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## Math Homework

This is a discussion on Math Homework within the General Off Topic+ forums, part of the QJ.net Forum Miscellaneous category; Math is awesome. What's even more awesome is math homework.... If anyone has any math homework problems, let me know... ...

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1. ## Math Homework

Math is awesome.
What's even more awesome is math homework....
If anyone has any math homework problems, let me know... I can help you out with it.... (My knowledge ranges from Algebra to Calculus 3)

2. Im a senior and im in Pre cal right now..Its not HARD..But i just don't feel like being there

3. you sound like your a pretty smart guy
I can do my own math homework tho lol
althought you might be able to help some lazy guy that doenst wnat to do his homework.
.....
whats a calculus???

4. Its a type of math class. GEEZ

5. Originally Posted by montrob
you sound like your a pretty smart guy
I can do my own math homework tho lol
althought you might be able to help some lazy guy that doenst wnat to do his homework.
.....
whats a calculus???
Calculus is split up into two branches:
Differential Calculus - Branch of Calculus where one studies the rate of change of functions...
Integral Calculus - Branch of Calculus where one studies areas and volumes of functions...
If you're in Pre-Calculus or Physics, chances are you've experienced some calculus but you probably have been doing stuff the hard way. (e.g. finding the minimum/maximum of a parabola; you have to go through the trouble of completing the square, or memorizing the -b/2a formula, while the traditional method is to use differential calculus and find out where on the curve the slope is 0..)
Calculus isn't just where someone studies rate of change of functions, however... there are many applications of it such as finding roots of a cubic, quartic or even a quintic (Not sure if I spelt it right :P) equation using differentials... (Newton's Method)
You could even find the sine value of any number without using your calculator's sine function... (Taylor Series)

As to your comment about me sounding like a smart guy... my only strength is math... all I do at school is ***** about math and pretty much do terribly in my other classes, so I enjoy doing math.. :P

Again, I'm open to any questions.

6. here's a simple question...
How do you calculate sine,cosine, and tangent functions without using a calculater. I only ones I know how to do are ones that form 30-60-90 and 45-45-90 triangles.

7. Math Blows,

Pie is NOT 3.14159
Pie R Good.

8. ok so whats 2+2x2?? if you get that then I think you are smarter than Einstein!!

btw....*burp*

Edit: I'm sorry Im just bored and have nothing to do...Im 17 and I know how to do math but I'm just doin nothin in my votech class right now so I'm just postin a lil bit...so....yeah

