Question

Anyone want to help me on Number 2? I have some ideas but I'm not honestly sure...

Answer 1

Sure

Answer 2

1st year Calculus.

Answer 3

@nincompoop ,@amistre64

Answer 4

@amistre64

Answer 5

solve for R, what do we get?

Answer 6

it's already solved for R.

Answer 7

Ohh I see.

Answer 8

\[R=\frac{ 1 }{ (\frac{ 1 }{ R1 } + \frac{ 1 }{ R2 })}\]

Answer 9

\[\frac1R=\frac{R_1+R_2}{R_1R_2}\]

\[R=\frac{R_1R_2}{R_1+R_2}\]

Answer 10

yes I did that next.

Answer 11

I can't see how they got a square root.

Answer 12

Multiply by conjugate perhaps?

Answer 13

well, lets find R^2 and see if that helps, then theres the restrictions they provide to test things

Answer 14

So square both sides?

Answer 15

if R^2 < R1R2/2 ..... we can start to make somme assumptions

Answer 16

Yeah but we just have R. I guess we square both sides to make it R^2.

Answer 17

yep

Answer 18

this is just an idea at the moment, with any luck it heads us in the right direction

Answer 19

Yep.

Answer 20

\[R^2=\frac{(R_1R_2)^2}{R_1^2+R_2^2+2R_1R_2}\]

Answer 21

Nice latex!

Answer 22

Yep I got your answer so far...

Answer 23

now can we compare this with R^2 setup with R1R2/2 to show that its less than it
\[\frac{(R_1R_2)^2}{R_1^2+R_2^2+2R_1R_2}<\frac{R_1R_2}{2}\]

Answer 24

How did you get R1R2/2 ?

Answer 25

if we cross multiply .... its stated as a criteria already: R < sqrt(R1R2/2) therefore R^2<R1R2/2 for some condition i believe, something greater than 1 maybe

\[2(R_1R_2)^2<R_1R_2({R_1^2+R_2^2+2R_1R_2})\]

Answer 26

Ohh I see. That is what we are trying to show.

Answer 27

1/3 = sqrt(1/9)

1/9 = 1/9 or maybe no restrictions

Answer 28

yes, we are trying to determine a reasonable approach, we know R^2 is equal to what we found, now we need to compare it to what is stated

Answer 29

\[2(R_1R_2)^2<{R_1^3R_2+R_1R_2^3+2(R_1R_2)^2}\]

since the R parts are >0 then when we add something to what we have, it gets bigger right?

Answer 30

But problem. We eliminated the R from the whole equation.

Answer 31

no we did not eliminate it, we used an equivalent statement for it

Answer 32

Okay. I just thought we had to match the statement given in the question. But yes this is perfectly valid.

Answer 33

we said: R = k, then used k as a substutite to work with since it has the same parts that we are trying to proof with

Answer 34

we could have replaced R by 1/R1 + 1/R2 to start with, but i just took the path i did because im old and stubborn i guess lol

Answer 35

It's fine :P .

Answer 36

Divide both sides by R1R2?

Answer 37

\[\color{red}{R^2}=\left(\frac{(R_1R_2)^2}{R_1^2+R_2^2+2R_1R_2}\right)<\left(\sqrt{\frac{R_1R_2}{2}}\right)^2\]

Answer 38

1/3 ^2 = 1/9

so when we are less than 1, the square is smaller .....

Answer 39

this might have to be worked into it somehow

Answer 40

I'm looking...

Answer 41

let me gather my idea together to see how valid it is

Answer 42

Factor the bottom back to what it was before maybe...?

Answer 43

show that:\[\frac{1}{R_1}+\frac{1}{R_2}<\sqrt{\frac{R_1R_2}{2}}\]

Answer 44

pfft, sqrt(2/R1R2) since 1/R is = to .... slight typo

Answer 45

\[\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}<\sqrt{\frac{R_1R_2}2}\]
thats the objective .... agreed?

Answer 46

I don't see a typo. You're just stating the thing we have to find.

Answer 47

yeah, just trying to get it subbed in at the start to work ot

Answer 48

Why the reciprocal?

Answer 49

Ohh wait nvm. I'm dumb.

Answer 50

we want to show that R < sqrt(....), but 1/R = 1/R1 + 1/R2

Answer 51

I get it.

Answer 52

yeah yeah :P .

Answer 53

we can take a value in the interval: 0<k<1, say k=1/4 .... sqrt(1/4) = 1/2 or (1/2)^2 < 1/2

Answer 54

so if R1 and R2 are small .... does this work out well for us

Answer 55

Define "small" .

Answer 56

between 0 and 1

Answer 57

.000000000000000000001 lol

Answer 58

Okay :P .

Answer 59

Maybe we can manipulate the LHS like we did before?

Answer 60

maybe

Answer 61

I'm trying afew things myself.

Answer 62

thats always good, since my ideas may or maynot be useful: show that IF
\[\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}<\sqrt{\frac{R_1R_2}2}\]
so we assume this to be true

1/3 < 1/2

1/9 < 1/4 still hold true for the squares

so my concerns at the moment seem to be without merit, square both sides and the inequality holds

Answer 63

Induction?

Answer 64

no need for induction, my concern was for some reason an interval between 0 and 1. had to make sure that i wasnt introducing a bad element

Answer 65

if 0 < a < b < 1 < c then squaring it all:

0 < a^2 < b^2 < 1 < c^2 the ineuality still holds true

Answer 66

That is true but how does that help us?

Answer 67

it simply means that we can work the squares to show that the inequailty is good
\[\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}<\sqrt{\frac{R_1R_2}2}\]
\[\frac{R_1R_2}{R_1+R_2}<\sqrt{\frac{R_1R_2}2}\]
\[(\frac{R_1R_2}{R_1+R_2})^2<(\sqrt{\frac{R_1R_2}2})^2\]
\[\frac{(R_1R_2)^2}{R^2_1+R^2_2+2R_1R_2}<\frac{R_1R_2}2\]
since R1 and R2 are positive, then we have no negative stuff to contend with and the sign issues are no problem
\[2(R_1R_2)^2<{R_1R_2}({R^2_1+R^2_2+2R_1R_2})\]
\[2(R_1R_2)^2<{R^3_1R_2+R_1R^3_2+2(R_1R_2})^2\]
subtracting 2(RR)^2 we get
\[0<{R^3_1R_2+R_1R^3_2}\]
which is true for any R1, R2 > 0 regardless of size

Answer 68

Wonderful! Thanks :D .

Answer 69

we can factor out the R1 R2 and divide off the excess to show that 0<R1R2 as well, ut how far you want to take it is up to you

Answer 70

I'm taking complex analysis and I can't do basic algebra...

Answer 71

i asked my discrete math teacher: do you know how much algebra i had to forget to learn this?

Answer 72

Haha xD .