C2 - Logarithms

Original post by Chittesh14

Is my answer right in terms of explanation or have I blundered somewhere.
The answer is 3 I got that :redface:

You say it cant be negative but you don't explain why. Use the cubic for that.

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Reply 181

Zacken

Original post by Chittesh14

Is my answer right in terms of explanation or have I blundered somewhere.
The answer is 3 I got that :redface:

You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check

a = \pm 3

, in the cubic, you'll find that

a=3

is the only valid solution.

Reply 182

Original post by Zacken

You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check

a = \pm 3

, in the cubic, you'll find that

a=3

is the only valid solution.

Oh lol. Thanks, yeah even if I put a = -3 in my other simultaneous equation, it doesn't work.

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Reply 183

Original post by RDKGames

You say it cant be negative but you don't explain why. Use the cubic for that.

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Thanks, I just used the second simultaneous equation lol it's shorter that way.

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Reply 184

Original post by Zacken

Original post by RDKGames

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I tried working out part b by using the argument of Z1 and Z2 and multiplying them Together and dividing them but the answers were always wrong until I actually found Z1Z2 and then worked out the argument. Does is rule only apply for modulus?

image.jpeg165.1KB

Reply 185

B_9710

Original post by Chittesh14

It can be easier to represent complex numbers in their exponential form.

\displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a}

Reply 186

Original post by Chittesh14

No. For the argument you add/subtract the individual arguments respectively rather than multiply/divide. You should find out what really is going on visually when two complex numbers multiply/divide in order to understand it, or just prove it to yourself algebraically.

I take it you are doing further maths if you're looking at complex numbers?

Reply 187

Original post by B_9710

It can be easier to represent complex numbers in their exponential form.

\displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a}

Sorry, I don't understand what you mean by the re^{ix} part. I understand the r as modulus. I understand x as a part in working out the argument, but I'm not too sure about e and i or whatever because maybe I haven't learnt it? :O.

Original post by RDKGames

Yeah, I was going to do that lol and actually do it by working out Z1 * Z2 and then doing the argument. That's what I did at the end, but I was too mad by then to actually explore it lol. So, it's like powers / exponents. In terms of the same power (argument), you add or subtract.

Yeah, FP1 - Complex Numbers lolz.

Reply 188

Original post by Chittesh14

That's in FP2 and is simply a different form of a complex number, don't worry about it.

Reply 189

Original post by RDKGames

You say it cant be negative but you don't explain why. Use the cubic for that.

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Original post by Zacken

You've got two simultaneous equations, you've only used one equation. It stands to reason that you need to use the other as well. Indeed, if you check

a = \pm 3

, in the cubic, you'll find that

a=3

is the only valid solution.

Original post by B_9710

It can be easier to represent complex numbers in their exponential form.

\displaystyle a+bi \equiv re^{ix}, \ r=\sqrt{a^2+b^2}, \ x=\arctan \frac{b}{a}

Nvm guys, sorry.

(edited 7 years ago)

Original post by Chittesh14

Nvm guys, sorry.

For c it's simply the case of showing that the angles from that -2 to the other other points add up to 90 :smile:

Or you can also do it by finding gradients, they should be negative reciprocal sod each other

Reply 191

Original post by RDKGames

For c it's simply the case of showing that the angles from that -2 to the other other points add up to 90 :smile:

Or you can also do it by finding gradients, they should be negative reciprocal sod each other

Lol. I just joined up the sides and it formed a right angled triangle lmfao. That's enough proof!!!
Yeah, I know that m₁m₂ = -1 rule for perpendicular lines lol.

Reply 192

Original post by Chittesh14

Lol. I just joined up the sides and it formed a right angled triangle lmfao. That's enough proof!!!
Yeah, I know that m₁m₂ = -1 rule for perpendicular lines lol.

"Yeah it looks right-angled enough.... QED!"

Try that in the exams :smile:

Reply 193

Original post by RDKGames

"Yeah it looks right-angled enough.... QED!"

Try that in the exams :smile:

Unfortunately, I might not get full UMS if I try it. So, fk the risk lol.

Reply 194

Original post by RDKGames

Original post by Zacken

Original post by B_9710

Hi guys, I'm just feeling a bit crazy XD I've done FP1 - Chapter 2 - Linear Interpolation. It's taking so long like 20 minutes per question or less idek (it feels long), or more. But, I may be doing something wrong or extra - maybe there is a shorter way of the method idk.

I'll post my working out to the question later, but I'll post the question now - can any of you just do the question and post your working out too - I might compare it before I post my working out too lol. I don't really know when I can post it lol that's the problem XD.

Question:

2. (a) Show that a root of the equation 5x³ - 8x² + 1 = 0 has a root between x = 1 and x = 2.
(b) Find this root using linear interpolation correct to one decimal place.

(edited 7 years ago)

Reply 195

Zacken

Original post by Chittesh14

2. (a) Show that a root of the equation 5x³ - 8x² + 1 = 0 has a root between x = 1 and x = 2.
(b) Find this root using linear interpolation correct to one decimal place.

Eh? This takes less than a minute. Denote your function by

f

, then since we have

f(1) = -2

and

f(2) = 9

, then since

f

is continuous, there exists a root

\alpha \in (1, 2)

.

Now, linearly interpolate

f

in the above interval, we have

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha = 13 \Rightarrow \alpha \approx 1.2 \end{equation*}

And then interpolate again in

(1, 1.2)

etc...

(edited 7 years ago)

Reply 196

Original post by Zacken

Eh? This takes less than a minute. Denote your function by

f

, then since we have

f(1) = -2

and

f(2) = 9

, then since

f

is continuous, there exists a root

\alpha \in (0, 1)

.

Now, linearly interpolate

f

in the above interval, we have

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha = 13 \Rightarrow \alpha \approx 1.2 \end{equation*}

That's not the way I'm taught it in the book. Also, 1.2 is not the correct answer either so idk lol. In the book, you draw out a sketch then use similar triangles to work out an approximation and keep on repeating the process until you get two successive approximations which round to the same answer when rounded to 1 d.p.

Reply 197

Original post by Zacken

Eh? This takes less than a minute. Denote your function by

f

, then since we have

f(1) = -2

and

f(2) = 9

, then since

f

is continuous, there exists a root

\alpha \in (0, 1)

.

Now, linearly interpolate

f

in the above interval, we have

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{2 - \alpha}{9} = \frac{\alpha - 1}{2} \Rightarrow 4 - 2\alpha = 9\alpha - 9 \Rightarrow 11\alpha = 13 \Rightarrow \alpha \approx 1.2 \end{equation*}

What you've done is found the next approximation to 1 d.p.
I done the process that my book showed me and then got 1. (18 recurring) and that rounds to 1.2 which is probably how you got that.

Reply 198