9. Originally Posted by Freehate
here's a simple question...
How do you calculate sine,cosine, and tangent functions without using a calculater. I only ones I know how to do are ones that form 30-60-90 and 45-45-90 triangles.
That really depends...
When you mean calculating sine, cosine and tangents without a calculator, you can do it with one of two ways...
First, you can be precise, or second you can be approximately near the value..
For instance, if you know the value of sin or cos 45 (pi/4 rad) and sin or cos 30 (pi/6 rad) then you can use the sum of angles formula to find the value of sin or cos 75 (pi/6 + pi/4 = 4pi/24 + 6pi/24 = 10pi/24 = 5pi/12 rad) assuming you knew the formula..
However, if you're just too lazy to use that formula, then you can do it another way-- Taylor Series...
Taylor series is a good approximation for a function using its derivatives and such...
The more terms you use for a Taylor series, the more accurate your answer will become...
A taylor series is like the following:
f(a) + f'(a)(x - a) + f''(a)(x - a)^2 / 2! + f'''(a)(x-a)^3 / 3! + f''''(a)(x-a)^4 / 4! ...
Where a is a point near where you want to approximate the value of the function..
However, most people are too lazy to expand those brackets and so they use MacLaurin Series which is a Taylor Series with x = 0:
f(0) + f'(0)x + f''(0)x^2 / 2! + f'''(0)x^3 / 3! + f''''(0)x^4 / 4! ....
So, how does this relate to sine and cosine? Let's say you wanted the sine of an angle... let's say pi/9..... which is approximately 3.1416/9 = 0.35... (Just take a guess here... you can be as close as you want..)
Now, we know that f(x) = sin x, so let's find its derivatives and come up with a Taylor Series of degree 4. (Degree 4 meaning we're going up to the 4th degree of x and 4th order in terms of derivatives)
So:
f(x) = sin x
f'(x) = cos x
f''(x) = -sin x
f'''(x) = -cos x
f''''(x) = sin x
So, we know sin 0 is 0, cos 0 is 1, thus we have:
f(0) = 0
f'(0) = 1
f''(0) = 0
f'''(0) = -1
f''''(0) = 0
So now we have:
sin x =approx= x - x^3 / 3!
Plugging in for 0.35 would give us 0.207...
Checking a calculator, this is PRETTY close.... the calculator gives 0.343 (3 d.p.) for the actual value and we get 0.343 (To 3 d.p.) using just the Taylor Series! Awesome! (Yes, geeky.. I know :P)
To show how increasing the degree of our Taylor Series would make the answer more accurate, let's be daring and go up a few more degrees...
f'''''(0) = 1
f''''''(0) = 0
f'''''''(0) = -1
Hmm... starting to see a pattern? We now have:
sin x = x - x^3 / 3! + x^5 / 5! - x^7 / 7!.....
See the pattern in sine approximations?
If you were to plug in 0.35 into this series, you'd find it's even MORE accurate to more decimal places..... which shows how amazing simple calculus is! (How do you think your calculator calculates sines and cosines? :P)
Anyway, I hope that helped some...

"ok so whats 2+2x2?? if you get that then I think you are smarter than Einstein!!"
6? :P
It must be one of those trick questions.. like "What is 2+2?" and then I'd say "4.." and you'd say "No, it's 22! HAHAH...."

So yeah.. I'm bored... any more questions are welcome.

10. ^^How long did it take you to write that? Geeeeeeeeez. I myself am in AA Math in 9th Grade (that's advanced), but I wouldn't consider myself a master at math.

11. Albert Einstein has entered the building!

Go be a Math teacher :)

12. wow, I got a little lost around the taylor series but the radians part I understood. I assumed thered be a way to figure it out using a unit circle but that is one long explanation. Im gonna ask my trig teacher about this later some time, after we come back from strike ;)

13. Originally Posted by Pseudo ANIMALISTIC
It must be one of those trick questions.. like "What is 2+2?" and then I'd say "4.." and you'd say "No, it's 22! HAHAH...."

So yeah.. I'm bored... any more questions are welcome.

well 1 + 1 = window

14. Originally Posted by Freehate
wow, I got a little lost around the taylor series but the radians part I understood. I assumed thered be a way to figure it out using a unit circle but that is one long explanation. Im gonna ask my trig teacher about this later some time, after we come back from strike ;)
Yeah, I did hear there was a way to do that but I never had a chance to do a Pre-Calc class... so my methods are weird.
Maybe the part you got lost with was the f(x), f'(x) part? (That's stuff from differential calc, if you haven't done the class yet..)

Albert Einstein has entered the building!

Go be a Math teacher :)
No thanks, don't think there are many chicks majoring in math.... or many chicks fancy math.

15. I haven't had calculus yet, Im only a junior. But it was something I always wondered since algebra I.

16. I'm taking FST (Functions, Statistics, and Trigonometry) after that I'm taking some calclus maybe..unsure tho

17. im taking pre-cal right now which is pretty much the same thing. Were doing standard deviations but I cheat and use my graphing calculator :icon_razz

