Inopeki Group Title Algebra, Turingtest :D 2 years ago 2 years ago

1. TuringTest Group Title

FOIL Inopeki :D (a+b)(c+d)

2. Inopeki Group Title

3. TuringTest Group Title

nice (x+3)(x-4)

4. Inopeki Group Title

x^2-4x+3x-12

5. TuringTest Group Title

good, simplify

6. Inopeki Group Title

x^2-1x-12

7. TuringTest Group Title

Good, (no need to write the 1 though) (3p+t)(2p-4t)

8. Inopeki Group Title

Oh right 6p^2+3p*-4t+t*2p+5t?

9. TuringTest Group Title

not 5t... everything else is right though

10. Inopeki Group Title

-5t

11. TuringTest Group Title

no... look closely

12. Inopeki Group Title

-5t^2?

13. TuringTest Group Title

-4t^2 :P

14. Inopeki Group Title

Huh :O

15. TuringTest Group Title

You don't see it? (-4t)(t)=-4t^2 look at the last terms in the parentheses

16. Inopeki Group Title

Ohhh, now i see it

17. TuringTest Group Title

so you have what then, before simplifying?

18. Inopeki Group Title

6p^2+3p*-4t+t*2p+4t^2?

19. TuringTest Group Title

-4t^2 :P

20. Inopeki Group Title

Oh right xd

21. TuringTest Group Title

can you simplify it?

22. Inopeki Group Title

6p^2+3p*-4t+t*2p-4t^2 6p^2+6p^2-4t^2-4t^2?

23. TuringTest Group Title

no... how did you the p^2 in the middle? we had 6p^2+3p(-4t)+t(2p)-4t^2 parentheses look cleaner btw, you should use them here what is 3p(-4t) and t(2p) ???

24. Inopeki Group Title

-12pt? 2pt?

25. TuringTest Group Title

right, so then we have...?

26. Inopeki Group Title

6p^2-12pt+2pt-4t^2

27. TuringTest Group Title

yes :D and how does that simplify?

28. Inopeki Group Title

I dont know actually, maybe 6p^2-14pt+pt-4t^2?

29. Inopeki Group Title

No

30. TuringTest Group Title

they are like terms in the middle -12pt+2pt=-10pt this goes for any combination of variables a^2bc+3a^2bc=4a^2bc wr^4dt^3+3wr^2dt+6wr^4dt^3=7wr^4dt^3+3wr^2dt notice in this last example we couldn't combine all the terms because one had different exponents. We can only combine when they are the exact same in exponents and number of variables.

31. Inopeki Group Title

But dont i need to multiply them to get -10pt?

32. TuringTest Group Title

no, just add the coefficients factoring is one way to see it... factor out 'pt' from the middle two terms: $6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2$

33. Inopeki Group Title

Oh wait, i get it now, it would me multiplication if it was -10pt^2

34. TuringTest Group Title

not necessarily. You can't multiply things unless the equation tells you too, or you do a trick like multiply both sides if you have -10pt^2+7pt you cannot simplify that...

35. Inopeki Group Title

Oh ok. But what i meant was that if it were to tell us to multiply it would look like this -10pt^2 and not this -10pt

36. TuringTest Group Title

if you multiplied you get $(2pt)(-12pt)=-24p^2t^2$

37. Inopeki Group Title

Oh right

38. Inopeki Group Title

Since you would be multiplying both p and t with itself

39. TuringTest Group Title

right

40. Inopeki Group Title

Whatever, were getting off track :)

41. TuringTest Group Title

not really, you need to get this down too. It's all related, and it's good to know where you're at. (2s^2+p)(s-3p^3)

42. TuringTest Group Title

hey I'm green! never been a mod for math before :)

43. Inopeki Group Title

Congratulations! You deserve it :)

44. TuringTest Group Title

thanks :) I can't delete things though, don't see the point oh well... (2s^2+p)(s-3p^3)

45. Inopeki Group Title

(2s^2+p)(s-3p^3) 2s^3+2s^3(-3p^3)+p(s)+3p^4

46. TuringTest Group Title

so far so good now try to simplify...

47. TuringTest Group Title

oh wait, small error

48. Inopeki Group Title

OH the -

49. TuringTest Group Title

2s^3+2s^3(-3p^3)+p(s)+3p^4 ^^ should be???

