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further pure 1

Original post by English-help


Have you found dydx\frac{dy}{dx} in terms of y and x??

Now just differentiate both sides again wrt x and multiply both sides by y.
(edited 7 years ago)
Original post by RDKGames
Have you found dydx\frac{dy}{dx} in terms of y and x??

Now just differentiate both sides again wrt x and multiply both sides by y.


Tbh i dont even understand the question , what is it trying to tell me to find out and how should i start?
Original post by English-help
Tbh i dont even understand the question , what is it trying to tell me to find out and how should i start?


Well the first part is asking for a stationary point. Stationary points are found when you set dydx=0\frac{dy}{dx}=0 for an expression.

You have y=xx,x{R:x>0}y=x^{-x}, \forall x \in \{ \mathbb{R} : x>0 \} and you need to differentiate this. Begin by rewriting this as ln(y)=ln(xx)=xln(x)\ln(y)=\ln(x^{-x})=-x\ln(x) and proceed with implicit differentiation on the LHS for dydx\frac{dy}{dx}
(edited 7 years ago)
Original post by RDKGames
Well the first part is asking for a stationary point. Stationary points are found when you set dydx=0\frac{dy}{dx}=0 for an expression.

You have y=xx,x{R:x>0}y=x^{-x}, \forall x \in \{ \mathbb{R} : x>0 \} and you need to differentiate this. Begin by rewriting this as ln(y)=ln(xx)=xln(x)\ln(y)=\ln(x^{-x})=-x\ln(x) and proceed with implicit differentiation on the LHS for dydx\frac{dy}{dx}


Ive done 9a! :/ I was talking about 9b?
Sorry , Well i did say 9b in the OP
http://mathsathawthorn.pbworks.com/f/FP1Jun07.pdf
I need help on q5^ Basically I have done n=1 and n=k , when i put n=k+1 inside the function , it gives me 2-(k+3)(1/2)^k+1
Idk what to do next though
Original post by English-help
Ive done 9a! :/ I was talking about 9b?
Sorry , Well i did say 9b in the OP
http://mathsathawthorn.pbworks.com/f/FP1Jun07.pdf
I need help on q5^ Basically I have done n=1 and n=k , when i put n=k+1 inside the function , it gives me 2-(k+3)(1/2)^k+1
Idk what to do next though


When you said "Tbh i dont even understand the question" I assumed you simply didn't know where to start on the question altogether. For 9b, as I said, just take your dydx=y(1+ln(x))\frac{dy}{dx}=-y(1+\ln(x)) and differentiate it again to find d2xdy2\frac{d^2 x}{dy^2}. Then multiply both sides by yy of the expression and tidy it up.

For Q5 if you start your inductive step with r=1k+1[r(12)r]=2(k+2)(12)k+(k+1)(12)k+1 \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac{1}{2})^{k+1} and up with 2(k+3)(12)k+12-(k+3)(\frac{1}{2})^{k+1} then you've basically proved it. You can turn it into 2[(k+1)+2](12)k+12-[(k+1)+2](\frac{1}{2})^{k+1} for clarity which shows that if the statement is true for n=kn=k then it is true for n=k+1n=k+1 and you just need to write an ending statement to that.
(edited 7 years ago)
Original post by RDKGames
When you said "Tbh i dont even understand the question" I assumed you simply didn't know where to start on the question altogether. For 9b, as I said, just take your dydx=y(1+ln(x))\frac{dy}{dx}=-y(1+\ln(x)) and differentiate it again to find d2xdy2\frac{d^2 x}{dy^2}. Then multiply both sides by yy of the expression and tidy it up.

For Q5 if you start your inductive step with r=1k+1[r(12)r]=2(k+2)(12)k+(k+1)(12)k+1 \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac{1}{2})^{k+1} and up with 2(k+3)(12)k+12-(k+3)(\frac{1}{2})^{k+1} then you've basically proved it. You can turn it into 2[(k+1)+2](12)k+12-[(k+1)+2](\frac{1}{2})^{k+1} for clarity which shows that if the statement is true for n=kn=k then it is true for n=k+1n=k+1 and you just need to write an ending statement to that.


But do you know , how did you get this step when doing n=k+1 , thats what i dont understand :/
r=1k+1[r(12)r]=2(k+2)(12)k+(k+1)(12)k+1 \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac{1}{2})^{k+1}
Original post by English-help
But do you know , how did you get this step when doing n=k+1 , thats what i dont understand :/
r=1k+1[r(12)r]=2(k+2)(12)k+(k+1)(12)k+1 \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac{1}{2})^{k+1}


Right so when you do it for n=1n=1 it works out.

