### Integration exercises

Calculus Made Easy has a couple of exercises at the end of each chapter. In the chapter introducing Integration, I thought I understood the lesson, but I can't make a dent on the exercises:

1) Find the ultimate sum of 2/3 + 1/3 + 1/6 + 1/12 + 1/24 + etc.

How do you solve this?

Michael B
March 6th, 2007 1:50pm

2/3 + 1/3 + 1/6 + 1/12 + 1/24.. = 1/3 (2 + 1 + 1/2 + 1/4 + 1/8..)

= 1 + 1/3(1/2 + 1/4 + 1/8 +...) = 1 + 1/3 = 4/3

since (1/2 + 1/4 + 1/8 +...) -> 1

LeMonde
March 6th, 2007 1:55pm

Actually, that's not integration, that's a discrete infinite sum.

2/3 + 1/3 + 1/6 + 1/12 + 1/24 + etc

2/3 + 1/3 = 1

multiply by 3 and by 1/3

1/6 + 1/12 + 1/24 + etc = 1/3 (1/2 + 1/4 + 1/8 +...)

1/2 + 1/4 + 1/8 + 1/16 + .... --> 1

you have 1/3 * (sum converge to 1) = 1/3

final sum is 1 + 1/3

LeMonde
March 6th, 2007 2:14pm

Each fraction in the series infinitesmally approaches 0.

के. जे.
March 6th, 2007 2:18pm

That is, in (1/2 + 1/4 + 1/8 +...)

के. जे.
March 6th, 2007 2:18pm

"multiply by 3 and by 1/3"

That part, where do you get that from?

...but their sums infinitesimally approach 1.

JoC
March 6th, 2007 2:22pm

This is a stupid exercise for a chapter introducing integration. It's there for math teachers to get off on seeing their gifted students figure this out by themselves.

It's inappropriate for a low level math course like calculus. Math is commonly hated because the teachers can't stop themselves from pulling shit like this, that makes most students feel stupid banging their head against the wall in frustration.

no label
March 6th, 2007 2:26pm

It's a geometric progression. Each successive term is half the previous term. Use the geometric progression formula:

Sum = a(1-r^n)/(1-r)

In this case set n to infinity giving

Sum = a/(1-r)

since r = 0.5 and 0.5^(infinity) = 0.

bon vivant
March 6th, 2007 2:26pm

> Each fraction in the series infinitesmally approaches 0.

more math!!! we need more math!!

What about this series .... 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + ....

Doesn't "Each fraction in the series infinitesmally approaches 0" here as well? Is the statement by itself particularly helpful in solving the puzzle?

strawberry snowflake
March 6th, 2007 2:27pm

Strawberry, yes. Read Bon Vivant's post.

के. जे.
March 6th, 2007 2:29pm

A harmonic series is not a geometric progression.

But a geometric progression *is* a geometric progression.

bon vivant
March 6th, 2007 2:31pm

..and the good doctor's post too!

के. जे.
March 6th, 2007 2:31pm

I know diddly-squat about things like 1/2 + 1/4 + 1/8 +...

I let the neighbors kid solve it. He is very smart.

He is twelve years old. He does all the computer and

programming I need for my business.

LeMonde
March 6th, 2007 2:32pm

Of course, strawberry's series is not a geometric progression, and Aaron was just pointing that out. I wholly agree with Aaron.

bon vivant
March 6th, 2007 2:33pm

> Each fraction in the series infinitesmally approaches 0.

Is not a statement about geometric progressions, but a more general sort. Hence my calling it out. :-P

strawberry snowflake
March 6th, 2007 2:37pm

The solution to this:

2/3 + 1/3 + 1/6 + 1/12 + 1/24 + etc

only strikes me because I can reason out:

"1/3 + 1/6 + 1/12 + 1/14 + ..."

approaches 2/3. You can see it on paper. Or Python:

d = 0

n = 3.0

for i in range(100):

d += 1.0/n

n *= 2.0

d

0.66666666666666663

Thusly,

2/3 + 2/3 = 1 1/3

I think I'm missing the point?

And generalising is vital to understanding.

के. जे.
March 6th, 2007 2:40pm

Infinite series, and summations or series, and convergent and non-convergent series are all important topics related to calculus and treated in calculus text books. That's how the question quoted in the first post came about.

In this case the answer is to recognise a geometric progression. The book should have geometric progression in the index, and should give a treatment of geometric progressions somewhere in the text.

bon vivant
March 6th, 2007 2:48pm

//summations *of* series//

Damn the missing edit feature.

bon vivant
March 6th, 2007 2:49pm

> And generalising is vital to understanding.

