Anonymous at Fri, 15 Nov 2024 01:49:28 UTC No. 16475531

I tried some GRE math subject test practice problems and found out I'm retarded

Anonymous at Fri, 15 Nov 2024 01:57:03 UTC No. 16475537

>>16475531
same desu. was a straight a math minor and I got 10th percentile on my math gre lol

Anonymous at Fri, 15 Nov 2024 02:02:29 UTC No. 16475539

>>16475537
It's brutal. I'm going to keep trying though.
I'm a HS math teacher right now and my students do really well, but I know I'm no genius. I could go on like this but don't want to stagnate.

Anonymous at Fri, 15 Nov 2024 02:16:50 UTC No. 16475545

My online multiple integrals calc 3 test where I put 6^3=192

Anonymous at Fri, 15 Nov 2024 03:06:32 UTC No. 16475594

>>16475519
I've been studying for qualifying exams in my EE dept and getting absolutely shit on by previous probability and information theory exams.

Anonymous at Fri, 15 Nov 2024 03:12:01 UTC No. 16475601

>>16475519
been slacking on precalc in a nutshell desu. Looks like a great little book tho, bless whoever put that in the wiki

Anonymous at Fri, 15 Nov 2024 03:13:14 UTC No. 16475603

Why is log_a(xy) = log_a(x) + log_a(y)?

Consider log_10(10^x):

log_10(10) = 1
log_10(10 * 10) = 2
log_10(10 * 10 * 10) = 3
log_10(10 * 10 * 10 * 10) = 4
log_10(10 * 10 * 10 * 10 *10) = 5

Add two together and you get the multiple.

Back to the first equation, xy = (a^log_a(x) ) (a^log_a(y) ) = a^(log_a(y)+log_a(x))

The definition says if a^z = xy, log_a(xy) = z, so log_a(xy) = og_a(y)+log_a(x)

Anonymous at Fri, 15 Nov 2024 04:16:58 UTC No. 16475673

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on [math](ab)^{1/x}=a^{1/x}b^{1/x}[/math]
[math]=\sqrt[fc]{x^{fb}x^{ec}}[/math]
And perform the final addition:
[math]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/math]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.

Anonymous at Fri, 15 Nov 2024 04:22:21 UTC No. 16475677

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on
[math](ab)^{1/x}=a^{1/x}b^{1/x}[/math]
[math]=\sqrt[fc]{x^{fb}x^{ec}}[/math]
And perform the final addition:
[math]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/math]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.

Anonymous at Fri, 15 Nov 2024 04:25:13 UTC No. 16475680

Define a positive rational exponent as:
[math]x^{a+b/c} = x^a x^{b/c} = x^a\sqrt[c]{x^b}[/math]

Then we consider exponent addition:
[math]x^{a+b/c}x^{d+e/f}[/math]
First add the whole numbers, no need for proof:
[math]=x^{a+d}x^{b/c}x^{e/f}[/math]
Then get the same denominator:
[math]x^{b/c}x^{e/f}=x^{fb/fc}x^{ec/fc}[/math]
The exponents remain equal during the last step:
[math]x^{eb/ec}=({\sqrt[ec]{x}})^{eb} = (({\sqrt[ec]{x}})^{e})^b = ({\sqrt[c]{x}})^{b} [/math]
Rewrite the numerator based on
[math](ab)^{1/x}=a^{1/x}b^{1/x} [/math]
[eqn]=\sqrt[fc]{x^{fb}x^{ec}}[/eqn]
And perform the final addition:
[eqn]=\sqrt[fc]{x^{fb+ec}}=x^{(fb+ec)/fc}[/eqn]
So by splitting the process into more believable steps we understand why exponent addition holds for rational powers.

Anonymous at Fri, 15 Nov 2024 06:03:36 UTC No. 16475752

>>16475519
I've been working on increasing the accuracy of a model of caffeine metabolism specific to myself so I can tweak more efficiently.
It's proceeding slowly, as I don't have good measurements of many of the variables, but it's a fun hobby project.

Anonymous at Fri, 15 Nov 2024 10:14:45 UTC No. 16475877

>>16475519
I’m currently researching unitary representations of classical groups. Trying to find surjective maps from a certain class of groups onto their subgroups. This is needed for the method of induced representations.

Anonymous at Fri, 15 Nov 2024 11:47:42 UTC No. 16475969

>>16475877
is it fun?

Anonymous at Fri, 15 Nov 2024 11:50:50 UTC No. 16475973

>>16475969
Yeah, I’m a group autist. But it’s more research and less problem solving. Basically 90% of it is digging through literature, 10% is figuring things out on your own.

Anonymous at Fri, 15 Nov 2024 20:31:10 UTC No. 16476767

Been doing boring problems but in a new way. Learned how to use an abacus, can add, subtract (into negatives too), multiply and divide now. Decided to refresh my math from the floor up, currently fucking around with arithmetic, with abacus and mental math practice.

Raphael at Sat, 16 Nov 2024 00:57:24 UTC No. 16477249

>>16475531
KEK

Raphael at Sat, 16 Nov 2024 00:57:55 UTC No. 16477251

>>16475973
That’s all subjects