[Monte314]

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[Arcani]

Alright, I got 7 of them, but I've got some last minute shopping to do so the other 3 will have to wait.

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Some shortcuts I used in the solutions. I'm in a bit of a rush, as I said, so I'm just going to state them out without proof.

i^2 = -1

z^c = exp(c*log(z)); this is a definition from complex analysis but it holds true for real numbers as well.

Log(z) = ln|z| + i*Arg(z) ; z is complex and Arg(z) is the principle value of the angle of z in the complex plane.

cos(theta) = (1/2)*(exp(i*theta)+exp(-i*theta))
cosh(theta) = (1/2)*(exp(theta)+exp(-theta))
cos(i*theta) = cosh(theta)

(n-1)! = gamma(n)
gamma(n+1) = n*gamma(n)



a) i^i = exp[i*Log(i)] = exp(i*[ln|i| + i*pi/2]) = exp(i*ln(1) - pi/2) = exp(-pi/2)

b) (-1)^i = exp[i*Log(-1)] = exp(i*[ln|-1| + i*pi]) = exp(i*ln(1) -pi) = exp(-pi)

c)cos(i) = (1/2)*(exp(i^2)+exp(-i^2)) = (1/2)*(exp(-1)+exp(1)) = cosh(1) = 1.5431 (I used a calculator to get the actual value)

d) ln(-1) = ln|-1| + i*pi = i*pi

I'm horrible with dealing with series and limits so, I'm not too sure about this.
e) Lim (x^x) as x--> 0+
Since we're approaching 0 from the right, I rewrote the series for x<1 as:
Lim (x^(-1/x)) as x--> infinity from this I just solved it as a power series with A_n = (1)^(1/n) the limit as n--> infinity = 1 = radius of convergence and Lim(x^x) as x-->0+ = 1
I graphed it and the graph seems to verify my result at least...

f) Haven't even tried it yet.

g) cos(theta) = 2
Since cosine only has a range from -1 to 1 in the real plane, theta must be imaginary. Define: theta= i*phi; cos(theta)=cosh(phi). Since the hyperbolic cosine has a range from 1 to infinity, cosh(phi) = 2 is within the range and plugging into a calculator I got phi = 1.317 so theta = i*1.317.

h) Haven't tried this one either.

i) I have very little experience with gamma functions but, I think I know enough to solve this one relying on the predetermined value of gamma(1/2)=sqrt(pi).

(1/2)! = (3/2 - 1)! = gamma(3/2) = gamma(1/2 +1) = (1/2)*gamma(1/2) = sqrt(pi)/2

j) Not really sure where to start on this one...

I'll try out the three I missed later.

Monte, in part j, is the omega outside the summation?

[Monte314]

Originally Posted by Arcani To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Alright, I got 7 of them, but I've got some last minute shopping to do so the other 3 will have to wait. ( Show Spoiler ) Some shortcuts I used in the solutions. I'm in a bit of a rush, as I said, so I'm just going to state them out without proof. i^2 = -1 z^c = exp(c*log(z)); this is a definition from complex analysis but it holds true for real numbers as well. Log(z) = ln|z| + i*Arg(z) ; z is complex and Arg(z) is the principle value of the angle of z in the complex plane. cos(theta) = (1/2)*(exp(i*theta)+exp(-i*theta)) cosh(theta) = (1/2)*(exp(theta)+exp(-theta)) cos(i*theta) = cosh(theta) (n-1)! = gamma(n) gamma(n+1) = n*gamma(n) a) i^i = exp[i*Log(i)] = exp(i*[ln|i| + i*pi/2]) = exp(i*ln(1) - pi/2) = exp(-pi/2) b) (-1)^i = exp[i*Log(-1)] = exp(i*[ln|-1| + i*pi]) = exp(i*ln(1) -pi) = exp(-pi) c)cos(i) = (1/2)*(exp(i^2)+exp(-i^2)) = (1/2)*(exp(-1)+exp(1)) = cosh(1) = 1.5431 (I used a calculator to get the actual value) d) ln(-1) = ln|-1| + i*pi = i*pi I'm horrible with dealing with series and limits so, I'm not too sure about this. e) Lim (x^x) as x--> 0+ Since we're approaching 0 from the right, I rewrote the series for x<1 as: Lim (x^(-1/x)) as x--> infinity from this I just solved it as a power series with A_n = (1)^(1/n) the limit as n--> infinity = 1 = radius of convergence and Lim(x^x) as x-->0+ = 1 I graphed it and the graph seems to verify my result at least... f) Haven't even tried it yet. g) cos(theta) = 2 Since cosine only has a range from -1 to 1 in the real plane, theta must be imaginary. Define: theta= i*phi; cos(theta)=cosh(phi). Since the hyperbolic cosine has a range from 1 to infinity, cosh(phi) = 2 is within the range and plugging into a calculator I got phi = 1.317 so theta = i*1.317. h) Haven't tried this one either. i) I have very little experience with gamma functions but, I think I know enough to solve this one relying on the predetermined value of gamma(1/2)=sqrt(pi). (1/2)! = (3/2 - 1)! = gamma(3/2) = gamma(1/2 +1) = (1/2)*gamma(1/2) = sqrt(pi)/2 j) Not really sure where to start on this one... I'll try out the three I missed later. Monte, in part j, is the omega outside the summation?