But your statement is generically false. Knowing that the terms in a sequence approach 0 does not imply their sum converges, so does not help the OP out particularly at all.

Really, this is like whoever throwing a bunch of epsilons and deltas around yesterday and calling it a proof. Nope. Sorry. Try again.

strawberry snowflake
March 6th, 2007 2:50pm

>> I think I'm missing the point?

Not at all. A summation is a summation. You've used addition (your For loop). The textbook wants alegebraic application. I've been corrected by some in the aviation field that an algorithmic loop does not suffice for some critical applications.

के. जे.
March 6th, 2007 2:50pm

Michael, your last post is correct.

Rick, try writing better English
March 6th, 2007 2:51pm

>> But your statement is generically false. Knowing that the terms in a sequence approach 0 does not imply their sum converges, so does not help the OP out particularly at all.

I know it does not imply. But the purpose of text book is to learn Calculus. Not solve problems. You may have a different (and 100% valid) approach to learn it by learning the 'How' first. And then the 'Why'. I try to go the other way round.

>>..Nope. Sorry. Try again.

I will.

के. जे.
March 6th, 2007 2:54pm

"Really, this is like whoever throwing a bunch of epsilons and deltas around yesterday and calling it a proof. Nope. Sorry. Try again."

That was me. And it's not a bunch of opsilons and deltas. It's exactly a definition of limit that can be applied to the derivitive.

Get an Analysis textbook.

Rick, try writing better English
March 6th, 2007 2:55pm

Well, what is vital to understanding is reducing the amount of already chewed information.

Presumably the OP understands that if the terms don't get smaller, his task is Sisyphean. Pointing out that the terms need to get smaller makes it seem that there is nothing else to know about the series, which isn't true. A teacher can make sins of omission, don't you think?

strawberry snowflake
March 6th, 2007 3:02pm

"""Not at all. A summation is a summation. You've used addition (your For loop). The textbook wants alegebraic application."""

I resorted to a for loop because that's how I know how to do summations.

What's the algebraic solution?

Michael, Bon Vivant has said it earlier.

"In this case the answer is to recognise a geometric progression. The book should have geometric progression in the index, and should give a treatment of geometric progressions somewhere in the text."

के. जे.
March 6th, 2007 3:04pm

> Get an Analysis textbook.

Was your statement a proof as you stated or a definition of the meaning of L or a or some property of f (continuity perhaps)? Or a statement of the intent to prove, yet with no follow up?

"For all men there exists a woman such that if the distance between man and woman is smaller than 'L' implies they are married" is not a proof.

Though in some cultures it may be a definition.

strawberry snowflake
March 6th, 2007 3:43pm

Ah, के. जे., perhaps you're correct, there is some merit to pointing out the obvious. I see from subsequent posts, that a man armed with but a for-loop will iterate thru every problem in the same way.

strawberry snowflake
March 6th, 2007 3:48pm

I see "LeMonde" got the answer right.

PHD GeeK
March 6th, 2007 3:56pm

# learning is fun

import math

class geomprog:

'''geomprog(a,r): initialize with scale factor and common ratio'''

def __init__(self,a,r):

self.a = a

self.r = r

def iterator(self):

n = 0

while 1:

yield self.termN(n)

n += 1

def termN(self,n):

if self.a is None:

raise ValueError,"set scale factor please"

if self.r is None:

raise ValueError,"set common ratio please"

return (self.a * math.pow(self.r,n))

if __name__ is '__main__':

gp = geomprog(1,1.01)

print '183rd term is: ',gp.termN(183)

for i in gp.iterator():

print i

correction:

if __name__ == '__main__':

"Was your statement a proof as you stated or a definition of the meaning of L or a or some property of f (continuity perhaps)? Or a statement of the intent to prove, yet with no follow up?"

∀ε>0 ∃δ>0 ∋ 0<|x-a|<δ⇒|ƒ(x)-L|<ε

This means the limit of f(x) at a is L.

f(x) is continuous at a when (I forgot, need to look it up)

1) The limit exists.

2) The limit = f(a)

There is no number that is as close as not but not closer to zero, just like there is not a real number called positive and negative infinity.

Rick, try writing better English
March 6th, 2007 7:46pm

I think what you posted is a condition on the existence of the limit. *If* it is true for a particular function f() and a particular value a, then the limit exists and is L.

The statement itself is not actually a proof. The proof that it is (or is not) true for any given function of your choice near some value a in the function domain involves work that is not shown in your posts.

bon vivant
March 6th, 2007 8:08pm