Math tricks for everyone

  • Thread starter agentredlum
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In summary: The reason for the question is that the integral is not zero, because the log should be evaluated at the limits of integration and ln(-1) is not defined. Therefore, the statement is not correct.In summary, the conversation started with a request to share math tricks from all areas of mathematics. A trick was shared involving a quadratic formula, followed by another trick involving the value of i^i. There was then a discussion about the validity of the first trick, and a proof was shared for the existence of two irrational numbers whose product is rational. However, the proof was incorrect as the integral used was not zero and the statement was not true.
  • #1
agentredlum
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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 occurred 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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  • #2
Well...no one is posting any tricks, that's 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.
 
  • #3
[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]
 
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  • #4
TylerH said:
[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, thanks for posting.
 
  • #5
Here's a picture.
 

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  • #6
TylerH said:
Here's a picture.

VERY NICE! I like it a lot! POST MORE PLEASE! THANK YOU!
 
  • #7
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
 
  • #8
TejasB said:
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, thanks for posting
 
  • #9
agentredlum said:
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.
 
  • #10
TylerH said:
[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.
 
  • #11
Char. Limit said:
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.
 
  • #12
Oh. All right then.
 
  • #13
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.
 
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  • #14
I just now noticed they're different. I must have been reading it as what I expected to be there. :)
 
  • #15
agentredlum said:
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.
 
  • #16
pwsnafu said:
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!
 
  • #17
Char. Limit said:
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

:biggrin: :cool:
 
  • #18
TylerH said:
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.
:biggrin:
 
  • #19
micromass said:
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.

HELLO MICROMASS:smile:

The integral is the area under the curve 1/(x^2 + 1)

This is defined for all real numbers. z is a 'dummy' variable of integration and is replaced by x when the definite integral is computed.

So the trick works for all real numbers x.
 
  • #20
All right, here's a simple trick, i hope I'm not the only one who finds this interesting...

Consider the matrix

[0 1]
[1 1]

Powers of this matrix generate 3 Fibbonacci numbers. The nth power generates F(n-1), F(n), F(n+1)

Example, when n=4 it generates 2,3,5 which are the 3rd, 4th and 5th Fibbonacci numbers, cool huh?
 
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  • #21
Not really a trick, but something to think about if you've never heard of fractional calculus:

for a general polynomial of the form [itex]x^{k}[/itex]

the nth derivative is given by the formula

[itex]\frac{d^{n}(x^{k})}{dx^{n}}[/itex]= [itex]\frac{k!x^{k-n}}{(k-n)!}[/itex]

Selecting n = 1/2, gives

[itex]\frac{d^{1/2}(x^{k})}{dx^{1/2}}[/itex]= [itex]\frac{k!x^{k-1/2}}{(k-1/2)!}[/itex]

Applying another derivative with n=1/2 gives

[itex]\frac{d^{1/2}\frac{(k!x^{k-1/2})}{(k-1/2)!}}{dx^{1/2}}[/itex]= [itex]\frac{k!(k-1/2)!x^{k-1}}{(k-1/2)!(k-1)!}[/itex] = [itex]\frac{k!x^{k-1}}{(k-1)!}[/itex]

which is the same as letting n=1 and doing a standard derivative.
 
  • #22
WOW! I never heard of fractional calculus but i googled it and it is a legitimate subject. Thank you for the post. PLEASE POST MORE!

Can you please post the calculation as a picture? My browser does not decode TeX and I would love to follow the calculation.

Thanx again
 
  • #24
came in here expecting 1+1 = 11
saw posts
everything went better than expect :D
 
  • #25
amd123 said:
came in here expecting 1+1 = 11
saw posts
everything went better than expect :D

Thank you. I invite you to post a trick. Any trick you find interesting, or something not well known like HALF derivatives. :smile:
 
  • #26
amd123 said:
came in here expecting 1+1 = 11
saw posts
everything went better than expect :D

1+1=11 in base 1. :tongue:
 
  • #27
TylerH said:
1+1=11 in base 1. :tongue:

YOU'RE RIGHT! COOL OBSERVATION!:cool:

1 + 1 = 110 in base 1 also or 101 or 1001 or 1000...01

or any combination that starts with 1 and has just one more 1 anywhere and all other digits 0
:smile:
 
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  • #28
agentredlum said:
YOU'RE RIGHT! COOL OBSERVATION!:cool:

1 + 1 = 110 in base 1 also or 101 or 1001 or 1000...01

or any combination that starts with 1 and has just one more 1 anywhere and all other digits 0
:smile:

Also, 1+1+1=111 in base 1.
 
