]show that \(f_n(x)\) is a polynomial of degree \(n+1\) with integer coefficients. Tricky question.

Meta-math
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
take derivative\[e^x=\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}\]multiply by x\[xe^x=\sum_{k=1}^\infty\frac{x^{k}}{(k-1)!}\]and take derivative like this n times and multiply by x again every time finally we will have something like this\[e^xP(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]
actually better to write\[e^xP_n(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]
lets prove by induction that degree of \(P_n(x)\) is n+1 \[P_1(x)=x+x^2\]lets say degree of \(P_n(x)\) is n+1 and prove that degree of \(P_{n+1}(x)\) is n+2\[P_{n+1}(x)=xe^{-x}[e^xP_n(x)]'=x(P_n^'(x)+P_n(x))\] which is from degree 1+n+1=n+2

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@mukushla can u show me the derivative of |dw:1348828878035:dw|
\[e^x(1+x)=\sum_{k=1}^\infty\frac{kx^{k-1}}{(k-1)!}\]
OH....... NOW, I got it.... THANX @mukushla

Not the answer you are looking for?

Search for more explanations.

Ask your own question