50. Inopeki Group Title

2s^3+2s^2(-3p^3)+p(s)-3p^4 ^^ plus that :)

51. TuringTest Group Title

good catch I missed the sign lol !!! yes excellent now can you simplify it

52. Inopeki Group Title

Thanks :D 2s^3+2s^2(-3p^3)+p(s)-3p^4 2s^3-6s^2p^3+sp-3p^4?

53. TuringTest Group Title

perfect last one: (a+b)(a-b)

54. Inopeki Group Title

a^2-ab+ab-b^2? a^2-2ab-b^2?

55. TuringTest Group Title

not quite, I don't think those terms in the middle add...

56. Inopeki Group Title

The simplification or the first one?

57. TuringTest Group Title

a^2-ab+ab-b^2 ^^^^^ what is -x+x ?

58. Inopeki Group Title

Oh right, 0

59. TuringTest Group Title

so (a+b)(a-b) is?

60. Inopeki Group Title

a^2-b^2?

61. pratu043 Group Title

Some kind of coaching class going on here?

62. TuringTest Group Title

And that is a very important formula @pratu Inopkie wants to learn Algebra, a good cause in my book :)

63. pratu043 Group Title

ok...

64. pratu043 Group Title

Shall I help?

65. TuringTest Group Title

Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)

66. TuringTest Group Title

@Inopeki We just showed that$(a+b)(a-b)=a^2-b^2$That relation is called the 'Difference of Squares' formula. It is used all the time, as you will see, as a quick way to factor anything of the form$a^2-b^2$

67. Inopeki Group Title

so a^2-b^2=(a+b)(a-b)?

68. pratu043 Group Title

Its an identity.

69. TuringTest Group Title

Yes, it's always true for that form. This is handy as you will see because if you remember how factoring helped us solve other equations, it will do the same for us here: For instance$x^2-4=0$we can factor this..$(x-2)(x+2)=0$which as I mentioned befor by the 'zero factor property' means that we need to solve$x-2=0$and$x+2=0$to find the zeros of our polynomial

70. Inopeki Group Title

But isnt (x-2)(x+2) 0 because the 2s negate?

71. TuringTest Group Title

foil it out and tell me yourself

72. Inopeki Group Title

x^2+2x-2x+-2*2 Well pellet

73. TuringTest Group Title

right which simplified too what?

74. Inopeki Group Title

x^2-4

75. TuringTest Group Title

so that's not always zero, is it? if x=1 then f(x)=-3, not zero what the expression f(x) is equal to here depends on what we put for x. There are in fact only two values for x that will make f(x)=0 Which should make sense by what I said yesterday: The order of the equation is 2, so the total multiplicity of its roots is 2 (fundamental theorem of algebra)

76. Inopeki Group Title

isnt that f(1)=

77. TuringTest Group Title

yes, sorry, bad notation on my part f(1)=-3 f(0)-4 etc.

78. Inopeki Group Title

But how does that work?

79. TuringTest Group Title

which part?

80. Inopeki Group Title

Functions, maybe im not ready for that subject

81. TuringTest Group Title

No you are, perhaps I haven't been clear enough whenever I write x^2-4 If I want to know where this graph hits the x-axis it is useful to call this a function f(x) here we have f(x)=x^2-4 Think about a graph: When the graph is toughing the x-axis it is at y=0, correct?

82. TuringTest Group Title

I just 'called' it a function, and made it graph-able...

83. Inopeki Group Title

Yeah, when x and y intersect its 0,0

84. Inopeki Group Title

oooh

85. Inopeki Group Title

f(x)=y+2x-3 So if we want to figure out y we need to plug in something for x

86. TuringTest Group Title

well, wait now, if you have a y in the function then it is not f(x) here's one thing that I bet is confusing you: f(x)=y when we write a line equation y=mx+b we are saying something about the graph: that the y coordinate is proportional to the x coordinate in this way. However this is rather simplistic. It is really more that we have a function f(x)=mx+b and we have chosen to graph it by letting the y-coordinate of our graph be f(x) they are effectively the same here, but a function can be graphed in many different types of coordinates, so y would not be used unless we are in Cartesian coordinates as we are now.

87. TuringTest Group Title

hence it's better to think of graphs in terms of functions, represented in Cartesian coordinates by y

88. Inopeki Group Title

What are cartesian coordinates?