18. Originally Posted by Freehate
I haven't had calculus yet, Im only a junior. But it was something I always wondered since algebra I.
Well, calculus kicks ass.
I got bored with math when I was in Algebra 1 so I used to do a little research on Calculus online-- it doesn't really take much to get into the subject...
Everything you're taking in Pre-Cal is pretty much an alternative way to do stuff that you'd do in Calculus...
For example, minimums, maximums, points of inflection and intermediate value theorem stuff.. (What's the point of that theorem anyway?)
Let's say you had an arbitrary function, let's say... f(x) = x^3 + 2x^2 + x + 1
To find the minimums and maximums of this function in Pre-Calc, you'd either graph it on your calculator and use its features to find the mins/maxs..... or you'd look at a graph and look closely and approximate the point... (This is a pain... I know I screw up alot when it comes to reading co-ordinates off a graph, like I'd say (2,3) instead of (2,4)...)
An alternative way to do it, is to use the concept of differentiation from calculus..
So, let's analyze this graph... here's a pic of it I got from gnuplot: (pic not available for the moment; I'll upload it in a bit once gnuplot compiles)

As you should know, or notice, at a maximum point or minimum point (relative minima and maxima, or relative extrema, that is) the slope of a line tangent to that point would be equal to 0.

Let's get back to the topic of differential calculus...
Differential calculus is mainly concerned with the rates of changes of non-linear functions... but recall from Algebra, the slope is essentially the graphical/geometric representation of "rate of change", in a sense that since a straight line, as you should know, is linear such that it has a constant slope, that the next plotted points would always follow a specific, constant pattern.
For example, consider the graph y = 3x + 2...
At x = 0, y = 2; at x = 2, y = 8... and so on; the x value is always multiplied by a constant and a constant is then added to it.... this is an example of constant rate of change; the independent variable, x, is never multiplied by itself, nor is it ever logarithmized, trigonometrized (Like my wording? :P) or exponentiated, rather, its change is only brought about by constant numbers.

Now, with a graph of a polynomial function of degree higher than 1 (or any other non-linear function, for that matter) you'll notice the slope is non-constant; it seems the slope gets steeper and steeper, or perhaps even.... the opposite of steeper... this is where calculus comes into play... so here's where I'll define the derivative. (Probably not in a formal way, but enough for you to understand how they work and why you use them)

General definition: A derivative is the rate of change of a function, f(x), at any x value of that function. (In other words, the instantaneous slope)

Numerical method:
-----------------------
Let's go to the function f(x) = x^2, and let's say you wanted to find the slope at point, hmm, say.... x = 3, or (3, 9).
Let's go to the slope formula you should have learnt in Pre-Algebra/Algebra 1:
m = delta y / delta x = (y2 - y1)/(x2 - x1)
Where delta means "change in", that is, the difference between two y or x values.
So if you wanted the slope at this point, (3, 9), using numerical means, you have to pick a point sufficiently close to (3,9) that lies on the graph of f(x) = x^2.... let's pick the point x = 3.000001, or (3.000001, 9.000006).
Now, calculating the approximate slope: (9.000006 - 9)/(3.000001 -3) = (0.00006)/(0.000001) = 6
Using the point slope formula,
(y - 9)/(x - 3) = 6
y = 9 + 6x - 18 = 6x - 9
And that's the slope of the line tangent to the graph f(x) = x^2 at point (3,9)....

That should pretty much cover the numerical method for calculating derivatives...
This gets a little boring sometimes, so I'll introduce the analytical method.

Analytical method
---------------------
OK, here I'll make a very very brief definition of a limit before I explain the analytical method...
Let's say you had a function, f(x) = x - 2.... the limit of that function as x approaches 3 is 1; that is, for a simple linear function you can quite simply plug in the x value which the function approaches.
The limit as x approaches c is denoted as:
lim
x->c
Now, let's say you had the function f(x) = [x^2 - 1]/(x+1); what's the limit of that function as x approaches -1? You can't directly plug it in because then you'd end up with the result f(x) approaches infinity; so when you're dealing with limits, always simplify what you're dealing with as much as possible before tackling it head-on, or you'll get an undesired answer.

Why did I have to introduce limits? Because it's very important to the analytic definition of derivatives, and if I did clarify the simplifying part you would have gone "Whoah, wait... is a derivative defined?"

