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April 10

Operations on PowerSeries

Can someone help explain how to do this.

Find the power series representaion for f(x) and specify the radius of convergence.

$f(x)={\frac {x}{(1+x)^{2}}}$

Thanks in advance! Azninvazn88 03:20, 10 April 2007 (UTC)[reply]

Maybe even how to do power series in general. Azninvazn88 03:33, 10 April 2007 (UTC)[reply]

I assume you want a Taylor series? Around what point? x=0? Do you know the Taylor series for

{\frac {1}{1-x}}

around x=0? If so, it's not too hard to manipulate it into the other one. --Spoon! 05:23, 10 April 2007 (UTC)[reply]

Thanks for you guys' help! I understand how to do it now. Azninvazn88 19:07, 10 April 2007 (UTC)[reply]

Logrithm? Solve for x in this?

I am tring to find formula to solve for x: (1+x)^6=4.5

This is to work out the eqivalent interest rate. Like the return on investment is 450% after 6 days. So the daily interest rate would be (1+x)^6 = 450%, but I don't know what to do.

This isn't homework its for a video game called Harvest Moon. I am trying to compare investments, like one investment is 450% in 6 days. Another is 1,000% in 10 days. I need the average daily rate to compare them. Like I know I could compare them using a general multiple, 6*10= 60 days. So 4.5^10=3,405,063 compared 10^6 = 1,000,000 , but I want a general per day rate so I can make comparisons between many many more investments, such as 4000% in 30 days, or 2200% in 16 days. etc.

Thanks for any help!--Dacium 05:02, 10 April 2007 (UTC)[reply]

First take the 6th root of both sides:

(1+x)^6 = 4.5

1 + x = (4.5)^(1/6)

Then subtract 1 from each side:

    x = (4.5)^(1/6) - 1

Post your answer and we'll check it for you. StuRat 05:12, 10 April 2007 (UTC)[reply]

Thanks! I don't know what I was thinking! I got 28.5% for 450% in 6 days and 25.9% for 100% in 10 days. The answers for 60 days come to about 3,405,000 and 1,000,000 yay!. Yeah I had a mental block I forget that I could multiple the powers 6 * 1/6 to get to power 1.... and 1/6 duh. I feel stupid now :-)--Dacium 05:29, 10 April 2007 (UTC)[reply]

Assuming you meant to say 1000% instead of 100%, your figures all check out. Good job ! StuRat 06:04, 10 April 2007 (UTC)[reply]

There are 6 solutions.

(1) x = -2.28

(2) x = -1.64 - 1.11i

(3) x = -1.64 + 1.11i

(4) x = -0.35 - 1.11i

(5) x = -0.35 + 1.11i

(6) x = 0.28

The really amazing thing is that a negative interest rate of -2.28 ( -228% ) can generate a positive return in 6 days!!! Cool!

202.168.50.40 23:37, 10 April 2007 (UTC)[reply]

You can get a positive return on a negative interest rate in any even number of days, so long as it's not -100% interest, which will give you a return of 0. Here is an example: suppose you invest $1.00 at -200% interest per day. After one day you have $1.00 * -200% + $1.00 = -$1.00. After another day you have $-1.00 * -200% + $-1.00 = $1.00. Also, if the interest is between -100% and 0% the return will be positive, but below what you put in. For example: suppose you invest $1.00 at -50% interest per day. After one day you have $1.00 * -50% + $1.00 = $0.50.— Daniel 01:50, 11 April 2007 (UTC)[reply]

Oh! Please don't stop there! You got my panties in a knot. Tell me what kind of returns I can expect with imaginary interest rate, both positive imaginary and negative imaginary interest rates. 202.168.50.40 02:55, 11 April 2007 (UTC)[reply]

Imaginary interest rates — or exponential eigenvalues in general — correspond to oscillations: cycles in the complex plane. Considering only the real part will yield sine and cosine waves. Complex exponents, being a combination of such oscillations with normal exponential growth or decay, yield growing or shrinking waves; a negative real part (like this -2.28) should yield a decay, but we are effectively using Euler's method here and the method is unstable for the chosen time step. --Tardis 22:03, 11 April 2007 (UTC)[reply]

A small correction to Daniel's comment: The interest rate must be less than -100%. Anything between -100% and 0% will give exponential decay. -- Meni Rosenfeld (talk) 16:33, 11 April 2007 (UTC)[reply]