89. TuringTest Group Title

the regular ones you know x-axis and y-axis

90. Inopeki Group Title

Yeah

91. TuringTest Group Title

there are other coordinate systems like polar, spherical, and cylindrical coordinates If we tried to graph our function in polar coordinates, y would have no meaning. however the zeros of the function are independent of how we choose to graph it. The math will be the same. Hence when we want to know the zeros of some polynomial we call it f(x), and that allows us to put numbers into it and see what happens f(x)=x^2-4 f(0)=0-4=-4 f(1)=1-4=-3 f(2)=4-4=0 so we found a zero simply by calling this a function and putting numbers into it.

92. Inopeki Group Title

One question, what do we gain from functions, what are the numbers on the RHS?

93. TuringTest Group Title

which numbers on the RHS?

94. Inopeki Group Title

Like f(0)=0-4=-4<--- ^ |

95. TuringTest Group Title

well that's the simplified form of f(0): f(0)=-4 check it by plugging in x=0 into the function

96. Inopeki Group Title

But if we know that f(0)=-4 why do we do it?

97. TuringTest Group Title

lol, why /don't/ we want to do that?! Almost all things in physics can be represented by function. If you want to know when the water pressure of a system will become too high, what trajectory to shoot your spaceship, or where a quantum particle will be at a certain time, you must analyse functions in this way.

98. Inopeki Group Title

Really?

99. Inopeki Group Title

Then teach me! This is important!

100. TuringTest Group Title

Really, though of course actual physics problems are usually (though not always) a bit trickier than just plugging in a number but yes, functions are instrumental to calculus, and therefor physics As I said earlier calculus is about analyzing functions... I guess we should start to talk about functions in a very detailed way.

101. Inopeki Group Title

102. TuringTest Group Title

the definition of a function for you will be: if you have something like y=f(x) (where f(x) is some arbitrary polynomial with x's in it) and for every x you put in you get exactly one number for y back, then y=f(x) is a function, and y is a function of x try to soak that in, I'll tell you more about it momentarily

103. Inopeki Group Title

So what you are saying is that if f(x)=y x=y?

104. Inopeki Group Title

Well not really

105. Inopeki Group Title

What you are saying is that if x is one and y is 2 and you add 1 to x you add one to y aswell?

106. TuringTest Group Title

no, just that we get only one value of y for each x for instance f(x)=y I defined that, whether or not it's a function however if for every x we put in we get only one y back, then f(x) is a function, and the variable y is a function of the variable x

107. TuringTest Group Title

scratch the *for instance :/

108. Inopeki Group Title

Gotta go eat dinner, brb

109. TuringTest Group Title

look at a graph like a circle (I know you haven't studied them much yet this way)|dw:1326043068666:dw|this is not a function because there is more than one y for each value of x... see you in a bit, I'm gonna have breakfast.

110. TuringTest Group Title

drawing a line from x to its corresponding y value you should know we need a vertical line|dw:1326043155512:dw|see how one x-coordinate corresponds to two different y values y1 and y2? But what about y=2x+5 ?|dw:1326043286670:dw|this graph has exactly one value of y for each value of x, so it is a function.

111. TuringTest Group Title

the idea above is what's called the 'vertical line test': if you can draw a vertical line at any point on the graph and intersect it twice, then it is not a function.

112. TuringTest Group Title

when you return look at this say we have a table of values for some y=f(x) and we want to know if f(x) is a function.... look at the table x |y=f(x) --------- -1 | 2 3 | 7 2 | 0 5 | 2 Is this a function? why or why nmot?

113. Inopeki Group Title

Its not. EXEPT the one that says 2 | 0<---------------< cause that means that there is only one value of x ^

114. Inopeki Group Title

The others seem to have two values of x

115. TuringTest Group Title

actually look at the x's each x has one y, so it is a function we have that f(5)=f(-1)=5 so two different x's give the same y, but that is okay the only thing against the rules would be to have something like this x | y ------ 1 | 3 2 | 5 1 | 4 because this means that f(1) has two values: f(1)=3 and 4 so this is not a function... look at the x to see which gives more than one f(x)... (I am using the terms f(x) and y here interchangeably) x | y ----- 2 | 0 3 | 1 5 | 6 3 | 6 1 | 0 function or not ant why?

