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Understand that involution means to multiply value with the same value so many times as involution number. So, understand that $1*1=1$

$t^2 * t^2$ would be 2t^2 (sorry, now understand that this is incorrect. Should be $t^{(2+2)}$ because rule is that for multiplication must sum the exponents).

So far understand.

But why $+ t^4$?

What the rule determines that need $+ t^4$?

If instead t use some number, then yes, all is correct. Just want to understand why $+ t^4$?

What would need to do if, for example $(1-t^2)^4$ or $(1-t^3)^2$ ?

enter image description here

Here is link to "textbook" https://estudijas.rtu.lv/pluginfile.php/287370/mod_resource/content/0/Uzdevumi/Uzdevumi_1.pdf The first page, the first example (1. piemērs. Aprēķināt / 1.example. Must calculate)

an4s
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user2360831
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    Uhh... $t^2 \times t^2 = t^4$. You might double check your algebra textbook. – Yuki.F Feb 23 '21 at 16:59
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    You should look up “rules of exponents”, “FOIL method”, and perhaps “binomial theorem” for more complicated examples – Benjamin Wang Feb 23 '21 at 17:03
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    There seem to be several things you are doing wrong. First, recall that $(a+b)^2 = (a+b)\times (a+b)$. Next, recall that $(a+b)\times (c+d) = a\times c + a\times d + b\times c + b\times d$. Third, recall that $x^a\times x^b = x^{a+b}$. You seem to have been multiplying in some cases and adding when variables were involved instead of multiplying, you seem to have been trying to "distribute exponentiation over addition" rather than correctly expanding as a product of sums, and other errors – JMoravitz Feb 23 '21 at 17:05
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    Do you know how to expand the square? That is, do you know what $(a + b)^2$ and $(a - b)^2$ expand to? If not, that might be something you should look into. – an4s Feb 23 '21 at 17:05
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    $(1-t^2)^2 = (1 - t^2)(1 - t^2) = 1(1 - t^2) - t^2(1 - t^2) = 11 - 1t^2 - t^21 - t^2(-t^2) = 1 - t^2 - t^2 + t^2t^2 = 1 - 2t^2 + t^4$. Because squaring is iterated multiplication, the distributive law, the distributive law, properties of 1 and sign arithmetic, simplification and the aforementioned fact that $t^2t^2 = t^4$. – leslie townes Feb 23 '21 at 17:05
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    As for "what to do for $(1-t^2)^4$" you should see the binomial theorem. It will be $(1-t^2)^4=1-4t^2+6t^4-4t^6+t^8$ – JMoravitz Feb 23 '21 at 17:07
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    See The Freshman's Dream for an explanation on why $(x+y)^n = x^n+y^n$ is wrong, why you can't just distribute exponentiation over addition. See exponentiation for why $x^a\times x^b = x^{a+b}$ and so why $(t^2)^2 = t^4\neq 2t^2$ – JMoravitz Feb 23 '21 at 17:10

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By the binomial formula we have $(1-t^2)^2=1-2 \times t^2+{(t^2)}^2=1-2t^2+t^{2 \times 2}=1-2t^2+t^4$.

jojobo
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