Math tricks for everyone


by agentredlum
Tags: math, tricks
agentredlum
agentredlum is offline
#1
Jun27-11, 02:39 PM
P: 460
I would like to have this thread dedicated to showing math tricks from all areas of mathematics. Hopefully the title has aroused your interest and you have an interesting trick you would like to share with everyone. Let me start by showing one of my favorite tricks, perhaps something that has not occured to many of you?

Start with a general quadratic, do not set it equal to zero, set it equal to bx+c

ax^2 = bx + c

multiply everything by 4a

4(ax)^2 = 4abx + 4ac

subtract 4abx from both sides

4(ax)^2 - 4abx = 4ac

add b^2 to both sides

4(ax)^2 - 4abx + b^2 = b^2 + 4ac

factor the left hand side

(2ax - b)^2 = b^2 + 4ac

take square roots of both sides

2ax - b = +-sqrt(b^2 + 4ac)

add b to both sides

2ax = b +-sqrt(b^2 + 4ac)

divide by 2a, a NOT zero

x = [b +- sqrt(b^2 + 4ac)]/(2a)

This quadratic formula works perfectly fine for quadratic equations, just make sure you isolate the ax^2 term BEFORE you identify a, b, and c

1) Notice that this version has 2 less minus signs than the more popular version
2) The division in the derivation is done AT THE LAST STEP instead of at the first step in the more popular derivation, avoiding 'messy' fractions.
3) In this derivation there was no need to split numerator and denominator into separate radicals
4) Writing a program using this version, instead of the more popular version, requires less memory since there are less 'objects' the program needs to keep track of. (Zero is absent, 2 less minus symbols)

I hope you find this interesting and i look forward to seeing your tricks.

The method of completing the square... multiplying by 4a and adding b^2 i learned from NIVEN AND ZUCKERMAN in their book ELEMENTARY NUMBER THEORY however it was an example they used on a congruence, they did not apply it to the quadratic formula.
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agentredlum
agentredlum is offline
#2
Jun29-11, 05:36 AM
P: 460
Well.....no one is posting any tricks, thats sad. I'll post another trick, hope it motivates some of you.

What is i^i and how to show what it is.

This trick uses Eulers famous identity.....e^(ix) = cos(x) + isin(x)

Notice that when x = pi/2

e^(ipi/2) = cos(pi/2) + isin(pi/2)

cos(pi/2) = 0 and sin(pi/2) = 1

e^(ipi/2) = i

Now use this for the base in i^i, don't use it for the exponent

[e^(ipi/2)]^i

When you raise a base with an exponent to another exponent, you multiply the exponents

(ipi/2)*i = -pi/2 since i*i = -1

Therefore i^i = e^(-pi/2)

This is a real number! What an AMAZING result, an imaginary base to an imaginary power can be real.
TylerH
TylerH is offline
#3
Jun29-11, 06:45 AM
P: 737
[tex]\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{x-i}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{x-i}{x+i} \right)[/tex]

agentredlum
agentredlum is offline
#4
Jun29-11, 06:53 AM
P: 460

Math tricks for everyone


Quote Quote by TylerH View Post
[tex]\int darctanx=\int \frac{dx}{x^2+1}=\int \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{iln \left( \frac{x-i}{x+i} \right)}{2}[/tex]
My browser did not decode so i can't see it...too bad. Can you rewrite it using normal keyboard symbols?

