is sinx a polynomial........?????????

- anonymous

is sinx a polynomial........?????????

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- schrodinger

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- anonymous

It can be written as an infinite series, so yes, a polynomial of infinite degree.

- anonymous

cant get u!!!!!!!

- anonymous

sinx isn't polynomial
which only works for sin(x), is to notice that sin(x) has infinitely many zeroes - no polynomial or rational function has this many zeros.

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- anonymous

Do you know what a taylor series is, brah?

- anonymous

no......

- anonymous

\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \]

- anonymous

When x is measured in radian.

- anonymous

Sin x is a trigonometric function.
A polynomial is in the form ax^n+bx^n-1+...+cx+d
A rational function is the quotient of two polynomials.
e^x is not a rational function, but it is an exponential function because it is in the form a^x where is a constant, and e is a constant.

- anonymous

Tian do you go to Cambridge? I do.

- anonymous

Use the taylor's polynomial for sin(x) to find the rates of convergence of the following
Taylor expansion of sin(x) about x=0 is
sin(x) = x - x^3 / 3! + x^5 / 5! - x^7 / 7! + ...
so sin(x) / x = 1 - x^2 / 3! + ...
In the limit that x goes to zero, [sin(x)/x] = 1 - (0)^2 / 3! + ... = 1
Second way...extra:) hehe
Consider a trigonometric circle with its center O and a tangent to this circle at A
Let M be any point on The ciricle, and project M on the x-asis and name it H
and let T be the tangent Thetha which is from A to T
http://img534.imageshack.us/img534/1252/…
As you realize in the picture
Area of Triangle OAT = half tan thetha
and Area of triangular sector OAM = thetha / 2
and Area of OHM : sin thetha = MH /OM (where OM =1 cz its a trigonometic circle )
==> MH = Sin thetha
==> Area of OHM = (sin theha cos theha) /2
Therefore we can conclude that Area of OHM is smaller than OAM which is smaller than triangle OAT
==> (sing thetha cos thetha) /2 < thetha/2 < (tan thetha) / 2
multiply this by ( thetha / sin thehta)
==> we get
cos theha < thetha / sin thetha < 1/cos thetha
let thetha tend to zero that is cos thetha = cos 0 = 1
==> 1- < lim tetha /sin thetha as thetha goes to zero < 1+
==> lim sin thetha / tjhetha as thjetha goes to 0 = 1
PS: Dont forget that lim sin thetha / thetha as thetha tends to zero = lim thetha / sin thetha as thetha tends to zero !
Hope i helped you :)

- anonymous

What. The. flutter.

- anonymous

Newton I hope so

- anonymous

why?

- anonymous

I thought you said it wasn't a polynomial?

- anonymous

Then copypasted some random stuff that is vaguely related at the very most.

- anonymous

yeah I think sinx isn't polynomial by my first argument

- anonymous

:@

- anonymous

the point is that ..............my doubt is not clear yet........clear it!!!!!

- anonymous

Whatever. I'm not going to argue semantics.

- anonymous

I think a polynomial should have a finite number of terms.

- anonymous

sinx isn't polynomial, sin(x) has infinitely many zeroes - no polynomial or rational function has this many zeros.

- anonymous

Actually, 0 is a polynomial and has infinitely many 0s...

- anonymous

All we have to do to settle this is to check the definition of a polynomial.

- anonymous

Sin x may not be a polynomial (in your baby, non-cantab definitions), but the reasoning given is incorrect.

- anonymous

the power must be a whole numbr

- anonymous

Whatever, I don't have time to argue with people who get their arguments of wikipedia and yahoo answers. this is like when 48/2(9+3) was asked. Yes, wolfram says something so it's right. People need to learn what REAL mathematicians do and stop this nonsense.

- anonymous

Polynomial has to be of finite degree, so I made a mistake (not that definitions change how I do mathematics, so it's irrelevant).
But the person who argued with me copied their argument DIRECTLY from yahoo answers and made some glaring mistakes. I don't have time for this pellet any more.

