Navigating Maths: Finding Answers to Limit Questions

[sadhu]

i am not used to so much of maths
but sometime it makes my life hell when i don't find the answer to my questions
here one such

Limit(n$$\rightarrow\infty$$){sum(1^p+2^p...n^p)/n^(p+1)}

where p is a constant

 

[CompuChip]

$$\lim_{n \to \infty}\left( \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}}\right) = \lim_{n \to \infty} \left( \frac{1^p}{n^{p+1}} + \frac{2^p}{n^{p+1}} + \cdots + \frac{n^p}{n^{p+1}} \right)$$
What can you say about all the terms separately?

 

[sadhu]

well i just an hour before solved the question
but let's see your method

answer is not another limit, but an simple expression

but tell me how did you write math expressions on reply,the way you have written

 

[CompuChip]

OK, glad you solved it.
To get the fancy math just type [tex] and [/ tex] (but without the space between / and tex) and write LaTeX code in between.

 

[sadhu]

thanks alot...

 

[Gib Z]

What was your answer? I know the answer if |p| is greater or equal to 1, and if p=0, but I can't be stuffed to fill in the gaps.

 

[sadhu]

the answer is true for all real values of p and it is
1/(P+1).

try to telescope the whole series ...

 

[Gib Z]

That answer is not correct..

eg for p=-2, the limit becomes $$\lim_{n \to \infty}\left( \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}}\right) = \lim_{n \to \infty} \left( \frac{1^p}{n^{p+1}} + \frac{2^p}{n^{p+1}} + \cdots + \frac{n^p}{n^{p+1}} \right) = \lim_{n\to \infty} \left( \frac{n}{1^2} + \frac{n}{2^2} + \frac{n}{3^2} ...\right )$$ which obviously diverges...

 

[sadhu]

sorry it was written their that p is the element of R+
i don't know what that mean

then i think it is real positive no.

 

[Gib Z]

O that means p is a Positive real number. I still don't see how the series telescopes, perhaps I am blind today :(

 

[sadhu]

try this stuff

(1+x)^p=??
taking x>=1
rearrange and buy a new telescope to see the sum:rofl:

 

[Gib Z]

I'm still lost :(

 

[sadhu]

ok
(1+x)^(p+1)=x^(p+1) *(1+1/x)^(p+1)

(1+x)^(p+1)-x^(p+1)=?/
do it taking x having different values from 0 to n
then add them

you will get
(1+x)^(p+1)-1=sum of like terms ...(p+1) * sum(x ^(p+1)) +etc terms

do you get it

 

[Gib Z]

Ahh ill try this tomorrow, my headache is killing me. I won't be able to say this in an hour, but =D - I haven't slept all YEAR!

 

[sadhu]

it is not only a bit complicated but a slight long but i don't get any other method except telescoping(diff. of consequetive terms) to do it without using any formulas or bernoulis method which i think is beyond the standard of a high school guy.

 

[Gib Z]

Bernoulli's method? I don't know what you mean, but I did immediately think about using Faulhaber's formula, which gives the sum of the first n k-th powered integers in terms of polynomials whose coefficients are the Bernoulli numbers. I also thought it wouldn't be suitable to apply anything that advanced here either. What exactly is Bernoulli's method?

EDIT: http://mathworld.wolfram.com/BernoullisMethod.html doesn't seem to relate :(

 

[sadhu]

what you said is mentioned as bernoulis method in my college library book .
well thanks for telling me that.

 

[Gib Z]

If you are interested, I do have another method =] It seems to be much easier as well =P

 

[sadhu]

well then please bring it up

i don't know why this post is not highlighted in the index...strange

 

[Gib Z]

It just so happens that when we wish to evaluate the integral $$\int^b_0 x^k dx$$, there are two main methods of doing it- Using the Fundamental theorem of Calculus, Or Integration by Riemann sums. The first method of course yields $$\frac{b^{k+1}}{k+1}$$.

Now when I proved this to myself the first time, I tried to use equal subdivisions of the integral, which previously had always worked for me. You'll see soon why I ran into a problem using that subdivision =] Instead I had to solve the more general integral, with lower bound a, with the subdivisions $q= \left(\frac{b}{a} \right)^{1/n}$.

Well anyway, the problem with equal subdivisions meant that when one approximated the integral with n upper Riemann sums, one ran into;

$$\frac{b}{n} \left( (\frac{b}{n})^k + ( \frac{2b}{n})^k + (\frac{3b}{n})^k ...(\frac{nb}{n})^k \right)$$. Taking out the common factor;

$$\frac{b^{k+1}}{n^{k+1}} ( 1^k + 2^k + 3^k + 4^k...+ n^k)$$.

Now taking limits as n goes to infinity, since we know from the other method that this method should yield the b^(k+1)/(k+1), your limit is k+1. yay

 

[sadhu]

your method seams to be really a good one

but can you explain or give link to riemann sums , i never know that

I am hoping that this method can be used to find sum of other series too...only if can understand it ,thanks for posting

 

[Gib Z]

Have you learned integration yet?

