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February 1

The units of an interest rate

Interest is calculated as $P_{new}=P*(1+r)^{t}$ , where $P_{new}$ is the new amount of money, $P$ is the principal, $r$ is the rate, and $t$ is the time elapsed. Since $P$ and $P_{new}$ are both measured in the same units, the term $(1+r)^{t}$ must be dimensionless. And since $t$ is an amount of time, $(1+r)$ must be raised to the power of $\mathbf {time} ^{-1}$ .So a 2% annual interest rate be expressed as ${\sqrt[{\mathbf {year} }]{1.02}}$ . This is a rather strange unit. How do you refer to this kind of unit, if at all? Rannovania (talk • contribs) 01:24, 1 February 2008 (UTC)[reply]

You are confused. t is not a measure of time but a measure of terms. If t is a measure of t then it would be in units of seconds. It is a measure of terms, ie how many terms. And terms are dimensionless value. 202.168.50.40 (talk) 03:14, 1 February 2008 (UTC)[reply]

It's probably better to think of the formula as

P_{new}=P*(1+r)^{(t/t_{0})}

, where

t_{0}

is the unit of time used for expressing the interest rate. Then with a 2% annual interest rate (compound) the formula comes out as

P_{new}=P*(1.02)^{Y}

where

Y

is the time in years. HTH, Robinh (talk) 08:23, 1 February 2008 (UTC)[reply]

Quadratic inequalities

I know you are supposed to solve quadratic inequalities like you do quadratic equations, except that at the end, when you get the two solutions, you use an inequality sign instead of the equals sign. My question is, when do you have a solution like 1 < x < 2 and when do you have a solution like x < 1 OR x > 2? —Preceding unsigned comment added by 165.21.155.94 (talk) 05:06, 1 February 2008 (UTC)[reply]

That's a good question, and the very rough answer is that you do it either graphically, or with a couple of test points. So if you have, say, (x-a)(x-b) > 0, then you use the fact that it's concave up to show that the inequality holds on the left branch (x < a, say) and the right branch (x > b) but not on the little in-between interval. For high school mathematics that's about as far as you'd need to go, but for something more rigorous you'd look at the continuity of the function to ground your argument. Confusing Manifestation(Say hi!) 05:37, 1 February 2008 (UTC)[reply]

Take

e:=ax^{2}+bx+c

as your equation, with solutions

x_{1}\leq x_{2}

If

a>0

, then

e\geq 0\Leftrightarrow (x\leq x_{1})or(x\geq x2)

If

a<0

, then

e\geq 0\Leftrightarrow x_{1}\leq x\leq x_{2}

Morana (talk) 05:41, 1 February 2008 (UTC)[reply]

sin^2 x

Is it (sin x)^2 or sin (x^2) or something else? —Preceding unsigned comment added by 165.21.155.91 (talk) 05:11, 1 February 2008 (UTC)[reply]

sin²x means (sinx)². Strad (talk) 05:12, 1 February 2008 (UTC)[reply]

Thanks! —Preceding unsigned comment added by 165.21.155.91 (talk) 05:15, 1 February 2008 (UTC)[reply]

But be warned that sin⁻¹ x doesn't mean (sin x)⁻¹, it means arcsin x. -- BenRG (talk) 14:32, 1 February 2008 (UTC)[reply]

e to the power of

Given y-1=4, solving for y, you would say you "add one to both sides" or something similar. Same thing with subtraction, multiplication and division. But what about ln(y) = 4? Is there a word that describes "e to the power of both sides" as an action? My calculus teacher says "exponentiate", but admits he made that word up. —Preceding unsigned comment added by 70.240.254.243 (talk) 21:37, 1 February 2008 (UTC)[reply]

I would say "take exponentials of each side" or something similar. The inverse is "take logs of each side". --Tango (talk) 21:46, 1 February 2008 (UTC)[reply]

I'd be willing to bet that wikt:exponentiate was not made up by the teacher. --LarryMac | Talk 21:50, 1 February 2008 (UTC)[reply]

"Exponentiate both sides" is exactly what I'd say. I'd consider it standard usage. --Anonymous, 00:13 UTC, February 2, 2008.

