Question

Find each point of discontinuity of f , and for each give the value of the point of discontinuity and evaluate the indicated one-sided limits.

Let f(x)=x^2+9/9-x^2

Answer 1

f is discontinuous at x=3 and x=-3
x->3^(-), then f(x)->+inf
x->3^(+), then f(x)->-inf
x->-3^(-), then f(x)->-inf
x->-3^(+), then f(x)->+inf

Answer 2

thanks for the help

Answer 3

hey myininaya do you have a sec to think about something?

Answer 4

maybe its not too hard right
my belly hurts

Answer 5

points discontinuity:
9-x^2=0 which gives x=3 & x=-3

Answer 6

what do you want me to think about

Answer 7

i don't know if it's hard or not but I can't see how to do it right now. I want to expand n!/(n-k)!

Answer 8

write it as a series possibly

Answer 9

ok let me think and stuff
doesn't sound too hard but it might be

Answer 10

i figured it would've been done already but couldn't find an expansion anywhere

Answer 11

i think i should assume n>k

Answer 12

yeah we can assume that

Answer 13

it's the term in front of x^n when you take the kth derivative:
(n)(n-1)(n-2)(n-3)...

Answer 14

n(n-1)(n-2)...(n-k+1)x^(n-k)

Answer 15

if we have
\[\frac{n!}{(n-8)!}=
\frac{n!}{(n-0)!(n-1)(n-2)(n-3)(n-4)....(n-8)}
=\frac{1}{(n-1)(n-2)(n-3)(n-4)....(n-8)}\]
so if we have
\[\frac{n!}{(n-k)!}
=\frac{n!}{(n-(k-k))!(n-(k-(k+1)))...(n-(k-3))(n-(k-2))(n-(k-1))(n-k)}\]
=\[\frac{1}{(n-(k-(k+1)))...(n-(k-3))(n-(k-2))(n-(k-1))(n-k)}\]{}\]

Answer 16

does this help?

Answer 17

my thingy got cutoff lol

Answer 18

(n-8)!=(n-8)(n-9)(n-10)...

Answer 19

yeah it got cut off

Answer 20

where did the (n)(n-1)(n-2).. terms come from

Answer 21

oh wait no i didn't include those terms

Answer 22

i need a representation of the polynomial created by multiplying terms together:

(n)(n-1)(n-2)=(n^2-n)(n-2)=n^3-3n^2+2n

Answer 23

so the first term always has coeff 1. the last term is (-1)^(k+1)(k-1)! the second term is
k(k-1)/2

Answer 24

so i need a general formula for whichever term of the polynomial i need

Answer 25

like what is the coefficient on n^3 when k =6?

Answer 26

\[\frac{5!}{(5-2)!}=\frac{5!}{5!(5-1)(5-2)}=\frac{1}{(5-1)(5-2)}=\frac{1}{4*3}=\frac{1}{12}\]
\[\frac{5!}{3!}=\frac{5*4*3*2*1}{3*2*1}=5*4=20\]
not equal im dumb :(

Answer 27

i think its a harder problem than it sounds at first I cant figure it out

Answer 28

1/12 not equal to 5!/(5-2)!
contradiction im my logic
which makes it non logic
which makes it retarded

Answer 29

haha its just a mistake

Answer 30

which makes me a retard

Answer 31

ive made some bad mistakes in front of a class before

Answer 32

ok sometimes it helps me to work on non general cases before i go to a general case

Answer 33

i hate making mistakes too though

Answer 34

k try looking at n*(n-1)(n-2)(n-3) if you want and see if you can find a nice form for the coefficients maybe

Answer 35

i know when all the terms are the same we can use combinatorics to find the coefficients

Answer 36

so maybe its some extension of that

Answer 37

so we want to write a summation right not
one those weird big pi symbols that mean multiply

Answer 38

summation where each part is a term of the polynomial

Answer 39

i think it will help me find a general solution to this:
\[\sum_{n=0}^{\infty}\frac{n^k}{e^{an}}\]

Answer 40

i know how to do it for specific k and a but I want a general solution. I think there will be a somewhat nice one

Answer 41

did you try putting it in wolfram
i don't think this will be nice looking

Answer 42

with k and a as variables or for a specific value of k and a?

Answer 43

i tried to get maple to solve it generally and it wouldnt do it

Answer 44

the specific solutions are nice though

Answer 45

i dont know how to type it into wolfram

Answer 46

http://www.wolframalpha.com/input/?i=n!%2F(n-k)!

Answer 47

what is a hurwitz lurch transcendent!?

Answer 48

thats the solution apparently haha

Answer 49

lol i have no clue
never heard of it

Answer 50

hmm maybe there isnt a nice solution to this

Answer 51

well depending on what a hurwitz lurch transcendent is i geuss

Answer 52

lol
that series was for n=0
it wasn't even a general case for all n but for k

Answer 53

i can't look at this anymore lol

Answer 54

haha im reading about hurwitz function on wikipedia it's horrible.

Answer 55

allright maybe i will try a different problem

Answer 56

good luck

Answer 57

thanks for looking at it. im off to bed talk to you later :)

Answer 58

i go to bed too :) goodnight