 

[sadhu]

yes i have but i was not taught reimann sum stuff , i told you that I am in high school.

I don't know what is reimann sum but still i can understand this stuff.

the confusion was because when i searched google , text i found mentioned about trapezoidal rule. etc which i know definitely is far beyond the level ,in which i am

again thanks for your idea, it is certainly extremely useful to all the cases

 

[Gib Z]

How did you learn integration without Riemann sums :( ? It should be in every single calculus book, the definition of the integral (at the elementary level) is in terms of Riemann sums..Did you learn integration independently or from school? If independent - Get a reliable textbook. I recommend Courant when your a bit more advanced, right now look through mathwonks thread in the Careers advice section, "Who wants to be a mathematician" for other books.
If from school, they really haven't been teaching things properly :(

This is in no way a poke at your learning by the way, it's just this needs to be fixed.

 

[sadhu]

actually I haven,t learned definite integrals yet ,this is my first year with calculus
,up till now I have done only indefinite integration .that is next year

but i do practice definite integrals in physics , so i know about them ,and what
you said Riemann sum i already know that but i was not knowing its name , and the name is not given in any books only method is there.

 

[Gib Z]

Indefinite integration is a concept brought AFTER definite integrals ...Are you sure your books didnt mention the name of the definition it was giving? Especially when its such a widely known one...

 

[sadhu]

well I think its indefinite which comes first and i am sure it wasnt there

 

[Gib Z]

Well then if your textbook didn't have it there, and you really think indefinite integration comes first, i recommend you brush up on your calculus, get a *good* textbook.

 

[sadhu]

well thanks for your advice , but I am not talking about only my book ,I have seen many magazines and books both in library and in book Stores and even told my friends about it .
all said that definite integration as a limit of a sum is there in next class.
and indefinite integration is more basic & elementary than definite integration.....may be priorities change from region to region

 

[CompuChip]

Of course, it's possible to introduce indefinite integration as anti-derivatives ("find a function whose derivative is a given function") and derive the integration rules from that (e.g. x^2 gives 2x when differentiated, so x should give x^2/2 + C when integrated). Definite integration can be set up by Riemann sums. I think it's a matter of choice which you learn first and which you call "more fundamental".
One could also say that definite integration is more basic and elementary, and indefinite integration is just definite integration with unknown boundary values

 

[Gib Z]

I think I'll open a thread about this, I really don't think that's a very good way to have it taught...unless of course integration is defined from the outset to be the inverse operation of differentiation, and then later on you find out it just so happens to also under the area under the curve.

EDIT: CompuChip has beaten me =]

EDIT: Sigh actually, forgive me please, It is definitely possible to teach things in a different order with different definitions, and though some ways may be harder and be more deceiving to the student about the development of calculus, its still the same. Sorry guys, just my personal taste here.

 

[sadhu]

area under the curve is definite integration , indefinite integration in not area under the curve it is area + value of function at initial point...however i just believe that there is no confusion in taking indefinite integration as an inverse function or operation as you see in cases of inverse operations we need not to define the operation in terms of other simpler operations

we can just define it as the inverse of well defined function.

like you doing division when multiple is known is just like doing integration (indefinite) when its derivative is known

and definite integrals can just use the concept of indefinite integration, and thus it is possible to reverse the relation and define the other in terms of first ...but in fundamental theorem of calculus regarding definite integration ,we cannot proceed unless we know the anti derivative of the function f(b)-f(a)

which itself uses the idea of something being like inverse operation...leaving the method of Riemann...

 

[Gib Z]

Could you perhaps reword the second half of your post? It's a bit confusing.

If we define the definite integral just to be the indefinite integral evaluated at certain points, and the indefinite integral to be the inverse of differentiation, how do we reunite this with the Fundamental theorem of Calculus, and the fact that it gives the area under a curve? Even if we can, this just seems so un-natural.

 

[sadhu]

you see that both are inter related and can be defined in terms of each other

but what important is which one is easier to whom
many authors prefer to introduce the concept of indefinite first and definite later, but reverse can not be wrongif function be f(x)

its anti derivative will be $$\int f(x)*dx$$=F(x)+C

how do we know F(x), clearly by considering its inverse nature with differentiation

now if we want to integrate it between b,a

=F(b)-F(a)this can be represented as
f(b)+c-(F(a)+c)

$$\int f(x)*dx$$ (x=b) - $$\int f(x)*dx$$ (x=a)
however the entire relation can be expressed ,reversing the role of definite with indefinite , still it will be alright
but what is common between both the definitions , it is the reversibility with differentiation as even in definite integrals we need to find F(x) i,e anti derivativewhich not done by any specific process ...

this what i mean to say that both definitions of integrals i.e the relation , is correct , and is just the matter of personal interpretation .......like log(1+x)=the log series

but this does not mean that log is defined in that form , the definition of log is different and it is just the relation....