Question

give an algebraic equation in which no coefficients or constant term is \[\pi\] but it has a root equal to \[\pi\]

Answer 1

any one here???

Answer 2

dx - c = 0
where d is the diameter
c i s the circumference of any circle

Answer 3

what say?

Answer 4

circumference itself contains pi

Answer 5

but good try

Answer 6

no it doesnt

Answer 7

what???

Answer 8

\[ds=rd \theta\]\[s=2 \pi r\]

Answer 9

are u denying this???

Answer 10

the value of pi is derived from circumference

Answer 11

do u agree the basic formula that i've written ??

Answer 12

pi is DEFINED as the ratio of circumference to the diameter.....that is how u derive the formula for circumference

Answer 13

obviously not

Answer 14

the people who made this formula did not use line integrals to calculate length..instead they defined the ratio of the circumference to the diameter as pi

Answer 15

the state ment u gave is for the class 9th or 10th standard

Answer 16

no.

Answer 17

it is historical

Answer 18

but it has no meaning..

Answer 19

then u give the answer

Answer 20

i need an equation like f(x)=c
where all coefficients and constants must be rational no.

Answer 21

where did u mention rational nos?

Answer 22

now i'm mentioning it.

Answer 23

see @himanish u are a very good student, tell me correctly do u think that what u have answered is acceptable??

Answer 24

yes

Answer 25

bt thats only when i didnt know abt the rational no thing

Answer 26

i don't think so. if the ans could be so easy then i would not post it here.

Answer 27

bhalo

Answer 28

think some different approach i think u can do it.

Answer 29

\[\pi = 3\sum_{n=0}^{\infty}[\left(\begin{matrix}2n \\ n\end{matrix}\right) / 16^{n}(2n+1)]\]

Answer 30

what to do?

Answer 31

this is pi

Answer 32

well but how can u write algebraic equation??

Answer 33

[ix - ln(-1)] = 0

Answer 34

the above series is algebraic

Answer 35

thats great..

Answer 36

not the second one

Answer 37

[ix - ln(-1)] = 0

is this right?

Answer 38

yes but it contains i

Answer 39

dont u use i in algebra?

Answer 40

u used \[e^{i \pi}=-1\]

Answer 41

ohk

Answer 42

is i rational???

Answer 43

[x - |ln(-1)| ] = 0

now?

Answer 44

|ln (-1) | is pi isnt it?

Answer 45

yes

Answer 46

so is it okay?

Answer 47

so which one is better the first or it??

Answer 48

this one is shorter

Answer 49

first one is wrong.

Answer 50

the series?

Answer 51

again here is some mistakes but i can ignore it

Answer 52

so the answer?

Answer 53

no dx-c=0 is wrong

Answer 54

x=|ln(-1)|
x=(2n+1)pi
x=pi when n=1

Answer 55

what was the answer u were lookin for?

Answer 56

its impossible
there is no polynomial function with a transcendental root

Answer 57

there are a lot of answers.

Answer 58

like?

Answer 59

\[1-x^2/2!+x^4/4!-....=1\]

Answer 60

http://en.wikipedia.org/wiki/Transcendental_numbers

pi cannot be a root to a function with all rational coefficients

Answer 61

what do u think @dumbcow about the above equation??

Answer 62

dipankar how does it come to pi?

Answer 63

put pi over there u will find RHS=LHS

Answer 64

THAT IS VALID FOR ANY X

Answer 65

you have an infinite series, a algebraic equation is defined as finite number of terms

Answer 66

sorry right hand side will be -1

Answer 67

this answer is invalid

Answer 68

why invalid??

Answer 69

you can always write pi as an infinite series

Answer 70

like i said above

Answer 71

6arcsin(0.5)

Answer 72

sin is not algebraic.

Answer 73

bt its expressible as an infinite algebraic series

Answer 74

any function can be expressed as an algebraic series

Answer 75

by taylor series

Answer 76

i know

Answer 77

so what about |ln(-1)|??

Answer 78

it can be expanded 2

Answer 79

its constant.lol

Answer 80

yes

Answer 81

http://en.wikipedia.org/wiki/Pi