The word is antilogarithm. Bo Jacoby (talk) 23:38, 1 February 2008 (UTC).[reply]

Technically, maybe, not I've never heard anyone use the word (and I'm currently studying maths at Uni). --Tango (talk) 23:43, 1 February 2008 (UTC)[reply]

And it's a noun, so if you do use it you have to say "take the antilogarithm". --Anonymous, 00:13 UTC, February 2, 2008.

Our article on Exponentiation uses the verb "exponentiated". hydnjo talk 02:21, 2 February 2008 (UTC)[reply]

Both "to exponentiate" and "to take the antilog" are operations with an unspecified base. The number e plays a special role, in particular when the argument is complex. I don't know another name for the operation exp than exponential function. I'd abbreviate that in the obvious way when speaking out loud, like when tutoring, as "now take the exps of both sides" and expect to be understood. --Lambiam 03:09, 2 February 2008 (UTC)[reply]

In higher level maths, the base is always e unless explicitly stated otherwise. I don't think you need to worry about the base. --Tango (talk) 13:45, 2 February 2008 (UTC)[reply]

To use exponentiate I think you’d have to say “exponentiate e to both sides”, because exponentating each side by e I believe would look like "ln(y)^e=4^e"

"take exponentials of each side" can also make sense though, because it’s commonly viewed as a function rather than an operation: "exp(ln(y))=exp(4)" which is "e^ln(y)=e^4". GromXXVII (talk) 12:20, 2 February 2008 (UTC)[reply]

That would be raising to the power e, not exponentiating, I don't think there is any ambiguity there. --Tango (talk) 13:45, 2 February 2008 (UTC)[reply]

Replying to Tango on antilogarithm. In the days when everything was done with published tables rather than calculators (not so long ago) there were tables of antilogarithms as well as logarithms and the phrases "take the log of both sides" and "take the antilog of both sides" were common currency. Both noun and verb forms were used. Terminology may well have changed now, I don't know. If you really want to know what geological era this was practiced, you will find that my age is displayed on my userpage. SpinningSpark 14:56, 2 February 2008 (UTC)[reply]

Using log tables to do arithmetic is very different to usings logs/exponentials in higher maths. (If nothing else, one uses base 10, the other base e and one uses numbers, the other algebraic expressions.) I wouldn't expect to use the same terminology in both. --Tango (talk) 17:17, 2 February 2008 (UTC)[reply]

Please do not be condescending. I know perfectly well the difference between arithmetic, pure, and applied mathematics. You are the one with a gap in your knowledge, which you highlighted yourself, I was merely politely giving you some information. You are wrong on just about every count. My mathematical tables (Knott, 1965, W&R Chambers) include both base 10 and base e. Likewise my slide-rule (a pre-computer analog calculating device) has scales for both log base 10 and log base e. In engineering calculations I have often (possibly more often) used base e for convenience. It is certainly not the case that mathematicians have a monopoly on base e while the rest of us grind out calcualtions in base 10. On the other hand there is no reason at all that algebraic expressions should not use log base 10 where appropriate; in engineering, many, in fact, do so. SpinningSpark 19:17, 2 February 2008 (UTC)[reply]

Ok, forget the difference in the bases then - the more important difference is that log tables are about the logs of numbers, not expressions. Exponentiating something generally refers to an expression (eg. "x"), and that is very different. As for engineers using base 10 - not when they're doing higher maths, they don't. Sure, decibels and things are defined in terms of base 10 logs, but that's not mathematics, that's just a convention. Base 10 might appear in the odd formula involving such units, but any time exponentials come up in relation to actual maths (eg. solving ODEs), they are base e. Oh, and don't complain about me being condescending and then do it yourself - I know what a slide rule is... I even own one, although I've no idea where I put it. --Tango (talk) 21:57, 2 February 2008 (UTC)[reply]

Just as you talk about the natural logarithm and the base ten logarithm, you may talk about the natural antilogarithm and the base ten antilogarithm. If ln(y) = 4, then take the natural antilogarithm on both sides and get y=antiln(4)=ln⁻¹(4)=exp(4)=e⁴. If sin(y)=a, then take the arcus-sine on both sides and get y=arcsin(a)=sin⁻¹(a). Bo Jacoby (talk) 21:11, 3 February 2008 (UTC).[reply]