- #1

rock4christ

- 34

- 0

mine is -b + or - the square root of b

^{2}-4ac all over 2a

the quadratic formula. I found it easier than factoring for finding my x's

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rock4christ
- Start date

- #1

rock4christ

- 34

- 0

mine is -b + or - the square root of b

the quadratic formula. I found it easier than factoring for finding my x's

- #2

FunkyDwarf

- 489

- 0

e^(i.pi) = -1

just because its so Goddamn freaky

just because its so Goddamn freaky

- #3

murshid_islam

- 442

- 17

i think it will be the favourite equation of many ppl around here.

- #4

rdt2

- 125

- 2

http://www.mech.gla.ac.uk/~rthomson/teaching/equcharts.htm

- #5

Gib Z

Homework Helper

- 3,352

- 6

Please, if it is your favourite, give Euler's Identity the respect it deserves:

[tex]e^{i\pi} + 1 = 0[/tex]

I probably wouldn't be able to state my favorite, but here's one I found VERY useful: [tex]\int^b_a f(x) dx = F(b) - F(a)[/tex] where F'(x)= f(x)

EDIT: Favorite equation of mine, here it is. I am a mathematician, but this is really beautiful.

[tex] F^{\rightarrow} = \frac{dp^{\rightarrow}}{dt}[/tex]. Simple, effective, and has withstood the test of time. It's still correct to this date.

[tex]e^{i\pi} + 1 = 0[/tex]

I probably wouldn't be able to state my favorite, but here's one I found VERY useful: [tex]\int^b_a f(x) dx = F(b) - F(a)[/tex] where F'(x)= f(x)

EDIT: Favorite equation of mine, here it is. I am a mathematician, but this is really beautiful.

[tex] F^{\rightarrow} = \frac{dp^{\rightarrow}}{dt}[/tex]. Simple, effective, and has withstood the test of time. It's still correct to this date.

Last edited:

- #6

neutrino

- 2,091

- 2

- #7

murshid_islam

- 442

- 17

is that the way Euler originally wrote it? i think i read somewhere that Euler origially wrote it as [tex]e^{i\pi} = -1[/tex] and not in the more beautiful way you or other ppl nowadays writes it.Please, if it is your favourite, give Euler's Identity the respect it deserves:

[tex]e^{i\pi} + 1 = 0[/tex]

Gib Z said:Favorite equation of mine, here it is. I am a mathematician, but this is really beautiful.

[tex] F^{\rightarrow} = \frac{dp^{\rightarrow}}{dt}[/tex]. Simple, effective, and has withstood the test of time. It's still correct to this date.

isn't it [tex]\overrightarrow{F} \propto \frac{d\overrightarrow{p}}{dt}[/tex]

and is it true for velocities close to the velocity of light?

Last edited:

- #8

uart

Science Advisor

- 2,797

- 21

Although I think it's definitely a bit too geeky to have a "favorite" equation I have to say that the first time I saw the following it equation it really impressed me.

[tex] \prod_{n=1}^\infty \frac{1}{1-(1/p_n)^a} = \sum_{n=1}^\infty 1/n^a[/tex]

where p_n is the n_th prime and a>1.

[tex] \prod_{n=1}^\infty \frac{1}{1-(1/p_n)^a} = \sum_{n=1}^\infty 1/n^a[/tex]

where p_n is the n_th prime and a>1.

Last edited:

- #9

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,624

- 8

No, its as Gib Z written it;isn't it [tex]\overrightarrow{F} \propto \frac{d\overrightarrow{p}}{dt}[/tex]

and is it true for velocities close to the velocity of light?

[tex]\vec{F} = \frac{d\vec{p}}{dt}[/tex]

And yes, if applied correctly, is valid in Special Relativity. However, note that F=ma is

- #10

Curious3141

Homework Helper

- 2,858

- 88

- #11

ranger

Gold Member

- 1,685

- 2

I'll go simple. My favs are Kirchoff's voltage and current law. I use 'em practically everyday.