116. Inopeki Group Title

Yeah

117. TuringTest Group Title

That's right FFM does the curriculum meet your approval? I doubt it with your standards :P

118. Inopeki Group Title

x | y ----- 2 | 0 =f(2)=2? im sorry, i dont really understand 3 | 1 5 | 6 3 | 6 1 | 0

119. TuringTest Group Title

f(2)=0 that means when x=2, y=0, yes?

120. Inopeki Group Title

Ooooh

121. TuringTest Group Title

that's all there is to that so is there any f(x) on the list that gives more than one number for a single x?

122. Inopeki Group Title

So, x | y ----- 2 | 0 = f(2)=0 3 | 1 = f(3)=1 5 | 6 = f(5)=6 3 | 6 = f(3)=6 1 | 0 = f(1)=0 So, no.

123. TuringTest Group Title

actually what is f(3) ?

124. Inopeki Group Title

Oh! 1 and 6!

125. TuringTest Group Title

exactly, so not a function :)

126. Inopeki Group Title

f(3)=1+6 so its not a function!

127. Inopeki Group Title

YEah :D

128. Inopeki Group Title

And that means that f(3) could be graphed in a circle?

129. TuringTest Group Title

right not 1+6=7 of course, it equal the set$f(3)=\left\{ 1,6 \right\}$ It doesn't mean it would be a circle but it would fail the vertical line test, watch...

130. Inopeki Group Title

Of course.

131. Inopeki Group Title

Y cant have to places

132. Inopeki Group Title

two*

133. Inopeki Group Title

:)

134. TuringTest Group Title

thanks FFM, praise form Caeser :) |dw:1326046327322:dw|now look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?

135. Inopeki Group Title

No way!

136. TuringTest Group Title

lol *Praise from Ceaser

137. TuringTest Group Title

no way to draw a vertical line and hit more than one point you say?

138. Inopeki Group Title

Not more than 3.

139. Inopeki Group Title

And there are 5 so..

140. TuringTest Group Title

more than 3? You only need to hit 2 points to show that this is NOT a function and here it is:|dw:1326046665544:dw|so again we show by this that f(3) has more than one value...

141. Inopeki Group Title

Yeah

142. TuringTest Group Title

are the following functions?|dw:1326046773020:dw|(I'll fill in different graph above)

143. Inopeki Group Title

f(x)=y?

144. TuringTest Group Title

yes f(x)=y because we have defined it that way now:|dw:1326046848239:dw|function?

145. Inopeki Group Title

NO

146. TuringTest Group Title

draw the vertical line that hits more than one point on the graph if that is true

147. Inopeki Group Title

No wait, there are no points

148. Inopeki Group Title

|dw:1326046981078:dw|

149. TuringTest Group Title

there are infinite points on a line.... a line intersects another line at a point|dw:1326047055394:dw|P is our intersect point can you draw a vertical line that hits the curve at more than one point?

150. Inopeki Group Title

Oh, so y is a vertical line?

151. Inopeki Group Title

Then this could be a function

152. Inopeki Group Title

|dw:1326047157974:dw| not function

153. Inopeki Group Title

But yours is

154. TuringTest Group Title

now you got the idea :D|dw:1326047282819:dw|function?

155. Inopeki Group Title

No.

156. TuringTest Group Title

show

157. Inopeki Group Title

|dw:1326047285288:dw|

158. TuringTest Group Title

just one line is sufficient, but yes :)||dw:1326047413725:dw|function?

159. Inopeki Group Title

Yes

160. TuringTest Group Title

good :) x | y ----- 1 | 3 2 | 3 3 | 3 4 | 3 function?

161. Inopeki Group Title

Yes

162. Inopeki Group Title

For all of them

163. Inopeki Group Title

But if it was 1|3 1|4 It would not

164. TuringTest Group Title

good x | y ----- 0 | 3 0 | 2 0 | 1 0 | 0 I think you already proved you know this, but function?

165. Inopeki Group Title

No

166. Inopeki Group Title

0 has 4 values of y

167. TuringTest Group Title

nice job, exactly slightly different: f(x)=x+3 function?

168. Inopeki Group Title

I guess that depends on what you plug in for x?

