Math Is Fun Forum

jtiger102

Member

Registered: 2008-04-03

Posts: 3

Help solving logarithmic equation with limit

I've been out of school too long and can't get this one...

If I remember correctly, the last term of

will work itself out to 0 as the series can be treated as a limit function.  In this case, as n goes to infinity, the resulting answer is 0.

I can't get the logarithmic portion though.

Many thanks for help!

Last edited by jtiger102 (2008-04-03 04:09:21)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Help solving logarithmic equation with limit

That's very interesting.  What mathematics have you seen?  Here is a very nonrigorous explanation:

This is -1.  It comes from mostly from the Taylor series expansion of sine, cosine, and e (Euler's number).

This sums to 1.  The easiest way to see it is that you are going half way to 1 with each term.  Half way between 0 and 1 is 1/2.  Well, thats the first term.  Halfway between 1/2 and 1 is 3/4, well, 1/2 + 1/4 = 3/4.  Halfway between 3/4 and 1 is 7/8.  And of course, 1/2 + 1/4 + 1/8 = 7/8.  You will keep going half way to 1, and if you do this nonstop, you will get within an arbitrary distance of 1.  This is what mathematicians mean they say when an infinite sum converges (in this case) to 1.

So the result is 0.002.

Now all of that wasn't really very interesting.  But what is interesting is that on a bulletin board in the applied mathematics building at my school, there is a check, written out to a cable company, with that expression on it.  The bottom note says, "What now *expletive*".

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

jtiger102

Member

Registered: 2008-04-03

Posts: 3

Re: Help solving logarithmic equation with limit

That is the same place I pulled the equation from... the check I mean.  I thought it was funny and I was just curious as to what the equation actually worked out to be.

It's been about 10 years since I even had to think about calculus and advanced math, so I just couldn't remember how to work it out.
It's a shame that my engineering math just hasn't had much of a workout since I graduated.

Just for my own knowledge, in the exponential term, what is the

?  I can figure out the expansion of

.  But I'm just not remembering what

is supposed to represent.

Sorry to be asking so many fundamental questions.

Thanks for the explanation!

Last edited by jtiger102 (2008-04-03 04:20:15)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Help solving logarithmic equation with limit

i is the square root of -1, an "imaginary" number (although it really exists).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

jtiger102

Member

Registered: 2008-04-03

Posts: 3

Re: Help solving logarithmic equation with limit

i is the square root of -1, an "imaginary" number (although it really exists).

So, if I'm doing this correctly... would this be the solution for the exponential term?

I'm assuming that the reason why it comes out to 1 is that you ignore all of the subsequent terms becuase they would all be imaginary.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Help solving logarithmic equation with limit

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Help solving logarithmic equation with limit

And

Last edited by Dragonshade (2008-04-03 05:18:17)

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Help solving logarithmic equation with limit

i is the square root of -1, an "imaginary" number (although it really exists).

So, if I'm doing this correctly... would this be the solution for the exponential term?

I'm assuming that the reason why it comes out to 1 is that you ignore all of the subsequent terms becuase they would all be imaginary.

Not quite jtiger.  You can't just ignore all of the terms with i in them, they must be added up as well.  It just so happens that they all cancel out.  The i needs to be squared / cubed / etc. as well.  That makes the series

Note that since i is the square root of -1, i squared is -1, which makes the third term of the series, (i^2 pi^2)/2! become (-pi^2)/2, which is a real number.  By extention, i^3 = -i, and i^4 = 1.  This means that every other term in the series, starting with the first, will be a real number, and every other term starting with the second will be imaginary.  We can separate the Taylor series into real and imaginary parts:

If you are so inclined you can try to work out the sums.  Just know that the left sum must equal -1 and the right sum must equal 0.

Last edited by TheDude (2008-04-04 02:03:31)

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Wrap it in bacon