- #12

3trQN

- 337

- 1

Hmm, i have a few favourites, in order of their cognitive bias from greatest to least:

[tex]a^2 + b^2 = c^2[/tex]

[tex]e^{ix} = cos(x) + i sin(x)[/tex]

[tex]F(t) = \frac{1}{\sqrt{(2\pi)}}\int^{\infty}_{-\infty} e^{itx}f(x)dx [/tex]

and finally Maxwell's equations, which i don't fully remember or understand but when i do they will be next in the list....

[tex]a^2 + b^2 = c^2[/tex]

[tex]e^{ix} = cos(x) + i sin(x)[/tex]

[tex]F(t) = \frac{1}{\sqrt{(2\pi)}}\int^{\infty}_{-\infty} e^{itx}f(x)dx [/tex]

and finally Maxwell's equations, which i don't fully remember or understand but when i do they will be next in the list....

Last edited:

- #13

Gib Z

Homework Helper

- 3,352

- 6

In fact, F=ma is incorrect for any velocity larger than zero :P as small as the error is.

BTW uart, its not too geeky for have a favorite equation :) and that's a nice relation you've chosen, perhaps you could prove the related Riemann hypothesis for me? :P

EDIT: Forgot to address this. Euler Originally wrote it as Murshid_islam states it, but on later realisation of the equations profound consequences, changed it to the new form. He was inclined to do so by many collegues and espically number theorists, who found beauty in equations that were equated to zero.

BTW uart, its not too geeky for have a favorite equation :) and that's a nice relation you've chosen, perhaps you could prove the related Riemann hypothesis for me? :P

EDIT: Forgot to address this. Euler Originally wrote it as Murshid_islam states it, but on later realisation of the equations profound consequences, changed it to the new form. He was inclined to do so by many collegues and espically number theorists, who found beauty in equations that were equated to zero.

Last edited:

- #14

quasar987

Science Advisor

Homework Helper

Gold Member

- 4,793

- 21

[tex]F(t) = \frac{1}{\sqrt{(2\pi)}}\int^{\infty}_{-\infty} e^{itx}f(x)dx [/tex]

Written as such, this is nothing. The cool-looking thing is that

[tex]f(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty}f(\chi)e^{-i\omega\chi}d\chi \right)e^{i\omega t}d\omega[/tex]

- #15

rock4christ

- 34

- 0

I also kinda like a^{n} + b^{n} =/= c^{n}

Fermat's theorem. was a pain to prove

Fermat's theorem. was a pain to prove

- #16

MeJennifer

- 2,007

- 5

Forgive my ignorance but why do some here consider Euler's identity so special?

To me it seems that Euler's identity is a trivial instance of Euler's formula.

Furthermore the presence of [itex]\pi[/itex] is not significant, it is only there if you decide to measure angles in radians.

To me it seems that Euler's identity is a trivial instance of Euler's formula.

Furthermore the presence of [itex]\pi[/itex] is not significant, it is only there if you decide to measure angles in radians.

Last edited:

- #17

Curious3141

Homework Helper

- 2,858

- 88

Forgive my ignorance but why do some here consider Euler's identity so special?

I don't know about others, but speaking for myself :

1) It's a breathtakingly simple looking result that beautifully ties up four mathematical constants (e, pi, 0 and 1) in a single equation (at least when you write it with the RHS equal to zero).

2) It allows one to define the logarithm of a negative number as a complex number, as I've already alluded to.

3) It is an important result that allows a quick proof of pi's transcendence via the Lindemann-Weierstrass theorem (actually the exp(2*pi*i) = 1 variant is the one used most often here).

To me it seems that Euler's identity is a trivial instance of Euler's formula.

Maybe trivial to derive (from Euler's formula, which in itself is a beautiful tie-up between exponentiation and trig and leads to the shortest possible proof of De Moivre's theorem), but hardly trivial in its significance.

Furthermore the presence of is not significant, it is only there if you decide to measure angles in radians.