Now assume it works for n=kn=k therefore out inductive assumption is that r=1k[r(12)r]=2(k+2)(12)k\displaystyle \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k

Next we must test it for n=k+1n=k+1

So we have r=1k+1[r(12)r]=r=1k[r(12)r]+(k+1)(12)k+1(k+1)thterm=2(k+2)(12)k inductive assumption+(k+1)(12)k+1\displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] = \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] + \underbrace{(k+1)(\frac{1}{2})^{k+1}}_{(k+1)^{th} \text{term}} = \underbrace{2-(k+2)(\frac{1}{2})^k}_{\text{ inductive assumption}} + (k+1)(\frac{1}{2})^{k+1}
Original post by RDKGames
Right so when you do it for n=1n=1 it works out.

Now assume it works for n=kn=k therefore out inductive assumption is that r=1k[r(12)r]=2(k+2)(12)k\displaystyle \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k

Next we must test it for n=k+1n=k+1

So we have r=1k+1[r(12)r]=r=1k[r(12)r]+(k+1)(12)k+1(k+1)thterm=2(k+2)(12)k inductive assumption+(k+1)(12)k+1\displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] = \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] + \underbrace{(k+1)(\frac{1}{2})^{k+1}}_{(k+1)^{th} \text{term}} = \underbrace{2-(k+2)(\frac{1}{2})^k}_{\text{ inductive assumption}} + (k+1)(\frac{1}{2})^{k+1}


So you get the get the Inductive assumption from n=k and then do n=k=1 which comes from the LHS? + then add both of them?
Original post by English-help
So you get the get the Inductive assumption from n=k

Yes we need to establish an assumption before we can proceed with the proof.

and then do n=k+1 which comes from the LHS? + then add both of them?


You test for n=k+1n=k+1 and to do so you need to make use of the assumption you made earlier. The only way to do this is to split the sum from r=1r=1 to r=k+1r=k+1 into 2 different terms where one can be replaced by the assumption (the one which goes from r=1 to r=k) and the other is whatever is left over to balance the sum (the (k+1)th term). So then you just add the two things and should end up with an expression for n=k+1n=k+1
(edited 7 years ago)
Original post by English-help
So you get the get the Inductive assumption from n=k and then do n=k=1 which comes from the LHS? + then add both of them?


To add to what RDK games has said, the whole structure of a proof by induction is to make some use of your assumption in the inductive step - that's where the real power of this type of proof comes from.

In this case, you have assumed that r=1k((12)rr)=2(k+2)(12)k \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) = 2 - \displaystyle \left(k+2\right)\left(\frac{1}{2} \right)^{k} .

So a nice way in which we can use our assumption here is if we partition the sum in our inductive step:

r=1k+1((12)rr)=r=1k((12)rr)+the (k+1)th term \displaystyle\sum_{r=1}^{k+1} \left( \left(\frac{1}{2}\right)^r r\right) = \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) + \text{the }(k+1)^{\text{th}} \text{ term} ,

which allows us to use the assumption by just substitution.
Original post by crashMATHS
To add to what RDK games has said, the whole structure of a proof by induction is to make some use of your assumption in the inductive step - that's where the real power of this type of proof comes from.

In this case, you have assumed that r=1k((12)rr)=2(k+2)(12)k \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) = 2 - \displaystyle \left(k+2\right)\left(\frac{1}{2} \right)^{k} .

So a nice way in which we can use our assumption here is if we partition the sum in our inductive step:

r=1k+1((12)rr)=r=1k((12)rr)+the (k+1)th term \displaystyle\sum_{r=1}^{k+1} \left( \left(\frac{1}{2}\right)^r r\right) = \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) + \text{the }(k+1)^{\text{th}} \text{ term} ,

which allows us to use the assumption by just substitution.

So you get the K+1 term when you put k+1 into the left hand side basically and the K term from when you do n=k? right?
Original post by English-help
So you get the K+1 term when you put k+1 into the left hand side basically and the K term from when you do n=k? right?


Yes, if I understand what you're saying correctly :smile:
Original post by crashMATHS
Yes, if I understand what you're saying correctly :smile:


Thank you:smile:

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