## Inopeki 3 years ago Algebra, Turingtest :D

1. TuringTest

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

2. Inopeki

3. TuringTest

nice (x+3)(x-4)

4. Inopeki

x^2-4x+3x-12

5. TuringTest

good, simplify

6. Inopeki

x^2-1x-12

7. TuringTest

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

8. Inopeki

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

9. TuringTest

not 5t... everything else is right though

10. Inopeki

-5t

11. TuringTest

no... look closely

12. Inopeki

-5t^2?

13. TuringTest

-4t^2 :P

14. Inopeki

Huh :O

15. TuringTest

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

16. Inopeki

Ohhh, now i see it

17. TuringTest

so you have what then, before simplifying?

18. Inopeki

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

19. TuringTest

-4t^2 :P

20. Inopeki

Oh right xd

21. TuringTest

can you simplify it?

22. Inopeki

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

23. TuringTest

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

-12pt? 2pt?

25. TuringTest

right, so then we have...?

26. Inopeki

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

27. TuringTest

yes :D and how does that simplify?

28. Inopeki

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

29. Inopeki

No

30. TuringTest

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

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

32. TuringTest

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

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

34. TuringTest

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

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

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

37. Inopeki

Oh right

38. Inopeki

Since you would be multiplying both p and t with itself

39. TuringTest

right

40. Inopeki

Whatever, were getting off track :)

41. TuringTest

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

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

43. Inopeki

Congratulations! You deserve it :)

44. TuringTest

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

45. Inopeki

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

46. TuringTest

so far so good now try to simplify...

47. TuringTest

oh wait, small error

48. Inopeki

OH the -

49. TuringTest

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

50. Inopeki

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

51. TuringTest

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

52. Inopeki

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

53. TuringTest

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

54. Inopeki

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

55. TuringTest

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

56. Inopeki

The simplification or the first one?

57. TuringTest

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

58. Inopeki

Oh right, 0

59. TuringTest

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

60. Inopeki

a^2-b^2?

61. pratu043

Some kind of coaching class going on here?

62. TuringTest

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

63. pratu043

ok...

64. pratu043

Shall I help?

65. TuringTest

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

66. TuringTest

@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

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

68. pratu043

Its an identity.

69. TuringTest

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

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

71. TuringTest

foil it out and tell me yourself

72. Inopeki

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

73. TuringTest

right which simplified too what?

74. Inopeki

x^2-4

75. TuringTest

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

isnt that f(1)=

77. TuringTest

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

78. Inopeki

But how does that work?

79. TuringTest

which part?

80. Inopeki

Functions, maybe im not ready for that subject

81. TuringTest

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

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

83. Inopeki

Yeah, when x and y intersect its 0,0

84. Inopeki

oooh

85. Inopeki

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

86. TuringTest

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

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

88. Inopeki

What are cartesian coordinates?

89. TuringTest

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

90. Inopeki

Yeah

91. TuringTest

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

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

93. TuringTest

which numbers on the RHS?

94. Inopeki

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

95. TuringTest

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

96. Inopeki

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

97. TuringTest

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

Really?

99. Inopeki

Then teach me! This is important!

100. TuringTest

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

102. TuringTest

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

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

104. Inopeki

Well not really

105. Inopeki

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

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

scratch the *for instance :/

108. Inopeki

Gotta go eat dinner, brb

109. TuringTest

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

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

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

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

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

114. Inopeki

The others seem to have two values of x

115. TuringTest

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

Yeah

117. TuringTest

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

118. Inopeki

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

119. TuringTest

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

120. Inopeki

Ooooh

121. TuringTest

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

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

actually what is f(3) ?

124. Inopeki

Oh! 1 and 6!

125. TuringTest

exactly, so not a function :)

126. Inopeki

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

127. Inopeki

YEah :D

128. Inopeki

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

129. TuringTest

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

Of course.

131. Inopeki

Y cant have to places

132. Inopeki

two*

133. Inopeki

:)

134. TuringTest

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

No way!

136. TuringTest

lol *Praise from Ceaser

137. TuringTest

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

138. Inopeki

Not more than 3.

139. Inopeki

And there are 5 so..

140. TuringTest

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

Yeah

142. TuringTest

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

143. Inopeki

f(x)=y?

144. TuringTest

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

145. Inopeki

NO

146. TuringTest

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

147. Inopeki

No wait, there are no points

148. Inopeki

|dw:1326046981078:dw|

149. TuringTest

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

Oh, so y is a vertical line?

151. Inopeki

Then this could be a function

152. Inopeki

|dw:1326047157974:dw| not function

153. Inopeki

But yours is

154. TuringTest

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

155. Inopeki

No.

156. TuringTest

show

157. Inopeki

|dw:1326047285288:dw|

158. TuringTest

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

159. Inopeki

Yes

160. TuringTest

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

161. Inopeki

Yes

162. Inopeki

For all of them

163. Inopeki

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

164. TuringTest

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

165. Inopeki

No

166. Inopeki

0 has 4 values of y

167. TuringTest

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

168. Inopeki

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

169. Inopeki

Actually, no

170. Inopeki

It is a function

171. Inopeki

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

172. TuringTest

Excellent!!!!!!!!

173. Inopeki

Yes! :D

174. TuringTest

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

175. Inopeki

Only if we know that 0=x

176. TuringTest

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

tricky one, but now you know ;)

178. Inopeki

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

179. TuringTest

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

|dw:1326048301595:dw|

181. Inopeki

Thats 0,0

182. TuringTest

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

Oh, right. |dw:1326048543616:dw|

184. Inopeki

Didnt really think of that

185. TuringTest

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

186. Inopeki

No

187. TuringTest

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

188. Inopeki

OH you mean that? Never mind lol

189. TuringTest

so is the graph indicative of a function or not?

190. Inopeki

Yes, it is

191. TuringTest

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

|dw:1326049009726:dw|

193. TuringTest

yes, so is it a function?

194. Inopeki

Yes.

195. TuringTest

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

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

197. TuringTest

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

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

199. Inopeki

but f(x) has endless possibilities

200. TuringTest

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

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

202. TuringTest

right :)

203. Inopeki

Ah, now i see!

204. TuringTest

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

205. Inopeki

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

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

x is the dependent one i think

208. TuringTest

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

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

210. Inopeki

exactly

211. TuringTest

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

212. Inopeki

Yeah, that sounds right.

213. TuringTest

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

214. Inopeki

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

215. TuringTest

216. Inopeki

r is the dependent and t is independent

217. TuringTest

exactly :)