Pi is a mathematical constant, it need have nothing to do with measuring any angle as far as (the "four constant") Euler's identity goes. As for Euler's original formula, well, it's understood that the trig ratios have to be evaluated with arguments in radians. There is nothing arbitrary in this, radian measure is also considered by most to be far more fundamental than any other commonly used unit of angle measure (degrees, grad, etc.) It's the same sort of "natural bias" (pun not intended) when one compares a natural log with a common one.

And if you want to treat trig functions as abstractions without any immediate reference to angles or triangles (as is often the case in analysis), then you should leave everything in radians, far more elegant that way.

Last edited:

- #18

DieCommie

- 157

- 0

Well, it contains unique numbers... The square root of negative one, pi, e, 1 and 0. All these are very special numbers and they can be equated. Who would have thought that raising e (crazy number with elegant properties) to the power of i (an imaginary number that doesn't quite make sense) times pi (the ratio of circumference/diameter) and add one (an obvious important number) and it all equals zero (a number humanity struggled to come to grips with)? wtf? <--thats what I think. Like it was said before, kind of creepy.

Last edited:

- #19

d_leet

- 1,077

- 1

Well, it contains unique numbers... The square root of negative one, pi, e, 1 and 0. All these are very special numbers and they can be equated. Who would have thought that raising e (crazy number with elegant properties) to the power of i (an imaginary number that doesn't quite make sense) times pi (the ratio of circumference/diameter) andsubtractone (an obvious important number) and it all equals zero (a number humanity struggled to come to grips with)? wtf? <--thats what I think. Like it was said before, kind of creepy.

That should be add.

- #20

Gib Z

Homework Helper

- 3,352

- 6

- #21

MeJennifer

- 2,007

- 5

Well, then I respectfully disagee with those.There is nothing arbitrary in this, radian measure is also considered by most to be far more fundamental than any other commonly used unit of angle measure (degrees, grad, etc.) It's the same sort of "natural bias" (pun not intended) when one compares a natural log with a common one.

In my opinion, more fundamental would be to use for instance the simple range [0, 1] or even better [-1,1]. To me to use of the term [itex]\pi[/itex] is just getting fancy, it is really completely insignificant to me.

- #22

Gib Z

Homework Helper

- 3,352

- 6

Not to mention, when using radian measure with certain taylor series, than integrating them, it gives series for pi. That does not happen for any other angle measure, and it wouldn't give a series for what the measure is based on either, incase you were thinking its cause pi is the radian measures base.

Just like previous posts, you just wuoldnt expect it! It leads to many beautilful results.

- #23

theperthvan

- 184

- 0

[tex]x^x(1+ln(x))[/tex]

- #24

Curious3141

Homework Helper

- 2,858

- 88

[tex]x^x(1+ln(x))[/tex]

That's not an equation unless you prepend 'd/dx(x^x) ='.

- #25

rock4christ

- 34

- 0

most useful equation and why(this will definitely depend on what you do)

least useful and why

most:grams/molar mass=mol

least: quadratic formula. I love it but its useless

- #26

arunma

- 927

- 4

Although I think it's definitely a bit too geeky to have a "favorite" equation I have to say that the first time I saw the following it equation it really impressed me.

[tex] \prod_{n=1}^\infty \frac{1}{1-(1/p_n)^a} = \sum_{n=1}^\infty 1/n^a[/tex]

where p_n is the n_th prime and a>1.

This, I must admit, is pretty awesome. I wonder how it's derived, especially since there's no obvious formula for calculating the nth prime.

- #27

Gib Z

Homework Helper

- 3,352

- 6

Why: Helps alot.

Least Useful - [tex]\sum^{\inf}_{n=1} n = \frac{-1}{12}[/tex]

Why: Its cool, but I've never found a use for it. Maybe when I do string theory >.<

- #28

Gib Z

Homework Helper

- 3,352

- 6

arunma- That equation is derived here: http://en.wikipedia.org/wiki/Riemann_zeta_function

- #29

AsianSensationK

- 187

- 0

for r > 0,

[tex]\Sigma{(\frac{1}{1+r})^n} = \frac{1-(\frac{1}{1+r})^n}{r}[/tex]

This is the multiplying factor for present value of an annuity with level payments. Simply marvelous in it's uses in the field of finance.