Anyway, let's go back to the slope formula, which we'll write as Sy/Sx for simplicity, where S denotes delta:
m = delta y / delta x = Sy/Sx
What I did, numerically, was I picked a point that was VERY close to x to approximate a slope with....
This value was so close, you could say it was pretty much the same as x itself, but NOT quite...
Thus, we could say that the difference between the close number I chose, and the x itself, is very close, or approaches, 0..
We can denote this with limits:

m = lim Sy/Sx = (y2 - y1)/Sx
Sx->0

Why did I include the y2 - y1? Let's switch to functional notation:
= lim (f(x2) - f(x1))/Sx
Sx->0

Let's remember what Sx means-- the difference between the two x values I chose; recall how x2 is x1 plus the difference between x2 and x1? We can now rewrite the limit as:

= lim [f(x1 + Sx) - f(x1)]/Sx
Sx -> 0

Or simply:

= lim [f(x + Sx) - f(x)]/Sx
Sx -> 0

And, from this formula, you can derive a formula from a function which allows you to determine the slope of that function at any point...

Let's go back to f(x) = x^2... and use this formula:

lim [(x+Sx)^2 - x^2]/Sx = [x^2 + 2xSx + Sx^2 - x^2]/Sx = (2xSx + Sx^2)/Sx = 2x + Sx
Sx -> 0
And now, since we've simplified, we can directly substitute 0 into Sx, giving:
2x
Now let's test it with the point x = 3 we used in the numerical method section...
2 times 3 is 6, isn't it? That's the exact answer we got with our numerical method!
Since this limit method is long and tedious, most people just use the differentiation rules, which are simple:
C & n are constant, x is the independent variable:
If f(x) is a function, f'(x) denotes its derivative which can be found by the following rules:

f(x) = Cx^n -> f'(x) = Cnx^(n-1) For example, f(x) = 2x^3, f'(x) = 2 * 3 * x^(3 - 1) = 6x^2
f(x) = C -> f'(x) = 0

So anyway, I'm too lazy to elaborate on non-polynomials or whatever, so I'll just go back to our initial question-- how to find the relative minimum and maximum of f(x) = x^3 + 2x^2 + x + 1...
Let's use the above short hand rules to come up with a formula to find the slope:
f'(x) = 3x^2 + 4x + 1
Now, since we know that the slope would = 0 at the minimum and maximum, (since they are stationary points) we'll set f'(x) = 0:
3x^2 + 4x + 1 = 0
And now, we solve for x to get the minimum/maximum:
x = -4 +/- root(16 - 12) / 6 = -4 +/- 2 / 6 = -2 +/- 1 / 3 = -1 or -1/3
Now, which is the minimum and which is the maximum? To tell, we'll use the SECOND DERIVATIVE RULE, which just means you differentiate the function again and plug in these values, (we denote this f''(x)), so....
f''(x) = 6x + 4
Now, let's try -1..
f''(-1) = -2
Since it's negative, that means that that point is a MAXIMUM, that it's concave down....
f''(-1/3) = 2
Since it's positive, that means that that point is a MINIMUM, that it's concave up...

So yeah, I hope you enjoyed my long post, blahblah.