Sorry for the trouble, thanx for posting.
TylerH
TylerH is offline
#5
Jun29-11, 06:59 AM
P: 737
Here's a picture.
Attached Thumbnails
snapshot1.png  
agentredlum
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#6
Jun29-11, 07:11 AM
P: 460
Quote Quote by TylerH View Post
Here's a picture.
VERY NICE! I like it a lot! POST MORE PLEASE!! THANK YOU!
TejasB
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#7
Jun29-11, 07:11 AM
P: 2
Dude your formula does not hold good for all quadratic expressions. Instead use
x= -b+(b^2-4ac)^1/2 or x= -b-(b^2-4ac)^1/2
agentredlum
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#8
Jun29-11, 07:22 AM
P: 460
Quote Quote by TejasB View Post
Dude your formula does not hold good for all quadratic expressions. Instead use
x= -b+(b^2-4ac)^1/2 or x= -b-(b^2-4ac)^1/2
give me an example where it fails please, thanx for posting
agentredlum
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#9
Jun29-11, 07:46 AM
P: 460
Quote Quote by agentredlum View Post
VERY NICE! I like it a lot! POST MORE PLEASE!! THANK YOU!
Wait a minute TylerH, you got the order wrong, change signs in both denominators, then it works.
Char. Limit
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#10
Jun29-11, 05:41 PM
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Quote Quote by TylerH View Post
[tex]\int_0^x darctanx=\int_0^x \frac{dx}{x^2+1}=\int_0^x \frac{i}{2(x-i)} - \frac{i}{2(x+i)} dx = \frac{i}{2} ln \left( \frac{x-i}{x+i} \right) \Rightarrow \forall x \in \Re, \: arctanx = \frac{i}{2}ln \left( \frac{x-i}{x+i} \right)[/tex]
Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
TylerH
TylerH is offline
#11
Jun29-11, 06:29 PM
P: 737
Quote Quote by Char. Limit View Post
Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
What should it be? Wikipedia says it's correct.
Char. Limit
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#12
Jun29-11, 06:41 PM
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Oh. All right then.
micromass
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#13
Jun29-11, 06:57 PM
Mentor
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No, Char is correct. The formula certainly doesn't hold in 0. So there must be some mistake somewhere. (The mistake being that complex integrals do not behave in the way you're describing)

The wikipedia is correct though.
TylerH
TylerH is offline
#14
Jun29-11, 07:01 PM
P: 737
I just now noticed they're different. I must have been reading it as what I expected to be there. :)
pwsnafu
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#15
Jun29-11, 08:32 PM
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P: 779
Quote Quote by agentredlum View Post
This is a real number! What an AMAZING result, an imaginary base to an imaginary power can be real.
Theorem: There exists 2 irrational numbers, a and b, such that ab is rational.

Proof: Consider sqrt{2}sqrt{2}. If this number is rational we are done. Suppose not. Define a=sqrt{2}sqrt{2} and b=sqrt{2}. Then ab = (sqrt{2}sqrt{2})sqrt{2} = sqrt{2}2 = 2, completing the proof.
agentredlum
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#16
Jun29-11, 11:33 PM
P: 460
Quote Quote by pwsnafu View Post
Theorem: There exists 2 irrational numbers, a and b, such that ab is rational.

Proof: Consider sqrt{2}sqrt{2}. If this number is rational we are done. Suppose not. Define a=sqrt{2}sqrt{2} and b=sqrt{2}. Then ab = (sqrt{2}sqrt{2})sqrt{2} = sqrt{2}2 = 2, completing the proof.
Can you send a picture of the proof the way TylerH did above? My browser does not decode TeX sorry. Thanks for posting. POST MORE PLEASE!
agentredlum
agentredlum is offline
#17
Jun29-11, 11:53 PM
P: 460
Quote Quote by Char. Limit View Post
Did you remember to write this as a DEFINITE integral? Because something tells me that 1/2 ln(-1) is not zero.
e^x = -1 is no problem...this is FAMOUS ...x=ipi

e^(ipi) = -1 so ln(-1) = ipi

Put ln(-1) in TI89 calculator in COMPLEX mode

agentredlum
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#18
Jun30-11, 12:02 AM
P: 460
Quote Quote by TylerH View Post
I just now noticed they're different. I must have been reading it as what I expected to be there. :)
You can edit your post. Go back and change signs. When it asks you for a reason, write wrong numbers.


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