- anonymous

it isn't a polynomial :)

- anonymous

a polynomial is : f(x) = x^2 + 3 and etc. This is not :)

- anonymous

WOW.. did somebody kill somebody?

- anonymous

lol, why , who's dead ._.

- anonymous

hopefully not me :P

- anonymous

sin(x) has infinitely many zeroes - no polynomial or rational function has this many zeros.
^_^

- anonymous

sooooo, it's not a polynomial lol . I hope you're not confused sid :)

- anonymous

INew lol , cool down :)

- anonymous

STOP quoting some nobody off yahoo answers. 0 is a polynomial and has infinitely many 0s.

- anonymous

._. I'm not , I'm talking

- anonymous

based on knowledge and experience. Hey it's okay to make mistakes, don't get all boiled up for a question, calm down

- anonymous

u guys r more confuse thn me............

- anonymous

LOL!

- anonymous

LOL

- anonymous

hold on sid :)

- anonymous

clear my doubt..........

- anonymous

we apologize for the vagueness, just a second ^_^

- anonymous

INew, first calm down lol

- anonymous

Sin(x) has an infinite degree which (apparently) makes it not a polynomial.
Also, it has infinite zeros, so can'y be a polynomial.
Source: Yahoo answers

- anonymous

eat ice cream ^_^

- anonymous

just because our answers sounded alike, doesn't mean we have copied them INew :)

- anonymous

many answers sound alike, and ofcourse they will since all lead to the same explanation and theorem

- anonymous

Maybe not yours, but the guy before copied 2 paragraphs directly, so please don't patronise me.

- anonymous

fight..........killl..............

- anonymous

oh come on, let's not take it that way :)

- anonymous

no sid, that's not my style, I take things calmly :)

- anonymous

that guy is still onlyn!!!!!

- anonymous

Alright now, sid, read INew's post and tell me if you understood it or not okay?

- anonymous

My first post was wrong (apparently). But I didn't make a mistake, I just never needed to know the exact definition.

- anonymous

and INew, don't waste your energy in useless arguments, preserve it for better things ^_^

- anonymous

what is 'Inew'

- anonymous

INew = INewton lol

- anonymous

oh........sry!!!

- anonymous

read the recent post made by INew :)

- anonymous

So Newton, Could you write the exact definition down?

- anonymous

and thank you for your help tian ^_^ ,all of you have been a great help to sid, but let's calm down now

- anonymous

so finally.........sinx is a polynomial or not....with reasons!!!

- anonymous

lol, sin x is not a polynomial since it tends to go to infinity. A polynomial = has a finite number of zeros.

- anonymous

It doesn't go to infinity? It has an infinite number of terms.

- anonymous

contradiction comes.........i m frustrated!!!!!!!!!!!!!

- anonymous

Example : \[f(x) = x^2 -1\]
= (x-1)(x+1) where the zeros are x = 1, -1 <-- finite numbers of zeros
but sin x keeps on oscillating b/w -1 and 1 and never stops, so that's why it's not a polynomial

- anonymous

sid dear, calm down , don't panic :)

- anonymous

f(x) = 0
INFINITE NUMBER OF ZEROS BUT POLYNOMIAL OF DEGREE ZERO.

- anonymous

LOL! INew shut the caps =D

- anonymous

come to india .............and kill me!!!!!!!

- anonymous

I'll slap you instead to wake u up

- anonymous

lol, hush now, we're going to explain it, if you don't get me, then INew will explain it, if you don't get her , then someone else, alright? now hush LOL

- anonymous

both of u ......as a unit...........give me the final nd the CORRECT answer!!!!!!!

- anonymous

wtf I'm not a her

- anonymous

SID! LOL

- anonymous

INEW, eat ICE CREAM and stay away from bananas!

