Maclaurin series problems

[sheilaw]

1. Deduce from the Maclaurin series for e^t that
1/x^x= the sum from 0 to infinite of [(-1)^n * (xInx)^n]/n!

2. use the fact that for any positive integers m and n
integral from 0 to 1 of x^m(Inx)^ndx=n!(-1)^n/(m+1)^(n+1) to show that integral from 0 to 1 of dx/x^x =sum form 1 to infinite of 1/n^n

For the #1 question, I found that e^x= the sum from 0 to infinite of x^n /n!, is infinite of x^n /n! =the sum from 0 to infinite of [(-1)^n * (xInx)^n]/n!?
then slove the x?
I dont no how to start #2, anyone help me? Thanks a lot

 

[daon]

\(\displaystyle \frac{1}{x^x} = x^{-x} = e^{ln(x^{-x})} = e^{-xlnx} = \sum_{n=0}^{\infty} \frac{(-xlnx)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(xlnx)^n}{n!}\)

For 2, Let m=n Then

\(\displaystyle \int_0^1 x^m(lnx)^ndx = \int_0^1 (xlnx)^ndx = (-1)^n\frac{n!}{(n+1)^{(n+1)}}\)

Dividing both sides by the numberator on the RHS:

\(\displaystyle \int_0^1(-1)^n \frac{(xlnx)^n}{n!}dx = \frac{1}{(n+1)^{(n+1)}}\)

Since these are equal we know:

\(\displaystyle \sum_{n=0}^{\infty} \int_0^1(-1)^n \frac{(xlnx)^n}{n!}dx = \sum_{n=0}^{\infty} \frac{1}{(n+1)^{(n+1)}}\)

The RHS sum's index can be incremented. Can you take it from here?

 

[sheilaw]

what's RHS stand for?

 

[daon]

Right-hand side.

 

[sheilaw]

i didnt get the part of Dividing both sides by the numberator on the RHS....why they will equals?

 

[daon]

i didnt get the part of Dividing both sides by the numberator on the RHS....why they will equals?

Back to the basics. \(\displaystyle n\in N\) is a constant at this point. That is, it is a set number, like 3, but in general stands for any specific positive number.

If we know, for instance, \(\displaystyle 2x^2 = 4\), then if we do equivilant, valid algebraic operations to both sides, they will remain equal. Thus \(\displaystyle \frac{2x^2}{2} = \frac{4}{2}\). This will work for any non-zero number \(\displaystyle n\), or any non-zero value whatsoever. That is, if \(\displaystyle g(n)\) is a function \(\displaystyle g\) evaluated at a number \(\displaystyle n\) and is non-zero, then \(\displaystyle \frac{2x^2}{g(n)} = \frac{4}{g(n)}\). This is why equations are often compared with balance-beams. You may remove or add the same weight to both sides of a balanced beam and it will remain balanced.



We have \(\displaystyle \int f(x)dx = \frac{(-1)^nn!}{\dots}\)

We can divide both sides by \(\displaystyle (-1)^nn!\)

\(\displaystyle \frac{\int f(x)}{(-1)^nn!} = \frac{\frac{(-1)^nn!}{\dots}}{(-1)^nn!}\)

\(\displaystyle \implies \int \frac{f(x)}{(-1)^nn!} = \frac{1}{\dots}\)

Again, n is aconstant. Constants may be moved into/out of integrals.

Final note: \(\displaystyle \frac{1}{(-1)^n} = (-1)^n\)

 

[sheilaw]

oh, thanks a lot
i know how to do this question~

 

[BigGlenntheHeavy]

\(\displaystyle \int_{0}^{1}x^{m}[ln(x)]^{n}dx \ = \ \frac{(-1)^{n}n!}{(m+1)^{(n+1)}}\)

\(\displaystyle For \ example: \ \int_{0}^{1}x^{7}[ln(x)]^{10}dx \ = \ \frac{14175}{33554432} \ = \ \frac{(-1)^{10}(10!)}{8^{11}}.\)

\(\displaystyle When \ I \ plugged \ the \ above \ integral \ into \ my \ trusty \ TI-89, \ it \ went \ into \ busy \ mode \ for \ about\)

\(\displaystyle \ at \ least \ two \ hours \ or \ more, \ before \ spitting \ out \ the \ answer. \ I \ then \ plugged \ in \ the \ RHS \ of \ the\)

\(\displaystyle \ equation, \ and \ it \ gave \ me \ the \ answer \ practically \ simultaneously.\)

\(\displaystyle The \ secret \ here, \ is \ when \ evaluating \ the \ integral, \ if \ you \ were \ demented \ enough \ to \ so \ attempt, \ is\)

\(\displaystyle to \ note \ when \ using \ integration \ by \ parts, \ (\int udv \ = \ uv \ - \ \int vdu), the \ uv \ always \ goes \ to \ zero,\)

\(\displaystyle leaving \ one \ to \ only \ worry \ about \ the \ end \ integral.\)

\(\displaystyle Note \ also \ that \ the \ exponent \ on \ x \ is \ constant, \ the \ problem \ being \ the \ exponent \ on \ [ln(x)].\)

\(\displaystyle Knowing \ that,\int_{0}^{1}x^{7}[ln(x)]^{10}dx \ = \ (\frac{-10}{8})(\frac{-9}{8})(\frac{-8}{8})(\frac{-7}{8})(\frac{-6}{8})(\frac{-5}{8})(\frac{-4}{8})(\frac{-3}{8})(\frac{-2}{8})(\frac{-1}{8})\int_{0}^{1}x^{7}dx.\)

\(\displaystyle This \ gives \ us \ \frac{(-1)^{10}(10!)}{8^{11}} \ = \ \frac{14175}{33554432}.\)

\(\displaystyle Hence, \ very \ clever.\)