19. : |

20. Code:
```                     -1
how would if find tan (opposite/adjacent) if I only have tan() and atan().```
I did the above, just so i could have superscript. ;:)

21. Nevermind, I just found out that's what atan is. :)

22. well im in 9th grade and im in algegra 2 OOOOOOOO
and if ur so smart, whats the square root of two?

23. Approximately 1.4142135623730951

but your teacher would probably like the answer "the square root of 2" because it's the most exact.

24. Originally Posted by Pseudo ANIMALISTIC
Math is awesome.
What's even more awesome is math homework....
If anyone has any math homework problems, let me know... I can help you out with it.... (My knowledge ranges from Algebra to Calculus 3)
Errm, are you any good with Boolean Algebra, Binary, Logic & Proof, Truth Tables etc? cos my lecturers and teachers who i learn from ain't the best of explainers!!!

25. Originally Posted by infmz
Errm, are you any good with Boolean Algebra, Binary, Logic & Proof, Truth Tables etc? cos my lecturers and teachers who i learn from ain't the best of explainers!!!
Sorry... those are my weak points..

well im in 9th grade and im in algegra 2 OOOOOOOO
and if ur so smart, whats the square root of two?
Consider yourself one of the lucky ones then...
I didn't get to do a formal Algebra 2 course until last year. (10th grade)
Anyway, to approximate the square root of 2 without a calculator, we could use Taylor series around x = 1 since it's pretty close to 2, first forming a function whose root is the square root of 2:
f(x) = x^2 - 2
But it won't really get us anywhere... (Recall the 3rd derivative will be zero, thus the taylor series will not be infinite and it will end up giving us the same function we started with...) so let's use something derived from Taylor series known as the Newton-Raphson method...
It's defined as:
x = x0 - f(x0)/f'(x0)
Through successive guessing of x0, the square root of 2, we will get closer and closer to the answer.
I'll pick one guess of the square root of 2... err... 1.5 let's say...
f(1.5) = 2.25 - 2 = 0.25 = 1/4
f'(x) = 2x
f'(1.5) = 3
Now, let's form the formula for this method..
x = 1.5 - (1/4)/(3) = 1.5 - 1/12 = 3/2 - 1/12 = 18/12 - 1/12 = 17/12...
Let's go onto the next iteration:
f(17/12) = (17/12)^2 - 2 = 289/144 - 288/144 = 1/144
f'(17/12) = 34/12 = 408/144
Thus a better approximation is:
x = 17/12 - (1/144)/(408/144) = 17/12 - 1/408 = (using a calculator at this point, too lazy to do the numbers) = (169 + 408)/408 = 577/408
And that's an approximation of the square root of 2 to about 4-5 decimal places...
It'd get more and more close to the square root of 2 by each iteration, but since we can't exactly define the square root of 2 in terms of finite decimals, let's leave it at that. :P

26. Originally Posted by Pseudo ANIMALISTIC
Sorry... those are my weak points..
No problem. Good to hear that your offering the help. You know of any sources or sites that can help me in the fields specified?

27. Originally Posted by Pseudo ANIMALISTIC
Sorry... those are my weak points..

Consider yourself one of the lucky ones then...
I didn't get to do a formal Algebra 2 course until last year. (10th grade)
Anyway, to approximate the square root of 2 without a calculator, we could use Taylor series around x = 1 since it's pretty close to 2, first forming a function whose root is the square root of 2:
f(x) = x^2 - 2
But it won't really get us anywhere... (Recall the 3rd derivative will be zero, thus the taylor series will not be infinite and it will end up giving us the same function we started with...) so let's use something derived from Taylor series known as the Newton-Raphson method...
It's defined as:
x = x0 - f(x0)/f'(x0)
Through successive guessing of x0, the square root of 2, we will get closer and closer to the answer.
I'll pick one guess of the square root of 2... err... 1.5 let's say...
f(1.5) = 2.25 - 2 = 0.25 = 1/4
f'(x) = 2x
f'(1.5) = 3
Now, let's form the formula for this method..
x = 1.5 - (1/4)/(3) = 1.5 - 1/12 = 3/2 - 1/12 = 18/12 - 1/12 = 17/12...
Let's go onto the next iteration:
f(17/12) = (17/12)^2 - 2 = 289/144 - 288/144 = 1/144
f'(17/12) = 34/12 = 408/144
Thus a better approximation is:
x = 17/12 - (1/144)/(408/144) = 17/12 - 1/408 = (using a calculator at this point, too lazy to do the numbers) = (169 + 408)/408 = 577/408
And that's an approximation of the square root of 2 to about 4-5 decimal places...
It'd get more and more close to the square root of 2 by each iteration, but since we can't exactly define the square root of 2 in terms of finite decimals, let's leave it at that. :P
.....nobody likes a bragger* spelling

28. How would you calculate exponents without a calculator or by multiplying x by itself y times?

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