169. Inopeki Group Title

Actually, no

170. Inopeki Group Title

It is a function

171. Inopeki Group Title

It doesnt show multiple values of y, just x+3

172. TuringTest Group Title

Excellent!!!!!!!!

173. Inopeki Group Title

Yes! :D

174. TuringTest Group Title

perfect observation :) every x we put in gives us only one y. y=0 is y a function of x?

175. Inopeki Group Title

Only if we know that 0=x

176. TuringTest Group Title

not so, plug in various x into the function f(x)=0, we get the table x | y ---- 0 | 0 1 | 0 2 | 0 3 | 0 and so on... and I think you would agree that this chart constitutes a function.

177. TuringTest Group Title

tricky one, but now you know ;)

178. Inopeki Group Title

Yes, that chart constitutes a function. But if there were no chart?

179. TuringTest Group Title

They are all in agreement, the chart, the graph, everything. we don't need x=0 to get one value of y back; we always get one value of y. The same value of y, namely zero. do you know what the graph of y=0 looks like?

180. Inopeki Group Title

|dw:1326048301595:dw|

181. Inopeki Group Title

Thats 0,0

182. TuringTest Group Title

yes but I don't want the point (0,0) y=0 is a line: it's actually the x-axis the line x=0 is the y-axis think about that for a minute

183. Inopeki Group Title

Oh, right. |dw:1326048543616:dw|

184. Inopeki Group Title

Didnt really think of that

185. TuringTest Group Title

exactly, and does the graph of y=0 pass the vertical line test?

186. Inopeki Group Title

No

187. TuringTest Group Title

are you saying that you can draw a vertical line that intersects it twice? show...

188. Inopeki Group Title

OH you mean that? Never mind lol

189. TuringTest Group Title

so is the graph indicative of a function or not?

190. Inopeki Group Title

Yes, it is

191. TuringTest Group Title

perfect, what about the line x=0? (standard cartesian notation here , don't get hung up on the names of variables, we could easily have them to t and p or whatever)

192. Inopeki Group Title

|dw:1326049009726:dw|

193. TuringTest Group Title

yes, so is it a function?

194. Inopeki Group Title

Yes.

195. TuringTest Group Title

be careful can we draw a vertical line that touches the graph at more than one point? what would the table of values look like?

196. Inopeki Group Title

No wait, this becomes f(x)=x right?

197. TuringTest Group Title

no, this would be f(x)|x=0 (read f of x such that x is zero) here f(x) takes on a range of values from negative to positive infinity: x | f(x) ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 etc. not a function, right?

198. Inopeki Group Title

I think i get it, its because there is no y?

199. Inopeki Group Title

but f(x) has endless possibilities

200. TuringTest Group Title

no this chart is the same as x | y ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 I told you that y and f(x) are interchangeable y is the name of the coordinate which we are using to represent f(x) so here they are the same the chart itself is of the same type as you saw before, we have one value of x corresponding to multiple (in this case infinite) values of y, so it is not a function.

201. Inopeki Group Title

Ohhh, but if you had y corresponding to multiple values of x it would be a function.

202. TuringTest Group Title

right :)

203. Inopeki Group Title

Ah, now i see!

204. TuringTest Group Title

let me try to clear a few things up... what is the dependent and what is the independent variable here?

205. Inopeki Group Title

The dependent variable is y because if its a function or not depends on y having multiple values corresponding to one value of x.

206. TuringTest Group Title

right, but I'm not sure about your reasoning, so what about f(y)=x=y^2+3y which is dependent and which is independent?

207. Inopeki Group Title

x is the dependent one i think

208. TuringTest Group Title

exactly, so here x is a function of y we could have therefor more than one y value for an x value in this case, but not more than one x value for each y, if it is to be a function. so really the rule is better stated this way: "a relation is a function if each value of the independent variable corresponds to exactly one value of the dependent variable" because we can change the names or swap positions of the variables

209. Inopeki Group Title

Because we can have more of y but not more of x

210. Inopeki Group Title

exactly

211. TuringTest Group Title

exactly, look at the new version of our rule above (in quotes)

212. Inopeki Group Title

Yeah, that sounds right.

213. TuringTest Group Title

so whenever we see f(t)=r for instance which is dependent and which ins independent?

214. Inopeki Group Title

We should move to another "thread" this one is lagging for me

215. TuringTest Group Title