Lookin' pretty good here...

On part j, you don't have to worry about summation...

...and enWTFp has already gotten part j.

[Monte314]

OK, only f, g, and h haven't been answered.

Any takers?

[rahdam]

( Show Spoiler )

Part G

Let t = theta for simplicity

cos t = 2 = (e^it + e^-it) / 2

2 = (e^it + e^-it) / 2

multiply by 2e^it

4e^it = e^2it + 1

0 = e^2it - 4e^it + 1

Let e^it = x

0 = x^2 - 4x + 1

x = ( 4 (+/-) sqrt(12) ) / 2

e^it = ( 4 (+/-) sqrt(12) ) / 2

it = ln ( ( 4 (+/-) sqrt(12) ) / 2 )

t = ln ( ( 4 (+/-) sqrt(12) ) / 2 ) * i

t ~ +/- 1.3169579i

[Monte314]

OK, Arcani and radham have both solved part g.

This means only f and h have not been solved.

Any suggestions?

[rahdam]

( Show Spoiler )

L'Hospital's Rule is likely involved.

Computationally, it's 0.5

Part f

rahdam added to this post, 82 minutes and 0 seconds later...

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sqrt(3) / 4 seems to be a reasonable guess for part H.
Let me append this:
The answer lies in the interval between 0.4358104 and 0.4358105
I'm not entirely sure of the significance (read: irrational statement) of this number.) Come to think of it, that interval, and my decimal answer, will both come out high, due to computational limits. So the answer could well be sqrt(3)/4, because the actual value of x is less than the value that I use to check.

[Monte314]

There are two solutions, both real:

(For Christmas, my wife got me this writing pad that has a wireless connection to the pen, so as you write it captures a digital image that you can upload, OCR, whatever. You see here the solution written in my own paw....)

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This means that only part f has not been solved....

Oops, radham has solved part f.

Th-th-th-that's all, folks!

And for those of you who didn't get part j:

Rotate it clockwise 90 degrees.

[graciela224]

I'm officially as dumb as Tom Cruise. Doesn't matter, apparently I can still be rich and famous =]

[TheLastMohican]

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	And for those of you who didn't get part j: Rotate it clockwise 90 degrees.

This warrants retaliation. Just sayin'.

[Rhan]

Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

There are two solutions, both real: (For Christmas, my wife got me this writing pad that has a wireless connection to the pen, so as you write it captures a digital image that you can upload, OCR, whatever. You see here the solution written in my own paw....) ( Show Spoiler ) To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. This means that only part f has not been solved.... Oops, radham has solved part f. Th-th-th-that's all, folks! And for those of you who didn't get part j: Rotate it clockwise 90 degrees.

Well radham solved the limit but he didn't quite demonstrated why and I'm not satisfied with that, so I solved it (pretty easy) an I posted it. For the oter ones (except the last one...nice one Monte!) I don't really know how to do them, well I'm in highschool so I couldn't learn them (neither heard of them)...yet.

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[Monte314]

Originally Posted by Rhan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Well radham solved the limit but he didn't quite demonstrated why and I'm not satisfied with that, so I solved it (pretty easy) an I posted it. For the oter ones (except the last one...nice one Monte!) I don't really know how to do them, well I'm in highschool so I couldn't learn them (neither heard of them)...yet. ( Show Spoiler ) To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Most excellent work, Rhan. This is how I solved it when I first encounterd it over 30 years ago.

This problem is actually more difficult for more experienced mathematicians, because they tend immediately to think of using L'Hospital's Rule (i.e., differential calculus), which is often the correct approach for handling indeterminate forms. But it turns out that's not the right approach here... this problem is best handled using algebraic methods, such as the one Rhan has applied.

Good work!

[Rhan]

Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Most excellent work, Rhan. This is how I solved it when I first encounterd it over 30 years ago. This problem is actually more difficult for more experienced mathematicians, because they tend immediately to think of using L'Hospital's Rule (i.e., differential calculus), which is often the correct approach for handling indeterminate forms. But it turns out that's not the right approach here... this problem is best handled using algebraic methods, such as the one Rhan has applied. Good work!

Thanks Monte! Well, I know L'Hospital rule very well as well, but in this case it would just make things more complex, this rule it's mostly applied on (let's say xi = infinite) xi/xi case (the sign of the xi doesn't matter) or on 0/0 and it shall be applied only if there isn't any algebric way of a lower grade of difficulty. That is a principle I always respect when dealing with this kind of problems.