  • #29
agentredlum said:
YOU'RE RIGHT! COOL OBSERVATION!:cool:

1 + 1 = 110 in base 1 also or 101 or 1001 or 1000...01

or any combination that starts with 1 and has just one more 1 anywhere and all other digits 0
:smile:

Wait a minute. Would base 1 allow us to use the digit 1? Base 2 does not allow 2, base 3 does not allow 3, etc.

Base 0 would put us in the uncomfortable position of having to calculate 0^0 in the left-most position.
:tongue2:
 
  • #30
agentredlum said:
Wait a minute. Would base 1 allow us to use the digit 1? Base 2 does not allow 2, base 3 does not allow 3, etc.

Base 0 would put us in the uncomfortable position of having to calculate 0^0 in the left-most position.
:tongue2:

Basically, Base 1 only allows one digit: 1. Base 2 allows 1 and 0. That's why 1+1+1=111 in base 1... actually, 1+1+1...+1 = 111...1 in base 1.
 
  • #31
Char. Limit said:
Also, 1+1+1=111 in base 1.

YOU'RE RIGHT TOO!

Thanks to all posting

I'm not sure the digit 0 would be allowed in base 0.:grumpy:
 
  • #32
No digits would be allowed in base 0. After all, the base number shows how many different "numbers" you can allow. Base 1 has one number: 1. Base 2 has two numbers: 0 and 1. The amount of numbers in base 0 is... the empty set!
 
  • #33
Char. Limit said:
Basically, Base 1 only allows one digit: 1. Base 2 allows 1 and 0. That's why 1+1+1=111 in base 1... actually, 1+1+1...+1 = 111...1 in base 1.

Oh, so I can't use 0 in base 1 :cry:

Too bad the symmetry is broken.:cry:
 
  • #34
Char. Limit said:
No digits would be allowed in base 0. After all, the base number shows how many different "numbers" you can allow. Base 1 has one number: 1. Base 2 has two numbers: 0 and 1. The amount of numbers in base 0 is... the empty set!

Good point about base 0. :smile:

I understand why we would pick 1 as the only allowable number for base 1...for consistency.

so symmetry is exchanged for consistency...i can live with that.:biggrin:
 
  • #35
agentredlum said:
Oh, so I can't use 0 in base 1 :cry:

Too bad the symmetry is broken.:cry:

You can use whatever symbol you want, but you can only use that symbol.
 
<h2>1. What are some examples of math tricks for everyone?</h2><p>Some examples of math tricks for everyone include the "11 Times Trick" for multiplying by 11, the "9 Times Trick" for multiplying by 9, and the "Finger Counting Trick" for multiplying by 9 or 6.</p><h2>2. How can math tricks benefit everyday life?</h2><p>Math tricks can benefit everyday life by making calculations faster and easier. They can also improve mental math skills and help with problem-solving in various situations.</p><h2>3. Are math tricks suitable for all ages?</h2><p>Yes, math tricks can be suitable for all ages. Some tricks may be more advanced and require a basic understanding of math concepts, but there are also simple tricks that can be taught to young children.</p><h2>4. Can math tricks be used in academic settings?</h2><p>Yes, math tricks can be used in academic settings. They can be helpful for students who struggle with math or for those who want to improve their mental math skills.</p><h2>5. Are there any downsides to using math tricks?</h2><p>One potential downside to using math tricks is that they may rely on memorization rather than understanding of mathematical concepts. This could lead to difficulties in more complex math problems that cannot be solved using a trick. It is important to also have a strong foundation in math principles.</p>

1. What are some examples of math tricks for everyone?

Some examples of math tricks for everyone include the "11 Times Trick" for multiplying by 11, the "9 Times Trick" for multiplying by 9, and the "Finger Counting Trick" for multiplying by 9 or 6.

2. How can math tricks benefit everyday life?

Math tricks can benefit everyday life by making calculations faster and easier. They can also improve mental math skills and help with problem-solving in various situations.

3. Are math tricks suitable for all ages?

Yes, math tricks can be suitable for all ages. Some tricks may be more advanced and require a basic understanding of math concepts, but there are also simple tricks that can be taught to young children.

4. Can math tricks be used in academic settings?

Yes, math tricks can be used in academic settings. They can be helpful for students who struggle with math or for those who want to improve their mental math skills.

5. Are there any downsides to using math tricks?

One potential downside to using math tricks is that they may rely on memorization rather than understanding of mathematical concepts. This could lead to difficulties in more complex math problems that cannot be solved using a trick. It is important to also have a strong foundation in math principles.

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