An Elegant Solution to a Complex Problem

Before trying to solve the problem I present here in this note, please make sure that you have read the complete note clearly.

Problem. Prove 2itan1(iab)=logaba+b2i\tan^{-1} \left(\frac{ia}{b}\right) = \log \left|\frac{a-b}{a+b}\right| where i=1i = \sqrt{-1} and a,bRa,b \in \mathbb{R} such that a±ba \neq \pm b

Proof. Ofcoure, you can prove it in a several ways, like simplifying using complex analysis or using series expansion or maybe some other method(s).

But I'll share another elegant solution to the above problem.

Consider the following integral

0x1x2a2dx=12a0x1xa1x+adx=12alogxax+a\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \frac{1}{2a}\int_0^x \frac{1}{x-a} - \frac{1}{x+a} \mathrm{d}x = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right|

We can also evaluate the above integral as 0x1x2a2dx=0x1x2+(ia)2dx=iatan1(ixa)\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \int_0^x \frac{1}{x^2+(ia)^2} \mathrm{d}x = \frac{i}{a}\tan^{-1} \left(\frac{ix}{a}\right)

Thus, iatan1(ixa)=12alogxax+a\frac{i}{a} \tan^{-1} \left(\frac{ix}{a}\right) = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| 2itan1(ixa)=logxax+a\Rightarrow 2i \tan^{-1} \left(\frac{ix}{a}\right) = \log \left|\frac{x-a}{x+a}\right| \qquad \qquad \qquad \square

Now, here I have a problem. When I tried to verify the above result on Mathematica, it gives me the negative result. Here's a screenshot of my code (in particular, I did it for a=1a=1 and for x<1|x| <1). I expected Mathematica to return true for all values.

Can somebody help me, where am I going wrong. Is there a flaw in the result that I have proven , or is my Mathematica code wrong?

Note by Kishlaya Jaiswal
4 years, 3 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

i solved the problem in the exact manner ! it was in Arihant level 2 complex problems right?

I too solved it through integration and later i realised that using demoures theorem (means eixe^{ix} was much easier ,

Mvs Saketh - 4 years, 3 months ago

Log in to reply

I derived this integral in some boring class and thought it was something great. Then I found out that using eiθe^{i\theta}, this thing is very elementary.

Raghav Vaidyanathan - 4 years, 3 months ago

Log in to reply

@Mvs Saketh @Raghav Vaidyanathan But did you notice the last line of the note.

Mathematica doesn't gives this result, can you tell me the flaw in my code?

Kishlaya Jaiswal - 4 years, 3 months ago

Log in to reply

I'm sorry, I do not know mathematica. But I do know that wolfram if not error free. Your result seems correct to me, but I am not able to see why you are getting false values.

Raghav Vaidyanathan - 4 years, 3 months ago

Log in to reply

@Raghav Vaidyanathan See my comment for an explanation.

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

@Raghav Vaidyanathan Yep, @Agnishom Chattopadhyay has already resolved the error. (Read our conversation below)

Kishlaya Jaiswal - 4 years, 3 months ago

Log in to reply

The fallacy is in the definite integrals.

In a similar way, one could prove that sin^2(x) = -cos^2(x)

0xsinxcosxdx=0x(sinx)d(sinx)=12sin2x \int^x_0{\sin x \cos x \, dx} = \int^x_0{(\sin x) d(\sin x) } = \frac{1}{2} \sin^2 x

0xsinxcosxdx=0x(cosx)d(cosx)=12cos2x \int^x_0{\sin x \cos x \, dx} = \int^x_0{-(\cos x) d(\cos x) } = -\frac{1}{2} \cos^2 x

    sin2x=cos2x \implies \sin^2 x = - \cos^2 x

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

Actually your last integral is incorrect. Let me explain

0xsinxcosxdx=01(cosx)d(cosx)=12(cos2x0x=12(cos2x1)=12(1cos2x)=12sin2x\int_0^x \sin x \cos x dx = \int_0^1 -(\cos x)d(\cos x) = -\frac{1}{2}\left(\cos^2 x\right|_0^x = -\frac{1}{2}\left(\cos^2 x - 1\right) = \frac{1}{2}(1-\cos^2 x) = \frac{1}{2} \sin^2 x

Kishlaya Jaiswal - 4 years, 3 months ago

Log in to reply

Haha, I know that already.

Sorry to confuse you, I don't think the problem is with the integral.

The problem is with the N function. Because it forces the software to convert both sides to numeric expressions, it is unlikely that they'll be the same.

I am not exactly sure why it happens but something like this might happen: Mathematica's calculation of one side may converge faster than the other. For example, if both sides are 2\sqrt{2}, one might converge to 1.41 and another to 1.414 after a certain number of iterations.

In computation, many a times a 0 and a small 'weed' error are often indistinguishable.

To make sure what I am talking about is not crap, run the following:

1
2
3
4
f[x_] := 2 I ArcTan[(I x)/1];
g[x_] := Log[Abs[(x - 1)/(x + 1)]];

Table[f[x/100] - g[x/100] // N, {x, 0, 99}]

See how small the values are?

Okay, let us resort to symbolic computation!

1
2
3
f[x_] := 2 I ArcTan[(I x)/1];
g[x_] := Log[Abs[(x - 1)/(x + 1)]];
Table[f[x/100] - g[x/100] == 0 // FullSimplify, {x, 0, 99}]

Run the code! Does it satisfy you?

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

@Agnishom Chattopadhyay Oh,yeah, now I see my flaw. Thanks

And also,

1
Solve[2*I*ArcTan[(*x)/1] - Log[Abs[(x - 1)/(x + 1)]] == 0,x]

gives me all real solutions.

A great thanks to you for pointing it out. ¨\ddot \smile

Kishlaya Jaiswal - 4 years, 3 months ago

Log in to reply

@Kishlaya Jaiswal You are Welcome. :)

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

Actually, there shouldn't be a fallacy because I found that both the sides of expression have same series expansion.

Also, if still there is fallacy, then can you please pin-point the wrong step.

And please do share the fallacy which proves that sin2x=cos2x\sin^2x = \cos^2x

Kishlaya Jaiswal - 4 years, 3 months ago

Log in to reply

I'm not sure of the exact fallacy but check the last comment again :)

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

With regards to your fallacy, that's not what you want. What you intended is

sinxcosxdx=sinxdsinx=12sin2x \int \sin x \cos x \, dx = \int \sin x \, d \sin x = \frac{1}{2} \sin ^ 2 x
sinxcosxdx=cosxdcosx=12cos2x \int \sin x \cos x \, dx = \int - \cos x \, d \cos x = - \frac{1}{2} \cos ^ 2 x

Hence, this would lead someone to claim that sin2x=cos2x \sin^2 x = - \cos ^2 x .


This is a common example used to represent this mistake, which is that fdx=F(x)+C \int f \, dx = F(x) + C , as opposed to fdx=F(x) \int f\, dx = F(x) . This appears numerous times, even on Brilliant. E.g. we've had to delete a lot of questions that say " Evaluate fdx \int f\, dx at x=0 x = 0 ", because they didn't make sense due to the +C + C which was forgotten.

Do you see how the mistake comes into play here?

Calvin Lin Staff - 4 years, 3 months ago

Log in to reply

Yep, the constant is -1/2 ;)

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

@bobbym none I'd like you to throw some light on this

Agnishom Chattopadhyay Staff - 4 years, 3 months ago

Log in to reply

Mr Chattopadhyay; Your later comments are on the right track. You should follow a discussion that phrontister and I had a long time ago...

bobbym none - 4 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...