- anonymous

lolzzzzzzz!!!!!!!!!!........sry 4 dat

- anonymous

In mathematics, a BABY polynomial is an expression of finite length, containing variables and constants. The Fundamental Theorem of Algebra says that a BABY polynomial of degree n can be written as the product of n linear factors.
Sin(x) is a taylor series of infinite length, and therefore not a BABY polynomial.

- anonymous

lol, I liked how you've put it "baby" polynomial, so sid do you get it now?

- anonymous

so sinx is not a polynomial bcos it has infinite roots!!!!!!!is it.......

- anonymous

yes :)

- anonymous

NO. It is of an infinite degree, with an infinite number of terms. NOT because of infinite roots.

- anonymous

same scenario as long as he got the picture lol

- anonymous

But by your logic, f(x) = 0 isn't a polynomial.

- anonymous

listen u guyz .....u may be a cambridgian or etc.etc........i m a scul going boy............new to maths..........dont kno so advance concepts...lyk taylor series.etc.etc

- anonymous

Protip: Taylor series isn't advanced.

- anonymous

INew, let's not get him into the series maze ._. from now, just simply explain it in a simple way LOL

- anonymous

I am taking taylor series next week

- anonymous

so, it's sorta advanced for him?

- anonymous

at some point u guyz would also have been as i m today!!!!!!dont forget........

- anonymous

Lol, don't worry :)

- anonymous

Alright, listen
If you plot a graph for sin x, you'll notice it oscillating between -1 and 1 and never stops, as in never dies, it goes to infinity
On the other hand, when you plot a simple function such as f(x) = x^2 or f(x) = ax + bx + c ( which is a polynomial ) you'll have a line, parabola or curve.
A polynomial is : (ax + bx + c), does sin(x) have this form? if yes, then it's a polynomial, if not, then it isn't.

- anonymous

did I make it easier? ^_^

- anonymous

so now you tell me, is it a polynomial? :)

- anonymous

|;

- anonymous

no ...it is not

- anonymous

excellent ^_^

- anonymous

x^2 goes to infinity.

- anonymous

here comes confusion

- anonymous

lol, wait, did you understand what I have said?

- anonymous

hold on INew

- anonymous

gud nyt!!!!!!!

- anonymous

sweet drms.............!!!!!1

- anonymous

lol good night

- anonymous

:(

- anonymous

INew, you're right, >_< my bad lol

- anonymous

what's with the long face? :) cheer up, he got it and you have given awesome "advanced" explanations, but it was too much for his level. I thought they were great ^_^

- anonymous

Lol. I was wrong to start with, so rather than admit it, I spent pretty much the rest of the argument crying and pointing out tiny mistakes. Pretty disappointing, really.

- anonymous

well we're all humans, and we all make mistakes that we learn from right? :)

- anonymous

"Who hasn't made a mistake, hasn't tried anything new" -Albert Einstein ^_^

- anonymous

:-)

- anonymous

when you face such situations, yelling and screaming won't solve the problem, calm down and take it with reason :)

- anonymous

promise me no more overreacting?

- anonymous

yeah right

- anonymous

Indeed. It's just, the first time I've been wrong, ever. Kind of a shock to the system. I'll try.

- anonymous

again, human LOL, you should see MY mistakes. I subtract instead of adding and the vice versa LOL

- anonymous

well, not anymore, just rarely happens now ^_^"

- anonymous

you make a mistake, fix it, period =P
just no more cursing or any of that stuff lol

- anonymous

We'll see. :D.
One thing before this topic dies, in case you didn't see my correction, there is a little error here:
'if you don't get me, then INew will explain it, if you don't get *her*'

- anonymous

^_^" my bad, aha..

- anonymous

you said that you were a she last time ._.

- anonymous

or were my eyes twisted x_x

- anonymous

Hmm. I think I may have said it (sarcasticaly :( ). Hence my pseudonym (and real name, actually, if you can see it).

- anonymous

oh yeah, LOL. My dearest apologies ^_^

- anonymous

anyhow, I'm off to bed now, good night :) don't let me catch you cursing or overreacting next time =P

- anonymous

lol, OK, bye

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