[tex]\Sigma{(\frac{1}{1+r})^n} = \frac{1-(\frac{1}{1+r})^n}{r}[/tex]

This is the multiplying factor for present value of an annuity with level payments. Simply marvelous in it's uses in the field of finance.

Last edited:

- #30

murshid_islam

- 442

- 17

isn't this thread about favourite "equations"? but what rock4christ mentioned is not technically an "equation". the two sides are not equal for [tex]n \geq 3[/tex].I also kinda like a^{n}+ b^{n}=/= c^{n}

Fermat's theorem. was a pain to prove

is that a Ramanujan summation? or am i confusing it with something else?Least Useful - [tex]\sum^{\inf}_{n=1} n = \frac{-1}{12}[/tex]

Last edited:

- #31

Gib Z

Homework Helper

- 3,352

- 6

and yes, that's Ramanujans Summation, or zeta(-1).

- #32

Gib Z

Homework Helper

- 3,352

- 6

O and not to mention, they can be equal, he didn't say that a, b and c had to be positive integers.

- #33

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

- 9,624

- 8

So small in fact that we can't measure it :tongue2: ; so one must ask, is F=ma incorrect if we can't detect any discrepancy between it and F=dp/dt?In fact, F=ma is incorrect for any velocity larger than zero :P as small as the error is

- #34

Gib Z

Homework Helper

- 3,352

- 6

- #35

uart

Science Advisor

- 2,797

- 21

But Uart's example was also interesting...

[tex] \prod_{n=1}^\infty \frac{1}{1-(1/p_n)^a} = \sum_{n=1}^\infty 1/n^a[/tex]

This, I must admit, is pretty awesome. I wonder how it's derived, especially since there's no obvious formula for calculating the nth prime.

Yes that's the thing that impressed me about this equation. It looks so unlikely and yet the proof is actually very simple, requiring nothing more than a binomial expansion and the fundamental theorem of arthimetic.

Start with the application of the binomial expansion, [tex](1 - x)^{-1} = 1 + x + x^2 + x^3 + x^4 +\ldots[/tex] to each of the product terms on the LHS of the original equation.

This gives the LHS of the orgiinal equation as,

[tex]\prod_{i=1}^\infty (1+(1/p_i^a)+(1/p_i^a)^2+(1/p_i^a)^3+ ...)[/tex]

Or if you prefer to put in some numbers its,

[tex](1 + \frac{1}{2^a} + \frac{1}{4^a} + \frac{1}{8^a} + \ldots)\, (1 + \frac{1}{3^a} + \frac{1}{9^a} + \frac{1}{27^a} + \ldots) \, (1 + \frac{1}{5^a} + \frac{1}{25^a} + \frac{1}{125^a} + \ldots) \ldots (1 + \frac{1}{p_k^a} +\frac{1}{p_k^{2a}} + \frac{1}{p_k^{3a}} + \ldots) \ldots[/tex]

Now for the interesting part. If you understand the fundamental theorem of arithmetic (uniqueness of prime factorization) and you stare at the above expansion for long enough you'll suddenly realize why it is equal to the infinite sum on the RHS of the original equation. Try it and see, it's quite a revalation if you haven't seen it before and a startling demonstration of the uniqness of prime factorization.

Last edited:

Share:

- Last Post

- Replies
- 2

- Views
- 227

- Last Post

- Replies
- 7

- Views
- 372

- Last Post

- Replies
- 5

- Views
- 305

- Last Post

- Replies
- 4

- Views
- 406

- Replies
- 20

- Views
- 861

- Last Post

- Replies
- 4

- Views
- 490

- Last Post

- Replies
- 0

- Views
- 664

- Last Post

- Replies
- 7

- Views
- 523

MHB
Trig equation

- Last Post

- Replies
- 6

- Views
- 605

- Last Post

- Replies
- 3

- Views
- 264